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Exact solutions for neutral particles in the field of a circularly polarized plane electromagnetic wave
Physics Letters A 342 (2005) 67–76 www.elsevier.com/locate/pla
Exact solutions for neutral particles in the field of a circularly polarized plane electromagnetic wave Qiong-Gui Lin China Center of Advanced Science and Technology (World Laboratory), P.O. Box 8730, Beijing 100080, People’s Republic of China 1 Department of Physics, Sun Yat-Sen University, Guangzhou 510275, People’s Republic of China Received 18 February 2005; received in revised form 19 May 2005; accepted 23 May 2005 Available online 31 May 2005 Communicated by P.R. Holland
Abstract The Schrödinger equation for a neutral particle with general spin and magnetic moment moving in the field of a circularly polarized plane electromagnetic wave is solved exactly. Simple cyclic solutions are presented for general spin, and more complicated ones for spin 1/2. Many properties of the solutions, including time evolution of the spin vector, nonadiabatic geometric phase, etc. are studied in detail. 2005 Elsevier B.V. All rights reserved. PACS: 03.65.Ca; 03.65.Vf Keywords: Circularly polarized electromagnetic wave; Schrödinger equation; Neutral particle; Exact solution
In quantum mechanics, it is difficult to solve the Schrödinger equation with a time-dependent Hamiltonian. Analytic solutions are possible only in a few simple cases. Some well studied examples involve particles with spin and magnetic moment moving in a time-dependent magnetic field [1–16]. Among these simple examples, neutral particles are of more interest since the problem is easier and the Schrödinger equation can be solved analytically in a rather general case [16]. In addition to the interest in its own right, the model also serves as an explicit example for the study of cyclic solutions and geometric phases [17–23], etc. In the above mentioned examples, the magnetic field is homogeneous in space. In this Letter we consider the Schrödinger equation for a neutral particle with general spin and magnetic moment moving in the field of a circularly polarized plane electromagnetic wave. Though the magnetic field depends on both the space and time
E-mail address: [email protected] (Q.-G. Lin). 1 Not for correspondence.
0375-9601/$ – see front matter 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.physleta.2005.05.051
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variables, the Schrödinger equation can be solved exactly by using a position- and time-dependent unitary transformation. The solutions are rather simple, but we do not find similar results in the literature. We choose the coordinates system such that the plane electromagnetic wave propagates along the positive z direction. A nonrelativistic neutral particle interacts only with the magnetic field through its magnetic moment. The magnetic field takes the form Bx = B0 cos(ωt − kz),
By = B0 sin(ωt − kz),
Bz = B,
(1)
where ω and k are both positive, B0 can be taken as positive by appropriately choosing the position of the origin, = ±1 where +1 (−1) corresponds to positive (negative) helicity. For more generality, we also include a homogeneous magnetic field B parallel to the z axis, where B may be either positive or negative. Consider a neutral particle with spin s (s = 1/2, 1, 3/2, . . .) and magnetic moment µ = µs/s, where s is the spin operator in the unit of h¯ , satisfying [si , sj ] = iij k sk (for s = 1/2, s = σ /2 in the sz representation and µ = µσ , where σ are Pauli matrices). The Schrödinger equation is ih¯ ∂t Ψ = H Ψ,
(2)
where the Hamiltonian takes the form µ p2 + V (x, y) − s · B, (3) 2M s where M is the mass of the particle, and for more generality again we include a potential V (x, y) in the Hamiltonian. The result for a pure plane electromagnetic wave can be obtained by setting V = 0 and B = 0. Let us make some remarks on the above magnetic field before solving the Schrödinger equation. Imagining that z is a fixed constant, this has the form of a rotating magnetic field, which has been discussed by many authors, as cited above. On the other hand, if t is a fixed constant, this has the form of a so-called helical magnetic field. A neutral particle moving in such a magnetic field has been studied in early days and exact solutions have been obtained [24]. Later it was studied again by other authors in connection with the problem of geometric phases, both theoretically [25] and experimentally [26]. However, the later works only gave approximate treatment. Charged particles moving in such a helical magnetic field are also studied by some authors [27–29]. The latter problem is of importance in free electron laser [30]. Unfortunately, exact solutions in the general case have not been available for the charged particles. To solve the Schrödinger equation (2), we make a position- and time-dependent unitary transformation Ψ = W (z, t)Φ, W (z, t) = exp −i(ωt − kz)sz . (4) H=
Similar transformations have been previously used (explicitly or equivalently) for particles moving in a rotating magnetic field [2,13–15] or a helical magnetic field [24]. Under this transformation, the original Schrödinger equation is reduced to a similar one for Φ: ih¯ ∂t Φ = Heff Φ,
(5)
with an effective Hamiltonian Heff = W † H W − hωs ¯ z.
(6)
= x, It is easy to show that ¯ z , and x W = px , y W = py , z W = pz + hks W † sW = sx cos(ωt − kz) − sy sin(ωt − kz), sx sin(ωt − kz) + sy cos(ωt − kz), sz , W † xW
W †p
W †p
W †p
(7)
which leads to Heff = Hxy + Hzs ,
(8a)
Q.-G. Lin / Physics Letters A 342 (2005) 67–76
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where Hxy =
1 2 p + py2 + V (x, y), 2M x
(8b)
1 µB0 µB 2 Hzs = (pz + hks sx − + hω ¯ z) − ¯ sz . 2M s s
(8c)
We see that Heff is time independent, so that Eq. (5) can be easily solved. We will use coordinate representation for the orbital part of the state vector, and use sz representation for the spin part, then both Ψ and Φ are (2s + 1)component column vectors, each component of which is a function of x and t. The result takes the form Heff Φ(x, t) = Ueff (t)Φ(x, 0), Ueff (t) = exp −i (9) t . h¯ With the obvious relation Ψ (x, 0) = W (z, 0)Φ(x, 0) = exp(ikzsz )Φ(x, 0), we obtain the solution to the original Schrödinger equation: Ψ (x, t) = U (t)Ψ (x, 0),
(10a)
where
Heff U (t) = W (z, t)Ueff (t)W (z, 0) = exp −i(ωt − kz)sz exp −i t exp(−ikzsz ). h¯ †
(10b)
The time evolution operator U (t) involves no chronological product, and thus is convenient for practical calculations. The next task is to find the eigenfunctions of the effective Hamiltonian. First, consider Hxy . If V = 0, the eigenvalues are h¯ 2 (kx2 + ky2 ) , 2M and the eigenfunctions are xy
Ekx ky =
ϕkx ky (x, y) =
(11)
exp[i(kx x + ky y)] , Lx Ly
(12)
where Lx and Ly are lengths for box normalization, and according to periodic boundary conditions, kx =
2πmx , Lx
ky =
2πmy , Ly
mx , my ∈ Z.
(13)
xy
If V = 0, we denote the eigenvalues as En and the eigenfunctions as ϕn (x, y). The specific form of these results are not important for subsequent discussions. For Hzs , we denote the eigenvalue as E zs and write the eigenfunction as u(z) =
exp(ikz z) χ, √ Lz
(14)
where χ is the spin part of the wave function, which is a (2s + 1)-component column vector independent of the x. In this form the orbital and spin motion is separated and χ satisfies the eigenvalue equation 2 2 2 h¯ 2 kz2 h¯ kkz µB µB0 h¯ k 2 sz + − − hω sx χ = E zs − χ. (15) ¯ sz − 2M M s s 2M
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Explicit solutions to this equation for general s are still not available at present. We will deal with the simplest case with s = 1/2 at the moment and return to the general case below. In this case sz2 = 1/4 and the above equation can be recast in the form h¯ 2 kz2 h¯ 2 k 2 zs χ, −µ hΩs (16) · nχ = E − − ¯ 2M 8M where µ = 1 (−1) if µ > 0 (< 0), n = (sin θ, 0, cos θ ), and Ω and θ are defined by h¯ 2 kkz 2 1/2 h¯ Ω = (2µB0 )2 + 2µB + hω − , ¯ M
(17)
and µ (2µB + hω 2|µ|B0 ¯ − h¯ 2 kkz /M) (18) , cos θ = . hΩ hΩ ¯ ¯ Note that Ω and θ depend not only on the model parameters, but also on the eigenvalue kz . Since s · n = exp(−iθ sy )sz exp(iθ sy ), the solutions of the Eq. (16) can be expressed in terms of χα0 , the eigenstate of sz , satisfying sz χα0 = (1/2)αχα0 , where α = ±. The results are sin θ =
Ekzsz α =
h¯ 2 kz2 h¯ 2 k 2 1 + − αµ h¯ Ω, 2M 8M 2
(19)
and χα = exp(−iθ sy )χα0 .
(20)
In conclusion, the eigenvalues of Heff are h¯ 2 kz2 h¯ 2 k 2 1 xy eff α + , Enk = E − ≡ E + hΩ, E ¯ nk µ nk n z z α z 2 2M 8M and the corresponding eigenfunctions are ψnkz α (x) = ϕn (x, y)
exp(ikz z) χα . √ Lz
(21)
(22)
It should be remarked that the above eigenvalues are not observable quantities because Heff is not a physical operator. Now we take Ψ (x, 0) = exp(ikzsz )ψnkz α (x)
(23)
as an initial state and study its time evolution. According to Eq. (10), the state at time t is eff eff Enk Enk exp(ikz z) zα zα Ψ (x, t) = exp −i exp −i(ωt − kz)sz χα . t W (z, t)ψnkz α (x) = exp −i t ϕn (x, y) √ h¯ h¯ Lz (24) We define the spin vector as (Ψ (x, t), sΨ (x, t))s (25) , (Ψ (x, t), Ψ (x, t))s where the subscript s indicates that the inner product is performed only in the space of the spin state. For the above state, we find v(x, t) = χα , W † (z, t)sW (z, t)χα . (26) v(x, t) =
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It is easy to show that (χα , sχα ) = (1/2)αn, and according to Eq. (7) we obtain 1 v(x, t) = α sin θ cos(ωt − kz), sin θ sin(ωt − kz), cos θ . (27) 2 Therefore, the spin vector rotates around the z axis with the same frequency as the magnetic field, though at a different angle θ with the z axis. It also propagates along the positive z direction with the same speed as the magnetic field, though there is a phase lag π when α = −. Return to Eq. (24), it is easy to see that this is a cyclic solution in the time interval [0, τ ] where τ = 2π/ω, because χα can be expressed as a linear combination of χ+0 and χ−0 . More specifically, we have Ψ (x, τ ) = eiδ Ψ (x, 0),
(28)
with the phase change eff Enk zα
τ − απ, mod 2π. h¯ It is easy to calculate the mean value of H in the state Ψ (x, t) with the result δ=−
1 eff H = Ψ (x, t), H Ψ (x, t) = Enk + α hω ¯ cos θ. zα 2 Thus the dynamic phase is 1 β =− h¯
τ H dt = −
eff Enk zα
0
h¯
τ − απ cos θ.
(29)
(30)
(31)
The nonadiabatic geometric phase is then γ = δ − β = −απ(1 − cos θ ),
mod 2π.
(32)
For any fixed x, v(x, t) is a periodic function of t, and it traverses a closed trace on a sphere of radius 1/2 in the time interval [0, τ ]. The solid angle subtended by this closed trace is obviously Ωv = α2π(1 − cos θ ),
mod 4π.
(33)
Therefore, we obtain the following relation 1 γ = − Ωv , mod 2π. (34) 2 For spin s = 1/2 and time varying magnetic fields (independent of the space variable), this is well known. Here we obtain it in a both time and space varying magnetic field. Next we consider a more general initial state Ψ (x, 0) = ϕn (x, y)
exp(ikz z) exp(ikzsz )χ, √ Lz
(35)
2 + c2 = 1. The corresponding state at time where χ = c+ χ+ + c− χ− with c+ and c− arbitrary except satisfying c+ − t is Enkz exp(ikz z) Ψ (x, t) = exp −i (36) t ϕn (x, y) √ exp −i(ωt − kz)sz exp(iµ Ωts · n)χ. h¯ Lz The spin vector in this state is v(x, t) = χ, exp(−iµ Ωts · n)W † (z, t)sW (z, t) exp(iµ Ωts · n)χ . (37)
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We denote u(t) = χ, exp(−iµ Ωts · n)s exp(iµ Ωts · n)χ = v 0 − (v 0 · n)n cos Ωt + (v 0 × n)µ sin Ωt + (v 0 · n)n, where we have introduced the notation v 0 = (χ, sχ), and used the formula [15,31] exp(−iφs · n)s exp(iφs · n) = s − (s · n)n cos φ + (s × n) sin φ + (s · n)n.
(38)
(39)
The spin vector is then given by v(x, t) = ux (t) cos(ωt − kz) − uy (t) sin(ωt − kz), ux (t) sin(ωt − kz) + uy (t) cos(ωt − kz), uz (t) . (40) This result is somewhat complicated, but the physical picture is clear. We recast u(t) in the form u(t) = v0⊥ cos Ωte1 − v0⊥ µ sin Ωte2 + v0 e3 ,
(41)
where e1 = v 0 − (v 0 · n)n /v0⊥ , (42) e2 = n × v 0 /v0⊥ , e3 = n, 2 . It is easy to show that {e , e , e } is an orthonormal set and constitutes a rightand v0 = v 0 · n, v0⊥ = v 20 − v0 1 2 3 handed frame. Eq. (41) is essentially the same as Eq. (38). However, in this form it helps us recognize the physical picture of the motion. Now to get the vector v(x, t), one just rotates v 0 around e3 = n through an angle −µ Ωt (positive angle corresponds to anti-clockwise rotation) to get u(t), and then rotates u(t) around the z axis through an angle (ωt − kz). Thus the phase (ωt − kz) of the plane electromagnetic wave always plays an important role in determining the direction of the spin vector. The resulted motion involves nutation as well as rotation, and is not periodic in general. Two cases with periodic v(x, t) are available. First, if the initial condition is such that v 0 = (1/2)αn, we have u(t) = (1/2)αn, and v(x, t) is given by Eq. (27). This has been discussed above. Second, if the parameters of the system and the eigenvalue kz are such that Ω/ω is a rational number, then both Ωt and ωt may simultaneously become integral multiples of 2π at some later time τ , and we have v(x, τ ) = v(x, 0), independent of the initial condition. The solution Ψ (x, t) is indeed cyclic in this case. Assume that Ω/ω = K2 /K1 , where K2 and K1 are both natural numbers and prime to each other, and τ satisfies ωτ = 2πK1 and Ωτ = 2πK2 . Because χ , a linear combination of χ+ and χ− , can also be expressed as a linear combination of χ+0 and χ−0 , we see from Eq. (36) that Ψ (x, τ ) = eiδ Ψ (x, 0), with the phase change δ=−
Enkz τ + (µ K2 − K1 )π, h¯
(43)
mod 2π.
It is not difficult to find that H = Ψ (x, t), H Ψ (x, t) = Enkz − µ h¯ Ω(v 0 · n) + hωu ¯ z (t).
(44)
Thus the dynamic phase is β =−
Enkz τ + (µ 2πK2 − 2πK1 cos θ )(v 0 · n). h¯
(45)
The nonadiabatic geometric phase is then γ = (µ K2 − K1 )π − (µ 2πK2 − 2πK1 cos θ )(v 0 · n),
mod 2π.
(46)
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At the present case, v(x, t) for a fixed x traverses a closed trace on a sphere of radius |v 0 | [it is obvious that |v(x, t)| = |u(t)| = |v 0 |] in the time interval [0, τ ]. The solid angle subtended by this closed trace is 1 Ωv = |v 0 |
τ 0
vx v˙y − vy v˙x dt. |v 0 | + vz
(47)
It is easy to show that
vx v˙y − vy v˙x = ux u˙ y − uy u˙ x + ω u2x + u2y .
(48)
Because uz = vz , and u2 = v 20 , we have u2x + u2y = v 20 − vz2 . Therefore, 1 Ωv = |v 0 |
τ 0
ux u˙ y − uy u˙ x ω dt + |v 0 | + uz |v 0 |
τ
|v 0 | − uz dt.
(49)
0
The second integral can be calculated easily, and the first is recognized as the solid angle subtended by the trace of u(t), which is very easy to calculate in the coordinate frame expanded by {e1 , e2 , e3 }. The final result is v0 · n v0 · n Ωv = −µ 2πK2 1 − (50) + 2πK1 1 − cos θ , mod 4π. |v 0 | |v 0 | Here the first term is due to the rotation around e3 = n, and the second is due to the further rotation around the z axis. Compared with Eq. (46), and notice that for spin 1/2 we always have |v 0 | = 1/2, we reach the result (34) once again. From Eqs. (38) and (40) we see that the spin vector at the initial time is given by v(x, 0) = (v0x cos kz + v0y sin kz, −v0x sin kz + v0y cos kz, v0z ).
(51)
It varies with z, and such states may be called twisted polarized states. Even the simpler case (23) have a similar initial spin vector [see Eq. (27)]. Thus such initial state may be difficult to prepare. However, in the special case c+ = cos(θ/2), c− = − sin(θ/2), we have χ = χ+0 and the initial state (35) becomes exp[i(kz + k/2)z] 0 (52) χ+ . √ Lz This is polarized along the positive z direction, and has a definite momentum along that direction. When V = 0 and ϕn (x, y) = 1/ Lx Ly it becomes even simpler. Such a state might be easy to realize in experiment. At later times, however, the state contains polarizations along both positive and negative z directions, since Ψ (x, t) can be expressed as Enkz Ωt Ωt exp[i(kz + k/2)z] 0 1 + iµ cos θ sin ϕn (x, y) Ψ (x, t) = exp −i χ+ t − i ωt cos √ 2 2 2 h¯ Lz Enkz Ωt exp[i(kz − k/2)z] 0 1 ϕn (x, y) + exp −i χ− . (53) t + i ωt iµ sin θ sin √ 2 2 h¯ Lz Thus an initial particle beam with a definite momentum and spin polarization in the positive z direction is split into two beams at later times, one with the original momentum and polarization and the other with a different momentum and the opposite polarization. It is easy to show that the operator Ψ (x, 0) = ϕn (x, y)
F = pz − hks ¯ z
(54)
commutes with H and thus is a conserved quantity. The above state is an eigenstate of F with eigenvalue h¯ kz at all times. In other words, a proper linear combination of pz and sz is the same for the two beams. The case with c+ = sin(θ/2), c− = cos(θ/2) and χ = χ−0 can be discussed similarly.
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Now we return to Eq. (15) for general s. A similar equation has arisen in solving the Schrödinger equation in a helical magnetic field, and the case s = 1 has been solved analytically [24]. For larger s, it is more difficult to solve it, because the secular equation for the eigenvalue is an algebraic equation of degree 2s + 1. However, some general results can be asserted without specific calculations. First, because the matrix on the left-hand side of Eq. (15) is a (2s + 1) × (2s + 1) Hermitian one, it has 2s + 1 real eigenvalues. We denote these eigenvalues as (h¯ 2 k 2 /2M)λi (i = 1, 2, . . . , 2s + 1), where λi depends on all model parameters (except those in V ) and kz , but is dimensionless. Second, we have 2s + 1 linearly independent eigenvectors, which will be denoted as ξi . Third, if ξi corresponds to a nondegenerate eigenvalue, it is real up to an overall phase factor, because both sz and sx are real matrices. In such an eigenvector, the mean value (ξi , sy ξi ) is imaginary because the matrix sy is pure imaginary. On the other hand, sy is Hermitian and the above mean value should be real. Therefore, (ξi , sy ξi ) = 0, and this does not depend on the representation employed since a mean value is representation independent. Though it is not clear whether some of the eigenvalues are degenerate, the first and second conclusions are sufficient for the subsequent discussions. On account of the above conclusions, the eigenvalues of Heff are now h¯ 2 kz2 h¯ 2 k 2 + λi , 2M 2M and the corresponding eigenfunctions are xy
eff = En + Enk zi
ψnkz i (x) = ϕn (x, y)
exp(ikz z) ξi . √ Lz
(55)
(56)
Now we take Ψ (x, 0) = exp(ikzsz )ψnkz i (x) as an initial state and study its time evolution. According to Eq. (10), the state at time t is eff Enk exp(ikz z) zi exp −i(ωt − kz)sz ξi . t ϕn (x, y) √ Ψ (x, t) = exp −i h¯ Lz
(57)
(58)
It is easy to find the spin vector in this state. We define v 0 = (ξi , sξi ), then the result reads v(x, t) = v0xy cos(ωt − kz + Λ0 ), v0xy sin(ωt − kz + Λ0 ), v0z , (59) 2 + v 2 , and Λ is determined by cos Λ = v /v where v0xy = v0x 0 0 0x 0xy and sin Λ0 = v0y /v0xy . If ξi corresponds to 0y a nondegenerate eigenvalue, then v0y = 0 and Λ0 is 0 or π . Once again we find that the spin vector rotates around the z axis, and propagates along the positive z direction, both in the same way as the magnetic field. It is easy to see that Eq. (58) is a cyclic solution in the time interval [0, τ ] where τ = 2π/ω, because ξi can be expressed as a linear combination of χm0 , the eigenstates of sz . More specifically, we have Ψ (x, τ ) = eiδ Ψ (x, 0), with the phase change eff Enk zi
τ − 2πs, mod 2π. h¯ The dynamic phase can be found to be δ=−
eff Enk zi
τ − 2πv0z , h¯ and the nonadiabatic geometric phase is then β =−
γ = −2πs + 2πv0z ,
mod 2π.
(60)
(61)
(62)
Q.-G. Lin / Physics Letters A 342 (2005) 67–76
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For a fixed x, v(x, t) is a periodic function of t, and it traverses a closed trace on a sphere of radius |v 0 | in the time interval [0, τ ]. The solid angle subtended by this closed trace is obviously v0z , mod 4π. Ωv = 2π 1 − (63) |v 0 | Therefore, we obtain the following relation γ = −|v 0 |Ωv + 2π |v 0 | − s , mod 2π.
(64)
In this case γ contains two terms. The first is the familiar one that is proportional to Ωv . The second is an extra term which depends on the initial spin state ξi . It cannot be omitted since |v 0 | − s is in general not an integer. Similar results have been encountered in other systems [15,16,32–34]. If s = 1/2, we have |v 0 | = 1/2, and the above result reduces to Eq. (34) as expected. In summary, we have solved the Schrödinger equation for a neutral particle with general spin and magnetic moment moving in the field of a circularly polarized plane electromagnetic wave, plus a uniform magnetic field and a two-dimensional potential. The time evolution operator is obtained by using a position- and time-dependent unitary transformation, which reduces the original equation to a similar one with a time-independent effective Hamiltonian. The time evolution operator involves no chronological product and is convenient for practical calculations. The eigenvalue problem of the effective Hamiltonian is solved analytically. Though some details of the spin part requires further study when s > 1/2, the knowledge attained is sufficient for our discussions. For all spin values we find simple cyclic solutions in which the spin vector rotates with the same frequency and propagates with the same speed and in the same direction as the magnetic field. For spin 1/2, more complicated cyclic solutions are presented. The nonadiabatic geometric phases for all the cyclic solutions are calculated. The relation between the nonadiabatic geometric phase and the solid angle subtended by the closed trace of the spin vector is obtained in each case. For spin 1/2, we also find that an initial particle beam with a definite momentum and spin polarization in the propagation direction of the plane electromagnetic wave is split into two beams at later times, one with the original momentum and polarization and the other with a different momentum and the opposite polarization. A proper linear combination of the momentum and spin along the original direction is the same for the two beams.
Acknowledgements This work was supported by the National Natural Science Foundation of the People’s Republic of China (grant number 10275098).
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Exact solutions for neutral particles in the field of a circularly polarized plane electromagnetic wave Qiong-Gui Lin China Center of Advanced Science and Technology (World Laboratory), P.O. Box 8730, Beijing 100080, People’s Republic of China 1 Department of Physics, Sun Yat-Sen University, Guangzhou 510275, People’s Republic of China Received 18 February 2005; received in revised form 19 May 2005; accepted 23 May 2005 Available online 31 May 2005 Communicated by P.R. Holland
Abstract The Schrödinger equation for a neutral particle with general spin and magnetic moment moving in the field of a circularly polarized plane electromagnetic wave is solved exactly. Simple cyclic solutions are presented for general spin, and more complicated ones for spin 1/2. Many properties of the solutions, including time evolution of the spin vector, nonadiabatic geometric phase, etc. are studied in detail. 2005 Elsevier B.V. All rights reserved. PACS: 03.65.Ca; 03.65.Vf Keywords: Circularly polarized electromagnetic wave; Schrödinger equation; Neutral particle; Exact solution
In quantum mechanics, it is difficult to solve the Schrödinger equation with a time-dependent Hamiltonian. Analytic solutions are possible only in a few simple cases. Some well studied examples involve particles with spin and magnetic moment moving in a time-dependent magnetic field [1–16]. Among these simple examples, neutral particles are of more interest since the problem is easier and the Schrödinger equation can be solved analytically in a rather general case [16]. In addition to the interest in its own right, the model also serves as an explicit example for the study of cyclic solutions and geometric phases [17–23], etc. In the above mentioned examples, the magnetic field is homogeneous in space. In this Letter we consider the Schrödinger equation for a neutral particle with general spin and magnetic moment moving in the field of a circularly polarized plane electromagnetic wave. Though the magnetic field depends on both the space and time
E-mail address: [email protected] (Q.-G. Lin). 1 Not for correspondence.
0375-9601/$ – see front matter 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.physleta.2005.05.051
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variables, the Schrödinger equation can be solved exactly by using a position- and time-dependent unitary transformation. The solutions are rather simple, but we do not find similar results in the literature. We choose the coordinates system such that the plane electromagnetic wave propagates along the positive z direction. A nonrelativistic neutral particle interacts only with the magnetic field through its magnetic moment. The magnetic field takes the form Bx = B0 cos(ωt − kz),
By = B0 sin(ωt − kz),
Bz = B,
(1)
where ω and k are both positive, B0 can be taken as positive by appropriately choosing the position of the origin, = ±1 where +1 (−1) corresponds to positive (negative) helicity. For more generality, we also include a homogeneous magnetic field B parallel to the z axis, where B may be either positive or negative. Consider a neutral particle with spin s (s = 1/2, 1, 3/2, . . .) and magnetic moment µ = µs/s, where s is the spin operator in the unit of h¯ , satisfying [si , sj ] = iij k sk (for s = 1/2, s = σ /2 in the sz representation and µ = µσ , where σ are Pauli matrices). The Schrödinger equation is ih¯ ∂t Ψ = H Ψ,
(2)
where the Hamiltonian takes the form µ p2 + V (x, y) − s · B, (3) 2M s where M is the mass of the particle, and for more generality again we include a potential V (x, y) in the Hamiltonian. The result for a pure plane electromagnetic wave can be obtained by setting V = 0 and B = 0. Let us make some remarks on the above magnetic field before solving the Schrödinger equation. Imagining that z is a fixed constant, this has the form of a rotating magnetic field, which has been discussed by many authors, as cited above. On the other hand, if t is a fixed constant, this has the form of a so-called helical magnetic field. A neutral particle moving in such a magnetic field has been studied in early days and exact solutions have been obtained [24]. Later it was studied again by other authors in connection with the problem of geometric phases, both theoretically [25] and experimentally [26]. However, the later works only gave approximate treatment. Charged particles moving in such a helical magnetic field are also studied by some authors [27–29]. The latter problem is of importance in free electron laser [30]. Unfortunately, exact solutions in the general case have not been available for the charged particles. To solve the Schrödinger equation (2), we make a position- and time-dependent unitary transformation Ψ = W (z, t)Φ, W (z, t) = exp −i(ωt − kz)sz . (4) H=
Similar transformations have been previously used (explicitly or equivalently) for particles moving in a rotating magnetic field [2,13–15] or a helical magnetic field [24]. Under this transformation, the original Schrödinger equation is reduced to a similar one for Φ: ih¯ ∂t Φ = Heff Φ,
(5)
with an effective Hamiltonian Heff = W † H W − hωs ¯ z.
(6)
= x, It is easy to show that ¯ z , and x W = px , y W = py , z W = pz + hks W † sW = sx cos(ωt − kz) − sy sin(ωt − kz), sx sin(ωt − kz) + sy cos(ωt − kz), sz , W † xW
W †p
W †p
W †p
(7)
which leads to Heff = Hxy + Hzs ,
(8a)
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where Hxy =
1 2 p + py2 + V (x, y), 2M x
(8b)
1 µB0 µB 2 Hzs = (pz + hks sx − + hω ¯ z) − ¯ sz . 2M s s
(8c)
We see that Heff is time independent, so that Eq. (5) can be easily solved. We will use coordinate representation for the orbital part of the state vector, and use sz representation for the spin part, then both Ψ and Φ are (2s + 1)component column vectors, each component of which is a function of x and t. The result takes the form Heff Φ(x, t) = Ueff (t)Φ(x, 0), Ueff (t) = exp −i (9) t . h¯ With the obvious relation Ψ (x, 0) = W (z, 0)Φ(x, 0) = exp(ikzsz )Φ(x, 0), we obtain the solution to the original Schrödinger equation: Ψ (x, t) = U (t)Ψ (x, 0),
(10a)
where
Heff U (t) = W (z, t)Ueff (t)W (z, 0) = exp −i(ωt − kz)sz exp −i t exp(−ikzsz ). h¯ †
(10b)
The time evolution operator U (t) involves no chronological product, and thus is convenient for practical calculations. The next task is to find the eigenfunctions of the effective Hamiltonian. First, consider Hxy . If V = 0, the eigenvalues are h¯ 2 (kx2 + ky2 ) , 2M and the eigenfunctions are xy
Ekx ky =
ϕkx ky (x, y) =
(11)
exp[i(kx x + ky y)] , Lx Ly
(12)
where Lx and Ly are lengths for box normalization, and according to periodic boundary conditions, kx =
2πmx , Lx
ky =
2πmy , Ly
mx , my ∈ Z.
(13)
xy
If V = 0, we denote the eigenvalues as En and the eigenfunctions as ϕn (x, y). The specific form of these results are not important for subsequent discussions. For Hzs , we denote the eigenvalue as E zs and write the eigenfunction as u(z) =
exp(ikz z) χ, √ Lz
(14)
where χ is the spin part of the wave function, which is a (2s + 1)-component column vector independent of the x. In this form the orbital and spin motion is separated and χ satisfies the eigenvalue equation 2 2 2 h¯ 2 kz2 h¯ kkz µB µB0 h¯ k 2 sz + − − hω sx χ = E zs − χ. (15) ¯ sz − 2M M s s 2M
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Explicit solutions to this equation for general s are still not available at present. We will deal with the simplest case with s = 1/2 at the moment and return to the general case below. In this case sz2 = 1/4 and the above equation can be recast in the form h¯ 2 kz2 h¯ 2 k 2 zs χ, −µ hΩs (16) · nχ = E − − ¯ 2M 8M where µ = 1 (−1) if µ > 0 (< 0), n = (sin θ, 0, cos θ ), and Ω and θ are defined by h¯ 2 kkz 2 1/2 h¯ Ω = (2µB0 )2 + 2µB + hω − , ¯ M
(17)
and µ (2µB + hω 2|µ|B0 ¯ − h¯ 2 kkz /M) (18) , cos θ = . hΩ hΩ ¯ ¯ Note that Ω and θ depend not only on the model parameters, but also on the eigenvalue kz . Since s · n = exp(−iθ sy )sz exp(iθ sy ), the solutions of the Eq. (16) can be expressed in terms of χα0 , the eigenstate of sz , satisfying sz χα0 = (1/2)αχα0 , where α = ±. The results are sin θ =
Ekzsz α =
h¯ 2 kz2 h¯ 2 k 2 1 + − αµ h¯ Ω, 2M 8M 2
(19)
and χα = exp(−iθ sy )χα0 .
(20)
In conclusion, the eigenvalues of Heff are h¯ 2 kz2 h¯ 2 k 2 1 xy eff α + , Enk = E − ≡ E + hΩ, E ¯ nk µ nk n z z α z 2 2M 8M and the corresponding eigenfunctions are ψnkz α (x) = ϕn (x, y)
exp(ikz z) χα . √ Lz
(21)
(22)
It should be remarked that the above eigenvalues are not observable quantities because Heff is not a physical operator. Now we take Ψ (x, 0) = exp(ikzsz )ψnkz α (x)
(23)
as an initial state and study its time evolution. According to Eq. (10), the state at time t is eff eff Enk Enk exp(ikz z) zα zα Ψ (x, t) = exp −i exp −i(ωt − kz)sz χα . t W (z, t)ψnkz α (x) = exp −i t ϕn (x, y) √ h¯ h¯ Lz (24) We define the spin vector as (Ψ (x, t), sΨ (x, t))s (25) , (Ψ (x, t), Ψ (x, t))s where the subscript s indicates that the inner product is performed only in the space of the spin state. For the above state, we find v(x, t) = χα , W † (z, t)sW (z, t)χα . (26) v(x, t) =
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It is easy to show that (χα , sχα ) = (1/2)αn, and according to Eq. (7) we obtain 1 v(x, t) = α sin θ cos(ωt − kz), sin θ sin(ωt − kz), cos θ . (27) 2 Therefore, the spin vector rotates around the z axis with the same frequency as the magnetic field, though at a different angle θ with the z axis. It also propagates along the positive z direction with the same speed as the magnetic field, though there is a phase lag π when α = −. Return to Eq. (24), it is easy to see that this is a cyclic solution in the time interval [0, τ ] where τ = 2π/ω, because χα can be expressed as a linear combination of χ+0 and χ−0 . More specifically, we have Ψ (x, τ ) = eiδ Ψ (x, 0),
(28)
with the phase change eff Enk zα
τ − απ, mod 2π. h¯ It is easy to calculate the mean value of H in the state Ψ (x, t) with the result δ=−
1 eff H = Ψ (x, t), H Ψ (x, t) = Enk + α hω ¯ cos θ. zα 2 Thus the dynamic phase is 1 β =− h¯
τ H dt = −
eff Enk zα
0
h¯
τ − απ cos θ.
(29)
(30)
(31)
The nonadiabatic geometric phase is then γ = δ − β = −απ(1 − cos θ ),
mod 2π.
(32)
For any fixed x, v(x, t) is a periodic function of t, and it traverses a closed trace on a sphere of radius 1/2 in the time interval [0, τ ]. The solid angle subtended by this closed trace is obviously Ωv = α2π(1 − cos θ ),
mod 4π.
(33)
Therefore, we obtain the following relation 1 γ = − Ωv , mod 2π. (34) 2 For spin s = 1/2 and time varying magnetic fields (independent of the space variable), this is well known. Here we obtain it in a both time and space varying magnetic field. Next we consider a more general initial state Ψ (x, 0) = ϕn (x, y)
exp(ikz z) exp(ikzsz )χ, √ Lz
(35)
2 + c2 = 1. The corresponding state at time where χ = c+ χ+ + c− χ− with c+ and c− arbitrary except satisfying c+ − t is Enkz exp(ikz z) Ψ (x, t) = exp −i (36) t ϕn (x, y) √ exp −i(ωt − kz)sz exp(iµ Ωts · n)χ. h¯ Lz The spin vector in this state is v(x, t) = χ, exp(−iµ Ωts · n)W † (z, t)sW (z, t) exp(iµ Ωts · n)χ . (37)
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We denote u(t) = χ, exp(−iµ Ωts · n)s exp(iµ Ωts · n)χ = v 0 − (v 0 · n)n cos Ωt + (v 0 × n)µ sin Ωt + (v 0 · n)n, where we have introduced the notation v 0 = (χ, sχ), and used the formula [15,31] exp(−iφs · n)s exp(iφs · n) = s − (s · n)n cos φ + (s × n) sin φ + (s · n)n.
(38)
(39)
The spin vector is then given by v(x, t) = ux (t) cos(ωt − kz) − uy (t) sin(ωt − kz), ux (t) sin(ωt − kz) + uy (t) cos(ωt − kz), uz (t) . (40) This result is somewhat complicated, but the physical picture is clear. We recast u(t) in the form u(t) = v0⊥ cos Ωte1 − v0⊥ µ sin Ωte2 + v0 e3 ,
(41)
where e1 = v 0 − (v 0 · n)n /v0⊥ , (42) e2 = n × v 0 /v0⊥ , e3 = n, 2 . It is easy to show that {e , e , e } is an orthonormal set and constitutes a rightand v0 = v 0 · n, v0⊥ = v 20 − v0 1 2 3 handed frame. Eq. (41) is essentially the same as Eq. (38). However, in this form it helps us recognize the physical picture of the motion. Now to get the vector v(x, t), one just rotates v 0 around e3 = n through an angle −µ Ωt (positive angle corresponds to anti-clockwise rotation) to get u(t), and then rotates u(t) around the z axis through an angle (ωt − kz). Thus the phase (ωt − kz) of the plane electromagnetic wave always plays an important role in determining the direction of the spin vector. The resulted motion involves nutation as well as rotation, and is not periodic in general. Two cases with periodic v(x, t) are available. First, if the initial condition is such that v 0 = (1/2)αn, we have u(t) = (1/2)αn, and v(x, t) is given by Eq. (27). This has been discussed above. Second, if the parameters of the system and the eigenvalue kz are such that Ω/ω is a rational number, then both Ωt and ωt may simultaneously become integral multiples of 2π at some later time τ , and we have v(x, τ ) = v(x, 0), independent of the initial condition. The solution Ψ (x, t) is indeed cyclic in this case. Assume that Ω/ω = K2 /K1 , where K2 and K1 are both natural numbers and prime to each other, and τ satisfies ωτ = 2πK1 and Ωτ = 2πK2 . Because χ , a linear combination of χ+ and χ− , can also be expressed as a linear combination of χ+0 and χ−0 , we see from Eq. (36) that Ψ (x, τ ) = eiδ Ψ (x, 0), with the phase change δ=−
Enkz τ + (µ K2 − K1 )π, h¯
(43)
mod 2π.
It is not difficult to find that H = Ψ (x, t), H Ψ (x, t) = Enkz − µ h¯ Ω(v 0 · n) + hωu ¯ z (t).
(44)
Thus the dynamic phase is β =−
Enkz τ + (µ 2πK2 − 2πK1 cos θ )(v 0 · n). h¯
(45)
The nonadiabatic geometric phase is then γ = (µ K2 − K1 )π − (µ 2πK2 − 2πK1 cos θ )(v 0 · n),
mod 2π.
(46)
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At the present case, v(x, t) for a fixed x traverses a closed trace on a sphere of radius |v 0 | [it is obvious that |v(x, t)| = |u(t)| = |v 0 |] in the time interval [0, τ ]. The solid angle subtended by this closed trace is 1 Ωv = |v 0 |
τ 0
vx v˙y − vy v˙x dt. |v 0 | + vz
(47)
It is easy to show that
vx v˙y − vy v˙x = ux u˙ y − uy u˙ x + ω u2x + u2y .
(48)
Because uz = vz , and u2 = v 20 , we have u2x + u2y = v 20 − vz2 . Therefore, 1 Ωv = |v 0 |
τ 0
ux u˙ y − uy u˙ x ω dt + |v 0 | + uz |v 0 |
τ
|v 0 | − uz dt.
(49)
0
The second integral can be calculated easily, and the first is recognized as the solid angle subtended by the trace of u(t), which is very easy to calculate in the coordinate frame expanded by {e1 , e2 , e3 }. The final result is v0 · n v0 · n Ωv = −µ 2πK2 1 − (50) + 2πK1 1 − cos θ , mod 4π. |v 0 | |v 0 | Here the first term is due to the rotation around e3 = n, and the second is due to the further rotation around the z axis. Compared with Eq. (46), and notice that for spin 1/2 we always have |v 0 | = 1/2, we reach the result (34) once again. From Eqs. (38) and (40) we see that the spin vector at the initial time is given by v(x, 0) = (v0x cos kz + v0y sin kz, −v0x sin kz + v0y cos kz, v0z ).
(51)
It varies with z, and such states may be called twisted polarized states. Even the simpler case (23) have a similar initial spin vector [see Eq. (27)]. Thus such initial state may be difficult to prepare. However, in the special case c+ = cos(θ/2), c− = − sin(θ/2), we have χ = χ+0 and the initial state (35) becomes exp[i(kz + k/2)z] 0 (52) χ+ . √ Lz This is polarized along the positive z direction, and has a definite momentum along that direction. When V = 0 and ϕn (x, y) = 1/ Lx Ly it becomes even simpler. Such a state might be easy to realize in experiment. At later times, however, the state contains polarizations along both positive and negative z directions, since Ψ (x, t) can be expressed as Enkz Ωt Ωt exp[i(kz + k/2)z] 0 1 + iµ cos θ sin ϕn (x, y) Ψ (x, t) = exp −i χ+ t − i ωt cos √ 2 2 2 h¯ Lz Enkz Ωt exp[i(kz − k/2)z] 0 1 ϕn (x, y) + exp −i χ− . (53) t + i ωt iµ sin θ sin √ 2 2 h¯ Lz Thus an initial particle beam with a definite momentum and spin polarization in the positive z direction is split into two beams at later times, one with the original momentum and polarization and the other with a different momentum and the opposite polarization. It is easy to show that the operator Ψ (x, 0) = ϕn (x, y)
F = pz − hks ¯ z
(54)
commutes with H and thus is a conserved quantity. The above state is an eigenstate of F with eigenvalue h¯ kz at all times. In other words, a proper linear combination of pz and sz is the same for the two beams. The case with c+ = sin(θ/2), c− = cos(θ/2) and χ = χ−0 can be discussed similarly.
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Now we return to Eq. (15) for general s. A similar equation has arisen in solving the Schrödinger equation in a helical magnetic field, and the case s = 1 has been solved analytically [24]. For larger s, it is more difficult to solve it, because the secular equation for the eigenvalue is an algebraic equation of degree 2s + 1. However, some general results can be asserted without specific calculations. First, because the matrix on the left-hand side of Eq. (15) is a (2s + 1) × (2s + 1) Hermitian one, it has 2s + 1 real eigenvalues. We denote these eigenvalues as (h¯ 2 k 2 /2M)λi (i = 1, 2, . . . , 2s + 1), where λi depends on all model parameters (except those in V ) and kz , but is dimensionless. Second, we have 2s + 1 linearly independent eigenvectors, which will be denoted as ξi . Third, if ξi corresponds to a nondegenerate eigenvalue, it is real up to an overall phase factor, because both sz and sx are real matrices. In such an eigenvector, the mean value (ξi , sy ξi ) is imaginary because the matrix sy is pure imaginary. On the other hand, sy is Hermitian and the above mean value should be real. Therefore, (ξi , sy ξi ) = 0, and this does not depend on the representation employed since a mean value is representation independent. Though it is not clear whether some of the eigenvalues are degenerate, the first and second conclusions are sufficient for the subsequent discussions. On account of the above conclusions, the eigenvalues of Heff are now h¯ 2 kz2 h¯ 2 k 2 + λi , 2M 2M and the corresponding eigenfunctions are xy
eff = En + Enk zi
ψnkz i (x) = ϕn (x, y)
exp(ikz z) ξi . √ Lz
(55)
(56)
Now we take Ψ (x, 0) = exp(ikzsz )ψnkz i (x) as an initial state and study its time evolution. According to Eq. (10), the state at time t is eff Enk exp(ikz z) zi exp −i(ωt − kz)sz ξi . t ϕn (x, y) √ Ψ (x, t) = exp −i h¯ Lz
(57)
(58)
It is easy to find the spin vector in this state. We define v 0 = (ξi , sξi ), then the result reads v(x, t) = v0xy cos(ωt − kz + Λ0 ), v0xy sin(ωt − kz + Λ0 ), v0z , (59) 2 + v 2 , and Λ is determined by cos Λ = v /v where v0xy = v0x 0 0 0x 0xy and sin Λ0 = v0y /v0xy . If ξi corresponds to 0y a nondegenerate eigenvalue, then v0y = 0 and Λ0 is 0 or π . Once again we find that the spin vector rotates around the z axis, and propagates along the positive z direction, both in the same way as the magnetic field. It is easy to see that Eq. (58) is a cyclic solution in the time interval [0, τ ] where τ = 2π/ω, because ξi can be expressed as a linear combination of χm0 , the eigenstates of sz . More specifically, we have Ψ (x, τ ) = eiδ Ψ (x, 0), with the phase change eff Enk zi
τ − 2πs, mod 2π. h¯ The dynamic phase can be found to be δ=−
eff Enk zi
τ − 2πv0z , h¯ and the nonadiabatic geometric phase is then β =−
γ = −2πs + 2πv0z ,
mod 2π.
(60)
(61)
(62)
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For a fixed x, v(x, t) is a periodic function of t, and it traverses a closed trace on a sphere of radius |v 0 | in the time interval [0, τ ]. The solid angle subtended by this closed trace is obviously v0z , mod 4π. Ωv = 2π 1 − (63) |v 0 | Therefore, we obtain the following relation γ = −|v 0 |Ωv + 2π |v 0 | − s , mod 2π.
(64)
In this case γ contains two terms. The first is the familiar one that is proportional to Ωv . The second is an extra term which depends on the initial spin state ξi . It cannot be omitted since |v 0 | − s is in general not an integer. Similar results have been encountered in other systems [15,16,32–34]. If s = 1/2, we have |v 0 | = 1/2, and the above result reduces to Eq. (34) as expected. In summary, we have solved the Schrödinger equation for a neutral particle with general spin and magnetic moment moving in the field of a circularly polarized plane electromagnetic wave, plus a uniform magnetic field and a two-dimensional potential. The time evolution operator is obtained by using a position- and time-dependent unitary transformation, which reduces the original equation to a similar one with a time-independent effective Hamiltonian. The time evolution operator involves no chronological product and is convenient for practical calculations. The eigenvalue problem of the effective Hamiltonian is solved analytically. Though some details of the spin part requires further study when s > 1/2, the knowledge attained is sufficient for our discussions. For all spin values we find simple cyclic solutions in which the spin vector rotates with the same frequency and propagates with the same speed and in the same direction as the magnetic field. For spin 1/2, more complicated cyclic solutions are presented. The nonadiabatic geometric phases for all the cyclic solutions are calculated. The relation between the nonadiabatic geometric phase and the solid angle subtended by the closed trace of the spin vector is obtained in each case. For spin 1/2, we also find that an initial particle beam with a definite momentum and spin polarization in the propagation direction of the plane electromagnetic wave is split into two beams at later times, one with the original momentum and polarization and the other with a different momentum and the opposite polarization. A proper linear combination of the momentum and spin along the original direction is the same for the two beams.
Acknowledgements This work was supported by the National Natural Science Foundation of the People’s Republic of China (grant number 10275098).
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