Solutions of nonrelativistic Schrödinger equation from relativistic Klein–Gordon equation

Physics Letters A 374 (2009) 116–122

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Physics Letters A www.elsevier.com/locate/pla

Solutions of nonrelativistic Schrödinger equation from relativistic Klein–Gordon equation Hosung Sun Department of Chemistry, Sungkyunkwan University, Suwon 440-746, Republic of Korea

a r t i c l e

i n f o

Article history: Received 21 July 2009 Received in revised form 15 September 2009 Accepted 2 November 2009 Available online 5 November 2009 Communicated by R. Wu PACS: 03.65.-w 03.65.Ge 03.65.Pm

a b s t r a c t The relativistic one-dimensional Klein–Gordon equation can be exactly solved for a certain class of potentials. But the nonrelativistic Schrödinger equation is not necessarily solvable for the same potentials. It may be possible to obtain approximate solutions for the inexact nonrelativistic potential from the relativistic exact solutions by systematically removing relativistic portion. We search for the possibility with the harmonic oscillator potential and the Coulomb potential, both of which can be exactly solvable nonrelativistically and relativistically. Though a rigorous algebraic approach is not deduced yet, it is found that the relativistic exact solutions can be a good starting point for obtaining the nonrelativistic solutions. © 2009 Elsevier B.V. All rights reserved.

Keywords: Klein–Gordon equation Harmonic oscillator potential Coulomb potential

1. Introduction There has been a growing interest in finding exact solutions of the relativistic one-dimensional Klein–Gordon equation. To understand relativistic effects in nuclear physics or chemistry it is important to obtain bound state solutions of the Klein–Gordon equation with mixed vector and scalar potentials. For various types of potentials, such as linear [1–3], exponential [2–5], Coulomb [6,7], Hulthén [8–11], Rosen– Morse [12,13], etc., the exact bound state solutions of the one-dimensional (or three-dimensional s wave) Klein–Gordon equation have been reported. The bound state solutions of the Klein–Gordon equation for more complex potentials or for one-dimensional potentials with centrifugal term are also investigated [8,14–19]. It is also reported that the one-dimensional Klein–Gordon equation can be exactly solved for shape invariant potentials [1–4,8,20]. For some potential the exact relativistic solutions can be obtained from the Klein–Gordon equation but the nonrelativistic Schrödinger equation is not necessarily solvable for the same potential. It will be very interesting if one finds an algebraic way of obtaining approximate (or exact) bound state solutions of the nonrelativistic Schrödinger equation from exact particle bound state solutions of the relativistic Klein–Gordon equation. Since the nonrelativistic Schrödinger equation is bosonic in nature, the relativistic bosonic particle Klein–Gordon equation is selected instead of the relativistic Dirac equation for fermions. In the present work we search for the possibility with the harmonic oscillator and the Coulomb potentials. Though the Klein–Gordon energy values at the nonrelativistic limit for the harmonic oscillator and the Coulomb potentials were investigated [6,18], extensive investigations on the transformation work have not been performed yet. In general the Klein–Gordon equation with mixed vector and scalar potentials cannot be mathematically transformed to the Schrödinger equation. Note that Chen et al. state [7], “. . . the possibility of getting approximate nonrelativistic solutions from relativistic ones should definitely be investigated more profoundly . . . ”. From now on we will simply call this subject “the transformation work” for convenience. For a spinless particle of rest mass m, the (1 + 1)-dimensional time-independent Klein–Gordon (KG) equation is

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H. Sun / Physics Letters A 374 (2009) 116–122

−h¯ 2 c 2

 2  2 Ψ (x) + mc 2 + S (x) Ψ (x) = E − V (x) Ψ (x)

d2 dx2

117

(1)

where E is the total energy of the particle, c is the velocity of light, and h¯ is the Planck constant divided by 2π . V (x) is called Lorentz vector potential that is the time-component of the (1 + 1)-vector potential [ A (t , x)]. S (x) is called Lorentz scalar potential that couples to the space–time scalar potential [ S (t , x)]. The vector potential couples to the energy while the scalar potential couples to the mass of the particle. The two couplings are independent and intrinsically different. Up to now it has been known that: (i) When the scalar potential energy (| S |) is larger than the vector potential energy (| V |), the KG equation always has analytical bound state solutions. (ii) Sometimes though the relativistic KG equation produces bound states, the corresponding nonrelativistic Schrödinger equation may not have bound states for the same potentials or vice versa. The simplest example is the case of S (x) = − V (x). In this case, the KG equation can have bound states but the Schrödinger equation has continuum states. The corresponding nonrelativistic Schrödinger equation is

h¯ 2 d2

Ψ nonrel (x) + W (x)Ψ nonrel (x) = E nonrel Ψ nonrel (x) (2) 2m dx2 where W (x) ≡ S (x) + V (x). In the nonrelativistic limit, the Schrödinger equation may be obtained from the KG equation when the two potential energies are small compared to the rest mass energy mc 2 , then the nonrelativistic energy is E nonrel ∼ = E − mc 2 and the wave nonrel ∼ function is Ψ (x) = Ψ (x). Note that the vector and scalar potentials are treated on a same footing, i.e. they are no more distinguishable −

in the Schrödinger equation. The transformation work is performed as follows. A Schrödinger equation that we desire to solve is set up. The two potentials are chosen. The sum of the scalar S (x) and vector V (x) potentials must be equal to the potential W (x) in the Schrödinger equation to be solved. Also the two potential functions must have a form for which the KG equation can be exactly solved for bound states. Finally the appropriate nonrelativistic limits of the KG solutions are taken to obtain the bound state solutions of the Schrödinger equation. The harmonic oscillator and the Coulomb potential systems are investigated as example. For these potentials the Schrödinger equation can be directly and exactly solved. In another words one knows the exact nonrelativistic solutions of the Schrödinger equation beforehand, which permits one to analyze the nonrelativistic solutions obtained from the relativistic KG equation without ambiguity. 2. Harmonic oscillator potential Suppose that one desires eigenenergies (or energies) E nonrel and eigenfunctions (or wave functions) Ψ nonrel (x) of the nonrelativistic Schrödinger equation with the potential W (x) ≡ S (x) + V (x) = 12 kx2 (k > 0), i.e. the bound state solutions of the following equation with a boundary condition of Ψ nonrel (−∞) = Ψ nonrel (∞) = 0,

−

h¯ 2 d2

1

2m dx2

Ψ nonrel (x) + kx2 Ψ nonrel (x) = E nonrel Ψ nonrel (x) (−∞ < x < ∞). 2

(3)

Let us start from the KG equation. Let S (x) = ax2 and V (x) = bx2 where a + b = 12 k, then the KG equation is

−h¯ 2 c 2

2 2   Ψ (x) + mc 2 + ax2 Ψ (x) = E − bx2 Ψ (x) (−∞ < x < ∞).

d2 dx2

(4)

Since the solutions obviously depend on the choice of a and b, the a = b and a = b cases are investigated separately. Case I. S (x) = V (x), i.e. a = b = 14 k: The relativistic KG equation is

−h¯ 2 c 2

2 2   1 2 1 2 2 Ψ ( x ) + mc + Ψ ( x ) = E − Ψ (x) (−∞ < x < ∞). kx kx 2

d2

4

dx

4

(5)

The equation may be rewritten as

−

h¯ 2 d2

1

2m dx

2



Ψ (x) + k 2

mc 2 + E



2mc 2

x2 Ψ (x) = ε Ψ (x)

(6)

−m c with ε = E 2mc . As shown above, the KG equation can always be reduced to a Schrödinger-type equation when S (x) = V (x) [18]. It is a 2 textbook knowledge to solve the above KG equation that mimics the Schrödinger equation for harmonic oscillator potential, but here one should solve it iteratively and consistently because the energy E appears in the potential function. Let the initial guess for E be E (0) . Solving Eq. (6) with a boundary condition of Ψ (−∞) = Ψ (∞) = 0, one obtains the energy-like term ε , 2

ε=

2 4

E 2 − m2 c 4 2mc 2



= h¯

k(mc 2 + E (0) )/2mc 2

1/2 

n+

m

1



2



=K

1 2

+

E ( 0)

1/2

2mc 2

(7)

where the quantum number n (the number of nodes in Ψ (x)) is 0, 1, 2, . . . and K is



K ≡ h¯

k

1/2 

m

n+

1 2

 .

(8)

Solving Eq. (7) for E, one obtains the first iterative energy E (1) ,



E

(1 )

= ± m2 c 4 + 2mc 2 K



1 2

+

E ( 0) 2mc 2

1/2 .

(9)

118

H. Sun / Physics Letters A 374 (2009) 116–122

Since only particle states have meanings in this work, the + sign must be chosen, i.e.,

E

(1 )

= mc

 2

1+

2K

1

mc 2

2

+

E ( 0) 2mc 2

(10)

.

The first iterative normalized wave function Ψ (1) (x) is

−1/2  (1) 1/4 −α (1) x2 /2  (1) 1/2   Ψ (1) (x) = 2n n!π 1/2 α e Hn α x with

(11)

+E α (1) = 2π m1/2k1/2 ( mc2mc )1/2 . H n (x) is a Hermite polynomial of degree n. 2 2

(0)

Inserting E (1) into E in Eq. (6) and solving it again, one obtains the second iterative energy E (2) ,

 2K (2 ) 2 E = mc 1+ mc

 1

+ 2 2

 1 2

1+

2K

1

mc 2

2

+

E ( 0)

(12)

2mc 2

and the second iterative wave function Ψ (2) (x),

−1/2  (2) 1/4 −α (2) x2 /2  (2) 1/2   Ψ (2) (x) = 2n n!π 1/2 α e Hn α x with

 (2 ) 1/2 1/2 1 α = 2π m k + 2

 1 2

(13)

1+

2K

1

mc 2

2

+

E ( 0) 2mc 2

.

After infinite number of iterations, one obtains the E (∞) that must be equal to the exact energy E, i.e.,

E = E (∞) = mc 2 y where

(14)

     

1 1 2K 2K 1 1 2K 1 E ( 0)

y=

1 + mc 2 2 + 2 1 + mc 2 2 + 2 · · · 1 + mc 2 2 + 2mc 2 .

Here (· · ·) denotes the infinite repetition of the term, 1 +

2K mc 2



1 2

+

1√ ···. 2

(15)

Note that the value of y becomes independent of the choice

of E (0) because there is the infinite number of terms embedded. The exact wave function is

−1/2  (∞) 1/4 −α (∞) x2 /2  (∞) 1/2   Ψ (x) = Ψ (∞) (x) = 2n n!π 1/2 α e Hn α x with

α (∞) = 2π m1/2k1/2



1 2

(16)

+ 12 y.

There is another way of obtaining the exact E. From Eqs. (14) and (15) the following equation can be obtained, i.e.,



2

mc y = mc

2

1+



2K mc 2

1 2

+

1 2

(17)

y.

Combining Eq. (14) with Eq. (17), one obtains the equation,

E 4 − 2m2 c 4 E 2 − 2mc 2 K 2 E + m4 c 8 − 2m2 c 4 K 2 = 0.

(18)

Therefore one can also obtain the exact E by solving the above quartic equation. At this point it is worthwhile to examine the customary method for solving Eq. (6). As many authors suggested [1–13], let us assume that the exact E is already known, i.e., E is a constant. Then Eq. (6) can be regarded as a Schrödinger equation with harmonic oscillator potential and its bound state energy ε is

ε=

E 2 − m2 c 4 2mc 2

 = h¯

k



mc 2 + E 2mc 2

m

 1/2  n+

1 2

 .

(19)

Rearranging it, one obtains a quartic equation for E to solve

E 2 − m2 c 4 2mc 2

 =K

1 2

+

E 2mc 2

1/2 .

(20)

Eq. (20) is identical with Eq. (18) obtained from the iterative procedure. Therefore this customary method is found to be adequate. However, in general, this method is valid only when the KG equation is exactly solvable.

H. Sun / Physics Letters A 374 (2009) 116–122

119

Now let us examine the nonrelativistic solutions. In the nonrelativistic limit where k  1, since



can be approximated as E ∼ = mc 2 1 +

2K mc 2

2K mc 2

 1 and y ∼ =



1+

2K , mc 2 2

the energy

∼ = mc 2 + K . Note that the energy E consists of the rest mass energy term mc and the other

term K which is independent of the velocity of light c. Therefore the nonrelativistic energy is

E

nonrel

 1/2   k 1 2 ∼ n+ = E − mc = K = h¯ m

(21)

2

and the nonrelativistic wave function, by further approximation of y ≈ 1, is

−1/2 1/4 −α x2 /2  1/2   Ψ nonrel (x) ∼ α e Hn α x = 2n n!π 1/2

(22)

with α = 2π m k . An alternate way of obtaining the nonrelativistic energy is to use Eq. (19) that can be rearranged as 1/ 2 1/ 2

 E − mc 2 = h¯

k

1/2  n+

m

1



mc 2 + E

−1/2 (23)

.

2mc 2

2

In the nonrelativistic limit, i.e. E nonrel ∼ = E − mc 2 , one obtains

E nonrel ∼ = h¯



k

1/2  n+

m

1

 (24)

2

which is identical with Eq. (21). As seen in Eqs. (21) and (22), we have obtained the exact nonrelativistic solutions. It is due to the fact that, in this case, the relativistic KG equation is exactly transformed to the nonrelativistic Schrödinger equation at the nonrelativistic limit. Case II. S (x) = V (x), i.e. a = b and a + b = 12 k: The relativistic KG equation is

− with



h¯ 2 d2 2m dx2

ε=

Ψ (x) +

a2 − b 2 2mc 2

 4

x +

mc 2 a + Eb



mc 2

x2 Ψ (x) = ε Ψ (x)

(−∞ < x < ∞)

(25)

E 2 −m2 c 4 . 2mc 2

This KG equation cannot be exactly solved because of the x4 term present. Furthermore, it does not have bound state solutions when b2 > a2 . As mentioned before, it verifies that the KG equation does not necessarily have bound state solutions when the vector potential energy (| V |) is larger than the scalar potential energy (| S |). Therefore, in case of a harmonic oscillator potential, the choice of S (x) = V (x) is not suitable. We have successfully derived the eigenenergies and eigenfunctions of the nonrelativistic Schrödinger equation from the relativistic Klein–Gordon equation. From this example one learns that the judicious choice of the two potentials, S (x) and V (x), is critical in the transformation work. 3. Coulomb potential q

The nonrelativistic Schrödinger equation with the potential W (x) = − x (q > 0) with a boundary condition of Ψ nonrel (0) =

Ψ

nonrel

(∞) = 0 is

−

h¯ 2 d2

q

2m dx2

Let S (x) = − 2 2

−h¯ c

qs x

Ψ nonrel (x) − Ψ nonrel (x) = E nonrel Ψ nonrel (x) (0 < x < ∞).

(26)

x

and V (x) = −

qv x

where q s + q v = q, then the KG equation is



d2

2

dx2

Ψ (x) + mc −

qs

2

 Ψ (x) = E +

x

qv

2

x

Ψ (x) (0 < x < ∞).

(27)

The relativistic KG equation may be rewritten to be a Schrödinger-type equation of the form, i.e.

−

h¯ 2 d2 2m dx2

 Ψ (x) +

q2s − q2v



2mc 2

1 x2

 Ψ (x) +

−mc 2 q s − Eq v



mc 2

1 x

Ψ (x) = ε Ψ (x)

(28)

−m c with ε = E 2mc . 2 Eq. (28) can be exactly solved and the bound state solutions are similar to those for the well known one-dimensional hydrogen atom. According to de Castro who reported the detailed solutions for the above KG equation [6], the exact energy for a particle state n with a boundary condition of Ψ (0) = Ψ (∞) = 0 is 2

2 4

 E = mc − 2

where



qs q v

(l + n + 1)2 h¯ 2 c 2

+

1−

q2s − q2v

(l + n + 1)2 h¯ 2 c 2

 1+

q2v

(l + n + 1)2 h¯ 2 c 2

−1

(n = 0, 1, 2, . . .)

(29)

120

H. Sun / Physics Letters A 374 (2009) 116–122

l=−

1 2

+

q2s h2 c 2

¯

−

q2v h¯ 2 c 2

1

+ .

(30)

4

n is the quantum number or the number of nodes in wave function. The constraints, l  − 12 , must be satisfied for Eq. (28) to have bound state solutions. The wave functions are

Ψ (x) = N zl+1 e −z/2 Ln2l+1 ( z)

(31)

√ 2

where N is a proper normalization constant. L n2l+1 ( z) is an associated Laguerre polynomial of degree n and z = h¯

−2ε x.

q

Case I. S (x) = V (x), i.e. q s = q v = 2 : The relativistic energy for a particle is

4(l + n + 1)2 h¯ 2 c 2 − q2

E = mc 2

4(l + n + 1)2 h¯ 2 c 2 + q2

(32)

.

In the nonrelativistic limit where q  1, Eq. (32) is reduced to



E∼ = mc 2 1 −



2q2 4(l + n

+ 1)2 h¯ 2 c 2

= mc 2 −

m

q2

h¯ 2(n + l + 1)2 2

(33)

.

Note again that the energy E is the sum of the rest mass energy (mc 2 ) and the c-independent term. The nonrelativistic energy is

E nonrel ∼ = E − mc 2 = − Since

ε=

E 2 −m2 c 4 2mc 2

mq2

(n = 0, 1, 2, . . .).

2h¯ 2 (n + l + 1)2

(34)

∼ = E nonrel , the wave function is

Ψ nonrel ( z) ∼ = Ψ (z) = N zl+1 e −z/2 Ln2l+1 ( z)

(35)

√

2 m q x. h¯ n+l+1

where z = Though l is zero from Eq. (30), the index l is kept for bookkeeping. The derived nonrelativistic energy (Eq. (34)) and the wave function (Eq. (35)) are identical with the exact eigenenergy and eigenfunction q of the Schrödinger equation for the − x potential as well known [21,22]. We have successfully derived the bound state solutions of the nonrelativistic Schrödinger equation from the relativistic KG equation.

 q v and q s + q v = q: Two particular cases are examined. When q s = q and q v = 0, the relativistic energy Case II. S (x) = V (x), i.e. q s = 

for a particle state n is E = mc 2 1 −

E nonrel ∼ = E − mc 2 = −

q2

(l+n+1)2 h¯ 2 c 2

mq2 2

2h¯ (l + n + 1)2

with l = − 12 +

q2 h¯ 2 c 2

+ 14 . The nonrelativistic energy (q  1) is

(36)

. mc 2

When q s = 0 and q v = q, the relativistic energy for a particle state n is E = 

1+

E nonrel ∼ = E − mc 2 = −

mq2 2

2h¯ (l + n + 1)2

q2

. The nonrelativistic energy is

(l+n+1)2 h¯ 2 c 2

.

(37)

Note that Eqs. (36) and (37) are identical with the nonrelativistic energy in Eq. (34). For the Coulomb potential there is more freedom in choosing S (x) and V (x). Unlike the harmonic oscillator potential, even the choice of S (x) = V (x) is suitable for deducing the nonrelativistic solution from the relativistic solution. As a matter of fact this finding is related to the type of potential. For the harmonic oscillator potential the choice of S (x) = V (x) always produces the term x4 that prohibits one from obtaining the exact solutions. For the Coulomb potential the choice of S (x) = V (x) produces the term − x12 that is solvable. Furthermore, the two terms − x12 and − 1x both go to zero when x goes to infinity so that the boundary condition of Ψ (∞) = 0 is satisfied even for the S (x) = V (x) choice. In conclusion, from this example, one learns that not only the type of potential but also the boundary condition must be carefully examined in the transformation work. 4. Liner potential Suppose one is interested in obtaining the analytical bound state solutions of the following Schrödinger equation,

−

h¯ 2 d2 2m dx2

Ψ nonrel (x) + W (x)Ψ nonrel (x) = E nonrel Ψ nonrel (x) (0 < x < ∞)

(38)

where the potential is infinite ( W (x) = ∞) when x < 0 and is linear (W (x) = ax (a > 0)) when x  0. Note that W (0) = 0. Then the boundary condition for bound states is Ψ nonrel (0) = Ψ nonrel (∞) = 0. This Schrödinger equation is not exactly solvable. Therefore we try to obtain the approximate solutions from the corresponding Klein–Gordon equation.

H. Sun / Physics Letters A 374 (2009) 116–122

121

First, let us examine the following Schrödinger equation with boundary condition of Φ nonrel (− A ) = Φ nonrel (∞) = 0,

−

h¯ 2 d2

Φ nonrel (x) + axΦ nonrel (x) = ωnonrel Φ nonrel (x) (− A < x < ∞)

2m dx2

where A is positive. Comparing Eq. (39) with Eq. (38), one immediately notices that the eigenenergy E nonrel is related to

(39)

ωnonrel as

E nonrel = ωnonrel + a A

(40)

and the eigenfunction Ψ nonrel (x) is

Ψ nonrel (x) = Φ nonrel (x − A ).

(41)

Therefore, if one finds the solutions of Eq. (39), the solutions of Eq. (38) will be obtained. The approximate solutions of Eq. (39) can be obtained from the corresponding Klein–Gordon equation. Let us start from the KG equation with S (x) = ax and V (x) = 0, then the KG equation is

−h¯ 2 c 2

d2 dx2

2  Ψ (x) + mc 2 + ax Ψ (x) = E 2 Ψ (x) (− A < x < ∞).

(42)

Introducing the new variable z = x + A with

A=

mc 2 a

(43)

,

Eq. (42) is transformed to a Schrödinger-type equation of the form,

−

h¯ 2 d2

1

2m dz

2

Ψ ( z) + 2



a2

 z2 Ψ ( z) = ε Ψ ( z)

mc 2

(0 < z < ∞)

(44)

2

E where ε = 2mc 2 . Note that now z spans from 0 to ∞, which is the same as x does in Eq. (38). Eq. (44) is the well-known Schrödinger equation for harmonic oscillator on the half-line (0 < z < ∞). Solving it with a boundary condition of Ψ ( z = 0) = Ψ ( z = ∞) = 0, one obtains the orthonormalized eigenfunctions,

−1/2 1/4 −α z2 /2    Ψ ( z) = 22n (2n + 1)!π 1/2 α e H 2n+1 α 1/2 z (n = 0, 1, 2, . . .) where

α=

ε=

√ a

2mc 2

E2 2mc 2

(45)

. H 2n+1 ( z) is a Hermite polynomial of degree 2n + 1. And the eigenvalues are

=

h¯ a mc

 2n +

3 2

 (n = 0, 1, 2, . . .).

(46)

Therefore, the relativistic eigenenergy E of Eq. (42) is



E = ± h¯ ca(4n + 3). In the nonrelativistic limit where the approximation of E nonrel ∼ = E − mc 2 is valid, the corresponding nonrelativistic eigenvalue by taking the particle state (+ sign),



ωnonrel ∼ = −mc 2 + h¯ ca(4n + 3).

(47)

ωnonrel is, (48)

Finally, using Eqs. (40) and (43), the eigenenergy E nonrel of the nonrelativistic Schrödinger equation (38) is obtained, i.e.,

E nonrel ∼ =



h¯ ca(4n + 3) (n = 0, 1, 2, . . .).

(49)

The dimensionless eigenenergy is defined as

E nonrel ∼ √ E nonrel (dimensionless) ≡ √ = 4n + 3. h¯ ca Using Eq. (49) one can evaluate E nonrel (dimensionless) which are 1.732 (n = 0), 2.646 (n = 1), 3.317 (n = 2), 3.873 (n = 3), . . . . We have numerically solved the Schrödinger equation (38) using the Numerov method. The exact numerical E nonrel (dimensionless) are 1.855757 (n = 0), 3.244607 (n = 1), 4.381670 (n = 2), 5.386613 (n = 3), . . . . One sees that the eigenenergy expression (49) is very crude. Nevertheless, we have shown how the nonrelativistic eigenenergy of Schrödinger equation can be analytically obtained from the relativistic Klein–Gordon equation, although it is not exact. We believe that we will be able to obtain a better approximate expression to E nonrel when the eigenfunction (Eq. (45)) is utilized. The details of the work will be reported elsewhere in the near future.

122

H. Sun / Physics Letters A 374 (2009) 116–122

5. Conclusions and discussion We have shown how the approximate (or exact) bound state solutions of nonrelativistic Schrödinger equation can be obtained from the exact solutions of the relativistic Klein–Gordon equation for the harmonic oscillator, Coulomb, and linear potentials. Though a general way of transforming the relativistic Klein–Gordon equation to the nonrelativistic Schrödinger equation is not found, some nontrivial requirements (or information on the transformation work) have been revealed. The following requirements must be met for the transformation. (i) Obviously the sum of the scalar and the vector potential functions in the Klein–Gordon equation must be equal to the potential function in the Schrödinger equation of interest. (ii) Theoretically there are infinite possibilities of choosing each of the scalar and vector potential functions. But the Klein–Gordon equation with the chosen potentials must be exactly solvable. (iii) The Klein–Gordon equation must have particle bound state solutions satisfying the same boundary conditions of the Schrödinger equation. (iv) The choice of S (x) = − V (x) generally makes the Klein–Gordon equation not to have bound state solutions. The choice of S (x) = V (x) seems to be suitable. The choice of S (x) = V (x) often makes the Klein–Gordon equation insoluble. But, depending on the type of potential functions and boundary conditions, the choice of S (x) = V (x) may be adequate. For this case, this choice is recommended instead of S (x) = V (x). (v) The relativistic particle state energy should be expressed in a certain analytical form in which the rest mass energy (mc 2 ) term must be separated from the other terms. And the other terms must not depend on the velocity of light c. When the above five requirements are met, it is possible to deduce the nonrelativistic bound state solutions of the Schrödinger equation from the corresponding relativistic Klein–Gordon equation. It is worth explaining the requirements (iv) and (v) in detail. In (iv), whenever it is applicable, the choice of S (x) = V (x) (instead of S (x) = V (x)) is recommended because of the following reasons. When S (x) = V (x) is chosen, the nonrelativistic Schrödinger equation of interest has a potential of S (x) + V (x) = 2V (x) and the relativistic Klein–Gordon equation is

−

h¯ 2 d2 2m

dx2

Ψ (x) +

mc 2 + E mc 2

V (x)Ψ (x) =

E 2 − m2 c 4 2mc 2

Ψ (x).

(50)

It can be considered as a Schrödinger-type equation with the potential of (mc 2 + E ) V (x)/mc 2 . When the potential V (x) is shape invariant, the constant × V (x) is also shape invariant. The sufficient condition for exact solubility for Schrödinger equation is that the potential must be shape invariant [22]. Therefore both the Schrödinger equation and the Klein– Gordon equation are always exactly solvable when V (x) is shape invariant. Therefore, in the case of S (x) = V (x), there is no need to solve the Klein–Gordon equation to obtain the solutions of Schrödinger equation because the Schrödinger equation itself can be directly and exactly solvable. As a matter of fact, the harmonic oscillator and the Coulomb potential functions studied in the present work are shape invariant so that we do not present any new potential for which the Klein–Gordon equation is exactly soluble but the Schrödinger equation is inexactly soluble. Simply we have taken the two potentials in order to investigate a simple way of obtaining solutions for the nonrelativistic Schrödinger equation from the relativistic Klein–Gordon equation. The above requirements seem to be restrictive. Nonetheless the ‘transformation work’ is worth investigating further. Further searches for the approximate analytical solutions of Schrödinger equation with inexactly solvable potential are under way. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22]

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