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 #

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.

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 probability 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

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.

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. Leth ¯Ω = 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

The time independent Schroedinger equation takes the following forms: p2 x ψ = Eψ, x< 0 . 2m

p2 e2B2x2 x + ψ = Eψ, 0

Consider the scattering problem for an electron incident from x < 0 and moving perpendicular to the barrier. For an incident wave exp(ikx) there will, in general be a transmitted wave T exp(iktx) and a reﬂected wave R exp(−ikx).

(b) The transmitted wave vector kt is determined by simple kinematics in terms of k and Bd. What is that relation? Solution Using the Schroedinger equation for x < 0, we ﬁnd E =h ¯2k2/2m. Using the Schroedinger equation for d

¯h2k2 e2B2d2 E = t + , 2m 2mc2 so 2 2 2 2 2 2 2 e B d 2 m ω d kt = k − = k − , 2 2 2 r ¯h c r ¯h where ω = eB/mc is the cyclotron frequency for the ﬁeld B. End Solution

(c) For a given barrier, you will find that, below a certain critical energy E0, kt is imaginary. What does this mean? Give a classical argument that leads to the same critical energy. Solution kt is imaginary whenhk ¯ < mωd. This means that the wave is exponentially killed off moving to the right of the strip which means it is completely reflected. Classically, the particle enters the strip and starts moving on a circle. If the strip is greater than the radius of the circle, the particle will complete a semi-circle and leave the strip in the direction opposite to which it entered. This will occur if mv = mωd. Since mv corresponds tohk ¯ , this happens at the same energy classically as quantum mechanically. End Solution

(d) What is the direction of the transmitted probability ﬂux? It is not along the x-axis! Solution Here we use the expression for j from problem 1. That expression is pretty tough to remember as it stands, but a good way to remember might be as follows. First of all

the probability density is ψ∗ψ. When we have a density and we want a current, we need to multiply by the velocity. With quantum mechanics, we have to replace v by p/m and when we throw in an A, it’s (p − eA/c)/m and ﬁnally we need to take the average with the operator operating on ψ and its complex conjugate operating on ψ∗. In any case,

1 ¯h e ¯h e j = ψ∗ ∇ + A ψ + ψ ∇ + A ψ∗ . 2m i c −i c

For x>d, the wave function is T exp(iktx) and the T is irrelevant for this discussion.

1 ¯hk j = ψ∗p ψ + ψp† ψ∗ = t . x 2m x x m 1 eBd eBd j = ψ∗ ψ + ψ ψ∗ = ωd. y 2m c c If φ is the angle the ﬂux makes with the x-axis, then

mωd mωd tan φ = = . ¯hkt ¯h2k2 − m2ω2d2

It is interesting to note that at the criticalp energy found in the previous part, the transmitted probability current density is parallel to the y-axis. End Solution

(e) Find the reflection and transmission coefficients in the limit d → 0 with Bd fixed. Solution In this case, the vector potential is a step function at x = 0. It’s 0 for x < 0 and Bd for x > 0. Since it’s a step, ψ and its derivatives must be continuous at x = 0. These conditions give 1 + R = T , and ik − ikR = iktT . These can be solved to give

k − k 2k R = t , T = . k + kt k + kt In terms of probability flux, the reflection and transmission coefficents are

k − k 2 4kk r = k|R|2/k = t , t = k |T |2/k = t , k + k t (k + k )2 t t

and of course, we use the kt determined in part (b).

End Solution

4. In lecture, we worked out the generating function for Legendre polynomials,

∞ l 1 F (x, µ) = x Pl(µ) = . 1 − 2xµ + x2 Xl=0 p (a) Use the generating function to demonstrate the following recursion relation among the polynomials,

(l + 1)Pl+1(µ) − (2l + 1)µPl(µ) + lPl−1(µ)=0 .

Hint: what happens if you diﬀerentiate the generating function and the series with respect to x? Solution

∂ 1 µ − x µ − x = = F (x, µ) . ∂x 1 − 2xµ + x2 (1 − 2xµ + x2)3/2 (1 − 2xµ + x2) Or p ∂F (1 − 2xµ + x2) =(µ − x)F . ∂x Or ∞ ∞ 2 l−1 l (1 − 2xµ + x ) lx Pl(µ)=(µ − x) x Pl(µ) . Xl=0 Xl=0 Now equate coeﬃcients of xl on both sides. This gives

(l + 1)Pl+1 − 2lµPl(µ)+(l − 1)Pl−1(µ) = µPl(µ) − Pl−1(µ) , which, after rearrangement, becomes

(l + 1)Pl+1(µ) − (2l + 1)µPl(µ) + lPl−1(µ)=0 .

End Solution

(b) Demonstrate the following recursion relation among the polynomials

′ ′ ′ Pl+1(µ) − 2µPl (µ) + Pl−1(µ) = Pl(µ) .

Hint: what else can you diﬀerentiate with respect to? Solution

Diﬀerentiate the generating function with respect to µ.

∂F x = F . ∂µ 1 − 2xµ + x2

Or ∞ ∞ 2 l ′ l (1 − 2xµ + x ) x Pl (µ) = x x Pl(µ) . Xl=0 Xl=0 Now equate the coeﬃcients of xl+1,

′ ′ ′ Pl+1(µ) − 2µPl (µ) + Pl−1(µ) = Pl(µ) .

Note that with the recursion relations of parts (a) and (b) you can generate many others! End Solution

∞ l 2 (−1/2) x Pl(0) = (1+ x ) Xl=0 −1 x2 −1 −3 x4 −1 −3 −5 x6 = 1 + + + + ··· , 2 1! 2 2 2! 2 2 2 3! From this we can tell immediately that all the odd l polynomials vanish at µ = 0. (We knew that anyway since they’re odd functions of µ.) For even l,

(l − 1)!! (l − 1)!! P (0) = (−1)l/2 =(−1)l/2 , l 2l/2(l/2)! l!!

where !! means double factorial—a factorial where we leave out every other factor. End Solution