Phys 268r Fall 2013
Problem Set 4  PART II  Solutions
B. Halperin
Solutions to Problem Set 4  Part II
Problem 5.
a)
We have from the definitions:
Z
dxdx
0
ψ
†
(
x
0
)
h
Ψ(
x
0
)
,
Ψ
†
(
x
)
ˆ
H
Ψ(
x
)
i
=
EX,
where
ˆ
H
=
h
vp
x
σ
z
+
{
p
x
,
~
f
(
s
)
} ·
~σ
+
V
(
x
) +
h
(
x
)
σ
x
i
.
We compute the commutator, showing
the first and third terms of
~
H
(the others follow similarly). For the third term, we have:
Ψ(
x
0
)
,
Ψ
†
(
x
)
V
(
x
)Ψ(
x
)
=
V
(
x
)
(
Ψ(
x
0
)Ψ
†
(
x
)Ψ(
x
)

Ψ
†
(
x
)Ψ(
x
)Ψ(
x
0
)
)
=
V
(
x
)
(
δ
(
x

x
0
)

Ψ
†
(
x
)Ψ(
x
0
) Ψ(
x
)

Ψ
†
(
x
)Ψ(
x
)Ψ(
x
0
)
)
=
V
(
x
)
(
δ
(
x

x
0
)Ψ(
x
)

Ψ
†
(
x
)Ψ(
x
0
)Ψ(
x
)

Ψ
†
(
x
)Ψ(
x
)Ψ(
x
0
)
)
=
V
(
x
)
(
δ
(
x

x
0
)Ψ(
x
) + Ψ
†
(
x
)Ψ(
x
)Ψ(
x
0
)

Ψ
†
(
x
)Ψ(
x
)Ψ(
x
0
)
)
=
V
(
x
)
δ
(
x

x
0
)Ψ(
x
)
.
For the first term (different because of the
p
x
), we have:
Ψ(
x
0
)
,
Ψ
†
(
x
)
vp
x
σ
z
Ψ(
x
)
=
v
(
Ψ(
x
0
)Ψ
†
(
x
)
p
x
σ
z
Ψ(
x
)

Ψ
†
(
x
)
p
x
σ
z
Ψ(
x
)Ψ(
x
0
)
)
=
v
(
δ
(
x

x
0
)

Ψ
†
(
x
)Ψ(
x
0
)
p
x
σ
z
Ψ(
x
)

Ψ
†
(
x
)
p
x
σ
z
Ψ(
x
)Ψ(
x
0
)
)
=
v
(
δ
(
x

x
0
)
p
x
σ
z
Ψ(
x
)

Ψ
†
(
x
)Ψ(
x
0
)
p
x
σ
z
Ψ(
x
)

Ψ
†
(
x
)
p
x
σ
z
Ψ(
x
)Ψ(
x
0
)
)
=
v
(
δ
(
x

x
0
)
p
x
σ
z
Ψ(
x
)

Ψ
†
(
x
)Ψ(
x
0
)
p
x
σ
z
Ψ(
x
)

Ψ
†
(
x
)
p
x
σ
z
Ψ(
x
)Ψ(
x
0
)
)
=
v δ
(
x

x
0
)
p
x
σ
z
Ψ(
x
)

Ψ
†
(
x
)
iδ
(
x

x
0
)
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 Fall '13
 BertrandI.Halperin
 Physics, Equals sign, plane wave, Ψ, B. Halperin, px σz