Physics 505 Homework No. 6 Solutions S6-1 1. Probability

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Physics 505 Homework No. 6 Solutions S6-1 1. Probability Physics 505 Homework No. 6 Solutions S6-1 1. Probability conservation (based on a problem in Schwabl). Recall that the Hamiltonian for a charged particle (charge e) of mass m in an electromagnetic field described by the potentials φ(x,t) and A(x,t) is, 1 e 2 H = p − A + eφ . 2m c Show that a wave function ψ which is a solution of Schroedinger’s equation with this Hamiltonian, satisfies the probability continuity condition, ∂ (ψ∗ψ) + ∇· j =0 , ∂t with j defined as ¯h 2ie j = ψ∗(∇ψ) − (∇ψ∗)ψ − Aψ∗ψ 2mi ¯hc 1 ¯h e ¯h e = ψ∗ ∇ − A ψ + ψ ∇ − A ψ∗ . 2m i c −i c Note that A is assumed to be real. Solution To start with, calculate −∇ · j using the j given above. −1 ¯h ¯h −∇ · j = (∇ψ∗) · (∇ψ) + (∇ψ) · (∇ψ∗) 2m i −i " e e + − A · (∇ψ∗) ψ − A · (∇ψ) ψ∗ c c ¯h ¯h + ψ∗∇2ψ + ψ∇2ψ∗ i −i e e + − ψ∗(∇· A)ψ − ψ(∇· A)ψ∗ c c e e + − ψ∗(A · ∇)ψ − ψ(A · ∇)ψ∗ . c c # Only the terms in the first line cancel! It’s tempting to get rid of the ∇· A terms, but we’re not necessarily in a static situation or the Coulomb gauge. Simplifying slightly, 1 ¯h −∇ · j = −ψ∗∇2ψ + ψ∇2ψ∗ 2m" i e +2 (ψ∗(A · ∇)ψ + ψ(A · ∇)ψ∗ + ψ∗ψ(∇· A)) . c # Copyright c 2012, Edward J. Groth Physics 505 Homework No. 6 Solutions S6-2 The Schroedinger equation is ∂ψ Hψ = i¯h . (1) ∂t For it’s complex conjugate, we might be tempted to write ∂ψ∗ Hψ∗ = −i¯h , ∂t but this only works when H = p2/2m + V . In this case, we need to be careful. ∗ 1 ¯h e 2 1 ¯h e 2 ∂ψ∗ (Hψ)∗ = ∇ − A + eφ ψ∗ = ∇ − A + eφ ψ∗ = −i¯h . 2m i c 2m −i c ∂t ! ! (2) Multiply the first equation by ψ∗, the second by ψ, and subtract. ∂ ∂ψ ∂ψ∗ (ψ∗ψ) = ψ∗ + ψ ∂t ∂t ∂t −i = (ψ∗Hψ − ψH∗ψ∗) ¯h −i 1 ¯h e 2 1 ¯h e 2 = ψ∗ ∇ − A ψ − ψ ∇ − A ψ∗ + ψ∗eφψ − ψeφψ∗ ¯h 2m i c 2m −i c ! 1 ¯h ¯h = − ψ∗∇2ψ + ψ∇2ψ∗ 2m i i " e e + ψ∗(∇· A)ψ + ψ(∇· A)ψ∗ c c e e + ψ∗(A · ∇ψ) + ψ(A · ∇ψ∗) c c e e + ψ∗(A · ∇ψ) + ψ(A · ∇ψ∗) c c ie2 ie2 + − ψ∗A2ψ + ψA2ψ∗ , ¯hc2 ¯hc2 # which simplifies slightly to ∂ 1 ¯h (ψ∗ψ) = −ψ∗∇2ψ + ψ∇2ψ∗ ∂t 2m" i e +2 (ψ∗(A · ∇)ψ + ψ(A · ∇)ψ∗ + ψ∗ψ(∇· A)) , c # which is the same as we previously obtained for −∇ · j. Copyright c 2012, Edward J. Groth Physics 505 Homework No. 6 Solutions S6-3 Some comments. p − eA/c = mv is the kinetic momentum operator. The extra terms with A in the current come about because we need to correct the canonical momentum to get the kinetic momentum. ψ∗ψ and j as defined above are the matter probability density and the matter proba- bility current. How would you write the charge probability density and charge probability current? (Answer: multiply them by e.) End Solution 2. Crossed E and B fields (based on a problem in Schwabl). Consider a particle with mass m and charge e moving in uniform B = Bez and E = Eex fields with E<B. Use the gauge A = Bxey. The Hamiltonian is 1 e 2 H = p − A − eEx . 2m c Find the eigenfunctions and eigenvalues for this Hamiltonian. Note that if you should find that the eigenfunctions involve standard functions which we already know about, you don’t have to write out each one explicitly. For example, if you should find (you won’t) that the eigenfunctions involve the Legendre polynomials Pl(y), you can just leave Pl(y) in your result rather than explicitly write out what Pl(y) is. Of course, you have to say that Pl stands for a Legendre polynomial of order l. Hint: you might start by reminding yourself what the classical motion looks like. Solution To start with the classical motion, there is no force in the z direction, so the z- component of velocity is constant. Note that if the particle has vy = −cE/B, then the magnetic force exactly cancels the electric force and the particle moves at constant velocity in the negative y direction. If the particle has vy different from above or a non-vanishing vx, then the motion is circular motion in the xy-plane (clockwise as viewed from the positive z-direction) with a velocity in the −y-direction at vy = −cE/B. (Often called E × B drift.) So, we are looking for eigenfunctions which have a constant vz (or pz) along with circular and linear motion in the xy-plane. Writing out the Hamiltonian, we have p2 1 e 2 1 H = x + p − Bx + p2 − eEx . 2m 2m y c 2m z We see that py and pz commute with the Hamiltonian, so our eigenfunctions can be eigenfunctions of py, pz, and H, simultaneously. Let pzψ =¯hkzψ and pyψ =¯hkyψ. Then the eigenvalue equation becomes 1 1 e 2 ¯h2k2 p2 + ¯hk − Bx − eEx ψ = E − z ψ = E′ψ, 2m x 2m y c 2m Copyright c 2012, Edward J. Groth Physics 505 Homework No. 6 Solutions S6-4 where we have used E to stand for the energy eigenvalue to distinguish it from E and E′ for the energy eigenvalue with the energy due to the z motion removed. Define ω = eB/mc and v0 = cE/B. Then the time independent Schroedinger equation becomes 1 1 p2 + (¯hk − mωx)2 − mωv x ψ = E′ψ, 2m x 2m y 0 Let’s work on the terms above involving x. 1 mω2 ¯hk v ¯h2k2 (¯hk − mωx)2 − mωv x = x2 − 2 y + 0 x + y . 2m y 0 2 mω ω m2ω2 ! Define ¯hk v x = y + 0 . 0 mω ω Then 2 2 2 1 2 mω 2 ¯h ky 2 (¯hky − mωx) − mωv0x = (x − x0) + 2 2 − x0 2m 2 m ω ! mω2 1 = (x − x )2 − mv2 − ¯hk v . 2 0 2 0 y 0 We can clean up the last two terms by adding and subtracting eEx0, 1 1 ¯hk v − mv2 − ¯hk v + eEx − eEx = − mv2 − ¯hk v + mωv y + 0 − eEx 2 0 y 0 0 0 2 0 y 0 0 mω ω 0 1 = mv2 − eEx . 2 0 0 Returning to the time independent Schroedinger equation, we have 1 mω2 1 ¯h2k2 p2 + (x − x )2 + mv2 − eEx ψ = E − z ψ, 2m x 2 0 2 0 0 2m So we see that the x motion is that of a harmonic oscillator centered at x0 with frequency ω. The energy eigenvalues are ¯h2k2 1 E =(n +1/2)¯hω + z + mv2 − eEx , n,ky ,kz 2m 2 0 0 withhk ¯ y = mωx0 − mv0 and (unnormalized) eigenfunction ikyy ikzz ψn,ky ,kz = ψn(x − x0)e e , where ψn is the harmonic oscillator eigenfunction with frequency ω and quantum number n. Copyright c 2012, Edward J. Groth Physics 505 Homework No. 6 Solutions S6-5 So, the motion has an energy corresponding to whatever kinetic energy we want in the z-direction, the kinetic energy due to the drift velocity in the y-direction and the kinetic energy due a harmonic oscillator at frequency ω in the x-direction. Since the center of oscillation is x0, there is also an electric potential energy −eEx0. Classically, the kinetic 2 2 energy due to the motion in the xy plane looks like mvx/2 + m(vy − v0) /2 where vx and vy are the velocity components of the circular motion relative to the center of the circle 2 2 2 which is moving with velocity −v0ey. Averaging over a period gives mr ω /2 + mv0/2 where r is the radius of the circular orbit. The cross term in the y-velocity averages to zero as does the variation in the electric potential energy as the particle oscillates in x. In the quantum case, the oscillator energy corresponds to mr2ω2/2. You might wonder where the drift velocity went in all this. Let ¯hΩ = En,ky ,kz , so Ω is the oscillation frequency of the wave function. Then the group velocity component in the y-direction is dΩ/dky. The only place ky appears in the energy is in x0. Carrying out the differentiation, we find vg = −cE/B = −v0, exactly the classical drift velocity. End Solution 3. Magnetic barrier. This problem appeared on the May, 2002 prelims. The Θ functions used below are defined so that Θ(x)=0if x < 0, Θ(x)=1if x > 0 and Θ(0) = 1/2. x ′ ′ In other words, Θ is the unit step function and Θ(x) = −∞ δ(x ) dx . Also you may find problem 1 in this assignment to be useful. R Consider a charged particle moving in the xy-plane subject to a magnetic field Bz = BΘ(x)Θ(d − x). The magnetic field is constant in a strip of width d and zero everywhere else. We will study the problem of scattering of plane waves from this “magnetic barrier.” (a) Write down the Schroedinger Hamiltonian for this problem. You have to choose a gauge for the vector potential—choose the gauge Ax = Az = 0, and also choose Ay = 0 for x < 0. Solution The vector potential is ∇ × A = B. We take Ay = 0 for x < 0, Ay = Bx for 0 ≤ x ≤ d and Ay = Bd for d<x.
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