Problem solutions, 23 November 20111

D. E. Soper2

University of Oregon23 November 2011

Derive Sakurai, equation (2.7.27)

For a particle with charge q in a vector potential ~A and a scalar potentialφ, both time independent, we take

H =1

2m~Π2 − qφ(~x) , (1)

where~Π = ~p− q ~A(~x) . (2)

Then

dxidt

= − i [xi, H]

= − i 1

m

∑j

[xi, pj] (pj − qAj)

=1

m

∑j

δij (pj − qAj)

=1

m(pi − qAi)

=1

mΠi .

(3)

1Copyright, 2011, D. E. Soper2soper@uoregon.edu

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In order to evaluate ∂~Π/∂t, it is useful to first note that

[Πi,Πj] = − q[pi, Aj]− q[Ai, pj]

= − q[pi, Aj] + q[pj, Ai]

= − q∑k

[pi, xk]∂kAj + q∑k

[pj, xk]∂kAi

= iq∑k

δik ∂kAj − iq∑k

δjk ∂kAi

= iq[∂iAj − ∂jAi]

= iq∑k

εijkBk .

(4)

We also have

[Πi, φ(~x)] = [pi, φ]

=∑j

[pi, xj]∂jφ

= − i∑j

δij ∂jφ

= − i∂iφ .

(5)

Using these results, we have

dΠi

dt= − i [Πi, H]

= − i 1

2m

∑j

{[Πi,Πj]Πj + Πj[Πi,Πj]} − iq[Πi, φ]

=q

2m

∑j,k

{εijkBkΠj + ΠjεijkBk} − q ∂iφ

=q

2m

∑j,k

εijk(ΠjBk +BkΠj)− q ∂iφ

=q

2m

{(~Π× ~B)i − ( ~B × ~Π)i

}+ q Ei

(6)

Note that we have ΠjBk + BkΠj here because Πj and Bk do not commute.

If we write ~Π as m~v according to our first equation of motion, this reads

md~v

dt=q

2

{(~v × ~B)− ( ~B × ~v)

}+ q ~E . (7)

2

This is the standard Lorentz force equation except that ~B and ~v are operatorsso we have to write them in the correct operator order.

Sakurai, problem 2.39 We have a hamiltonian For a particle with chargeq in a vector potential ~A and a scalar potential φ, both time independent,we take

H =1

2m~Π2 , (8)

where~Π = ~p− q ~A(~x) (9)

and the magnetic field ~B = ~∇ × ~A is in the z direction and has a constantvalue B. Using

[Πi,Πj] = iq∑k

εijkBk , (10)

which was proved in the previous problem, we have

[Πy,Πz] = 0 ,

[Πz,Πx] = 0 ,

[Πx,Πy] = iqB .

(11)

We see from this that [Πz, H] = 0, so that we can diagonalize Πz and H atthe same time. Thus we can look for states

∣∣n, pz, ξ⟩ where H∣∣n, pz, ξ⟩ =

En

∣∣n, pz, ξ⟩ and Πz

∣∣n, pz, ξ⟩ = pz∣∣n, pz, ξ⟩.

Here ξ is another label for the state but the value of ξ does not affect theenergy. In this problem, we do not learn anything about ξ, but we need toinclude it because the solution for eigenvalues En, pz might not be unique.In fact, one can show that there is another operator that commutes with Hand pz, so that we do need another label. There is more than one way tochoose this operator. We do not study this issue in this problem.

Now let’s make this look like a harmonic oscillator by defining

a =1√2qB

(Πx + iΠy) ,

a† =1√2qB

(Πx − iΠy) ,

N = a†a .

(12)

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We note that

N =1

2qB(Πx + iΠy)(Πx − iΠy)

=1

2qB(Π2

x + Π2y − i[Πx,Πy])

=1

2qB(Π2

x + Π2y + qB)

=1

2qB(Π2

x + Π2y)−

1

2

=m

qBH − 1

2mΠ2

z −1

2

(13)

Thus

H =qB

m

(N +

1

2

)+

1

2mΠ2

z . (14)

Let us choose states∣∣n, pz, ξ⟩ that are eigenstates of Πz and N with

N∣∣n, pz, ξ⟩ = n

∣∣n, pz, ξ⟩ ,

Πz

∣∣n, pz, ξ⟩ = pz∣∣n, pz, ξ⟩ .

(15)

Then

H∣∣n, pz, ξ⟩ =

[qB

m

(N +

1

2

)+

p2z2m

] ∣∣n, pz, ξ⟩ (16)

This gives the eigenvalues of H as soon as we know what the eigenvalues ofN and Πz are. There is no restriction on the eigenvalues pz of Πz, so p2z/(2m)can be any positive number.

To find the eigenvalues of N , we note that

[a, a†] = − i

qB[Πx,Πy] = 1 . (17)

Thus

Na = a(N − 1) ,

Na† = a†(N + 1) .(18)

That is, applying a to an eigenstate of N lowers the eigenvalue by 1, whileapplying a† raises the eigenvalue by 1. Just as with the harmonic oscillator,we derive that the eigenvalues of N are non-negative integers, 0,1,2,. . . .

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