Charged Particle in a Magnetic Field

Transcription

1 Chapter 11 Charged Particle in a Magnetic Field The theory of the motion of a charged particle in a magnetic field presents several difficult and unintuitive features. The derivation of the quantum theory does not require the classical theory; nevertheless it is useful to first review the classical theory in order to show that some of these unintuitive features are not peculiar to the quantum theory, but rather that they are characteristic of motion in a magnetic field Classical Theory The electric and magnetic fields, E and B, enter the Lagrangian and Hamiltonian forms of mechanics through the vector and scalar potentials, A and φ: E = φ 1 c B = A. A t (11.1a) (11.1b) (The speed of light c appears only because of a conventional choice of units.) The potentials are not unique. The fields E and B are unaffected by the replacement A A = A + χ, φ φ = φ 1 χ c t, (11.2) where χ = χ(x,t) is an arbitrary scalar function. This change of the potentials, called a gauge transformation, has no effect upon any physical result. It thus appears, in classical mechanics, that the potentials are only a mathematical construct having no direct physical significance. The Lagrangian for a particle of mass M and charge q in an arbitrary electromagnetic fields is L(x, v, t)= Mv2 2 qφ(x,t)+ q v A (x,t), (11.3) c

2 308 Ch. 11: Charged Particle in a Magnetic Field where x and v = dx/dt are the position and velocity of the particle. The significance of (11.3) lies in the fact that Lagrange s equation of motion, [ ] d L L =0 (α =1, 2, 3), (11.4) dt v α x α leads, after an elementary calculation, to the correct Newtonian equation of motion, Mdv/dt = q(e + v B/c). From the Lagrangian, we can define the canonical momentum, p α = L/ v α. For a particle in a magnetic field it has the form p = Mv + q c A. (11.5) Since p, like A, is changed by a gauge transformation, it too lacks a direct physical significance. However, it is of considerable mathematical importance. Lagrange s equation (11.4) can be written as dp α /dt = L/ x α. Hence it follows that if L is independent of x α (or in other words, if L is invariant under a coordinate displacement of the form x α x α + a α ), then it is the canonical momentum p α that is conserved, and not the more intuitive quantity Mv α. The Hamiltonian for a particle in an electromagnetic field is H = v p L = Mv2 + qφ(x,t), (11.6) 2 with the terms involving A canceling out of the final expression. Since the magnetic force on a moving particle is perpendicular to the velocity of the particle, the magnetic field does no work and hence does not enter into the expression for the total energy H. How then can the Hamiltonian generate the motion of the particle, which does depend upon the magnetic field, when the magnetic field apparently does not enter into (11.6)? The answer lies in the fact that Hamiltonian is to be regarded as a function of position and momentum, not of position and velocity. Hence it is more proper to rewrite (11.6) using (11.5) as H = 1 ( p q ) 2 2M c A + qφ. (11.7) Hamilton s equations, dp/dt = H/ x and dx/dt = H/ p, then yield the familiar Newtonian equation of motion. Two important results from this classical theory, which also hold in the quantum theory, are the relation (11.5) between velocity and canonical momentum, and the fact that the apparently more complicated Hamiltonian (11.7) is

3 11.2 Quantum Theory 309 really just equal to the sum of kinetic plus potential energy. One should also remember that in the presence of a magnetic field the momentum p is not an observable quantity, but nevertheless it plays an important mathematical role Quantum Theory It was shown in Sec. 3.4 that the requirement of Galilei invariance restricts the possible external interactions of a particle to a scalar potential and a vector potential. Since the coupling of the particle to the electromagnetic field is proportional to the particle s charge q, the generic form of the Hamiltonian (3.60) should be rewritten as ( H = P q ) 2 c A /2M + qφ (11.8) in this case. (The factor 1/c is present only because of a conventional choice of units.) Here P is the momentum operator of the particle. The vector and scalar potentials, A = A(Q,t)andφ = φ(q,t), are operators because they are functions of the position operator Q. Their dependence (if any) on t corresponds to the intrinsic time dependence of the fields. (Here we are using the Schrödinger picture.) The velocity operator, defined in units of by (3.39), is V α = i ( )[ i { [H, Q α]= P α q } ] 2 2M c A α,qα = 1 ( P α q ) M c A α, (α = x, y, z). (11.9) As was the case for the classical theory, the momentum P is mathematically simpler than the velocity V. But the velocity has a more direct physical significance, so it is worth examining its mathematical properties. The commutator of the position and velocity operators is [Q α,v β ]=i M δ αβ. (11.10) Apart from the factor of M, this is the same as the commutator for position and momentum. However the commutator of the velocity components with each other presents some novelty: [V x,v y ]= [P x,p y ] M 2 + ( q ) 2 [Ax,A y ] Mc q M 2 c {[A x,p y ]+[P x,a y ]}.

4 310 Ch. 11: Charged Particle in a Magnetic Field The first and second terms vanish. The remaining terms can be evaluated most easily by adopting the coordinate representation (Ch. 4), in which a vector is represented by a function of the coordinates, ψ(x), and the momentum operator becomes P α = i / x α.thuswehave [V x,v y ]ψ = i = i = i { q M 2 A x c q M 2 c { y y A x + x A y A y ( ) ( )} Ax Ay + ψ y x q M 2 c ( A) q zψ = i M 2 c B zψ. } ψ x Since this result holds for an arbitrary function ψ, it may be written as an operator equation, valid in any representation: [V x,v y ]=i ( q/m 2 c) B z.this may clearly be generalized to [V α,v β ]=i q M 2 c ε αβγ B γ, (11.11) where ε αβγ is the antisymmetric tensor [introduced in Eq. (3.22)]. The commutator of two components of velocity is proportional to the magnetic field in the remaining direction. Heisenberg equation of motion The velocity operator (11.9) is equal to the rate of change of the position operator, calculated from the Heisenberg equation of motion (3.73). Similarly, an acceleration operator can be calculated as the rate of change of the velocity operator. The product of mass times acceleration may naturally be regarded as the force operator, M dv α dt = i M [H, V α]+m V α t. (11.12) (For simplicity of notation we shall not distinguish between the Schrödinger and Heisenberg operators. This equation should therefore be regarded as referring to the instant of time t = t 0 when the two pictures coincide.) To evaluate the commutator in (11.12), it is useful to rewrite the Hamiltonian (11.8) as H = 1 2MV V + qφ. Thuswehave

5 11.2 Quantum Theory 311 [H, V α ]= 1 2 M β [V 2 β,v α]+q [φ, V α ] = 1 2 M β {V β [V β,v α ]+[V β,v α ] V β } + q M [φ, P α] = 1 ( ) q 2 i (V β ε βαγ B γ + ε βαγ B γ V β )+ q Mc M i ( φ) α β,γ ) = 1 2 i ( q Mc β,γ ε αβγ ( V β B γ + B β V γ )+i ( ) q ( φ) α. M The last term of (11.12) has the value M V α / t = (q/c) A α / t. Combining these results and writing (11.12) in vector form, we have M dv dt = 1 2 ( q c ) (V B B V)+qE. (11.13) This is just the operator for the Lorentz force, the only complication being that the magnetic field operator B (and also the electric field operator E) isa function of Q, andsob does not commute with V. Coordinate representation The Hamiltonian (11.8) may be written as H = P P (q/c)(p A + A P)+(q/c)2 A A + qφ. 2M The difference between P A and A P can be determined by the action of these operators on an arbitrary function ψ(x): P A ψ = i Aψ = i A ψ i ψ ( A). Since ψ is an arbitrary function, this may be written as an operator relation, P A A P = i diva, (11.14) which holds in any representation. (It is always possible to choose the vector potential so that diva = 0, and this is often done.) The general form of the Hamiltonian in coordinate representation is H = 2 2M 2 + i q i q A + Mc 2Mc (diva)+ q2 2Mc 2 A2 + qφ. (11.15)

6 312 Ch. 11: Charged Particle in a Magnetic Field In interpreting this expression, it should be remembered that, in spite of the apparent complexity, the sum of the first four terms is just the 1 kinetic energy, 2 MV 2. Sometimes the first term is described as the kinetic energy, and the next three terms are described as paramagnetic and diamagnetic corrections. That is not correct, and indeed the individual terms have no distinct physical significance because they are not separately invariant under gauge transformations. For many purposes, it is preferable not to expand the quadratic term of the Hamiltonian, but rather to write it more compactly as H = 1 2M ( i q c A ) 2 + qφ. (11.16) Gauge transformations The electric and magnetic fields are not changed by the transformation (11.2) of the potentials. On the basis of our previous experience, we may anticipate that there will be a corresponding transformation of the state function that will, at most, transform it into a physically equivalent state function. Since the squared modulus, Ψ(x,t) 2, is significant as a probability density, this implies that only the phase of the complex function Ψ(x,t) canbe affected by the transformation. (This is similar to the Galilei transformations, which were studied in Sec. 4.3.) The Schrödinger equation, ( 1 2M i q ) 2 c A Ψ+qφΨ=i t Ψ, (11.17) is unchanged by the combined substitutions: A A = A + χ, φ φ = φ 1 χ c t, Ψ Ψ =Ψe i(q/ c)χ, (11.18a) (11.18b) (11.18c) where χ = χ(x,t) is an arbitrary scalar function. It is this set of transformations, rather than (11.2), which is called a gauge transformation in quantum mechanics. That the transformed equation 1 2M ( i q c A ) 2 Ψ + qφ Ψ = i t Ψ (11.17 )

7 11.2 Quantum Theory 313 is equivalent to the original (11.17) can be demonstrated in two steps. First, on the right hand side of (11.17 ) the time derivative of the phase factor from (11.18c) exactly compensates for the extra term introduced on the left hand side by the scalar potential (11.18b). Second, it is easily verified that ( i q ) ( i c A e i(q/ c)χ Ψ=e i(q/ c)χ qc ) A Ψ (11.19) since the gradient of the phase factor on the left hand side compensates for the extra term in the vector potential introduced by (11.18a). Hence it follows that (11.17 ) differs from (11.17) only by an additional phase factor on both sides of the equation, and so the original and the transformed equations are equivalent. From (11.19) it follows that the average velocity, ( Ψ V Ψ Ψ P M qa ) Ψ, Mc is invariant under gauge transformations, whereas the average momentum Ψ P Ψ is not. For this reason, the physical significance of a result will usually be more apparent if it is expressed in terms of the velocity, rather than in terms of the momentum. We can also show that the eigenvalue spectrum of a component of velocity is gauge-invariant, even though the form of the velocity operator, (P/M qa/m c), depends on the particular choice of vector potential. Suppose that ψ(x) is an eigenvector of V z, ( Pz M qa ) z ψ(x) =v z ψ(x). (11.20) Mc Consider now another equivalent vector potential, A = A+ χ. From (11.19) and (11.20) we obtain ( ) ( Pz M qa z e i(q/ c)χ ψ(x) =e i(q/ c)χ Pz Mc M qa ) z ψ(x) Mc = v z e i(q/ c)χ ψ(x). (11.21) Thus the operators P z qa z /c and P z qa z /c musthavethesameeigenvalue spectrum. Probability current density The probability current density J(x,t) was introduced in Sec. 4.4 through the continuity equation, divj +( / t) Ψ 2 = 0, which expresses the conservation of probability. In the presence of a nonvanishing vector potential, the

8 314 Ch. 11: Charged Particle in a Magnetic Field expressions (4.22) and (4.24) are no longer equal, and it is the latter that is correct: { [ P J(x,t)=Re Ψ (x,t) M qa(x,t) ] } Ψ(x,t). (11.22) Mc (Proof that this expression satisfies the continuity equation is left for Problem 11.2.) It is apparent from (11.19) that this expression for J(x,t) is gauge-invariant Motion in a Uniform Static Magnetic Field In this section we treat in detail the quantum theory of a charged particle in a spatially homogeneous static magnetic field. Only the orbital motion will be considered, and any effects of spin or intrinsic magnetic moment will be omitted. Throughout this section the magnetic field will be of magnitude B in the z direction. There are, of course, many different vector potentials that can generate this magnetic field. Some of the following results will depend only upon the magnetic field, whereas others will depend upon the particular choice of vector potential. Although the vanishing of the electric field requires only that the combination (11.1a) of scalar and vector potentials should vanish, we shall assume that the vector potential is static and that the scalar potential vanishes. Energy levels The most direct derivation of the energy levels can be obtained by writing the Hamiltonian (11.8) in terms of the components of the velocity operator (11.9): H = H xy + H z, with H xy = 1 2 M(V x 2 + V y 2)andH z = 1 2 MV z 2. Since B x = B y = 0, it follows from (11.11) that V z commutes with V x and V y. Hence the operators H xy and H z are commutative, and every eigenvalue of H is just the sum of an eigenvalue of H xy and an eigenvalue of H z. By introducing the notations γ =( q B/M 2 c) 1/2,V x = γq and V y = γp, we formally obtain H xy = 1 2 ( q B/Mc)(P 2 +Q 2 ) with Q P P Q = i. (Note that Q and P are only formal symbols and do not represent the position and momentum of the particle.) These equations are isomorphic to (6.7) and (6.6) for the harmonic oscillator (Sec. 6.1), and therefore the eigenvalues of H xy must be equal to (n ) q B/Mc, wheren is any nonnegative integer. The eigenvalue spectrum of H z is trivially obtained from that of V z. The spectrum of V z P z /M qa z /M c has been shown to be gauge-invariant. Because the magnetic field is uniform and in the z direction, it is possible to

9 11.3 Motion in a Uniform Static Magnetic Field 315 choose the vector potential such that A z = 0. Therefore the spectrum of V z is continuous from to, like that of P z. Thus the energy eigenvalues for a charged particle in a uniform static magnetic field B are E n (v z )= (n ) q B + 1 Mc 2 Mv2 z, (n =0, 1, 2,...). (11.23) This result is independent of the particular vector potential that may be used to generate the prescribed magnetic field. This form for the energies is easily interpreted. The motion parallel to the magnetic field is not coupled to the transverse motion, and is unaffected by the field. The classical motion in the plane perpendicular to the field is in a circular orbit with angular frequency ω c = qb/mc (called the cyclotron frequency), and it is well known that periodic motions correspond to discrete energy levels whose separation is ω c. If we want not only the energies but also the corresponding state functions, it is necessary to choose a particular coordinate system and a particular form for the vector potential. Solution in rectangular coordinates Let us choose the vector potential to be A x = yb, A y = A z = 0. One can easily verify that A =0andthat A = B is in the z direction. The Hamiltonian (11.8) now becomes H = (P x + yqb/c) 2 + P 2 y + P 2 z 2M. (11.24) It is apparent that P x and P z commute with H (and that P y does not), so it is possible to construct a complete set of common eigenvectors of H, P x,and P z. In coordinate representation, the eigenvalue equation H Ψ = E Ψ now takes the form 2 2M 2 Ψ i q Mc By x Ψ+ q2 B 2 2Mc 2 y2 Ψ=EΨ. (11.25) Since Ψ can be chosen to also be an eigenfunction of P x and P z, we may substitute Ψ(x, y, z) = exp{i(k x x + k z z)}φ(y), (11.26)

10 316 Ch. 11: Charged Particle in a Magnetic Field thereby reducing (11.25) to an ordinary differential equation, 2 d 2 φ(y) 2M dy 2 + qbk [ x q 2 B 2 ] yφ(y)+ Mc 2Mc 2 y M (k2 x + kz) 2 E φ(y) =0. (11.27) The term linear in y can be removed by shifting the origin to the point y 0 = k x c/qb. The equation then takes the form 2 d 2 [ ] φ(y) Mω 2 2M dy 2 + c (y y 0 ) 2 E φ(y) =0, (11.28) 2 where ω c = qb/mc is the classical cyclotron frequency, and E = E 2 kz 2/2M is the energy associated with motion in the xy plane. This is just the form of Eq. (6.21) for a simple harmonic oscillator with angular frequency ω = ω c, whose eigenvalues are E = ω(n+ 1 2 ),n=0, 1, 2,... Thus the energies for the charged particle in the magnetic field must be E = ω c (n )+ 2 kz 2/2M, confirming the result (11.23). The function φ(y) is a harmonic oscillator eigenfunction, of the form (6.32). Apart from a normalization constant, the eigenfunction (11.26) will be Ψ(x, y, z) = exp{i(k x x+k z z)} H n {α(y y 0 )} exp { 12 } α2 (y y 0 ) 2, (11.29) with α =(M ω c / ) 1/2 =( q B/ c) 1/2 ), and y 0 = k x c/qb. Here H n is a Hermite polynomial. It is useful to define a characteristic magnetic length, a m = α 1 = ( ) 1/2 c. (11.30) q B In terms of this length, the center of the Hermite polynomial in (11.29) is located at y 0 = (q/ q )k x a 2 m. The interpretation of this state function is far from obvious. The classical motion is a circular orbit in the xy plane. But (11.29) does not reveal such amotion,thex dependence of Ψ being an extended plane wave, while the y dependence is that of a localized harmonic oscillator function. The x and z dependences of Ψ are the same; nevertheless the energy E is independent of k x while k z contributes an ordinary kinetic energy term. These puzzles can be resolved by considering the orbit center coordinates.

11 11.3 Motion in a Uniform Static Magnetic Field 317 Fig viewer). An orbit of a charged particle (q >0) in a magnetic field (directed toward the Orbit center coordinates Consider a classical particle moving with speed v in a circular orbit of radius r, as shown in Fig The magnetic force is equal to the mass times the centripetal acceleration, qvb/c = Mv 2 /r. The angular frequency, ω c v/r = qb/mc, is independent of the size of the orbit. The equations for the orbital position and velocity of the particle are of the forms x x 0 = r cos(ω c t + θ), y y 0 = r sin(ω c t + θ), v x = ω c r sin(ω c t + θ), v y = ω c r cos(ω c t + θ). (11.31) (These equations are correct for both positive and negative charges if we take ω c to have the same sign as the charge q.) Hence the coordinates of the orbit center are x 0 = x + v y /ω c and y 0 = y v x /ω c. We conclude this brief classical analysis with the seemingly trivial remark that the orbit center coordinates are constants of motion. Let us now, by analogy, define quantum-mechanical orbit center operators, X 0 and Y 0, in terms of the position operator and the velocity operator (11.9): X 0 = Q x + V y ω c, Y 0 = Q y V x ω c. (11.32) It is easy to verify using (11.10) and (11.11), that if the x and y components of the magnetic field vanish then [H, X 0 ]=[H, Y 0 ]=0. (11.33) (This is another case in which it is simpler to express the Hamiltonian in terms of the velocity than in terms of the momentum.) Thus the orbit center

12 318 Ch. 11: Charged Particle in a Magnetic Field coordinates are quantum-mechanical constants of motion, a result that is independent of the particular choice of vector potential. It is not possible to construct eigenfunctions corresponding to a definite orbit center because the operators X 0 and Y 0 do not commute. A simple calculation yields [X 0,Y 0 ]= i c qb = i q q a2 m. (11.34) In accordance with (8.27) and (8.31), there is an indeterminacy relation connecting the fluctuations of the two orbit center coordinates: X 0 Y a2 m. It is possible to construct common eigenfunctions of H and X 0,orofH and Y 0, but not of all three operators. For the particular vector potential A x = yb, A y = A z = 0, the orbit center operators become X 0 = Q x + cp y /qb, Y 0 = cp x /qb. Thus it is apparent that the energy eigenfunction (11.29) is also an eigenfunction of Y 0 with eigenvalue y 0 = c k x /qb. This result illustrates the nonintuitive nature of the canonical momentum in the presence of a vector potential. The reason why the energy eigenvalue of (11.27) does not depend on k x is that in this case the momentum component k x does not describe motion in the x direction, but rather position in the y direction! Roughly speaking, we may think of the state function (11.29) as describing an ensemble of circular orbits whose centers are distributed uniformly along the line y = y 0. (That this picture is only roughly accurate can be seen from the quantum fluctuations in the orbit size, as evidenced by the exponential tails on the position probability density in the y direction.) Degeneracy of energy levels Even for fixed n and v z, the energy eigenvalue (11.23) is highly degenerate. Although the degree of degeneracy must be gauge-invariant, it is easier to calculate it for the particular coordinate system and vector potential corresponding to (11.29). These degenerate energy levels (for fixed n and k z ) are often called Landau levels, after Lev Landau, who first obtained the solution (11.29). With k z held constant, the problem is effectively reduced to two dimensions. For convenience, we assume that the system is confined to a rectangle of dimension D x D y and subject to periodic boundary conditions. The allowed values of k x are k x =2πn x /D x, with n x =0, ±1, ±2,... Now the orbit center coordinate, y 0 = (q/ q )k x a 2 m = (q/ q )a2 m 2πn x/d x, must lie in the range 0 <y 0 <D y. In the limit as D x and D y become large, we may ignore any

13 11.3 Motion in a Uniform Static Magnetic Field 319 problems associated with orbits lying near the boundary, since they will be a negligible fraction of the total. In this limit the number of degenerate states corresponding to fixed n and k z will be D x D y /2πa 2 m. This result suggests a simple geometrical interpretation, namely that each state is associated with an area of magnitude 2πa 2 m in the plane perpendicular to the magnetic field. The quantity Φ 0 =2π c/q = hc/q is a natural unit of magnetic flux. In a homogeneous magnetic field B, the area 2πa 2 m encloses one unit of flux. Thus the degeneracy factor of a Landau level is simply equal to the number of units of magnetic flux passing through the system. Orbit radius and angular momentum It is possible to obtain a more direct description of the circular orbital motion of the particle than that contained implicitly in the state functions of the form (11.29). We may confine our attention to motion in the xy plane, since it is now apparent that the nontrivial aspect of the problem concerns motion perpendicular to the magnetic field. From the position operators, Q x and Q y, and the orbit center operators, X 0 and Y 0, we construct an orbit radius operator, r c : rc 2 =(Q x X 0 ) 2 +(Q y Y 0 ) 2. (11.35) From (11.32) we obtain rc 2 = ω 2 c (Vx 2 + V y 2 ), and hence the transverse Hamiltonian satisfies the relation H xy 1 2 M(V 2 x + V 2 y )= 1 2 Mω2 cr 2 c, (11.36) a relation that also holds in classical mechanics. From the known eigenvalues of H xy, which are equal to ω c (n+ 1 2 ), we deduce that the eigenvalues of r2 c are a 2 m(2n + 1), with n =0, 1, 2,... The degeneracy of the energy levels is due to the fact that the energy does not depend upon the position of the orbit center. Since the operators X 0 and Y 0 commute H xy but not with each other, it follows that the degenerate eigenvalues of H xy form a one-parameter family (rather than a two-parameter family, as would be the case if the two constants of motion, X 0 and Y 0, were commutative). To emphasize the rotational symmetry of the problem, we introduce the operator R 0 2 = X Y 0 2, (11.37) whose interpretation is the square of the distance of the orbit center from the origin. The degenerate eigenfunctions of H xy can be distinguished by the eigenvalues of R 0 2. [These will not be the particular functions (11.29).]

14 320 Ch. 11: Charged Particle in a Magnetic Field The set of three operators {X 0 /a m,y 0 /a m,h 1 2 (X 0/a m ) (Y 0/a m ) 2 } are isomorphic in their commutation relations to the position, momentum, and Hamiltonian of a harmonic oscillator (see Sec. 6.1). Hence the eigenvalues of H are equal to l with l =0, 1, 2,... Thus the eigenvalues of R 0 2 are equal to a 2 m(2l +1), (l =0, 1, 2,...). Suppose that the system is a cylinder of radius R. Since the orbit center must lie inside the system, we must impose the condition R 2 0 R 2. If we ignore any problems with orbits near the boundary, since they will be a negligible fraction of the total in the limit of large R, then the degeneracy factor of an energy level (the number of allowed values of l) is equal to 1 2 (R/a m) 2 = πr 2 /2πa 2 m. This agrees with our previous conclusion that each state is associated with an area 2πa 2 m. The orbital angular momentum in the direction of the magnetic field is L z = Q x P y Q y P x = M(Q x V y Q y V x )+(q/c)(q x A y Q y A x ). It will be constant of motion if we choose the vector potential to have cylindrical symmetry. Therefore we take the operator for the vector potential to be A(Q) = 1 2 B Q, which has components ( 1 2 BQ y, 1 2 BQ x, 0). Thus we obtain L z = M(Q x V y Q y V x )+ q 2c (Q2 x + Q 2 y) = qb 2c (R 0 2 r c 2 ). (11.38) [The second line is obtained by using (11.32) to eliminate the velocity operators.] It is apparent that the angular momentum is indeed a constant of motion, but it is not independent of those already found. Recall that r c 2 is proportional to the energy of transverse motion, and that R 0 2 is an orbit center coordinate that distinguishes degenerate states. Those degenerate states could equally well be distinguished by the orbital angular momentum eigenvalue m. It is now apparent that the angular momentum can have a very unintuitive significance in the presence of a magnetic field. If we consider it to vary at fixed energy, it has little to do with rotational motion, but is instead related to the radial position of the orbit center. Suppose the radius R of the system becomes infinite. Then for fixed energy (fixed r c 2 ) the allowed values of angular momentum will be bounded in one direction and unbounded in the other, since R 0 2 is bounded below. (If R 0 2 is fixed at its minimum value, and the energy and angular momentum are allowed to vary together, then the angular momentum plays a more familiar role.) It is possible to solve the Schrödinger equation directly in cylindrical coordinates, verifying in detail the results obtained above, and also obtaining

15 11.4 The Aharonov Bohm Effect 321 explicit eigenfunctions (Problem 11.6). But the interpretation of those eigenfunctions as physical states would be very obscure without knowledge of the relation (11.38) The Aharonov Bohm Effect In classical electrodynamics, the vector and scalar potentials were introduced as convenient mathematical aids for calculating the electric and magnetic fields. Only the fields, not the potentials, were regarded as having physical significance. Since the fields are not affected by the substitution (11.2), it follows that the equations of motion must be invariant under that substitution. In quantum mechanics these changes to the vector and scalar potentials must be accompanied by a change in the phase of the wave function Ψ. The theory is then invariant under the gauge transformation (11.18). Because of its classical origin, it is natural to suppose that the principle of gauge invariance merely expresses, in the quantum mechanical context, the notion that only the fields but not the potentials have physical significance. However, Aharonov and Bohm (1959) showed that there are situations in which such an interpretation is difficult to maintain. They considered an experiment like that shown in Fig. 11.2, which consists of a charged particle source and a double slit diffraction apparatus. A long solenoid is placed perpendicular to the plane of the figure, so that a magnetic field can be created inside the solenoid while the region external to the solenoid remains field-free. The solenoid is located in the unilluminated shadow region so that no particles will reach it, and moreover it may be surrounded by a cylindrical shield that is impenetrable to the charged particles. Nevertheless it can be shown that the interference pattern depends upon the magnetic flux through the cylinder. Let Ψ (0) (x,t) be the solution of the Schrödinger equation and boundary conditions of this problem for the case in which the vector potential is everywhere zero. Now let us consider the case of interest, in which the magnetic field is nonzero inside the cylinder but zero outside of it. The vector potential A will not vanish everywhere in the exterior region, even though B = A =0 outside of the cylinder. This follows by applying Stokes s theorem to any path surrounding the cylinder: A dx = ( A) ds = B ds =Φ. Ifthe flux Φ through the cylinder is not zero, then the vector potential must be nonzero on every path that encloses the cylinder. However in any simply connected region outside of the cylinder, it is possible to express the vector potential as the gradient of a scalar, from the zero-potential solution by means of a gauge transformation, Ψ = Ψ (0) e i(q/ c)λ.

16 322 Ch. 11: Charged Particle in a Magnetic Field Fig The Aharonov Bohm experiment.charged particles from the source a pass through the double slit.the interference pattern formed at the bottom screen depends upon the magnetic flux through the impenetrable cylinder. This technique will now be applied to each of the (overlapping) regions L and R shown in Fig In region L, which contains the slit on the left, the wave function can be written as Ψ L =Ψ (0) L ei(q/ c)λ1,whereψ (0) L is the zeropotential solution in region L, andλ 1 =Λ 1 (x,t)= A dx, with the integral taken along a path within region L. Since A =0insideL, thevalueof this integral depends only upon the end points of the path, provided, of course, that the path remains within L and does not cross the cylinder of magnetic flux. A similar form can be written for the wave function in the region R, which contains the slit on the right. At the point b, in the overlap of regions L and R, the wave function is a superposition of contributions from both slits. Hence we have Ψ(b) =Ψ (0) L ei(q/ c)λ1 +Ψ (0) R ei(q/ c)λ2. (11.39) Here Λ 1 = A dx with the path of integration running from a to b through region L, andλ 2 = A dx with the path of integration running from a to b through region R. The interference pattern depends upon the relative phase of the two terms in (11.39), e i(q/ c)(λ1 Λ2). But (Λ 1 Λ 2 ), the difference between the integrals along paths on either side of the cylinder, is equivalent to an integral around a closed path surrounding the cylinder, A dx =Φ. Therefore the interference pattern is sensitive to the magnetic flux inside of the cylinder, even though the particles never pass through the region in which the magnetic field is nonzero! This prediction, which has been experimentally verified, was very surprising when it was first announced. Several remarks about this effect are in order. First, the relative phase of the two terms of (11.39) is (e iqφ/ c ). If the magnetic flux Φ were quantized