Schrödinger Equation II: Three Dimensions

6 Schrodinger Equation II: Three Dimensions

We are now ready to study the three-dimensional world. In this chapter, we extend the Schrodinger equation to three dimensions, and we investigate the properties of the solutions for cases in which the potential energy is spherically symmetric. A most important property of any solution is its angular momentum, so we shall spend considerable time on the quantum mechanical description of angular momentum. We shall then use our knowledge to solve three-dimensional counterparts to some of the problems of Chapter 5, and to study some features of scattering of particles from a spherically symmetric potential. 187

188

SCHRODINGER EQUATION II: THREE DIMENSIONS

6.1

EXTENSION TO THREE

OF THE SCHRODINGER DIMENSIONS

EQUATION

In three dimensions, the total energy of a particle is (p + p + pl)l2m + V(x, y, ζ)· We have seen (Section 5.6) that the Schrodinger equation may be constructed by replacing p by the operator (h/i)(d/dx). There is no essential difference between the χ coordinate and the y and ζ coordinates, so we do the same for p and p , replacing them by (h/i)(dldy) and (h/i)(d/dz) respectively, to obtain the Schrodinger equation in three dimensions: 2

2

x

x

y

2

h

2

Id

2

d

2

d\ 2

= in — dt

(1)

As before, the wavefunction φ for a definite energy may be written as a product of a space-dependent part and a time-dependent p a r t : φ(χ, y ζ , 0= w ( x , y, 9

z)e~

iEtlh

Particle in a Box. To illustrate how the three components of momentum fit together, let us find the wavefunction for a particle which is confined to the inside of a rectangular box. This apparently artificial example has a fruitful application to the theory of electrons in metals, as we shall see in Section 11.4.

Fig. L "Box" containing a particle whose potential energy is zero inside and infinite outside it. Let the walls of the box be the planes χ = 0, χ = α, y = 0, y = b, ζ = 0, ζ = c, and assume that the potential energy is zero inside the box and infinite outside the box (see Fig. 1). As in the one-dimensional square well (Section 5.4), the function u must go to zero when the potential step is infinite, so u = 0 at the walls of the box.

6.1

EXTENSION OF THE SCHRODINGER EQUATION18

9

We shall solve the problem by the standard technique of separation of vari­ ables, which is applicable to a great variety of partial differential equations. We assume that u(x,y,z) may be written as a product of three functions: X(x), a function of χ only, Y(y), a function of y only, and Z(z), a function of ζ only: u(x,y,z)=X(x)-Y(y)-Z(z) Equation(1 ) now becomes, for the region inside the box, ft - — (X"YZ 2

+ XY"Z

+ XYZ")

= EXYZ

where Χ" Ξ d X l d x , Y"=d Y\dy , and Z" = d Z/dz . sides of Eq. (2) by the product X YZ, we obtain 2

2

2

h

2

2

2

IX"

- 2 ^ { χ

Y" +

Ύ

Z"\ Υ )

+

2

(2) If we divide both

„ =

Ε

( 3 )

Suppose we evaluate the left side of Eq. (3) at two different values of x, leaving y and ζ fixed. When χ changes, the only term which could possibly change is the first term, but since the sum of the three terms is the constant E, the first term must remain unchanged also. But this term is a function of χ only; if it does not change when χ is varied, it must never change; i.e., it is equal to a constant, and we may write γ = - * *

(4a)

2

Similarly, we may deduce that the second term equals another constant: Y" f = - k

(4b)

2 y

and finally that Z" γ =

(4c)

~kl

Thus we have completed the separation of variables, obtaining three ordinary differential equations, one for each coordinate. The constants are written in this form because we know that they should be negative; a negative constant makes each equation equivalent to the equation of simple harmonic motion, whereas a positive constant would lead to an exponential solution which cannot satisfy the boundary condition that u be zero at each wall. Thus the solutions to Eqs. (4) are X= A s i n ^ - x ) Y=

Bsm(k 'y)

Ζ =

Cun(k -z)

y

z

190

SCHRODINGER EQUATION II: THREE DIMENSIONS

We may now evaluate k , k , and k by using the boundary conditions, which become, in terms of X, Y, and Z : x

X(0) = 0,

7(0) = 0,

y

z

Z(0) = 0,

X(a) = 0,

Y(b) = 0,

Z(c) = 0

We have already satisfied the first three conditions by choosing the sine rather than the cosine for our three functions. The last three conditions are satisfied if k a = n n, k b = n n, kc = nn x

x

y

y

z

z

where n , n , and n are integers. The complete solution to the problem is therefore x

y

z

. η πχ ψ(χ y, z, t) = constant χ sin a

. n ny . η πζ sin sin b c

χ

y

9

ζ

e

_

iFt/h lc,t/

where E=!^(ki+k +k ) 2

2

y

z

In other words, we have three independent wavenumbers, k k and k which take the place of the single wavenumber k that we had in the onedimensional case; these wavenumbers determine the momentum components: X9

p = hk , 2

2

x

6.2

2

x

p = hk , 2

2

y

SPHERICALLY SYMMETRIC AND ANGULAR MOMENTUM

2

y

p = hk 2

2

y9

z9

2 z

POTENTIALS

The example of the particle in a box was particularly simple because the potential energy was constant in a region whose boundaries were parallel to the coordinate axes. We cannot usually expect such simple situations in nature, although we can use the above result to make a fairly good approximation to the behavior of electrons in a metal (see Chapter 11). A situation which is not quite so simple, but still manageable, arises when the potential energy is a function of only one of the three coordinates. A very important special case is that of a central force field, in which the potential is a function only of r (in spherical coordinates, r, 0, and φ). F o r example, this is true for any two particles which interact by the Coulomb force only, provided that we use the center-of-mass coordinate system, in which r is the distance between the two particles. Thus the theory developed in this section is an essential part of the theory of the hydrogen atom, where the* dominant force is the Coulomb force between electron and proton. However, we shall first develop the theory in a form that applies to all central forces, before taking up the hydrogen atom in detail in Chapter 7.

6.2

SPHERICALLY SYMMETRIC POTENTIALS19

1

Fig. 2. Spherical and rectangular coordinates of a point P. We begin by writing the Schrodinger equation in spherical coordinates, to take advantage of the fact that the potential energy is a function of r only, and to pave the way for the separation of variables. By referring to Fig. 2, we see that the spherical coordinates may be written in terms of cartesian coordinates as r = x+ 2

C O S f l

y + z

2

2

2

"r"(x +/ + 2

2)/

(6)

2 12

tan φ = χ The space derivatives in Eq. (1) may be converted to derivatives with respect to r, Θ, and φ by using the 'chain rule": 4

du

du dr

cu δθ

du δφ

δχ

dr δχ

δθ δχ

δφ δχ

with corresponding equations for du/dy and duldz. The necessary partial deriva­ tives can be shown, with the aid of Eqs. (6), to be δτ . , — = sin θ cos φ; δχ

— = sin 0 sin φ; dy

— = cos 0 δζ

— = - cos θ cos φ\ δχ r

— = - cos θ sin φ; dy r

δθ

δφ

1 sin φ

δφ

1 cos φ

δχ

r sin 0 '

δγ

r sin θ '

ι

— = sin 0 δζ r 1 ο δζ δ

=

192

SCHRODINGER EQUATION II: THREE DIMENSIONS

To find d uldx obtaining 2

one simply substitutes du/dx

2

flr

u _ Γ d dx 2 I 2

fdu\] UJC

/

dr^ dx

J'

. \_d_( [βθ\

for u into the chain rule,

βιΛΊ Sx ) \

0Θ_ ,\j_(du\]

dx

θφ \ dx ) \ dx

[θφ

and one finds d uldy and d u/dz in corresponding fashion. It is understood that the first partial derivatives are already expressed in terms of spherical coordinates and derivatives with respect to those coordinates. Thus 2

&

2

2

2

- ο ί η Λ Λ Λ ο Λ ^

u

— dx

ι

1

Λ.

£k

du

sin

sin θ cos φ —— + — cos θ cos φ - ^ r dr r dd

-

φ

du

— - £ χ r sin θ dφ

Completing these operations for all three coordinates, one finally obtains the time-independent Schrodinger equation in spherical coordinates:

We can now separate the variables by writing Μ(Γ,0,Φ) as a product of two functions, one a function of r only, the other a function of θ and φ: u(r, θ, φ)=

R(r)Y(0

φ)

9

Making this substitution in Eq. (7), and multiplying both sides by r /w, we obtain 2

r d

2

,

n x

1 Γ 1

d ( .

dY\

1

n

δ Υ1

2mr

2

2 τ / /

χ π

(8) The dependence on the variables θ and φ is all contained in the second term in Eq. (8). Therefore, if θ or φ is varied, the other terms in the equation cannot change, and we conclude that the second term in the equation must be a constant, according to the same reasoning used in Section 6 . 1 : 1 Γ 1

d I . dY\ n

1

d Y~\ 2

or Q Y = O L Y

(9)

6.2

SPHERICALLY SYMMETRIC POTENTIALS19 3

where we have defined the operator Ω as _

J

d_( .

_

sin 0 50I

d\

<

1_ W sin 0
2

The eigenvalue equation (9) must be obtained from the Schrodinger equa­ tion every time that the potential energy depends on r only; the eigenfunctions of (9) must describe the angular dependence of the wavefunction in all such cases. Before attempting to find these eigenfunctions, let us inquire into their physical meaning, and the meaning of the operator Ω and its eigenvalue(s) a. We can obtain a clue to the meaning of the operator Ω by writing the kinetic energy in spherical coordinates as 1

where L is the square of the total angular momentum of the particle. (L = r χ p. ) Equation (10) tells us that the time independent Schrodinger equation should be 2

Γ2

where p a n d I are operators. If we compare this equation with Eq. (7), we see that they agree, provided that 2

2

V? ,

2 ρ

·

u

χ

~7~d?

=

(m)

and L

u

=

-^roFe[ -o)-^W sme

2

( 1 2 )

d

(Notice that the operator for p is not simply —ft d /dr , as it would be if r were a Cartesian coordinate like x, y, or z.) Comparison of Eq. (12) with the definition of the operator Ω shows that L = ή Ω . We may also deduce the above result directly from the definition of L . SinceL = r χ p,the operators for the rectangular components ofL should be 2

2

2

L*=yp*-*p>

2

2

2

=

- {yir - T ) it

z

z

y

I t i s eas y t o verif y tha t Eq . (10 ) i s th e correc t expressio n fo r T i n term s o f L. T h e velocit vecto r ca n b e resolve d int o t w o c o m p o n e n t s , v, an d v , whic h ar e respectivel y paralle l an perpendicula r t o th e radiu s vecto r r . The n T= mv}jl + mv\\2, an d L = (mv r) s o tha v\=- - L /m r . Substitutin g fo r v int o T y i e l d s Γ = - mvf/2+ L /2mr , whic h i s equivalen t t Eq . (10) , wit h p — mv . 1

±

2

2

±

2

2

2

2

±

r

r

2

y

y d t o

194

SCHRODINGER EQUATION II: THREE DIMENSIONS

L.-xp,-yp --lh(XL—y£ldy x

which become, in spherical coordinates, L = ih\sin φ ^ - 4- cot θ cos φ 4~r) \ cO c
L

y

i d o\ = — /Λ cos φ — - cot θ sin φ — 1 \ cO οφΙ

(13)

as you may verify yourself by using the chain rule and the transformation equations given above. Equations (13) may now be used to compute L = L + L 4- L ; if you work it out, you will find, in agreement with our previous result, that Lr ~ / ι Ω . Having determined the meaning of the operator Ω, we can proceed to find the eigenvalues the values of α which yield an acceptable function Y when we solve Eq. (9). These eigenvalues will tell us the possible results of a measure­ ment of I}. To imd them, we complete the separation of variables as follows. We write 2

2

2

2

2

Y((K Φ) -

Θ(0)Φ(φ)

Using the definition of Ω, we may write Eq. (9) as Φ

cte\

d /

; T n 7 i ^ (

s

^ ^ )

m

Θ
+

s

^

^

=

~

a

0

a

>

or, after multiplication by (sin 0)/ΘΦ, 2

de\

d I ,

1 .

R 7o( M) i>W~m0

an0

ASM0

+

1 ί/ Φ 2

( , 4 )

By the same reasoning used in our previous separation of variables, we deduce thaiI he only -dependent term in the equation must e^ual a constant:
constant χ Φ (15) Λφ A very simple condition on the wavefunction tells us what this constant must be. We require that the wavefunction be single valued—that there be just one value of ψ for a given point in space at a given time. (Although it gives the right answer in this case, this requirement is not strictly necessary. See Problem 8.) As a consequence, 2

6.2

SPHERICALLY SYMMETRIC POTENTIALS19

5

Φ(φ + 2 π ) = Φ(ψ) since the angle φ + 2π describes the same points in space as the angle φ. We can satisfy this condition if we write Eq. (15) as — αφ

= -

2 m

0

(16)

where m is an integer, so that the equation becomes identical to the standard equation of simple harmonic motion, whose solution has period In and may be written as Φ = Ae * 2

3

im

where A is an arbitrary constant which may be chosen to normalize the wavefunction. Notice that this solution is an eigenfunction of the L operator, for 2

L - Φ = -ih

d ^- Φ = -ih dφ

d — Ae + = πιΗΦ dφ im

and the eigenvalue is mh. Therefore, according to the eigenvalue postulate (Section 5.6), the only possible results of a measurement of a component of angular momentum are integral multiples of ft. It seems that after all that work we are right back where we started, with the Bohr condition! However, our next step, the investigation of the eigenvalues of L , will show that there is a significant difference between the results of the Bohr theory and those of the present theory. Continuing our search for the eigenvalues a, we substitute from Eq. (16) into Eq. (14) to eliminate Φ, and obtain an equation in the single variable Θ: 2

sin θ d

ι.

_ d&\

~

.

0

Λ

Equation (17) appears formidable because of the sine functions scattered through it, but these may be eliminated by the substitution cos θ = χ. With the function ^(Λ:) replacing the function Θ(0), the equation becomes d_ dx which may be solved by assuming y to be a power series in x. The details of

This is the standard notation, which could possibly lead to confusion between the quantum number m and the mass m, but the meaning should always be clear from the context. Of course there are two independent solutions of a second-order equation, but the other solution, e- * becomes the same as the first solution when we consider negative values of m as well as positive values. 2

3

im

y

196

SCHRODINGER EQUATION II: THREE DIMENSIONS

the general solution need not concern us h e r e ; the important feature is that the solution is finite for all values of 0—that is, for all values of χ between — 1 and + 1—only when the constant α is of the form 4

a = /(/+l) where / is an integer or zero, and / > \m\. Thus Eq. (9) becomes Ω7=/(/+

l)Y

but we found that the L operator is equal to h Ω, so that 2

2

L Y(0, φ) = /(/ + l)ft Y(0, φ) 2

2

In other words, the eigenvalues of the L operator are / (/ + 1)Λ , so that these values: 0, 2# , 6/i , 12/i ,.. .are the only possible results of a measurement of /A Contrast these values with the prediction of the Bohr theory, where one expects that L could have, for example, the value Λ , which is simply the square of an allowed value of a component of L . We shall see in Section 6.3 that the value of fi is excluded because the uncertainty principle prevents our knowing the exact direction ofL ; if L is known to be exactly equal to ft, then there must be another nonzero component of L , making L greater than ft . We shall also see that the mysterious-looking eigenvalues /(/ + l)ft can be deduced from a straightforward analysis of how L is determined experimentally. 2

2

2

2

2

2

2

2

z

2

2

2

2

Eigenfunctions of L . The eigenfunctions of L are the products of the functions Θ and Φ, solutions of Eq. (17) and Eq. (16), respectively, cor­ responding to possible combinations of the integers / and m. The functions Φ, as we have seen, depend only on m, but the functions Θ depend on / and on the absolute value of m, because Eq. (17) contains m as well as /. Thus we write the eigenfunctions of L as 2

2

2

2

where the functions P} ^ (cos 0), found by solving Eq. (17), are called associated Legendrepolynomials. The functions Yj, (0, φ) are called spherical harmonics, and are well known in classical theory; the first few are: m

w

The method is very similar to that used for the solution of the radial part of the Schrodinger equation for the hydrogen atom, which is fully worked out in Appendix C. For details of the solution of Eq. (17), see, for example, R. Leighton, "Principles of Modern Physics," pp. 166-171. McGraw-Hill, New York, 1959. 4

6.2

SPHERICALLY SYMMETRIC POTENTIALS ^2,2 =

197

sin 0 e * 2

\32π]

2i

/ 1 5 \ 1/2 1—1 sin 0 cos θ β

1/2

ίφ

3\ /3\

1/2

1 / 2

cos 0 1

/

Yz.a

—

/15\

2

0-1)

2

1 / 2

15 \

^2,-2

(3 c o s

1 / 2

s i n 2 θΛ^-2ίψ e 2

(32^

(The minus signs in front of Y and Y , i are not essential, and they do not always appear in tables of spherical harmonics. They are included here in order to simplify the matrix representation of the angular momentum operators, to be developed in the problems of Chapter 7.) Notice that the θ dependence of Y _ is the same as that of Y , because the function Θ depends on \m\ rather than on m itself. The numerical factors shown in the above functions are simply normalization factors derived from a requirement that the integral of | y ( 0 . φ)\ over an entire sphere be equal to unity: U1

2

ltfn

m

lt

2

i>m

Γ o

J

Γ\Υ , \ ύηθάθάφ

= 1

2

J

ι ηι

ο

(18)

This requirement enables one to interpret I ^ J as a probability density for the particle's angular coordinates. In other words, if a particle has a wavefunction whose angular dependence is y , the quantity 2

/ m

Γf V Φι

(

,J

2

sine άθάφ

0ι

is equal to the probability of finding the coordinates θ and φ of the particle between the limits Φι <φ < φ

2

and

θ <θ

<θ

ί

2

when one measures the position of the particle. Because of the exponential form of the φ dependence, the absolute square | 7 | is a function of θ only. A polar plot of | ^ / | versus θ can give us some insight into the connection between spherical harmonics and the angular momentum of a particle. Figure 3 shows polar plots of | 7 | > 1*2, il > d | Y 2 | - Each of these functions represents a particle with the same value of L , but the values of L are 0, and +2h, respectively. As L increases, the 2

2

i m

>ιη

2

2

2 > 0

2

t2

2

2

2

a n

198

SCHRODINGER EQUATION II: THREE DIMENSIONS θ=

0 °0

= 0 °0

= 0 °

Fig. 3. Polar plots of some probability densities \

Yi, \ m

2

as functions ofΘ.

probability density shifts from 0 = 0 (the ζ axis) toward 0 = 90°. F o r a given value of r, an orbit in the xy plane (0 = 90°) must correspond to the greatest value of L classically as well as quantum mechanically; and of course a particle with 0 = 0 is located on the ζ axis and must have L = 0. Let us recapitulate the somewhat tedious operations of this section, in order to bring out the essential points. z9

z

1.

A straightforward transformation into spherical coordinates yields a time-independent Schrodinger equation of the form d - r — ( m ) + ΩΜ =

2mr

2

2

2

[ £ - F(r)]u

where Ω is an operator involving derivatives with respect to 0 and φ. 2.

By means of the definition L =r χ p , and substitution of the operators for p ,p , and p , expressions for the operators L L L and L can be found, and the operator L is found to be equal to h Q. 2

x

y

z

X9

2

3.

y9

Z9

2

Separation of the variables in the Schrodinger equation, by means of the substitution w(r, 0, φ) = R(r)Y(9. φ) yields the eigenvalue equation ΩΥ = aY. T o be acceptable as a factor in the wavefunction, the functions Y(0 φ) must have a φ dependence of the form e * where m is a positive or negative integer, or zero; these functions are then eigenfunctions of the L operator, with eigenvalues mh. Furthermore, the eigenvalue α must have the form /(/ + 1), where / is an integer greater than or equal to \m\. Thus each function Y is an eigenfunction of both L and L , and it is characterized by two indices, / and m: 9

im

t

9

z

2

z

L y,, (0,4>) = / ( / + i ) f i r 2

2

m

1 > m

( 0 , ψ)

and

L YU9 2

9

φ) = ηΛΥ β> ΐ9Μ

Φ)

These functions Y (0, φ) are extremely important because they form the angular part of the wavefunction every time the potential energy is spherically />m

6.3

MEASUREMENT OF ANGULAR MOMENTUM

199

symmetric. Because the L operator involves only θ and φ, the complete wavefunction u(r, θ, φ) = R(r)Y (9, φ) is also an eigenfunction of L and L as well as an eigenfunction of energy. Thus a particle may always be in a state which has a definite value of L as well as a definite value of energy. This fact, of course, has its classical basis in the fact that angular momentum is conserved when a particle moves under the influence of a central force. It is left as an exercise for the reader to verify that each of the Y listed in this section is indeed a solution of Eq. (14), with α = /(/ + 1). 2

2

lm

z

2

5

lm

6.3

MEASUREMENT

OF ANGULAR

MOMENTUM

a. T h e o r y . In the preceding section, we showed t h a t the functions are eigenfunctions of the L operator. They are not eigenfunctions of L or L , a fact you can easily verify by applying either of these operators [Eq. (13)] to any of the Yi, . Except for the trivial case of y ,o, the operation fails to yield a constant multiple of the original Yi, . But there cannot be an essential difference between the ζ axis and the χ or y axis. We can measure a component ofL with respect to any direction in space, so for any particular value of L , the eigenvalues of L and L must be the same as those of L , even though the eigenfunctions are different. For example, if / = 1, then L can be4- ft, 0, or —ft, so the same must be true of L or L . The fact that the eigenfunctions of L are not also eigenfunctions of L or L means that one can measure only one component ofL at a time; when we write the wavefunction as an eigenfunction of L , we are in effect defining the ζ axis as the axis along which we have chosen to measure a component of L . The method of measurement of L will be discussed in Section 6.3b. Let us now turn to L , to inquire into the verification of those seemingly peculiar eigenvalues /(/ + l)ft . We can imagine the following thought experiment to determine the value of L : We assemble in free space a large number of systems that all have the same value of L , but for each of which the direction ofL is completely undetermined. The systems are otherwise identical. We measure L for each of these systems, verifying that L is always an integer times ft. We define the integer / as the maximum observed value of L /n. Each of the possible values of L ( + /ft, + (/ - 1)/?,... - lit) must occur with equal frequency, because in the absence of an

Υι,,Λ^,Φ)

z

x

y

m

0

m

2

x

y

z

z

x

y

z

x

y

z

z

2

2

2

2

z

z

z

z

There is a possibility that eigenfunctions corresponding to different values of L may accidentally have the same energy eigenvalue. In this case, one could form an energy eigenfunction from a superposition of eigenfunctions with different values of L , thereby creating an eigenfunction of energy which is not an eigenfunction of ZA However, such situ­ ations do not actually occur in nature; the nearest thing to such a situation occurs in the hydrogen atom, for which eigenfunctions of different L value can have almost the same energy—to one part in 10 . 5

2

2

2

4

200

SCHRODINGER EQUATION II: THREE DIMENSIONS

external force field, there is no energy difference between states with different values of L . From these values, we can easily compute ( L ) . There are 21 + 1 equallylikely values of L , so we sum these values and divide by 21 + 1, obtaining z

2

z

( L

z

2

)

=

2

(21

+ l)"

1

hΣ

ι

2

m

2

=

4-l)"

2(21

1

ι

hΣ

m

2

m = -l

2

= 1

m

The series sum is easily shown by mathematical induction to be equal to /(/ + 1)(2/ + l ) / 6 , so L = 1(1 + 1)Λ /3. Since there is no preferred direction, we know that z

2

2

( L / ) = {W)

=

( L

z

2

)

.

We also know that L is the same for each system, so 2

L

=

2

( L

2

)

=

( L

x

2

+

( V )+

)

{Lz )

=

2

3 ( L

Z

2

)

=

1(1

+ l)ft . 2

Thus we have deduced the eigenvalues of L from a statistical analysis of the observed values of L . Having measured L and L , we can visualize our knowledge of the angular momentum vector as follows; L lies somewhere on a cone whose axis is the z a x i s a n d whose apex angle is equal to c o s ( L / | L | ) = c o s ( m / [ / ( / + 1 ) ] ) (see Fig. 4). We assume that all azimuthal angles for L are equally likely, be2

z

2

z

- 1

_1

1/2

Z

Fig. 4. Visual represent­ ation of the angular momentum vectorL .

cause we do not have any additional information about L or L ; the eigenfunction Y provides no additional information because | Y \ is independent of φ. In many introductions to the subject, the theory of angular momentum is left at this point, because the eigenvalues have been determined and no further information can be obtained. Let us examine more carefully the reason why no further information can be obtained, and let us investigate what happens if we do try to measure two components of angular momentum. Study of these points will give us much more insight into the meaning of superposition x

t

m

y

t

m

2

6.3

MEASUREMENT OF ANGULAR MOMENTUM20

1

in quantum theory, and will also show how the uncertainty principle affects all of these measurements. In attempting to measure two components of angular momentum, we must first measure one component and then measure the second component. (In part b of this section we shall describe how this is done experimentally.) The measurement of the first component defines our ζ axis; so we must give a different label to the second axis. Let us call it the χ axis, and let us therefore determine the eigenfunctions of the L operator. This would be a rather tedious chore if we attempted to compute these eigenfunctions directly from expression (13) for L , but fortunately there is a much simpler way to do it. We can convert eigenfunctions of L into eigenfunctions of L simply by rotating the coordinate system. This is done most easily if we write the Y, in rectangular coordinates first; then, if we rotate the coordinate system by 90° about the y axis, ζ is replaced by χ and χ is replaced by — z, leaving y unchanged (see Fig. 5). Let us perform these operations for the x

x

z

x

#m

- y

Fig. 5. A 90° rotation about the y axis causes the χ axis to replace the ζ axis. The ζ coordinate becomes the χ coordinate, and the χ coordinate becomes the —z coordinate.

three functions with / = 1. In rectangular coordinates, we have /3\ = - 1 — I 1

Y

i t l

1/2 / 2 /

-χ \ 1/ 1 / 22

/3\ (sin θ cos φ 4- i sin θ sin φ) = - I — I

x

+ iy —-—

(19) / 3\ Y

t

1 / 2

_ = j— I t

\ 8 π /\ο7Γ

/ 3\

1 / 2

χ - iy

(sin Θ cos φ — j sin 0 s i n φ) = I — I /

r

202

SCHRODINGER EQUATION II: THREE DIMENSIONS

After the substitutions χ -> - ζ , y y, and ζ -> χ, these functions become the eigenfunctions of L which we denote as Y : x

1>mjc

We may think of the operation which generated the Y as simply a relabelling. The functions Y have exactly the same spatial distribution as the functions Y ; only the coordinates used to describe this distribution have changed. Thus these functions are still eigenfunctions of L , with / = 1; and their respective eigenvalues of L are the same as the L eigenvalues of the original functions. As an exercise, you may verify by direct application of the L operator (13) and the L operator that these functions do indeed have the eigenvalues which we claim. N o w suppose that we have measured L and L , finding that / = 1 and m = 1; that is, I? = /(/ + l)ft = 2ft , and L = mh = +ft.W h a t will happen when we now attempt to measure L ? Since / = 1, there are three possible results: L = +ft,0, or —ft.But the wavefunction's angular part is Y before the measurement, so the wavefunction is not an eigenfunction of L . Can we make any prediction at all about the values of L which are likely to result from the measurement under these circumstances? Certainly. Just as we can Fourier analyze a wave packet t o determine the probabilities of obtaining various values of momentum in a measurement (even though the wave packet is not an eigenfunction of momentum), we can analyze the function Υ and determine it to be a superposition of various eigenfunctions of L . That is, we can write 1>mjc

itTnx

1 > m

2

x

z

2

x

2

z

2

2

z

x

x

ltl

x

x

1Λ

x

Yi,i="Yuix

+

+ cY -

l>YL0

lt

X

(21)

ix

and the coefficients a, b, and c should then determine the probabilities of obtaining the results +ft, 0, and —ft, respectively, when L is measured. As before, the square of the amplitude determines the probability; that is, \a\ equals the probability of obtaining t h e result +ft, just as |φ(Α:)| equals the probability of obtaining a value of ρ equal to hk in a momentum measurement. Before performing a computation of a, b, and c for this example, let us try to deduce the probabilities from a knowledge of the expectation values. We know that L + L = L - L , which equals ft in this case. The χ and y directions are equivalent, so ( L ) = (L ) = ft /2. There are only two possible values for x

2

2

2

x

2

2

2

y

2

z

2

x

2

y

2

6.3

MEASUREMENT OF ANGULAR MOMENTUM20

3

L : either ft or 0. Therefore, to have the proper expectation value, L must equal zero half of the time, and h half of the time. When L is ft , L is either +ft or - f t , with equal probability, so each of these probabilities must be i. That means that the three probabilities are i, i, and i, for obtaining eigenvalues of +ft, 0, and - f t , respectively. This result is confirmed when we compute a, b, and c. Substitution from Eqs. (18) and (19) into Eq. (20) yields 2

2

2

x

2

2

x

-

x

+

(y = ( z ~ (y) + r g

r

frV2* r

+

2

x

c(-z-ry) r

This equation must hold for all values of x, y, or z, so we may find a, b, and c by equating the coefficients of x, y, and ζ in turn, obtaining fcV2 = 1 ,

a + c = 1,

and

α - c = 0

with the final result that a

=

h

b

— ρ ,

=

and

c=

i

or \a\ = i, 2

\b\ = h 2

and

| c | = J, 2

in agreement with the previous result. This sort of procedure is of such general use in quantum theory that it is worthwhile to generalize it by stating a third fundamental postulate (to go with the two postulates stated in Section 5.6).

POSTULATE 3. Any acceptable wavefunction φ can be expand­ ed in a series of the eigenfunctions of the operator corresponding to any dynamical variable. The probability of finding a particular value of a dynamical variable as a result of a measurement on a system with normalized wavefunction ψ is equal to the sum of the absolute squares of the coefficients of the corresponding normalized eigenfunctions in the expansion of the function ψ.

In other words, any set of eigenfunctions, like the spherical harmonics, forms a " complete s e t " of functions, so that by combining members of the set in various ways one can form any other function which is reasonably well behaved—that is, any function which satisfies the requirements of continuity, etc., for an acceptable wavefunction. Of course, one could always prove mathematically that any particular set of functions forms a complete set; the

204

SCHRODINGER EQUATION II: THREE DIMENSIONS

postulate is a physical one, regarding the completeness of any set of functions which form the eigenfunctions of an operator for a dynamical variable. The statement of the postulate regarding the sum of the squares of the coefficients recognizes the possibility that more than one independent eigenfunction may correspond to the same eigenvalue. Further discussion of the spherical harmonics will illustrate this point. If the angular dependence of the wavefunction is contained in a factor / ( 0 , φ), one can always write / ( 0 , φ) as a superposition of Y with constant coefficients c : Um

lttn

Σ ci. YU0>4>) () = -i Each term in the series (22) has a particular pair of values for the variables L and L , and the probability | c | , with a given value of / and m, is the probability of finding the pair of values /(/ + l)ft and raft, respectively, when one measures L and L . F o r a given /, there are 21 + 1 terms (with ra = /, ra = / — 1 , . . . , ra = — /), and the probability of finding the corresponding value for L , without regard to the value of L , is thus equal to the sum of the squares of the coefficients c in those 21 + 1 terms, or Σ = -ι ki,ml We have interpreted the squares of the coefficients as probabilities by analogy to the fact that the intensity of a wave is the square of its amplitude. This analogy is not optional; it is a strict mathematical necessity, if we are to be consistent with what we have done before, in particular in our treat­ ment of expectation values (Section 5.6). F o r example, we write the expecta­ tion value of L as Λθ,φ)=

Σ

22

m

1= 0

m

2

2

2

i m

2

2

z

2

2

z

ι

Z m

2

η

2

= Γ

f > ( 0 , tf>)L /(0, φ) sin 0 άθ άφ

2

2

o p

whenever / ( 0 , φ) is the angular part of the wavefunction. Substitution of the expansion (22) into this integral yields

<*< >= Σ Σ Γ f * '* ° p «.- >- 2

c

L,ML',M' 0 J

J

y

L

2

c

Y

s i nθ ά θ d

*

0

and performing the indicated operation with the L operator, we obtain 2

< ' >= Σ Σ

Γ

L

.τ * Kl + IK

A **?

L ML\M' 0 J

J

T

2

m

Υι, M sin 0 άθ άφ

0

To evaluate the integral, one makes use of the fact that the Y , like the sine and cosine, are all orthogonal to one another. This means that / > m

Γ

fY* . m

Y

lt m

sin θάθάφ = 0

6.3

MEASUREMENT OF ANGULAR MOMENTUM20 5

unless / ' = / and m' = m. Thus every term of the infinite series vanishes, except the terms for which / ' = / and m' = m. F o r the nonvanishing terms, we use the fact that the Y are normalized, and we finally obtain ltM

(i ) = ZKJ'0 + i)i 2

2

l,m

= ZKi + W

2

2

Σ

k..J

2

I Thus it is necessary to identify the quantitym=Yf-I _ | c | as the probability of obtaining the result L = /(/ + l)ft , because the average value of L is by definition equal t o the sum of the possible results of the measurement, weighted by the probability of each result. We can take advantage of the fact that the Y are orthogonal and nor­ malized, t o find a general formula which gives the coefficients c in the expansion of any given function / ( 0 , φ). The procedure is simply to multiply both sides of Eq. (22) by a particular spherical harmonic Y* >, and then to integrate both sides of the equation over all angles. (This is a perfect analogy to the procedure used in finding the coefficients in a Fourier series (Section 4.3); here we integrate over the surface of a sphere, which is analogous t o integrating over a complete period for the sine and cosine series.) Because of the orthogonality of the Y , all terms except one drop out of the series, and we obtain the formula 2

m=

2

z

/ i W

2

2

ltTn

l%m

tm

/ i m

(23) which enables us to find the coefficient for any given indices /' and rri. Let us now explore further implications of our discussion of the measure­ ment of L and L . We assumed that we had measured L , and we computed the probabilities of various results of a subsequent measurement of L . N o w suppose that we go ahead and measure L . Can we then say that we know both L and L l N o , we cannot, because after the measurement of L , the angular part of the wavefunction becomes one of the functions Yi rather than one of the functions Y / . I t therefore consists of a super­ position of different eigenfunctions of L , so that L is no longer known with certainty. Thus we conclude that the process of measuring L must disturb L ; in part b of this section we shall discuss the physical mechanism which causes this disturbance. In quantum theory one encounters many similar cases in which measure­ ment of one variable disturbs another variable so that one cannot know the value of both variables simultaneously. There is a general rule for knowing whether or not a pair of variables can be measured simultaneously: If it is possible t o measure one variable without disturbing the value of another 2

x

z

x

x

z

x

x

ttnx

>w

z

z

x

z

206

SCHRODINGER EQUATION II: THREE DIMENSIONS

variable, the operators O commute: that is,

and O

a

representing the two variables must

b

oov=oov a

b

b

a

for any wavefunction Ψ, where we assume that the operator on the right is applied to Ψ first, and the other operator is then applied to the result. To prove this we note first that, if the value of one variable is to be unchanged when a second variable is measured, it must be possible to expand any eigen­ function of the first variable in a series of eigenfunctions of both variables. Since any wavefunction may be written as a series of eigenfunctions of the first variable, and each of the latter wavefunctions may be written as a series of eigenfunctions of both variables, we see that any wavefunction may be written as a series of eigenfunctions of both variables. Suppose that T is one such eigenfunction: 6

t i

ΟΨ

υ

=

u

= O bjV

α

αΨ ι

υ

and Then OO ¥ x

a

b

a

lJ

=

ab W

=

ab ¥

l

J

iJ

and O Oa ¥ =--O a W x

b

ij

b

l

lJ

x

i

j

ij

so that Ο Ο„Ψ = Ο Ο Ψ that is O and O commute when applied to any of the eigenfunctions But in that case, these operators must commute when applied to any wavefunction Ψ, for we may expand Ψ as α

a

υ

ι>

α

υ

b

T

J

so that Ο Ο,Ψ α

=

2Σ^α ^Ψ ι

υ

=

Ο ΟΨ ι>

α

ι J

It is also possible to prove that the commutation of these operators is a sufficient as well as a necessary condition for the corresponding variables to be simul­ taneously measurable. An example of a pair of commuting operators is the energy operator Η for a spherically symmetric potential and the operator L . Η may be written as (Section 6.2) 2

2t 2m

2 2mr

Suppose that O Ψ = ΑΨ, where A is a constant, so that Ψ is an eigenfunction of variable α. Since any function can be expanded in eigenfunctions of variable b, we may write Ψ = Χ ο Ψ where the Ψ are eigenfunctions of variable b. After we make a measurement of variable b, the wavefunction of the system must be one of the Ψ . But if we can make the measurement without disturbing the value of variable a, then this wavefunction (one of the Ψ ) must be an eigenfunction of variable a as well as variable b. Thus for this type of measurement, all of the Ψ are eigenfunctions of a as well as b. 6

a

η

η 6

η 6

η 6

η1>

ηΰ

6.3

MEASUREMENT OF ANGULAR MOMENTUM20

7

L clearly commutes with this operator, because L , being a function of θ and φ only, must commute with p , which is a function of r only. Therefore both the energy and the total angular momentum may be known simultaneously for a particle in a spherically symmetric potential (as we have already assumed). On the other hand, the operators L and L clearly do not commute. In fact, you may verify from Eqs. (13) that 2

2

2

x

LL x

z

=

z

L L -ihL z

x

y

b.Experiment . Direct evidence of quantization of angular momentum was obtained in 1922 by Stern and Gerlach. They made use of the fact that a circulating electron in an atom forms a tiny loop of electric current, upon which a magnetic field can exert a force. The influence of a magnetic field upon a current loop is most simply described in terms of the magnetic dipole moment of the loop, defined as a vectorμ perpendicular to the plane of the loop, whose magnitude is equal (in mks) to I A, where / is the current and A is the area of the loop. (A magnetic dipole so defined behaves in a magnetic fieldΒ just as an electric dipole behaves in an electric field; that is, there is a torque on the dipole equal toμ χ Β , which tends to alignμ parallel to B , and the potential energy of this dipole is equal to — μ · Β . For further details see any text on electricity and magnetism. ) From this definition one can easily show that an orbiting electron has a momentμ which is proportional to its orbital angular momentum L , and in mks unitsμ = —eL/2m. Therefore, if the values of a component ofL are quantized, the same must be true for a component ofμ , and the experiment of Stern and Gerlach, which showed that μ is quantized, is a verification of the quantization of L . Stern and Gerlach saw the effect of the momentμ by passing a beam of atoms through an inhomogeneous magnetic field (see Fig. 6). In such a field a magnetic dipole feels a net force whose ζ component is 7

8

9

ζ

z

dB

z

(Fig. 6(b) indicates the origin of this net force.) Therefore each atom of the beam was deflected in passing through the magnet; the deflection of the beam was observed by catching the atoms on a photographic plate. The atoms were O. Stern and W. Gerlach, Z. Phys. 8, 110; 9, 349 (1922). For example, Chapter 10 of Ε. M. Purcell, Electricity and Magnetism," Vol. 2 of the Berkeley Physics Course. McGraw-Hill, New York, 1965. In the actual experiment of Stern and Gerlach, the angular momentum resulted from internal spin of the electron rather than from its orbital motion; but the general idea is the same, because all kinds of angular momentum are quantized. We shall discuss spin in Sec­ tion 7.3. 7

8

9

44

208

SCHRODINGER EQUATION II: THREE DIMENSIONS

Β

Β

(b)

Fig. 6. (a) Schematic view of Stern-Gerlach apparatus. Atomic beam travels along sharp pole piece, which produces a field stronger near the north pole (N). As a result, field lines converge and are not exactly vertical, (b) Edgewise view of a current loop in theΒ field of (a); current is into paper at right. Because theΒ field lines are not parallel, the forces F on opposite sides of the loop are not in opposite directions; each has a downward component, so there is a net downward force on the loop.

found in distinct clusters, indicating that the angular momentum possessed only discrete values rather than the continuous range of values which one would expect classically. Let us now try to understand, on the basis of the mechanism of the SternGerlach experiment, why we cannot measure one component ofL without disturbing the other components. Suppose we do two consecutive SternGerlach experiments, one with theΒ field along the ζ axis, and a second one with theΒ field along the χ axis. Let L = 2h (that is, / = 1) throughout the experiments. As a result of passing through the B field, the beam splits into three beams, which can be identified with the values L — -\-h, L = 0, and L = —h, respectively. We can split off the beam in which L = H-ft and send it through the B field, to measure the value of L for each atom of this beam. We know from the preceding theoretical discussion that the beam should split into three components, with one-fourth of the atoms having L = +ft, one-half having L = 0, and one-fourth having L = —ft. Consider the atoms for which L = Η-ft (and for which we already found L to be +ft). Theory tells us that L is no longer equal to +ft for all of these atoms. What happened to change L while we were measuring L l We can 2

2

z

z

z

z

z

x

x

x

x

x

x

z

z

z

x

6.3

MEASUREMENT OF ANGULAR MOMENTUM

209

understand classically that the measurement of L must change L to some extent. The magnetic fieldΒ causes a torque equal toμ χ Β on each atom, and by classical mechanics x

2

torque = dL/dt Sinceμ =

—eL/2m,

In the measurement of L ,Β is in the χ direction. If we assume thatL initially has components L = +h and L = + then x

x

z

wherej is a unit vector in the y direction. ThusL In a small time St, the change inL is

acquires a y component.

As time goes on, the vectorL precesses about the χ axis; the tip of theL vector describes a circle about the χ axis. In time t, the angle through which the tip ofL moves is \SL/L \ = (e/2m)B t (see Fig. 7). The angular velocity of the tip of theL vector is ω = eBJlm, as this point moves on a circle centered on the χ axis. X

x

Ζ

5L

y

X

Fig. 7.

Precession of the tip of the L vector in time δί.

210

SCHRODINGER EQUATION II: THREE DIMENSIONS

This discussion, assuming as it does that L and L are known initially, is entirely based on classical theory. It shows that measurement of L does disturb L , even in classical theory, for as L precesses, L must change. But according to classical theory, we still retain knowledge of L after the measure­ ment of L , because the amount of precession, and hence the amount of change in L , may be precisely calculated from the knowledge of the initial conditions. On the other hand, when we consider the limitations imposed by the un­ certainty principle, we find that we cannot design a Stern-Gerlach experiment to measure L in such a way that we can predict the resulting change in L . Let us examine this statement closely. In order to predict the change in L , we must know the field B which acts on each atom. But because of the presence of the field gradient dBJdx, and because we cannot know the JC coordinate of each atom precisely, we do not know B for each atom. Thus each atom's angular momentum vector pre­ cesses by an uncertain amount, and we lose our knowledge of L . It remains to be seen that the precession is sufficiently great that we are justified in saying that our knowledge of L is completely lost. Let us make a rough calculation of the uncertainty in the precession angle for the above case, in which L was originally + ft, and L is 2ft . As mentioned above, measurement of L splits the beam into three p a r t s ; the L = 0 part is undeflected, while the L = + ft part is deflected, acquiring a momentum of p = F t = μ (δΒ /6χ)ί, where / is the time required to traverse the magnetic field. If this component is to be distinguishable from the L = 0 component, p must be greater than twice the uncertainty in p demanded by the uncer­ tainty principle. Thus, if the beam initially had a spread of ±δχ in its χ co­ ordinate, x

z

x

z

z

z

x

z

x

z

z

x

x

z

2

2

2

z

x

x

x

x

x

χ

χ

x

x

x

Px > 2δΡχ=

N$x

so that p ^tSx>h

(24)

x

The product (3Β /δχ)δχ is the uncertainty in the value of B for an atom in the beam, the uncertainty resulting from the spread of the beam over an area in which the field is not uniform. There must be a corresponding uncertainty in the rate of precession of L . Since ω = eB /2m, χ

x

x

δω =

eJBx

2m 2m

dx

δχ

6.4

THE THREE-DIMENSIONAL HARMONIC OSCILLATOR

211

The angle of precession is then uncertain by an amount

But from Eq. (24), with μ

= (eh/2m),

χ

we have

The precise value of 1 rad should not be taken too seriously, because it resulted from a somewhat arbitrary assumption about the value of p re­ quired to separate the beams, and we have not justified our classical approach to the measurement process. But we do see that the basic uncertainty principle for position and momentum, combined with classical laws, predicts an uncertainty in the direction ofL which justified our statement that the knowledge of L is destroyed by measurement of L . x

z

6.4

x

AN EXAMPLE: HARMONIC

THE THREE-DIMENSIONAL OSCILLATOR

The potential energy of a spherically symmetric three-dimensional harmonic oscillator may be written r)= \Kr

2

= \γηω\χ +

y+

2

z)

2

2

The time-independent Schrodinger equation is therefore Γ

h

2

Id

2

d\

2

(a?

which may. " be2 ^written

d

+

1

2

+

T?

a?)

+

2

2 /

m

(

°

(

2

x+

(H + H + H )u = Eu x

where

y

z

2

y

+

2

z

)

/|

\

u=

E

u

(25)

212

SCHRODINGER EQUATION II: THREE DIMENSIONS

H , H , and H are simply the energy operators (or Hamiltonian operators) for one-dimensional oscillation along the respective axes. We know, there­ fore, from the result of Section 5.7, that x

y

2

H u (x) x

= (q +±)ha>u (x)

q

q

where q is an integer and u (x) is one of the eigenfunctions found in Section 5.7: q

u (x) =(J^-

αχ^%- '

(a = mcolh)

αχ2 2

q

(26)

and a similar equation holds for the variables y and z. We can construct an eigenfunction of Eq. (25) simply by taking the product of the eigenfunctions of H , H , and H : x

y

z

u (x,y,z)

= u (x)u (y)u (z)

qst

q

s

(27)

t

where q, s, and t are integers, and the functions on the right are defined in Eq. (26). Application of the energy operator then yields (H + H + H )u (x, x

y

z

qs(

>>, z) = ( ? + j + s + i + i+\)fi^u (x qst

y, z)

y

(28)

so that u (x,y,z) is indeed an eigenfunction of the energy operator, and the energy levels are QSt

E = (n+ i)tw

(n = q + s + t)

n

where η is any positive integer, or zero. There are three linearly independent eigenfunctions for η = 1: w , w , and w . We see here an example of degeneracy, independent eigenfunctions with the same eigenvalue are said to be degenerate. Level E is sixfold de­ generate, for there are six eigenfunctions for η = 2 : w oo> w > t4 , u , w , and w . Level E is tenfold degenerate, for there are ten ways to combine the integers q, r, and s so that they total three. It is interesting to examine the angular momentum of these eigenfunctions. T o do this, we examine the angular dependence of each function and try to express it as a combination of spherical harmonics; according to postulate 3, this should be possible. We start with η = 0; there is one eigenfunction with energy eigenvalue E = %hcoi 1 0 0

0 1 0

001

2

2

1 0 1

011

020

002

l i 0

3

0

"ooo

= o(x)u (y)uo{z) u

0

=

e~

ar2/2

This eigenfunction is spherically symmetrical, so its angular dependence, a constant, can be expressed as the single spherical harmonic Y , with total angular momentum of zero. (Clearly any spherically symmetrical wavefunc­ tion has total angular momentum zero, because the L operator yields zero when it operates on a spherically symmetrical function.) 0 t 0

2

6.4

THE THREE-DIMENSIONAL HARMONIC OSCILLATOR21 3

The three eigenfunctions for η = 1 may be written, according to Eqs. (26) and (27), as " l o o=

«ι(χ )« ω«ο(ζ)=

( £"

0

A X Y - ' ^ E - ^ E - "

=

2

'

2

-2axe-°

r2/2

« ο ί ο=

-2aye-" '

« o o i=

-2aze-"

2 2

212

The angular dependence of these wavefunctions is all contained in the respective factors x, y, and z. But E q s . (19) show that χ may be written in terms of the sum of the two spherical harmonies Y , and Y, _, because t

so that r /8π\ / * - ( ) J

5

2

+

T

and where Ν is a normalization constant. Therefore w is an eigenfunction of L , with / = 1, composed of equal parts of m = + 1 and m = — 1; a measure­ ment of L for a particle whose wavefunction is w would yield either L = +ft or L = —ft with equal probability. In a similar way, we find that 2

1 0 0

z

z

1 0 0

r

«bio=/(rXyi

B

i-yi.-i )

and Woo i = ^ (

r

)^i,o

where the functions F(r) and g(r) contain the appropriate normalization constants as well as the factors r and E ~ . Thus the three linearly inde­ pendent eigenfunctions of the energy operator for η = 1 are all eigenfunctions of L , with / = 1, b u t only one of them, w , is also an eigenfunction of L ; the other two are mixed states, as far as L is concerned. (It should be clear that u is an eigenfunction of L and w is an eigenfunction of L .) It is often more convenient to work with functions which are eigenfunctions of L . W e can construct linear combinations of u , u , and w oi which are linearly independent and are eigenfunctions of L , as follows: A

R

2

/

2

2

0 0 1

z

z

100

x

0 1 0

y

z

l 0 0

0 1 0

0

z

"α =

w

i o o + "Όιο

u

= w oi

u

= u

b

c

0

100

-

/wqio

(« = 1,

/ = 1,

m =

(n=l,

I = 1,

m = 0)

/ =

m

(« =

1,

1,

=

+1) -1)

214

SCHRODINGER EQUATION II: THREE DIMENSIONS

The functions u , u , and u are simultaneous eigenfunctions of energy, of L , and of L , with eigenvalues indicated by the quantum numbers given above. Of course, they all have the same η and / quantum numbers because these numbers are characteristic of the three original functions u , w , and w . The factor / comes in with w because Y ±i contains the factor χ ± iy. It becomes a little more interesting if we go through the same procedure for η = 2. The six linearly independent eigenfunctions for η = 2 contain quadratic terms such as x , y , z , xy xz, yz, as well as terms which are spherically symmetrical. In order to construct six independent functions as superpositions of spherical harmonics, we need six spherical harmonics. We cannot use any with / > 2, for these contain cubic or higher order terms. Neither can we use / = 1, for these contain only linear terms, which are absent from the η = 2 eigenfunctions. This leaves us with the five functions for 1 = 2 plus the function for / = 0. Thus when η = 2, L may be either 2(2 + \)h = 6ft , or zero, and L may be +2ft, h, 0, - f t , or — 2h. Just as we did for η = 1, one may construct new energy eigenfunctions which are eigen­ functions of L and L , by using appropriate linear combinations of the six functions u , u , u , w , w , and w . Similarly, for η = 3, the ten eigenfunctions are linear combinations of the seven spherical harmonics for 1=3 and the three for / = 1; for η = 4 there are fifteen eigenfunctions, which are linear combinations of the nine spherical harmonics for / = 4, the five for 1=2, and the one for / = 0, etc. 2

a

b

c

z

l 0 0

0 1 0

2

010

001

it

2

2

9

2

2

2

2

2

2

200

6.5

020

THE RADIAL

002

110

101

011

EQUATION

We have not yet dealt with the radial part of Eq. (8). Let us eliminate the angular part of that equation by substituting the eigenvalue /(/ + 1) for the operator Ω. We obtain - £ ^ ( r * )+

/(/+l )=

^ [ E - V W l

(29)

Consider first the case / = 0. We may then rewrite Eq. (29) as {rR) = - ψ ΐ Ε -

V(r)](rR)

(30)

Equation (30) is identical to the one-dimensional Schrodinger equation, Chapter 5, Eq. (4), except that the variable is r rather than χ and the eigenfunction is rR(r) rather than u(x). We may use Eq. (30) to find the wavefunction for a free particle with / = 0. We let V(r) = 0 and we obtain d , 2mE, ^ ( r * ) = - - ^ ) 2

x

x

6.5

THE RADIAL EQUATION

215

whose solution is /2m£\ (withfc=(^)

r * = e*»'

1 / 2

\ )

so that the wavefunction, neglecting the normalization factor, is u{r, 0, φ) = Λ ( Γ ) 7 , ( 0 , φ) ο

ο

oce /r

(31)

±ikr

since Υ , (θ φ) is a constant. If we multiply w(r, θ, φ) by the time dependent factor e~ \ we see that the solution e* /r represents a wave traveling outward from the origin (in the positive r direction), and the solution e' /r represents a wave traveling inward toward the origin. To obtain a solution that is acceptable in a region that includes the origin, we must combine these waves to form a standing wave (see Sec. 5.2) of the form (sin kr)/r. Equation (30) may also be used to find the spherically symmetric (/ = 0) solutions for any spherically symmetric potential energy. For example, if we let V(r) be the harmonic oscillator potential : 0 0

9

i0i

ikr

ikr

10

ma> r 2

V(r) =

2

we find that corresponding to each solution of the one-dimensional harmonic oscillator equation (Section 5.7) there is a spherically symmetric solution for the three-dimensional harmonic oscillator. There is only one difference; the lowest eigenvalue is £ ftco in the three-dimensional case, rather than i δ ω . How can this be, when the equations are identical? Suppose we put Ε = \ hu> in Eq. (30), to see what happens. The product rR(r) should then be the same function as the eigenfunction u found for the one-dimensional case; that is, since u (x) = then , n rR(r) = e~ 0

a x 2 / 2

0

e

v

n

ar

m

2

1 2

However, this function is not acceptable for rR(r), because there is another boundary condition, that rR(r) must vanish at the origin. This condition is necessary t o ensure that the expectation values of all dynamical variables remain finite, when they are computed by the rules of Section 5.6. There­ fore we rule out the eigenvalue %ha> for the three dimensional case, in agreement with the result of Section 6.4. (Notice also that the abovementioned functions of the form e /r can only be used in regions of space that exclude the origin.) ±ikr

In this paragraph, unlike the previous paragraph, ω is the classical oscillator frequency, and is not equal to the frequency of the time-dependent part of the wavefunction. 1 0

216

SCHRODINGER EQUATION II: THREE DIMENSIONS

It is easy to see that Eq. (30) gives us the same answer that we found in Section 6.4 for the eigenvalue \\ιω also. In this case the product rR(r) should be the same function as the eigenfunction w for the one-dimensional case, so, since t

u (x)

=

xe-

rR(r) =

re-

l

ax2/2

ar2/2

Therefore R(r) = e~ ' , which is the eigenfunction w we found in Section 6.4 to correspond to an energy of \\ιω. (Remember that w(r, θ, φ) = R(r) because the wavefunction is spherically symmetric.) a 2/2

0 0 0

9

Th e Centrifuga l Potential . If we rewrite Eq. (29) with / Φ 0, we can obtain an equation of the same form as Eq. (30):

Equation (32) is also of the same form as the one-dimensional Schrodinger equation, if we consider the potential energy to be the sum

2mr

2

The additional term is called the centrifugal potential; it is analogous to the fictitious potential energy which appears when one uses a rotating coordinate system in classical mechanics. To see that this is reasonable, notice that this energy is equivalent to l}/2mr . As we saw in Section 6.2, the kinetic energy in spherical coordinates is 2

L

2

2m

2

2mr

so that P V £L = = E~ <
(33)

£

(34)

2

E

V

2

while in one dimension, = E-V{x)

Therefore, to write a radial equation in the form of the one-dimensional equation, it makes sense to replace the right-hand side of Eq. (34) by the right-hand side of Eq. (33), replacing the p operator by the p operator. x

r

6.6

6.6

THE THREE-DIMENSIONAL SQUARE WELL21

THE THREE-DIMENSIONAL

SQUARE

7

WELL

As an example of the application of the radial equation, let us consider the three-dimensional counterpart to the " square w e l l " treated in Section 5.4. The potential is V(r)=+V (r>a) V(r) = 0 (r < a) 0

When / = 0, the radial equation (30) can be written, for r < a, as d —

2mE —

2

( R) r

-k\rR)

where we have again used the definition k = 2mElh . The solution is 2

2

rR(r) = A cos kr + Β sin kr

(35)

or u(r, 0, φ) = R(r) A cos kr

Β sin kr

=+ r r We write u in this form rather than the exponential form, because u must be finite at r = 0, and we can ensure this by making Λ = 0. Therefore , Bsinfcr u(r 0, φ) = (r < a) r Λ

l v

9

Outside the well, the radial equation is

F o r a bound state, (E — V ) is negative, so the solution must be a decaying exponential, as in Section 5.4: 0

rR = Ce~*

r

(r > a)

(36)

with α defined as (2m(V — £ ) / f t ) , as before. N o w we can proceed as in Section 5.4, matching the wavefunction and its derivative at r = a. Because the functions involved are the same, the result must be exactly as in Section 5.4 for the odd parity solution, because rR, the function analogous to u(x), is the odd function sin A t ; the boundary condition that rR = 0 at r = 0 eliminates the even solution. Therefore the energy levels are given by the solutions of Eq. (36), Chapter 5: 2

1/2

0

11

sin ka=

11

(β = k + α ) 2

βα

Se eth e discussio na t th ebotto m o fpag e 215 .

2

2

(37)

218

SCHRODINGER EQUATION II: THREE DIMENSIONS

Because we are restricted to the odd solutions, it is possible that no bound state exists at all, for a shallow well (or a narrow well). (Remember that the lowest energy solution for the one-dimensional case was an even parity state.) The lowest-energy odd solution has - < ka < π 2 and since ka < βα [to satisfy Eq. (37)], we see that if βα < π/2, that is, if 2mV a /h < (π/2) , there is no bound state. There is one bound state if π/2 < βα < 3π/2, two if 3π/2 < βα < 5π/2, and so on. If / Φ 0, the presence of the centrifugal potential makes the radial equation a bit more complicated: 2

2

2

0

or (rR) = - k\rR)

ar

+ ^±-^ r

(rR)

(r < a)

(38)

and - (

r

R

) = - ^ [ E - V

0

^ - ^ ^ \ ( r R )

or £ (rR) = - ( / a ) ( r K ) + dr

(rR)

2

r

z

z

(r > a)

(39)

where k and a are defined as before. With the addition of the centrifugal potential, the effective well is as shown in Fig. 8. The well is effectively narrower, because the particle must stay away from the origin if it is to have some angular momentum. Therefore we might expect fewer bound levels for successively higher values of /, in a given well. The functions R(r) which are solutions of Eq. (38) or Eq. (39) are not hard to work out for a given /; they belong to a class of functions called spherical Bessel functions ji(kr) and spherical N e u m a n n functions n (kr)} 12

3

t

Notice that Eq. (39 ) is identical to Eq. (38) , except that k is replaced by ia. Obviously, differential equations of the form of Eqs. (38 ) and (39 ) were studied and solved long before quantum theory was invented. Professor Bessel died in 1846 . The general formula for ji(x) is ji(x) = ( 7 τ / 2 χ ) . / ! *)(.*), where J + + (x) is the ordinary Bessel function, defined by the infinite series x [ χ i" 12

13

1/2

(

+

it

n

2T(/ z

+1 )1

2· 4 · 6(2, i+

}

2

4

2(2n+ 2 )^ 2 · 4(2n + 2)(2n+ 4 )

^

2)(2n + 4)(2n+ 6 ) ' +

6.6

THE THREE-DIMENSIONAL SQUARE WELL21

/( / + 1)h

9

2

2m r

2

Fig. 8. The effective potential well which ap­ pears in the radial equation, when one adds the centrifugal potential to a square-well potential

(Of course, for Eq. (39) the argument of the functions is iar rather than kr.) For each value of / there are two linearly independent solutions—a spherical Bessel function and a spherical N e u m a n n function. The first three of each are Jo(kr) =

sin kr kr

sin kr Ji(fcr) = (kr)

cos kr kr

2

3 cos kr

[(k-U

-

SMKR

h(kr)=

(kr)

2

c o s kr

kr cos kr

M,(fer)

>

i

k

kr

2

Γ n

sin kr

(kr) r

)

=

3

[ - ^

1 "I

,

3sinfcr

rr\ ° -lkrT

+

C

Skr

We have already seen that j (kr) and n (kr), as given above, are solutions of the radial equation (38) when / = 0, and you may easily verify that the other j and η functions given here are also solutions of Eq. (38) with the appropriate values of /. 0

0

220

SCHRODINGER EQUATION II: THREE DIMENSIONS

The functions j\(kr) are important aside from their application to the special case of a square-well potential. If a particle is completely free and has a definite angular momentum, its radial equation is Eq. (38) for all r, and the radial part of its wavefunction must therefore be ji(kr) for all r. (Consider a —> <*>). Thus, in the expansion of a plane wave in a series of eigenfunctions of L , the coefficients are multiples of the ji(kr) (see Section 6.7, Eq. (49)). 2

Fig. 9. Graphs of jo(kr), j (kr), andj\ (kr). Vertical scale is 4 units per inch for j , and0.4 units per inch for j and y'io . (Plotted from values tabulated in Tables of Spherical Bessel Functions, National Bureau of Stan­ dards, Columbia Univ. Press, New York, 1947.) 5

0

0

5

Figure 9 shows a few of the ji(kr). Notice how the function is " p u s h e d a w a y " from the origin as /increases. In general, for / Φ 0,ji(kr) is maximum at kr just greater than /, and is quite small for kr < I. This is the behavior that one should expect; a particle with momentum ρ = hk and angular momentum \L\ = [/(/ + l ) f i ] « I h should not be found closer to the origin than r « l/k, because 2

1 / 2

|L |= | r χ p |< |r||p |=

ΓΛΛΓ

so Ih < rhk

or

r > Ilk

To find the energy levels for the square well, one proceeds in much the same way as before. The solution for a given / for r < a must be j (kr), because the functions n (kr) go to infinity at r = 0. F o r r > a, we require a solution which goes to zero as r - ^ oo, and we can obtain such a solution from a linear combination of ji(iar) and n (i
t

t

cos(/ar) + ι sin(/ar) = e

i(iar)

=

e~

ar

6.6

THE THREE-DIMENSIONAL SQUARE WELL22

1

w e construc t th e combinatio n h\ \iar)

= j (i
l

t

in^iar)

whic h i s calle d a spherica l Hanke l functio n o f th efirst k i n d .

14

I nparticular ,

and

Thes e function sclearl y hav e th erigh t behavio r a s r -> oo, s o w e us e the m for th e regio n r> a. Thus , fo r example , th e complet e solutio nfo r / = 1 i s R(r) = Aj^kr) (

r<

a)

R(r) = BhYXiar) (

r> a )

wher e A an d Βar earbitrar y constant s t o b e determine d b y th e boundar y condition s an d normalization . W e ma y eliminat e A an d Β b y th erequiremen t that th e rati o (dR/dr)/R b e continuou s a t r= a.Therefor e

Ji(ka) or

hYXiza)

d_ [si n kr _ co sfcrl sin ka co

sfca

(ka) ~ ~

ka

2

d_Γ

_ / J _1

\ 1

(αα)

\αα

/

or 2 si n ka fe a 2

3

2 co sfea si n ka

+ — Γ - 32 —

+

ka

sin ka co (ka) 2

a

sfca ka

1

2 a


1

( 2

1 4

T

2

3

1 (oca)2

oca

Ther e i sa spherica lHanke lfunctio n o fth e secon d kind , H (wit hargumen t IAR)o r Eq . (38 ) (wit hargumen t KR):

oc a

2

\ whic h i s als oa solutio n o f Eq . (39 )

Thisfunctio n i slinearl y independen t o f A ,bu t w e d ono t us ei t becaus e H \IAR) contain sth efacto r , s otha t an y wavefunctio ntha t contain s HI \IAR) woul d blo w u p a s R - » oo. Yo u ma y wis ht o verif y tha t JUICER), whic hi sa superpositio n o f HI \IAR)an d HF \IAR) goe st oinfinit y a s R —• oo. (Se e theasymptoti c expressio nfo r JT(KR) give ni n Sectio n 6.7. ) (,)

I2

T

E

+

A

R

(2

N

2

F

222

SCHRODINGER EQUATION II: THREE DIMENSIONS

After multiplication by -tf, and performing the division on each side, we have -2 +

\_\ka)

ka

J

[aa

\
or cot ka \ka) \ka)

ka

\oca) ' W

oca oca

(40)

Equation (40), together with the condition that (

K

A

Y

+

« A Y

I

=

-

2

^

(4i)

may be solved graphically to find the possible values of k, and hence the energy levels, for / = 1. EXAMPLE PROBLEM 1. for / = 1 when

Use Eq. (40) to show that there is no bound state

Solution. For ka < π cot ka < Ilka. Therefore the left side of Eq. (40) is negative for these values of Ka. But the right side of Eq. (40) is always posi­ tive, so there is no solution for ka < π. But Eq. (41) tells us that 15

2mV a

2

0

2

so that, when ka ^ π—that is, when there is a solution—we have 2mV a

2

0

^

_

2

h

2

or nh 2

Va> 2

0

2

2m Q.E.D.

We can have a solution with ka = π. In that case, cot ka -> oo, cca = 0, and 2mV ft

2

{

2

Remember that α was defined to be a real, positive quantity; if it were not, our solution which goes as e~ would not behave properly at infinity. 1 5

ar

6.7

SCATTERING OF PARTICLES22

3

A second bound state appears when

~ F ~ ~

( 2 π )

Notice that the first bound level for / = 1 appears at a value of V a which is four times the value required in order that there be a bound level for / = 0. 2

0

6.7

SCATTERING SPHERICALL

OF PARTICLES Y SYMMETRIC

FROM A POTENTIAL

To conclude our general discussion of three-dimensional problems, we consider the three-dimensional counterpart of the problem of transmission of particles past a potential barrier (Section 5.5). The problem, of course, has a completely different character in three dimensions, for the particles which strike the barrier (or well) can go off in any direction, instead of simply being transmitted or reflected. Suppose that we send a " b e a m " of particles—a plane wave of the f o r m Ψ = Ae —along the ζ axis toward a " scattering center," and suppose that the potential energy is nonzero over a limited region, r < a, surrounding the scattering center (see Fig. 10). The density of particles in the beam is ΙΨ-nJ = l ^ | , and the intensity of the beam—the number of particles cross­ ing a unit area in a unit of time—is, as usual, the product of particle density and particle velocity, or v\A\ . 16

ikz

ιηο

2

2

2

Fig. 10. Scattering of a plane wave from a scat­ tering center, resulting in a spherical scattered wave. The interaction which produces the scattered wave occurs only in the region r
i(at

in writing the

224

SCHRODINGER EQUATION II: THREE DIMENSIONS

A certain fraction of the particles will be scattered, forming a wave which emanates from the scattering center, while the remainder continue on as the plane wave. At large r, the scattering potential and the centrifugal potential both go to zero, so we know from Eq. (31) that the radial part of the scattered wave is e Ir. (We use e rather than e^ , because we want a wave which is traveling outward, toward positive r.) Therefore we write the scattered wave as ψ = Α/(θ, 4)e /r, where the factor A expresses the fact that the scattered wave amplitude is proportional to the incident wave amplitude. The particle density in the scattered wave is \A\ \f(0, (j))\ /r , and the intensity of the scattered wave is v\A\ \f(0^)\ /r , in the region of large r. M k r

+ i k r

kr

ikr

8<5

2

2

2

2

2

2

Scatterin g Cros s Section . Given a scattering potential, we shall see that it is straightforward, although perhaps tedious, to compute / ( 0 , φ) and hence to determine the scattered wavefunction. In practice, one has the more difficult task of deducing the scattering potential after measuring | / ( 0 , φ) \ . To measure | / ( 0 , φ)\ , one places a detector at a point Ρ (Fig. 10) where it can detect particles in the scattered wave but not those in the incident w a v e . The number N of particles seen by the detector per unit time is equal to the prod­ uct of the intensity of the scattered wave and the area dA of the detector, so 2

2

17

d

\v\A\

| / ( 0 , φ)\ ~*

2

2

dA

But dA/r is just the solid angle άΩ subtended by the detector at the scatter­ ing center, and v\A\ is the intensity / of the incident beam, so 2

2

i n c

N = 1/(0, Φ Λ Ι)ηο di2 (άΩ = sin 0 άθ άφ) 2

d

or where the differential da is defined as ^

_ N 7

d

N u m b e r of particles/sec striking detector

i n c

N u m b e r of particles/sec-cm in incident beam

^

2

The ratio άσ/άΩ, which has the dimensions of an area, is called the differential cross section for scattering from the scattering center. Its integral over a sphere is the total scattering cross section a: a = jda

=

άΩ =

fj f

| / ( 0 , φ)\ sin 0 άθ άφ 2

The incident beam cannot be a pure plane wave, because it has a finite width, but we assume that the width is much greater than the wavelength, so that we do not have to worry about diffraction effects. The detector is placed in the region where the incident wave is zero. 1 7

6.7

SCATTERING O F PARTICLES

225

Thus by counting particles, one can determine | / ( 0 , φ)\ experimentally for all values of 0 and φ. The problem then is to fit the experimental results to a theoretical determination of / ( 0 , φ) on the basis of an assumed potential function. We shall see one way to go about this in this section. There is a physical significance to the fact that σ has the dimensions of an area. If the scattering were classical—for example, that of a uniform stream of bullets bouncing off a piece of armor plate—the integral of da as defined by Eq. (42) would be just the cross sectional area of the plate; that is, the total number of bullets scattered per second would be the product of the area of the plate and the number per unit area per second striking the plate (assuming the plate's area to be smaller than the cross sectional area of the stream of bullets). In quantum theory, however, the cross section, although still an area, is not directly given by the area covered by the scattering potential, because diffraction effects complicate the situation. 2

Partial Waves. So far the discussion has been quite general. But this is not the place for a full exposition of scattering t h e o r y ; we simply wish to show one treatment as an application of the concepts of angular momentum and the solutions of the Schrodinger equation which have been developed in this chapter. Because we assume the scattering potential to be spherically symmetric, we can simplify the analysis in two ways. First, there is symmetry with respect to rotation about the ζ axis, so the φ dependence disappears, and / ( 0 , φ) becomes / ( 0 ) . Second, angular momentum is conserved, so that we can de­ compose the incoming wave into components (partial waves) with different /-values, and can then calculate the scattering of each component separately. We begin by writing the complete wavefunction in the region of large r: 18

Ψ -+ A je

ifcz

+ /(0)

(r -+ oo)

(43)

This is to be compared with the actual solution of the Schrodinger equation, which in general is a linear combination of spherical harmonics, each multi­ plied by a radial factor R (r). Since there is no φ dependence, we have t

Ψ = Σ * / * * Μ Λ ( ο ο 8 0)

( 4 4 )

P,(cos 0) is identical to Pj°(cos 0) and is called the Legendre polynomial of order /. We determine the scattering amplitude J ( 0 ) by matching expression (43) to the limit of expression (44) as r goes to infinity. In order to find this limit, 19

Fo r a mor e complet e discussio n o fscatterin g theory , se ean y standar d tex t o nquantu m mechanics , for example , L .I .Schiff ,*'Quantu m Mechanics, "3r d Edition . McGraw-Hill , Ne w York , 1968 . Excep t fo r a normalizin gfactor . Se e Eq . (48 )an d Sectio n 6.2 . 1 8

1 9

226

SCHRODINGER EQUATION II: THREE DIMENSIONS

we make use of the fact that the scattering potential V(r) is zero at large r (r > a), so that Ri is a linear combination of solutions of Eq. (37), namely R (r) = cos SJtikr)

— sin d^^kr)

t

(45)

(r > a)

The sum of the squares of the coefficients in this combination must be 1, so that Ri(r) will be normalized (as j and n are). Hence the two constant coefficients are not independent and may be written as we have done, expressing the fact that for r > a the solution for a given / value is completely specified by the single parameter δ*, called the phase shift of the /th partial wave. The name 'phase shift" is appropriate for reasons which will become clear in a moment. We may now find the limit of Ri(r) as r — »oo , because the limits of the functions ji(kr) and tii(kr) are known to be given by t

t

4

singer -

^

kr

(r —• co)

°i - i)

c n,(kr)-> - ·

\

kr l kr

so that s i n

R,(r) -> cos <5,

( "T) k r

kr

. ,

+ sin δ

ν

c o s

( -T) k r

kr

which reduces to

Λ |

( ) Γ

sin {kr — y + ^ ^

(r - oo)

(46)

The quantity δ thus appears as a phase factor in the asymptotic limit of Ri(r). This phase factor is a convenient measure of the effect of the scattering potential on the wavefunction; if the scattering potential is zero, then δ = 0, because if V(r) = 0 for all r, the same solution R (r) must hold all the way in to r = 0, and thus R (r) cannot contain the function n^kr), which goes to infinity at r = 0. The effect of a potential on the /th partial wave at large distances is a shift in phase by the amount <5/.We shall see that there is a simple expression for the scattering cross section σ in terms of the various phase shifts δ . ι

ι

t

t

;

6.7

SCATTERING OF PARTICLES

227

Now we can equate expression (43) to the limit of expression (44) as r— > oo , obtaining 20

s i n g e r +

JKR

e

f(9)—

ikz

+

=

Σ * i / ( p

c

o

s

(> 47

——r — 1

β)

1

r kr In order to solve this equation for / ( # ) , we must find the coefficients b and to do that we expand the left side in a series of Legendre polynomials, in order to match the two sides term by term. We know that this expansion is possible; the angular dependence of any wavefunction may be expressed as a series of spherical harmonics, which reduce to the Legendre polynomials when m = 0 (and m is zero in this case because the function e = e has no φ dependence). Thus we write u

ikz

i k r c o s e

e =

ffliW^iicosfl) /=o The coefficients a (r) may be evaluated in the same way as the coefficients in a Fourier series, by using the fact that ikz

t

P,(cos fl)P (cos Θ) djcos Θ) = j r

2

(48)

^27TT

J se=-i CO

( ί

~

}

The result is e

ikz

= f (2/ + l)i'Mkr)PtcQs

Θ)

(49)

1= 0

Substitution of this expression for e

into Eq. (47) yields

ikz

ji -i) kr

Σ (2/ + (=o

= Σ b, —

v

•

ln\ IKR

E

~ ' P/(cos Θ) fcr

r

1

— r

Λ(α> 8 θ)

where we have replaced the function ji(kr) by its limit as r - > oo. If we re­ write the sine functions in exponential form [using the identity sin χ = (e — e ~ ) / 2 / ] , the last equation becomes ix

ix

MlikfW)

+ V (2/ + l y e - ^ P X c o s 0)) - e "

I

i=

= e

ikr

f 1 =0

f ( 2 / + l ^ V ^ P ^ c o s 0)

J

0

e " " " ' V ' P i i c o s 0) - e "

i f c r

ifcr

f *>/ e

i , n /

/=0

V

i 5 ,

P

z

(cos 0)

1=0

Th econstan t / I o f Eq . (43 ) i s n o longe r needed , becaus ei t ca n b e "absorbed "int oth econstan t coefficients b .

2 0

t

228

SCHRODINGER EQUATION II: THREE DIMENSIONS

The constants b are found by equating the coefficients of e~ sides, obtaining

ikr

t

b = (21 +

l)i e l

x

iSl

We may then find f(9) by equating the coefficients of /(Θ) = (2iky Σ

on the two

e: ikr

(2/ - \)(e *>- l )P (cos Θ)

1

2i

(

or /(»)

= lΣ

(

+

2 ί

Ι)**

s

i

n

«

ι *ι (<*> 8

0

)

Therefore

= (p)lI(2/+l> sin<5 P (cos0)| W l

/

2

i

(50)

and σ =

— dSl (21 + 1) sin <5 P^cos Θ) sin θ dfj Z

the cross products in the summation having vanished because of the orthogonal­ ity of the Pj(cos Θ) [Eq. (48)]. Equation (48) may now be used to evaluate the integrals, with the final result that

(51) Κ

1= 0

Thus we have reduced the problem of finding σ to one of finding the phase shifts produced by the potential in the various angular momentum compo­ nents of the incident plane wave. This is a remarkable simplification, but it appears that we are still faced with a formidable problem, because Eq. (51) contains an infinite series. To analyze a scattering experiment, we must assume a scattering potential V(r), use it to calculate all of the δ use Eq. (51) to find σ, and then compare this with the observed σ. We can repeat the process using several different assumed potentials, to see which one gives the best fit to the data. But if we have to compute an infinite number of phase shifts in order to use Eq. (51), there is not much point in commencing the calcu­ lation. ΐ9

6.7

SCATTERING OF PARTICLES22

9

There are many situations, fortunately, in which only one or two of the phase shifts are nonzero. T o understand this fact, we may visualize the de­ composition of the incident plane wave into partial waves as follows (Fig. 11):

Fig. 11. Illustration of the decomposition of a beam of particles into separate annular beams, each with a definite angular momentum about the origin. Particles passing through the shaded area have angular momentum between Ih and (/ + l)h.

The plane wave is a stream of particles of m o m e n t u m ρ = hk moving parallel to the ζ axis. If the particles were localized, the angular momentum of a given particle about the origin would be \L\ = pd = hkd, where d is the dis­ tance of that particle from the ζ axis. Thus in this semiclassical view, particles with angular m o m e n t u m between Ih and (/ + \)h pass through an annulus whose radius lies between l/k and (/ + l)/k. [Notice that the number of particles passing through such an annulus is proportional to its area, which is proportional to ( 2 / + 1); since these particles contribute to the total cross section in proportion to their number, we see why there is a (2/ + 1) factor in each term of the series in Eq. (51).] If Ilk is greater than the radius a at which the scattering potential becomes zero, these particles cannot be influ­ enced by the potential, and the corresponding phase shift b must be zero. 9

t

230

SCHRODINGER EQUATION II: THREE DIMENSIONS

So if ka < 1, even nonlocalized particles with / = 1 are but slightly influenced by the potential, and σ is given to good accuracy by the first term in Eq. (52): 4π σ = ρ sin δ

(ka < 1)

2

0

(52)

Of course, if one is trying to find out something about the shape of the scattering potential, one does not use incident particle energies so low that ka < 1. Obviously one cannot measure the shape of something by doing an experiment whose result is determined by only one parameter; a physician cannot determine the condition of a patient simply by weighing h i m . T o put it another way, a particle whose de Broglie wavelength is A cannot respond t o details of dimensions much less than λ in the scattering potential. So it is often necessary to analyze experiments in which several phase shifts are involved, in order to explore details of various potentials. Study of Fig. 12 may give you a better idea of the connection between the scattering potential and the phase shifts. Because the radial equation is mathematically identical to the one-dimensional Schrodinger equation, we can sketch the function rR (r) which is mathematically analogous to u(x) guided by the same principles we used in Section 5.4 in discussing Fig. 3 of Chapter 5. In the region where the effective potential—the scattering potential plus the centrifugal potential—is greater than the particle's total energy £ , the function rR (r) curves away from the axis. There is a point of inflection at r = r , where Ε = 1(1+ \)h /2mr + V(r) ; for r > r , the total energy is greater than the effective potential, and the function curves toward the axis, eventually becoming sinusoidal as / · - • oo. The region r < r is the classically forbidden region, and the wavefunction does not amount to much in this region. Therefore if the scattering potential cuts off at a value of r much less than r = r , it can have very little influence on the wavefunction, and the phase shift must be zero for the particular / value in question. The present argument gives the same criterion as the semiclassical view illustrated in Fig. 11, for determining which partial waves contribute to the cross section. We may find r by setting K equal to £ , obtaining 21

t

9

9

t

2

2

c

c

c

c

c

eff

1(1+

l)fe

2

2mr

2

2

2m

2

or

hk

ικι+»ν _.ι ΐ2

r

k

~i

Sometimes the physician can tell that you are " o u t of shape," but he must at least know your height as well as your weight. 2 1

6.7

SCATTERING O F PARTICLES 231

(a)

(b)

Fig. 12. Connection between scattering potential and phase shift, (a) Total energy E, centrifugal potential V t, and effective potential V ff (dotted line); V f=V + K(r), where V(r) is the scattering poten­ tial, which is negative in this case, (b) The product rR (r) versus r, for the case when V(r) = 0. In this case Ri(r) = j (kr). (c) The product rR (r) versus r for the effective potential shown in (a). At larger r, this func­ tion has the same form as the function plotted in (b), but it is shifted toward smaller r, because the whole curve has been "pulled in " toward the origin by the attractive potential, the point of inflection having moved from r , where Ε = V , in to r , where Ε = V . (Shift shown is &t/k rather than δ| itself, because the functions are plotted versus r rather than kr.) The phase shift is positive, by definition, in this case; remember that the asymptotic form of Ri(kr) was written (sin (kr -(In/2) + Sft/kr.If V(r) were posi­ tive, the curve would be pushed away from the origin, and the phase shift δ, would be negative. cen

e

ef

cent

t

t

t

c

cent

a

eff

232

SCHRODINGER EQUATION II: THREE DIMENSIONS

In order for a given partial wave to contribute, r for that wave must be not much larger than the radius a at which V(r) cuts off. That is, l/k <, a or / ;$ ka for a wave of a given / value to be able to contribute appreciably t o the cross section. (Also see the discussion associated with Fig. 9.) c

EXAMPLE PROBLEM 2. Find σ and do/dQ for the scattering of particles from a "perfectly r i g i d " sphere (infinitely repulsive potential) of radius a, for the case ka <^ 1. Solution.

The potential energy is V(r) = + o o V(r) = 0

(r < a) (r > a)

Therefore the wavefunction must vanish at r = a, just as it did in the onedimensional square well with rigid walls. For r > a, the radial function must be a combination of j (kr) and n (kr): 0

0

R(r) =

sin(kr+

δ) 0

kr

because none of the partial waves for / > 0 should be expected to contribute to the scattering in this case. We find the phase shift <5 by using the boundary condition that R(r) = 0 at r = a: 0

sin(fca + δ ) 0

ka so that ka + δ ο = 0 and δ ο = -ka. 47i

= 0

Equation (52) gives the cross section: An π ρ (-ka)

a = ρ un (-ka) 2

2

= Απα

2

Because / = 0, the scattered wave is isotropic. Therefore άσ dh

=

σ ~An

2 =

a

Notice that σ is just four times the geometrical cross section which the sphere presents to the beam. It is instructive to see how small ka must be so we can safely neglect the / = 1 wave. For this wave, the radial function is [from Eq. (45)] R (r) = cos SJ^kr) x

- sin ^ n (kr) x

(r ^ a)

6.7

SCATTERING OF PARTICLES

233

Setting R(a) equal to zero, we have tan δ

=

χ

n^ka) sin(fca) — (ka) cos(fca) - c o s ( k a ) - (ka) sin(fca) Using the series expansions of sine and cosine, we obtain (ka)

\

3

tan δ

, L

(ka)

2

=

χ

-{i-V 'r r "~ + e

t e

e )

+

which becomes

tan^ =--γν

when we retain only the lowest-order terms in ka in numerator and denominator. Using Eq. (51) to find σ , we have a = ρ ( s i n <5 + 3 s i n ί 0 » 4 π α j l + ^ y ^ j 2

2

2

0

and the error resulting from neglect of the / = 1 term is seen to be about (ka)*/3 χ 100 percent. The effect of the / = 1 term shows up at smaller k values in the differential cross section, which is given by Eq. (50) as

= JL |e«a

αΩ

k

i 8

S

n

+ 3^ .

Q

s

= ^ {sin <5 + 3e ~* 2

i(di

3 e

-i(*i-*>)

s i n

δ ι

s

n

C

os 0|

2

sin ^ sin δ cos 0

o)

0

+

i ^

0

i 5 n

o s θ + 9 s i n δ c o s 0} 2

o C

2

ί

= ~ {sin 5 + 6 sin <5 sin <5 c o s ^ — δ ) cos 0 + 9 s i n <5 c o s 0} A: 2

2

0

X

0

0

2

X

With the substitutions sin δ = — ka, cos <5 = 1, sin δ = tan ^ and cos <5j = 1, we have 3 0

0

χ

^ 1 L , „ (ka) _ (/"0 Λ s

-

6 Λ 2 {(*»)» + tta L - L cos 0 + 9 L _ c o 2/ s 0j 2

p

» a { l + 2(/ca) cos 0} 2

2

=—(ka) /3, 3

234

SCHRODINGER EQUATION II: THREE DIMENSIONS

Thus at 0 = 0 or 180°, the percent error in do/dQ caused by neglecting the / = 1 term is 2(ka) χ 100 percent—considerably larger than the error in σ. If we begin to bombard a scattering center with low energy particles, and then increase the energy, we see a difference between the forward and back­ ward differential cross sections much sooner than we see the effect of the / = 1 contribution to the total cross section. 2

PROBLEMS 1.

Evaluate the three lowest energy levels for an electron in a " b o x " whose dimensions are 3.0 Ax 3.0 Ax 4.0 A. Give the degeneracy of each level.

2.

Verify by direct substitution that Υ (Θ,φ) is a solution of Eq. (9), with the ex­ pected eigenvalue ( a =1(1 + 1) = 2).

3.

For a particle in a state for which / = 1 and m = 1, compute (L ) from the three-dimensional definition of expectation value:

ΧΛ

directly

2

x

(L )

=

2

x

fj f

ltf L

2

0

using expression (13) for the L values of L and L l

Y

U1

sin θ d6 άφ ,

operator. Is the result consistent with the

x

2

z

4.

Using Eqs. (13), verify that L L

5.

Suppose that we have a beam of particles in which L = 2h (1 = 1) and L = +h. What will be the result of measuring the angular momentum L , about an axis z' which makes an angle of a with the ζ axis? Write ex­ pressions in terms of a for the probability of finding L , to be equal to 0, and —h, respectively. (Hint: Write the three / = 1 eigenfunctions of L , in rectangular coordinates x, y, and ζ by rotating the coordinate system as we did in finding the eigenfunctions of L in Section 6.3. Then write the appropriate spherical harmonic as a linear combination of these eigenfunc­ tions.)

x

y

— LL y

=

x

ihL . z

2

2

z

z 9

z

z

x

6.

Extend the discussion of Section 6.3a to the case / = 2: By a rotation of axes, find an expression for K ,2j-> the eigenfunction of L with eigenvalue +2ft. Then express Y ,2x linear combination of the functions F , m Finally, from the coefficients in this expression, deduce the probabilities of the various possible results of a measurement of L on a system for which L = +2h and L = 6ft . 2

a s

x

a

2

2

z

2

2

x

7.

Use the probabilities found in Problem 6 to compute the value of (L ) for that system. Explain why (Lf) should be equal to h . 2

Z

2

PROBLEMS23 5

(a) Equation (17) requires that / be an integer because m is an integer. If m were a half-integer, Eq. (17) would yield half-integer values of / as well. Thus there exist spherical harmonics such as Yi /2,i/2 = constantx

(sin 0 e'*) ' = θ ( 0 ) Φ(φ) 1 2

Verify by direct substitution that θ ( 0 ) for this spherical harmonic is a solution to Eq. (17) with m = V2, and that the eigenvalue / equals /(/ + 1) = V2V/2 + 1) =

(b) We ruled out solutions such asKi ,i/2 because they are not singlevalued. However, the wavefunction itself is not observable—the ob­ servable is I ψ | — s o it is unduly restrictive to require φ itself to be single valued. But we can rule out solutions of noninteger m in another way, illustrated by the following example: The functions Y and ^1/2,-1/2 are the only two solutions for / = 2. Therefore it should be possible to find two eigenfunctions of L that are linear combinations of these two solutions, just as we found three eigenfunctions of L for / = 1 (Eqs. 20). Find one of these functions, Yi/ ,i/2x, in the same way that we found Y in Eqs. (20), namely, by writing Yy in cartesian coordinates and making the substitutions χ —> - ζ and ζ —> χ. Show that Κ 1/2,1/2χ cannot be written as a linear combination of Κ 1/2,1/2 and Y 1/2,-1/2. Thus the functions with / = 2 do not behave properly (they violate Postulate 3) and they must be discarded. /2

2

1 / 2 a / 2

x

x

2

l t l x

2>1/2

One can generate all of the spherical harmonics from the formula γ

=

1(21 + ! ) ( / +

|m I ) !

1

I d

V - 1*1

Μ 4ττ (/ - jm I ) ! 2 l\ (sin 0)I - 1 \d cos θ) (sin 0 ) e + Use this formula to generate Y , and verify that it is a solution of Eq. (17) with the correct eigenvalue. Show that the sum of the probability density functions l

l m

2/

im

3t2

+2

Χ m =

\Υ,.η,(θ,φ)\

2

-2

I

ι

is spherically symmetric. (The general result, that

ί

m = -I

Y ,m t

2

is spherically symmetric for any value of /, was proven by A. Unsold, Annalen der Phys. 82 , 355 (1927).) Consider an anisotropic three-dimensional harmonic oscillator, with a dif­ ferent frequency of oscillation along each axis, so that the potential energy is

236

SCHRODINGER EQUATION II: THREE DIMENSIONS

K(x, y, ζ) = - ( ω χ 2

2

+ wy 2

+

2

y

ωζ) 2

2

Write an expression for the energy levels, in terms of ω , ω , and ω . Find the four lowest energy levels, for the case ω = ω = 2ω /3, and deter­ mine the degeneracy of each level. Do you expect the energy eigenfunc­ tions to be eigenfunctions of L ? Explain. χ

υ

υ

χ

ζ

ζ

2

12.

Construct the six linear combinations of

#200» 020> 002> 110» 101» and w

w

u

M

uon which are eigenfunctions of L and L . (See Section 6.4.) Develop the analysis of Section 6.4 for the case η = 2, as follows: Write the function u as a linear combination of eigenfunctions of L and and thereby find the probabilities of obtaining various values of L and L in a measurement on a system with that wavefunction. Use your results to compute (L ) and (L ) for this system. Do you expect to find (L ) = 3(L ) in this case? 2

z

13.

2

200

2

:

2

2

2

14.

Verify that j (kr)

15.

Find the lowest energy level for an electron trapped in a spherically sym­ metric well for which V = 0 when 1 A < r < 4 A, and V is infinite everywhere else. (Hint: The ground-state wavefunction must be a super­ position of j (kr) and n (kr), and it must be zero where V is infinite.)

x

0

16.

and j (kr)

2

as given in Section 6.6 are solutions of Eq. (38).

2

0

Find the probability that the electron of the previous problem will be found within a spherical shell of inside diameter 3.9 A and outside diameter 4.0

A. 17.

One can generate all of the spherical Bessel functions by means of the formula

Show that this is true by using the formula t o find ji(kr) and j (kr). find/ (A:r) and show that it is a solution of Eq. (37) for / = 3. 2

Also

3

18. Use the first three partial waves (/ = 0, / = 1, and / = 2) to compute σ and da/dQ for scattering from a perfectly rigid sphere when ka = Compare with the result that you would obtain by using only one (/ = 0) or two (/ = 0, / = 1) partial waves. 19. A particle of mass m is scattered from a spherically symmetric potential V(r), (see figure) where

PROBLEMS23

7

(0
(a
2a)

(r>2a)

2a

Assume that the incident particle has very low energy, so that ka <^ 1, where k is the wavenumber for r > 2a. Find σ and da/dQ for the case where (2mV /h ) = π/2α. (Hint: Since Ε <ζ V the wavenumber for r
0

09

1

2

[A sin k r x

kr

(r
t

Β sin(fc r + δ') 2

k r

(a
< 2a)

2

C sin(fcr + δ) kr

(r > 2a)

and one can eliminate the arbitrary constants A B and C, and find <5, by using the boundary conditions on the wavefunction at r = a and r = 2a.) 9

9

20. If <5 = 180° and all the other phase shifts are quite small, the cross section is very small for incident particles of just the right energy, and σ becomes much larger for particles of slightly different energy, because sin <5 , which provides the sole contribution to σ increases whether δ increases or decreases. A sharp minimum in σ for the scattering of electrons from rare-gas atoms (called the Ramsaucr-Townsend effect) was explained in this way by Niels Bohr. Show that this effect can occur with an attractive potential but not with a repulsive potential. 0

2

0

9

0