Anisotropic Harmonic Oscillator in a Static Electromagnetic Field

Transcription

1 Commun. Theor. Phys. Beijing, China pp c International Academic Publishers Vol. 38, No. 6, December 15, 00 Anisotropic Harmonic Oscillator in a Static Electromagnetic Field LIN Qiong-Gui Department of Physics, Sun Yat-Sen Zhongshan University, Guangzhou 51075, China China Center of Advanced Science Technology World Laboratory, P.O. Box 8730, Beijing , China Received September 18, 001; Revised July 4, 00 Abstract A nonrelativistic charged particle moving in an anisotropic harmonic oscillator potential plus a homogeneous static electromagnetic field is studied. Several configurations of the electromagnetic field are considered. The Schrödinger equation is solved analytically in most of the cases. The energy levels wave functions are obtained explicitly. In some of the cases, the ground state obtained is not a minimum wave packet, though it is of the Gaussian type. Coherent squeezed states their time evolution are discussed in detail. PACS numbers: Ge, 4.50.Dv Key words: anisotropic harmonic oscillator, electromagnetic field, exact solutions, coherent states, squeezed states 1 Introduction The anisotropic harmonic oscillator potential is of physical interest in quantum mechanics, since it may describe the motion of, say, an electron in an anisotropic metal lattice. With external electromagnetic fields imposed on, such models may also be useful in semiconductor physics. [1,] For an isotropic harmonic oscillator potential a homogeneous static magnetic field the Schrödinger equation can be easily solved in the cylindrical coordinates. If the harmonic oscillator potential is anisotropic, the problem is not so easy. Nevertheless, it can be solved analytically because the Hamiltonian is quadratic in the canonical coordinates momenta. There exist quite general formalisms for solving systems with such quadratic Hamiltonians, [3] the above problem represents a typical example. [4] In Ref. [4] only the isotropic case is considered as an example for illustrating the general method. The anisotropic case with a homogeneous static magnetic field was studied in Refs. [5] [6]. In these works the magnetic field was arranged to point to one of the rectangular axis directions. In this paper we will solve the problem in a somewhat easier formalism. A homogeneous static electric field is also included, though its effect is trivial. We will also deal with a case the magnetic field has a more general direction. This seems not having been considered before. Coherent squeezed states of these systems will be discussed in some detail. In terms of these concepts the time evolution of some wave packets can be discussed very conveniently. In the next section we consider a charged harmonic oscillator with arbitrary frequencies ω x, ω y ω z in three rectangular axis directions. A homogeneous static magnetic field in the z direction a homogeneous static electric field in an arbitrary direction are imposed to the system. The problem can be easily reduced to one equivalent to the case without an electric field by making some coordinate transformation. In Sec. 3 we diagonalize the reduced Hamiltonian give the energy eigenvalues. Although this has been previously done, the method used here is different seems more convenient, especially for the subsequent discussions of coherent squeezed states. The wave functions are explicitly worked out in Sec. 4. In Sec. 5 we discuss some special cases, some of which may need different hlings, or even cannot be solved. In Sec. 6 we consider a somewhat different case, ω z = 0 E z = 0 E z is the z component of the electric field, but the magnetic field has an x or y component, in addition to the z component. To our knowledge, this was not considered before. It is solved analytically by the formalism of Sec. 3. In Sec. 7 we discuss the coherent squeezed states their time evolution in detail. A brief summary is given in Sec. 8. Reduction of the Hamiltonian Consider a charged particle with charge q mass M, moving in an anisotropic harmonic oscillator potential, a homogeneous static magnetic field B = Be z, a homogeneous static electric field E = E x e x + E y e y + E z e z, e x, e y e z are unit vectors of the rectangular coordinate system, B, E x, E y E z are all constants. The stationary Schrödinger equation is H T Ψx = E T Ψx, 1 The project supported by National Natural Science Foundation of China Foundation of the Advanced Research Center of Sun Yat-Sen University qg lin@163.net

2 668 LIN Qiong-Gui Vol. 38 the Hamiltonian is H T = 1 p q M c A 1 + Mω xx + ωyy + ωzz qe x, E x = E x x + E y y + E z z. Since the magnetic field points to the z direction, one may take A x A y independent of z, A z = 0. Then the Hamiltonian can be decomposed as H T = H xy + H z, H xy = 1 M [ p x q c A x + p y q ] c A y + 1 Mω xx + ω yy qe x x + E y y, 3 H z = 1 M p z + 1 Mω zz qe z z. 4 Therefore the total wave function can be factorized as Ψx = ψx, yzz, ψx, y Zz satisfy the following equations H xy ψx, y = E xy ψx, y, 5 H z Zz = E z Zz 6 with E xy + E z = E T. Obviously, H z can be recast in the form H z = 1 M p z + 1 Mω zz z 0 q Ez Mωz, 7 z 0 = qe z /Mωz, so that equation 6 can be easily solved with the following results, En z 3 = n hω z q Ez Mωz, n 3 = 0, 1,,..., 8 Z n3 z = ψn ωz 3 z z 0, n 3 = 0, 1,,..., 9 we have quoted the stard wave functions for harmonic oscillators ψnx ω Mω 1/ = n n! exp Mω Mω π h h x H n h x, n = 0, 1,, Therefore our main task is to solve Eq. 5. We take the gauge A x = 1 Bỹ, A y = 1 B x, 11 x = x x 0, ỹ = y y 0, x 0 = qe x /Mω x, y 0 = qe y /Mω y, then H xy can be written as H xy = 1 M p x + p y + 1 Mω 1 x + 1 Mω ỹ ω B Lz q Ex Mωx q Ey Mωy, 1 L z = xp y ỹp x, ω B = qb/mc, ω 1 = ωx + ωb, ω = ωy + ω B. 13 Note that ω B may be either positive or negative, depending on the signs of q B. Now equation 5 can be written as Hψx, y = Eψx, y, 14a H = 1 M p x + p y + 1 Mω 1 x + 1 Mω ỹ ω B Lz, 14b E = E xy + q Ex Mωx + q Ey Mωy. 15 Equation 14 has the same form as that for an anisotropic harmonic oscillator moving on the xy plane under the influence of a homogeneous static magnetic field perpendicular to the plane, except that x, y are replaced by x, ỹ. The main feature is that the reduced Hamiltonian is quadratic in the canonical variables. Though this equation has been studied by some authors, we will solve it in a somewhat different simpler way in the next two sections. Here we point out that the calculations below will be simpler if one takes the gauge A x = Bỹ, A y = 0 or A x = 0, A y = B x. However, we prefer the above gauge 11, since in this gauge it would be convenient to compare the results with known ones when ω x = ω y, or ω 1 = ω. 3 Diagonalization of the Reduced Hamiltonian If ω x = ω y, equation 14 can be easily solved in the cylindrical coordinates. In the general case this does not work, so other methods are necessary. We define a column vector X = x, p x, ỹ, p y τ, 16 τ denotes matrix transposition, then the reduced Hamiltonian 14b can be written as H = 1 Xτ HX = 1 X HX, 17 H is a c-number matrix which we take to be symmetric. We do not write down H here since this is easy. If one can diagonalize H then the Hamiltonian would become a sum of two one-dimensional Hamiltonians for harmonic oscillators or something similar repulsive harmonic oscillators or free particles. However, a crucial point here is to preserve the commutation relation [X α, X β ] = hσ y αβ, α, β = 1,, 3, 4, 18 after the linear transformation that diagonalizes H, Σ y = diagσ y, σ y, σ y is the second Pauli matrix. Therefore the needed transformations are the so-called symplectic ones. [3,4] Here we will transform the Hamiltonian into a form expressed in terms of raising lowering operators, so the formalism is somewhat different. Let us proceed as follows.

3 No. 6 Anisotropic Harmonic Oscillator in a Static Electromagnetic Field 669 Because H is quadratic in the canonical variables X, it is easy to show that [ih, X] = hωx, 19 Ω = iσ y H 0 is another c-number matrix. As both H iσ y are real, so is Ω. Our first step is to diagonalize Ω, so we write down it here, 0 1/M ω B 0 Mω1 0 0 ω B Ω = ω B 0 0 1/M. 1 0 ω B Mω 0 The characteristic polynomial for this matrix is Since detλi Ω = λ 4 + bλ + c, a b = ω x + ω y + 4ω B, c = ω xω y. b b 4c = ω x ω y +8ω Bω x +ω y +ω B 0, 3 we have two real roots for λ. If ω x = 0 or ω y = 0, one of the two roots is zero, the following discussions are not valid. We will return to this case latter. Currently we assume that both ω x ω y are nonzero, then c > 0 < b, both roots for λ are negative. Therefore the above characteristic polynomial has four pure imaginary roots {λ 1, λ, λ 3, λ 4 } = { iσ 1, iσ 1, iσ, iσ }, 4 b + 1/, b 1/ σ 1 = σ =, 5 σ 1 σ > 0. The equal sign appears when ω B = 0 ω x = ω y. Because Ω is not symmetric, right eigenvectors left ones are different. We define two left eigenvectors row vectors u i corresponding to the eigenvalues iσ i i = 1, by u i Ω = iσ i u i, i = 1,, 6 then the other two are u i, corresponding to the eigenvalues iσ i i = 1,. Similarly, the right eigenvectors column vectors are v i vi i = 1,, satisfying the equations Ωv i = iσ i v i, i = 1,, 7 the complex conjugate ones. From these eigenvalue equations it can be shown that u i v j = u i v j = 0, i, j = 1,, 8a u i v j = u i v j = 0 for σ i σ j this should be checked individually when σ 1 = σ, see below. By appropriately choosing the normalization constants we have u i v j = u i v j = δ ij, i, j = 1,. 8b The specific expressions for the above eigenvectors are not necessary in this section. However, a relation between the left right eigenvectors is very useful in the following. Using Eqs. 0, 6, 7, it is easy to show that v i Σ y u i, so we choose v i = Σ y u i, i = 1,. 9 Note that the condition 8b only determines the product of the normalization constants in u i v i. With this relation both the constants are fixed up to a phase factor. Also note that equations 8b 9 lead to u i Σ y u i = 1 no summation over i on the left-h side. The left-h side of this equation is real which can be easily shown, but not necessarily positive. Actually it can be shown by using the eigenvalue equation that u i Σ y u i = u i Σ y Hu i Σ y /σ i. Thus the sign of this quantity depends on the matrix H, in general equation 9 should be replaced by v i = ɛ i Σ y u i, ɛ i = ±1. For the present case, however, equation 9 is sound. Now we define a 4 4 matrix Q by arranging the column vectors in the following order Using Eq. 8 it is easy to show that Q = v 1, v 1, v, v. 30 Q 1 = u τ 1, u τ 1, u τ, u τ τ. 31 With the help of the eigenvalue equations 6 7, we have Q 1 ΩQ = diag iσ 1, iσ 1, iσ, iσ. 3 Therefore the matrix Ω is diagonalized. Using the relation 9, it is not difficult to show that Q = Σ z Q 1 Σ y, 33 Σ z = diagσ z, σ z, σ z is the third Pauli matrix. This relation is very useful in the following. Next we will diagonalize the Hamiltonian. With the above results this is easy. We define two operators a i = u i X/ h, i = 1,. 34a Their hermitian conjugates are a i = u i X/ h, i = 1,. 34b Using Eqs. 18, 8, 9 it is easy to show that the nonvanishing commutators among these operators are [a i, a j ] = δ ij, i, j = 1,. 35 Therefore they are similar to the raising lowering operators for harmonic oscillators. We define a column vector A = a 1, a 1, a, a τ, 36

4 670 LIN Qiong-Gui Vol. 38 then equation 34 can be written in the matrix form The inverse is thus A = Q 1 X/ h. 34 X = hqa, 37 X τ = X = ha Q. 38 Substituting these into Eq. 17, using Eqs. 0, 3, 33, it is rather easy to show that H = 1 ha ΣA, 39 Σ = diagσ 1, σ 1, σ, σ. Using the commutation relations 35 this becomes H = hσ 1 a 1 a hσ a a Therefore the Hamiltonian is diagonalized. More specifically, it becomes the sum of two one-dimensional harmonic oscillator Hamiltonians. The energy levels are readily available E n1n = hσ 1 n hσ n + 1, n 1, n = 0, 1,, When ω B = 0, we have σ 1 = maxω x, ω y σ = minω x, ω y, these energy levels reduce to known results, as expected. Substituting these into Eq. 15 we obtain the energy levels for the motion on the xy plane En xy 1n = hσ 1 n hσ n + 1 q Ex Mωx q Ey Mωy, n 1, n = 0, 1,,... 4 The total energy is the sum of E xy n 1n E z n 3 given by Eq. 8. The wave functions on the xy plane will be worked out in the next section. The method employed in this section is similar to that used in Ref. [7], which dealt with a charged particle moving in a rotating magnetic field also studied in Refs. [8] [9]. However, we present here a relation between the right left eigenvectors, which leads to the very useful relation 33, so that the calculation here seems more straightforward easier. The method can be easily extended to systems with more degrees of freedom. 4 The Wave Functions in the Coordinate Representation The eigenstates of the Hamiltonian 40 corresponding to the eigenvalues 41 are n 1 n = 1 n1!n! a 1 n1 a n 00, n 1, n = 0, 1,,..., is the ground state satisfying a 1 00 = a 00 = The task of this section is to work out these states in the coordinate representation, that is, the wave functions ψ n1n x, y = xy n 1 n. 45 For this purpose the left eigenvectors are necessary. These are given by u i = K i imσ i σ i ω y ω B, σ i ω y, Mω B σ i + ω y, iω B σ i, i = 1,, 46a their complex conjugates, the normalization constants are K i = {Mσ i [σ i ω y + 4ω Bω y]} 1/. 46b When ω B = 0, {σ i i = 1, } = {ω x, ω y }. For ω x, the last two components of u i vanish. For ω y, the first two components of u i vanish. Thus the orthogonality is ensured. If the further condition ω x = ω y holds which yields σ 1 = σ, one may take the limit ω y ω x after everything is worked out. First we should work out the ground state. In the coordinate representation, equation 44 takes the form ξ ij x j i hη ij j ψ 00 x, y = 0, i = 1,, 47 x 1 = x, x = y similarly for x 1 x, j = / x j = / x j, we have defined two matrices u11 u 13 ξ = ξ ij =, u 1 u 3 u1 u 14 η = η ij =, 48 u u 4 u iβ i = 1,, β = 1,, 3, 4 is the β-th component of u i. Suppose that ψ 00 x, y = N 0 exp[ sx, y], sx, y = x i Λ ij x j / h, 49 S is a symmetric matrix whose elements are complex numbers, N 0 is a normalization constant. Substituting this into Eq. 47 we obtain Λ = iη 1 ξ. 50 This can be easily worked out. The elements are Λ 11 = Mω xσ 1 + σ ω x + ω y hλ x, 51a Λ = Mω yσ 1 + σ ω x + ω y hλ y, 51b Λ 1 = Λ 1 = imω Bω x ω y ω x + ω y i hλ xy, 51c

5 No. 6 Anisotropic Harmonic Oscillator in a Static Electromagnetic Field 671 the relation σ 1 σ = ω x ω y has been used. Therefore the ground-state wave function is ψ 00 x, y = N 0 exp 1 λ x x 1 λ yỹ iλ xy xỹ, 5a the normalization constant is λx λ [ y Mσ1 + σ ω x ω y ] 1/ N 0 = =. 5b π π hω x + ω y When ω B = 0, we have λ x = Mω x / h, λ y = Mω y / h, λ xy = 0. On the other h, if ω x = ω y, we have ω 1 = ω, σ 1 = ω 1 + ω B, σ = ω 1 ω B, λ x = λ y = Mω 1 / h, λ xy = 0. These are all expected results. It is remarkable that the uncertainty relation for the above ground state is x p x = y p y = h 1 + λ 1/ xy h λ xλ y. 53 The equal sign holds only when B = 0 or ω x = ω y. Therefore the ground state is in general not a minimum wave packet, though it is of the Gaussian type. It seems that this is not noticed in the previous literature. For any function F x, y, it is easy to show that ξ ij x j i hη ij j F x, y = i h exp[s x, y]η ij j {exp[ s x, y]f x, y}, i = 1,. 54 According to Eq. 43, using the above relation we obtain n1+n i h 1 ψ n1n x, y = N 0 K n1 1 n1!n! Kn exp λ x x + 1 λ yỹ iλ xy xỹ [σ 1 ω y x iω B σ 1 y ] n1 [σ ω y x iω B σ y ] n exp λ x x λ yỹ Some Special Cases In this section we discuss some special cases some of the parameters are zero. i If ω z = 0 E z = 0, the motion in the z direction is free, E z 0 is continuous. ii If ω z = 0 but E z 0, the motion in the z direction is that in a homogeneous electric field, which is also well known, E z is also continuous, can take any real value. [10] iii In the preceding sections we have assumed that both ω x ω y are nonzero. In this case, if any one of E x or E y is zero, it can be substituted into the above results directly no modification to the formalism is needed. However, if one of ω x ω y is zero, the situation is different. We would deal with ω y = 0 in the following. The other case could be discussed in a similar way. iv If ω y = 0 but E y 0, the Hamiltonian could not be reduced to a quadratic form, to our knowledge the problem could not be solved analytically. v If ω y = 0 E y = 0, we choose the gauge A x = 0, A y = Bx, let ψx, y = 1 π expikyφx, 56 then φx satisfies the equation [ 1 M p x + 1 M ω 1x x k ] φx = E x φx, 57 ω 1 = ω x + 4ω B, x k = qe x + k hω B M ω 1 E x = E xy + qe x + k hω B M ω 1, 58 h k M. 59 The energy levels wave functions for Eq. 57 are readily available, so the final results are E xy n = h ω 1k 1 qe x + k hω B n M ω 1 + h k M, n 1 = 0, 1,,..., k, +, 60 the corresponding wave functions are ψ n1kx, y = 1 π expikyψ ω1 n 1 x x k. 61 If further ω x or E x or both are zero, the results can be obtained from the above ones directly. vi Finally, if both ω x ω y are zero, one can appropriately choose the coordinate axes of the xy plane such that E y = 0. Thus it is a special case of the above case v. 6 A More General Magnetic Field In this section we set ω z = 0 E z = 0, consider a magnetic field with an x or y component but not both in addition to the z component. To our knowledge this was not considered previously. Without loss of generality we take B = B e x + Be z. 6 The gauge convenient for the present case might be A x = 0, A y = B x = Bx x 0, A z = B y, 63 x 0 is the same as defined before. In this gauge the total Hamiltonian takes the form H T = 1 M p x + p y + 1 M p z Mω By + 1 M ω 1 x + 1 Mω yy qe y y ω B xp y q Ex Mωx, 64

6 67 LIN Qiong-Gui Vol. 38 ω B ω 1 are the same as defined before, ω B = qb /Mc. p z is obviously a conserved quantity, so we make the factorization Ψx = 1 π expikzψx, y. 65 The Schrödinger equation 1 is then reduced to the form 14a now H = 1 M p x +p y+ 1 M ω 1 x + 1 M ω ỹ ω B xp y, 66 E = E T + q Ex Mωx + qe y + k hω B M ω h k M. 67 In these equations ỹ = y y k, which is different from the previous one, ω = ω y + 4ω B, y k = qe y + k hω B M ω. 68 Now the reduced Hamiltonian H is quadratic, different from Eq. 14b only in the last term. Thus the problem can be solved in much the same way as before. We only give the results here. The energy levels are E T n 1n k = hσ 1 n hσ n + 1 q Ex Mωx qe y + k hω B M ω + h k M, n 1, n = 0, 1,,..., k, +, 69 σ 1 σ are given by Eqs. 5, b, 3, but with ω y replaced by ω. The wave functions are given by Ψ n1n kx = 1 π expikzψ n1n kx, y, 70 ψ n1n kx, y has the same form as Eq. 55 with ω y replaced by ω every, including in the expressions for λ x, λ y, K 1 K, but here λ xy = Mω B ω hω x + ω. 71 Also note that ỹ depends on k, which is the reason why ψ n1n kx, y has the subscript k. The orthonormal relation is Ψ n 1 n k xψ n 1n kx d x = δk k δ n1n 1 δ n n. 7 By the way, the wave functions 61 satisfy a similar orthonormal relation. As before, the ground state obtained here, though being of the Gaussian type, is not a minimum wave packet, except when B = 0 then B = B e x. 7 Coherent Squeezed States Coherent squeezed states are useful objects in quantum mechanics quantum optics. These are widely discussed in the literature. In the presence of magnetic fields, some discussions can be found in Ref. [11]. There are also examples on other applications of such states. [1] However, such states for anisotropic harmonic oscillators in the presence of magnetic fields are not discussed previously, to the best of our knowledge. In this section we will discuss the definition, the properties the time evolution of such states for the case treated in Secs The discussions can be easily extended to the case of Sec. 6. Since the motion in the z direction is separable is essentially that of a usual harmonic oscillator, we only discuss the coherent squeezed states on the xỹ plane which is a simple translation of the xy plane, their time evolution is governed by the Hamiltonian H given by Eq. 14a or 40 the additional constants in H xy has only trivial consequence for the time evolution. The definitions used below are natural generalizations of those for a single harmonic oscillator. [13 15] We define a unitary displacement operator Dα 1, α = D 1 α 1 D α, D i α i = expα i a i α i a i, i = 1,, 73 the α i s are complex numbers. For an arbitrary state ϕ one can define a corresponding displaced state ϕ, α 1 α D = Dα 1, α ϕ = D 1 α 1 D α ϕ. 74 It is easy to show that D α 1, α a i Dα 1, α = a i + α i, i = 1,, 75 so if ϕ = 00 is the ground state we have a i 00, α 1 α D = α i 00, α 1 α D, i = 1,. 76 Thus the coherent states may be defined as the displaced ground state 00, α 1 α D. In terms of the original variables the displacement operator takes the form Dα 1, α = exp i ī h xd i p D i exp h pd i x i exp ī h xd i p i, 77 x D i = i hα j η ji c.c., p D i = i hα j ξ ji c.c.. 78 If the wave function for the state ϕ is ϕ x, ỹ, then the one for the displaced state ϕ, α 1 α D is ϕ D α 1α x, ỹ = exp i ī h xd i p D i exp h pd i x i ϕ x x D, ỹ y D. 79 Thus we see the reason why the operator Dα 1, α is called a displacement operator. From this Eq. 5 it is easy to obtain the wave function for the coherent states. Obviously the displaced state the original one have the same shape in the configuration space, thus the

7 No. 6 Anisotropic Harmonic Oscillator in a Static Electromagnetic Field 673 uncertainty X α in ϕ, α 1 α D is the same as that in ϕ. This can also be easily shown by using Eqs Now consider the time evolution of the displaced states. It is easy to show that e iht/ h a i e iht/ h = expiσ i ta i, i = 1,. 80 If the state at the initial time t = 0 is ψ0 = ϕ, α 1 α D, then the state at the time t is ψt = ϕt, α 1t α t D = D 1 α 1t D α t ϕt, 81 α it = exp iσ i tα i ϕt = e iht/ h ϕ. Therefore if ϕt is known, ψt can be obtained by a time-dependent displacement. A simple special case is ϕ = n 1 n, that is ψ0 = n 1 n, α 1 α D, 8 a displaced number state. In this case [ ϕt = exp i n σ 1 t i n + 1 ] σ t n 1 n, [ ψt = exp i n σ 1 t i n + 1 ] σ t n 1 n, α 1t α t D. 83 This means that ψt is also a displaced number state, except that the displacement parameters are timedependent, a time-dependent phase factor is gained. The position velocity of the wave packet ψt, x, ỹ = xỹ ψt is characterized by the expectation values x c αt = ψt X α ψt. Using Eq. 37 it is easy to show that x c αt = h [exp iσ j tα j v jα + c.c.]. 84 By straightforward calculations one can show that they satisfy the following equation ẋ c αt = Ω αβ x c βt, 85 the eigenvalue equation 7 has been used, at t = 0 they give the same results as those obtained by using Eqs On the other h, the equations of motion for X α governed by the Hamiltonian H according to classical mechanics are Ẋ α = {X α, H} PB = Ω αβ X β, 86 the Poisson bracket {X α, H} PB turns out to be of the same form as the commutator [X α, H]/i h because H is quadratic in X α. The above two equations mean that the center of the wave packet moves like a classical particle. In fact, it can be shown that this holds for any wave packet in any quadratic system. For the displaced number states, some more specific properties should be emphasized. First, the shape of the wave packet keep unchanged with time. Second, the center of the wave packet is oscillating, but the motion is in general not periodic except when σ /σ 1 is a rational number. Next we discuss the squeezed states. We define a unitary squeeze operator Sζ 1, ζ = S 1 ζ 1 S ζ, 1 S i ζ i = exp ζ ia i a i 1 ζ i a i a i, i = 1,, 87 the ζ i s are complex numbers. For an arbitrary state ϕ the corresponding squeezed state may be defined as ϕ, ζ 1 ζ S = Sζ 1, ζ ϕ = S 1 ζ 1 S ζ ϕ. 88 The following equation is useful for subsequent calculations S ζ 1, ζ a i Sζ 1, ζ = a i cosh ρ i + a i eiφi sinh ρ i, i = 1,, 89 ρ i = ζ i φ i = arg ζ i. Unlike the displaced state, it is difficult to obtain the explicit wave function for the squeezed state in terms of the wave function for the original one. However, in the squeezed number state n 1 n, ζ 1 ζ S, it can be shown by using Eq. 89 that X α = 0, 90 X α = { hn j + 1[ v jα cosh ρ j + Rev jαe iφj sinh ρ j ]} 1/. 91 Comparing with the results for ζ 1 = ζ = 0, we see that the center of the squeezed state is the same as that of the original one, but the uncertainties are different. Thus the states are indeed squeezed. Now consider the time evolution of the squeezed states. If the state at the initial time t = 0 is ψ0 = ϕ, ζ 1 ζ S, then the state at the time t is ψt = ϕt, ζ 1t ζ t S = S 1 ζ 1t S ζ t ϕt, 9 ζ it = exp iσ i tζ i ϕt = exp iht/ h ϕ. Therefore if ϕt is known, ψt can be obtained by a time-dependent squeeze. A simple special case is ϕ = n 1 n, that is ψ0 = n 1 n, ζ 1 ζ S, 93 a squeezed number state. In this case [ ϕt = exp i n σ 1 t i n + 1 ] σ t n 1 n, [ ψt = exp i n σ 1 t i n + 1 ] σ t n 1 n, ζ 1t ζ t S. 94 This means that ψt is also a squeezed number state, except that the squeeze parameters are time-dependent, a time-dependent phase factor is gained. Though it is difficult to obtain the wave function ψt, x, ỹ = xỹ ψt explicitly, in these states it can be shown that X α t = 0, 95

8 674 LIN Qiong-Gui Vol. 38 t X α = { hn j + 1[ v jα cosh ρ j + Rev jα e iφj iσjt sinh ρ j ]} 1/. 96 These results mean that the center of the wave packet is at rest, but the uncertainties are oscillating. Thus the shape of the wave packet changes with time apparently. As before, the motion is in general not periodic except when σ /σ 1 is a rational number. 8 Summary In this paper we have studied a charged anisotropic harmonic oscillator moving in a homogeneous static electromagnetic field. Several configurations of the electromagnetic field are considered. One of these configurations has been studied in the literature. However, the formalism used here seems more convenient. We have studied the coherent squeezed states of these systems in some detail. In terms of these concepts the time evolution of some wave packets can be discussed very conveniently. References [1] U. Merkt, J. Huser, M. Wagner, Phys. Rev. B [] K.D. Zhu S.W. Gu, Phys. Lett. A [3] M. Moshinsky P. Winternitz, J. Math. Phys [4] R. Bogdanovic M.S. Gopinathan, J. Phys. A [5] N. Kokiantonis D.P.L. Castrigiano, J. Phys. A [6] M.A.Z. Habeeb, J. Phys. A [7] B. Mielnik D.J. Fernández C., J. Math. Phys [8] Q.G. Lin, Phys. Rev. A [9] Q.G. Lin, J. Phys. A [10] L.D. Lau E.M. Lifshitz, Quantum Mechanics, 3rd ed., Pergamon, Oxford [11] H.Y. Fan, Phys. Lett. A ; H.Y. Fan, Y.H. Wang, Y. Fan, Commun. Theor. Phys. Beijing, China ; H.Y. Fan Y. Fan, Commun. Theor. Phys. Beijing, China [1] H. Jing, Y.A. Han, J.L. Chen, Y.X. Miao, Chin. Phys. Lett [13] M.M. Nieto, Phys. Lett. A [14] G.J. Ni S.Q. Chen, Advanced Quantum Mechanics, Fudan Univ. Press, Shanghai 000 in Chinese. [15] X.L. Ka, Advanced Quantum Mechanics, 1st ed., Higher Education Press, Beijing 1999; nd ed., Higher Education Press, Beijing 001 in Chinese.