Physics 215C Homework #1 Solutions - KITP - University of ...

Physics 215C Homework #1 Solutions
Richard Eager
Department of Physics
University of California; Santa Barbara, CA 93106
Shankar 20.1.1
Derive the continuity equation
where P = ψ † ψ and j = cψ † αψ
∂P
∂t + ∇ · j = 0
The Dirac equation takes the equivalent forms
The conjugate equation is
i ∂|ψ〉
∂t
= ( cα · P + βmc 2) |ψ〉
(
∂ψ
∂t = −cα · ∇ − i )
βmc2 ψ
∂ψ †
∂t
=
(−cα · ∇ + i βmc2 )
ψ †
∂P
∂t
∂ψ
= ψ†
∂t + ∂ψ†
∂t ψ
= −c ( ψ † α · ∇ψ + (α · ∇ψ † )ψ )
∇ · j = c ( ∇ψ † αψ + ψ † α · ∇ψ )
Adding both equations together yields the desired result
∂P
∂t + ∇ · j = 0
Both terms in the Hamiltonian, cα · P and βmc 2 are Hermitian, so where did
the relative minus sign come from? To show that P is a Hermitian operator you
need to integrate by parts. If you are working with the L 2 inner product you
can drop boundary terms, but when working locally to derive the continuity
equation we don’t integrate by parts and get a relative minus sign.
1

2
Show that the probability current j of the previous exercise reduces in the nonrelativistic
limit to Eq.(5.3.8) [which is the same as Sakurai Eq.(2.4.16)].
( ) 0 σ
The probability current j = cψ † αψ and α = We can write the wave
(
σ 0
χ
function ψ = in terms of its relativistic and non-relativistic components,
Φ)
Φ and χ respectively.
In the non-relativistic limit (20.2.13)
j = c ( ( ( )
χ † Φ †) 0 σ χ
σ 0)
Φ
(0.1)
= c ( χ † σΦ + Φ † σχ ) (0.2)
Φ ≈ σ · π
2mc
j = χ † ˆp
2m χ + ( ˆp
2m χ† )χ
which is the non-relativistic current (5.3.8)
j =
from the identification ˆp = −i∇.
Shankar 20.2.1
2mi (ψ∗ ∇ψ − ψ∇ψ ∗ )
Show that
where π = P − qA c .
π × π = iq
c B
π × π = P × P − qA c × P − P × qA c + qA c × qA c
= − q (A × P + P × A)
c
= iq
c
(A × ∇ + ∇ × A)
The simplest way to manipulate operators is to act on a test function ψ
Therefore
iq
c
iq
(A × ∇ + ∇ × A) ψ = (A × ∇ψ + ∇ × (ψA))
c
= iq
c
= iq
c
= iq
c Bψ
(A × ∇ψ + (∇ψ)A + (∇ × A)ψ)
(∇ × A) ψ
π × π = iq
c B
2

Shankar 20.1.1
Solve for the 4 spinors w that satisfy Shankar Eq. 20.3.3. You may assume that
the 3- momentum ⃗p is along the z-axis. Normalize them to unity, and show that
they are mutually orthogonal.
Equation (20.3.3) is
Ew = (α · p + βm)w
In terms of the relativistic and non-relativistic components,
( ( ) (
E − m −σ · p χ 0
=
−σ · p E + m)
Φ 0)
The ( solutions ( are given in equations (20.3.7) and (20.3.8). Choosing the basis
1 0
and for Φ and letting p be in the z direction,
0)
1)
w 1,3 =
⎛
⎞
p/(±E − m)
⎜
⎝
0
1
0
⎟
⎠ w 2,4 =
⎛
⎞
0
p/(±E − m)
⎜
⎟
⎝ 0 ⎠
1
Orthogonality of the spinors is easy to see using E 2 = p 2 + m 2 .
5
The five terms in 20.2.28 are
P 2
2m
V
− P 4
8m 3 c 2
iσ · P × [P, V ]
4m 2 c 2
P · [P, V ]
4m 2 c 2
Hermitian
Hermitian
Hermitian
Hermitian
anti-Hermitian
Recall that ˆP is a Hermitian operator, the potential V is assumed to be
Hermitian (for conservation of probability) and the Pauli matrices σ are Hermitian.
The commutator [X, Y ] of two Hermitian operators is anti-Hermitian
since [X, Y ] † = (XY ) † − (Y X) † = Y † X † − X † Y † = Y X − XY = −[X, Y ].
The cross product of two Hermitian (vector) operators is again Hermitian since
(X × Y ) † = X † × Y † = X × Y.
3