ALL 12 CHAPTERS COVERED
SOLUTIONS MANUAL
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0.1 Solutions Ch. 1 . . . . . . . . . . . . . . . . . . . . . . . . . iii
0.1 Solutions Ch. 1
1.
(i) The orthonormality of the states is demonstrated as follows
1 0
< 1 j = 1
= 1; <
1i 0 1 j 2i = 1 0 = 0. Similarly one
0 1
can show < 2 j 1 i = 0 and < 2 j 2i =1
(ii) The column matrix can be written as
a 1 0
=a +b
b 0 1
(iii) The outer products j i i h jj give the following matrices
1 1 0 1 0 1
j 1i h 1j = 1 0 = ;j 1i h 2j = 0 1 = ;
0 0 0 0 0 0
0 0 0 0 0 0
j 2i h 1j = 1 0 = ;j 2i h 2j = 0 1 =
1 1 0 1 0 1
0
(iv) The j i i s satisfy completeness relation from the following relation
,iv Contents
P 1 0
i j ii h ij = j 1i h 1j +j 2i h 2j = =1
0 1
(v) write
a b 1 0 0 1 0 0 0 0
A= =a +b +c +d = aj 1i h 1j +
c d 0 0 0 0 1 0 0 1
b j 1i h 2j + c j 2i h 1j + d j 2i h 2j
(vi)
Aj 1i = +j 1i and A j 2i = j 2i
Constructing the matrix elements from the above relation and using ortho-
normality we …nd
h 1j A j 1i h 1j A j 2i 1 0
fAg = =
h 2j A j 1i h 2j A j 2i 0 1
2.Start with the relation
AA 1 = 1
Take the derivative with respect to
d
AA 1 = 0, therefore
d
dA 1 dA 1 1
A + A = 0; multiplying on the left by A and then moving
d d
the second term to the left gives
dA 1 dA 1
= A 1 A
d d
3.
For an operator A;
y y 1
AA 1 = 1;therefore AA 1 = 1 or A 1 = Ay
1 + iK 1 1
U= = (1 + iK) (1 iK) = (1 iK) (1 + iK)
1 iK
the last step follows from the fact that K 0 s commute among themselves
Therefore,
y
y 1 1
U y = (1 iK) (1 + iK) = (1 iK) (1 + iK) since K is Her-
mitian, and h i
1 1 1
and U U y = (1 + iK) (1 iK) (1 iK) (1 + iK) = (1 + iK) (1 + iK) =
1
One can write
eiC=2 1 + i tan C=2)
eiC = iC=2 =
e 1 i tan C=2)
and identify
K = tan C=2)
One can also show that
U = eiC = cos C + i sin C
, 0.1 Solutions Ch. 1 v
If U = A + iB then identifying
A = cos C, B = sin C we note that A and B commute.
4.
Let U be a unitary operator diagonalizing A, so that
AD = U AU y
is a diagonal matrix. Then
T r(A) = T r(AD )
det(A) = det(AD )
Similarly by expanding eA in powers of A we get
det(eA ) = det(eAD ) = e(AD )11 e(AD )22 e(AD )33 :::::: = eT r(AD ) = eT r(A)
5. P P
T r [j i h j] = < n j i h j n >= h j n >< n j i = h j i
n n
6.
A = j ih j + j ih j + j ih j + j ih j
In the matrix form it can be written (take j i = j1i ; j i = j2i to imple-
ment matrix notation)
1
fAg =
Let a be the eigenvaliues, then
2
(1 a) ( a) j j = 0
The solutions are
q
2 2
(1 + ) (1 ) + 4j j
a=
2
(i) = 1;q = +1
2
2 4j j 2
a= =1 j j
2
= 1;q = 1
2
a= 1+j j
(ii) = i; = +1
a = 2,0
= i; p = 1
a= 2
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0.1 Solutions Ch. 2 . . . . . . . . . . . . . . . . . . . . . . . . . iii
0.1 Solutions Ch. 2
1.
j i=j i+ j i
Since h j i 0
2
[h j + h j] [j i + j i] = h j i + h j i + h j i + j j h j i 0
(1)
To obtain the inequality we need, let
h j i
=
h j i
This choice eliminates the …rst two terms on the right hand side of (1).
We obtain in place of (1)
2
h j i
h j i+ h j i 0
h j i
Thus
2
h j i h j i jh j ij
2.
We start with
2
h j i h j i jh j ij
Let
,iv Contents
j i = A j i, j i = B j i
then
2 2 2
h j ( A) j i h j ( B) j i h j ( A) ( B) j i
We can write
( A) ( B) = 12 [ A; B] + 21 f A; Bg
We note that if two operators C, and D are Hermitian, then if F is their
commutator
F = CD DC
then
F y = Dy C y C y Dy = DC CD = F
Thus F must be purely imaginary
while if G is their anticommutator
G = CD DC
then
Gy = Dy C y + C y Dy = DC + CD = G
Therefore, G is purely real
Hence
2 2 2
jF + Gj = jF j + jGj
Therefore,
2 2 2
jh j ( A) ( B) j ij = 41 jh j [ A; B] j ij + 14 jh j f A; Bg j ij
1 2
4 jh j [ A; B] j ij
This implies then that
2 1 2
j( A) ( B)j 4 j[ A; B]j
Since
[ A; B] = [A; B]
therefore.
2 1 2
j( A) ( B)j 4 j[A; B]j
3. In the above problem we put B = H then
2 1 2
j( A) ( H)j 4 j[A; H]j
However,
dA
i~ = [A; H]
dt
Thus
dA
j( A) ( H)j ~2
dt
We note that ( H) = E and de…ne
1 1 dA
=
t A dt
This leads to
E t ~2
4.
(i) Here
p2
H= + Kr
2m
, 0.1 Solutions Ch. 2 v
From uncertainty relation this gives
~2
E= + Kr
2mr2
To …nd the minimum, we write
@E ~2
= +K =0
@r mr3
Therefore, the minimum value r0 is given by
~2
r03 =
mK
At this minimum 1
~2 3 ~2 3
E= + K r0 = K
2mr02 2 mK
(ii) In this problem
p2 Ze2
H=
2m r
Uncertainty relation gives
~2 Ze2
E= 2
2mr r
@E ~2 Ze2
= + =0
@r mr3 r2
and
~2
r0 =
Zme2
1 Ze2
E=
2 r0
5. For an operator A to be Hermitian, we must have A = Ay ; therefore
y
h jAj i = h jAj i :That is
Z Z
3
d r A = d3 r A (1)
If A = i~@=@r then let us consider the left hand side (LHS)of (1) which
after partialZintegration is of the form
Z Z
LHS = d3 r i~ @@r = d dr r2 i~ @@r
Z Z Z Z h i
= i~ d @
dr @r r2 = i~ d dr r2 @@r + 2r (2)
We note that because of the second term on the right hand side of (2)
( i~@=@r) is not Hermitian.
Let
@ a
A = i~ +
@r r
The
Z new LHS is now Z Z
3 @ a @ a
d r ( i~) ( + ) = d dr r2 (i~) +
Z Z @r r @r r
@ 2 2 a
= d dr ( i~) @r r + (i~) r
r
,vi Contents
Z Z
a
= d dr ( i~) r2 @@r + 2r + (i~) r2
Z Z r
h i
2@
= d dr ( i~) r @r + ( i~) [2r ra ]
Z
@ (2 a)
= d3 r ( i~) ( + )
@r r
Therefore, in order for A to be Hermitian a = 1:Thus
@ 1
A = i~ +
@r r
6
r r
D=p + p
r r
Now
r
p
r
@ x @ y @ z 1 x 1 y 1 z
= i~ + + = i~ +x + +y + +z
@x r @y r @z r r r3 r r3 r r3
x@ y@ z@
i~ + +
r @x r @y r @z
r x@ y@ z@
p = i~ + +
r r @x r @y r @z
r r
D=p + p
r r
2
r 3 x@ y@ z@
= i~ +2 + +
r3 r r @x r @y r @z
2 @
= ( i~) +2
r @r
1 @
= 2 ( i~) +
r @r
Thus 12 D is the same as the operator in the previous problem. It is, of
course, Hermitian
7.
We must take D in this problem
X as
D = 12 [p r + r p] = 12 (pj xj + xj pj )
(i) [D; xi ] = Dxi xi D
Consider
pj xj xi xi pj xj = pj xi xj xi pj xj = (xi pj i~ ij ) xj xi pj xj = i~xi
Similarly, one can show that
xj pj xi xi xj pj = i~xi
Therefore,
[D; xi ] = i~xi
(ii) [D; pi ] = Dpi pi D
, 0.1 Solutions Ch. 2 vii
pj xj pi pi pj xj = pj (pi xj + i~ ij ) pj pi xj = i~pi
Following the same steps as in (i), we get
[D; pi ] = i~pi
(iii)
X [D; Li ] = DLi Li D
ibc [Dxb pc xb pc D]
Consider
pj xj xb pc xb pc pj xj = pj xb xj pc xb pc pj xj
= (xb pj i~ bj ) xj pc xb pc pj xj = xb pj xj pc i~xb pc xb pc pj xj
= xb pj xj pc i~xb pc xb pc pj xj
= xb pj ( jc + pc xj ) i~xb pc xb pc pj xj = 0
Therefore,
[D; Li ] = 0
To prove the relation
ei D=~ xi e i D=~ = e xi (1)
We use the relation
1
eA Be A = B + [A; B] + 2! [A; [A; B]] + :::::(2)
Thus
2
ei D=~ xi e i D=~ = xi + i~ [D; xi ] + 2!
1 i
~ [D; [D; xi ]] + ::::::(3)
We have already found that :::
[D; xi ] = i~xi
Thus the right hand side of (3) is
2
xi + i~ [D; xi ] + i~ [D; [D; xi ]] + ::::::
2 2
= xi + i~ ( i~xi ) + 2!1 i
~ ( i~) xi + :::::
2
= xi + xi + 2! xi + :::::
2
= (1 + + 2! + :::::)xi
= e xi
This proves relation (1)
8.
Start with [x; p] = i~
(i) x; p2 = [x; p] p + p [x; p] = 2i~p
(ii) x2 ; p = x [x; p] + [x; p] x = 2i~x
(iii) x2 ; p2 = x2 ; p p + p x2 ; p = 4i~xp
9.
[x; F (d)] j i = xF (d) j i F (d)x j i = x [F (d) j i] F (d) [x j i] = x [F (d) j i]
(x d) [F (d) j i] = dF (d) j i
Hence
[x; F (d)] = dF (d) (1)
h d j x j d i = h j F y xF j i (2)
However, from the above relation
xF F x = dF = F d
Therefore,
F y xF x = d i.e. F y xF = x + d
, viii Contents
and (2) gives
h dj x j di = h j x + d j i = h j x j i + d
10. The double commutator [[x; H0 ] ; x] can be written as
[[x; H0 ] ; x] = 2xH0 x H0 x2 x2 H0 (1)
The diagonal elements are given by
hi j[[x; H0 ] ; x]j ii = 2 hi jxH0 xj ii 2Ei i x2 i (2)
where we have used the fact that x is Hermitian. To determine the …rst
term on the right hand side of (2), we insert the complete set of jni
P P 2
hi jxH0 xj ii = hi jxj ni hn jH0 xj ii = jhn jxj iij En (3)
n n
where we have used the relation hnj H0 = En hnj :The second term on the
right hand side of (2), is similarly given by
P P 2
i x2 i = hi jxj ni hn jxj ii = jhn jxj iij (4)
n n
From (2), (3) and (4) we obtain
P 2
hi j[[x; H0 ] ; x]j ii = 2 (En Ei ) jhn jxj iij (5)
n
Let us calculate the left hand side of (1) in a di¤erent way. First we note
that
px
[x; H0 ] = i~ (6)
m
and, similarly,
i~ ~2
hi j[[x; H0 ] ; x]j ii = hi j[px ; x]j ii = (7)
m m
Equating (5) and (7) we have
P 2 ~2
(En Ei ) jhn jxj iij =
n 2m
10. consider the matrix element
hij [[H; eik r ]; e ik r
X] jii = hij He e
ik r ik r
eik r He ik r e ik r Heik r +
ik r ik r ik r ik r
e e H jii = fhij He jni hnj e jii hij eik r H jni hnj e ik r jii
ik r
hij
Xe H jni hnj eik r jii + hij e ik r jni hnj eik r H jiig
= fEi hij eik r jni hnj e ik r jii En hij eik r jni hnj e ik r jii En hij e ik r jni hnj eik r jii+
Ei hijXe ik r jni hnj eik r jiig
2 2
= (Ei En ) hij eik r jni + (Ei En ) hij e ik r jni (1)
However,
2 2 2
hij e ik r jni = jhij [cos (k r) i sin (k r)] jnij = jhij cos (k r) jnij +
2 2
jhij sin (k r) jnij = hij eik r jni
Therefore, (1) can be written X as 2
hij [[H; eik r ]; e ik r ] jii = 2 (Ei En ) hij eik r jni (2)
One can also write the operator on the left hand side of (2) as
[[H; eik r ]; e ik r ] = 2H eik r He ik r e ik r Heik r (3)
We use for the second two terms on the right hand side of (3)
1
eA Be A = B + [A; B] + 2! [A; [A; B]] + :::::(4)
SOLUTIONS MANUAL
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0.1 Solutions Ch. 1 . . . . . . . . . . . . . . . . . . . . . . . . . iii
0.1 Solutions Ch. 1
1.
(i) The orthonormality of the states is demonstrated as follows
1 0
< 1 j = 1
= 1; <
1i 0 1 j 2i = 1 0 = 0. Similarly one
0 1
can show < 2 j 1 i = 0 and < 2 j 2i =1
(ii) The column matrix can be written as
a 1 0
=a +b
b 0 1
(iii) The outer products j i i h jj give the following matrices
1 1 0 1 0 1
j 1i h 1j = 1 0 = ;j 1i h 2j = 0 1 = ;
0 0 0 0 0 0
0 0 0 0 0 0
j 2i h 1j = 1 0 = ;j 2i h 2j = 0 1 =
1 1 0 1 0 1
0
(iv) The j i i s satisfy completeness relation from the following relation
,iv Contents
P 1 0
i j ii h ij = j 1i h 1j +j 2i h 2j = =1
0 1
(v) write
a b 1 0 0 1 0 0 0 0
A= =a +b +c +d = aj 1i h 1j +
c d 0 0 0 0 1 0 0 1
b j 1i h 2j + c j 2i h 1j + d j 2i h 2j
(vi)
Aj 1i = +j 1i and A j 2i = j 2i
Constructing the matrix elements from the above relation and using ortho-
normality we …nd
h 1j A j 1i h 1j A j 2i 1 0
fAg = =
h 2j A j 1i h 2j A j 2i 0 1
2.Start with the relation
AA 1 = 1
Take the derivative with respect to
d
AA 1 = 0, therefore
d
dA 1 dA 1 1
A + A = 0; multiplying on the left by A and then moving
d d
the second term to the left gives
dA 1 dA 1
= A 1 A
d d
3.
For an operator A;
y y 1
AA 1 = 1;therefore AA 1 = 1 or A 1 = Ay
1 + iK 1 1
U= = (1 + iK) (1 iK) = (1 iK) (1 + iK)
1 iK
the last step follows from the fact that K 0 s commute among themselves
Therefore,
y
y 1 1
U y = (1 iK) (1 + iK) = (1 iK) (1 + iK) since K is Her-
mitian, and h i
1 1 1
and U U y = (1 + iK) (1 iK) (1 iK) (1 + iK) = (1 + iK) (1 + iK) =
1
One can write
eiC=2 1 + i tan C=2)
eiC = iC=2 =
e 1 i tan C=2)
and identify
K = tan C=2)
One can also show that
U = eiC = cos C + i sin C
, 0.1 Solutions Ch. 1 v
If U = A + iB then identifying
A = cos C, B = sin C we note that A and B commute.
4.
Let U be a unitary operator diagonalizing A, so that
AD = U AU y
is a diagonal matrix. Then
T r(A) = T r(AD )
det(A) = det(AD )
Similarly by expanding eA in powers of A we get
det(eA ) = det(eAD ) = e(AD )11 e(AD )22 e(AD )33 :::::: = eT r(AD ) = eT r(A)
5. P P
T r [j i h j] = < n j i h j n >= h j n >< n j i = h j i
n n
6.
A = j ih j + j ih j + j ih j + j ih j
In the matrix form it can be written (take j i = j1i ; j i = j2i to imple-
ment matrix notation)
1
fAg =
Let a be the eigenvaliues, then
2
(1 a) ( a) j j = 0
The solutions are
q
2 2
(1 + ) (1 ) + 4j j
a=
2
(i) = 1;q = +1
2
2 4j j 2
a= =1 j j
2
= 1;q = 1
2
a= 1+j j
(ii) = i; = +1
a = 2,0
= i; p = 1
a= 2
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0.1 Solutions Ch. 2 . . . . . . . . . . . . . . . . . . . . . . . . . iii
0.1 Solutions Ch. 2
1.
j i=j i+ j i
Since h j i 0
2
[h j + h j] [j i + j i] = h j i + h j i + h j i + j j h j i 0
(1)
To obtain the inequality we need, let
h j i
=
h j i
This choice eliminates the …rst two terms on the right hand side of (1).
We obtain in place of (1)
2
h j i
h j i+ h j i 0
h j i
Thus
2
h j i h j i jh j ij
2.
We start with
2
h j i h j i jh j ij
Let
,iv Contents
j i = A j i, j i = B j i
then
2 2 2
h j ( A) j i h j ( B) j i h j ( A) ( B) j i
We can write
( A) ( B) = 12 [ A; B] + 21 f A; Bg
We note that if two operators C, and D are Hermitian, then if F is their
commutator
F = CD DC
then
F y = Dy C y C y Dy = DC CD = F
Thus F must be purely imaginary
while if G is their anticommutator
G = CD DC
then
Gy = Dy C y + C y Dy = DC + CD = G
Therefore, G is purely real
Hence
2 2 2
jF + Gj = jF j + jGj
Therefore,
2 2 2
jh j ( A) ( B) j ij = 41 jh j [ A; B] j ij + 14 jh j f A; Bg j ij
1 2
4 jh j [ A; B] j ij
This implies then that
2 1 2
j( A) ( B)j 4 j[ A; B]j
Since
[ A; B] = [A; B]
therefore.
2 1 2
j( A) ( B)j 4 j[A; B]j
3. In the above problem we put B = H then
2 1 2
j( A) ( H)j 4 j[A; H]j
However,
dA
i~ = [A; H]
dt
Thus
dA
j( A) ( H)j ~2
dt
We note that ( H) = E and de…ne
1 1 dA
=
t A dt
This leads to
E t ~2
4.
(i) Here
p2
H= + Kr
2m
, 0.1 Solutions Ch. 2 v
From uncertainty relation this gives
~2
E= + Kr
2mr2
To …nd the minimum, we write
@E ~2
= +K =0
@r mr3
Therefore, the minimum value r0 is given by
~2
r03 =
mK
At this minimum 1
~2 3 ~2 3
E= + K r0 = K
2mr02 2 mK
(ii) In this problem
p2 Ze2
H=
2m r
Uncertainty relation gives
~2 Ze2
E= 2
2mr r
@E ~2 Ze2
= + =0
@r mr3 r2
and
~2
r0 =
Zme2
1 Ze2
E=
2 r0
5. For an operator A to be Hermitian, we must have A = Ay ; therefore
y
h jAj i = h jAj i :That is
Z Z
3
d r A = d3 r A (1)
If A = i~@=@r then let us consider the left hand side (LHS)of (1) which
after partialZintegration is of the form
Z Z
LHS = d3 r i~ @@r = d dr r2 i~ @@r
Z Z Z Z h i
= i~ d @
dr @r r2 = i~ d dr r2 @@r + 2r (2)
We note that because of the second term on the right hand side of (2)
( i~@=@r) is not Hermitian.
Let
@ a
A = i~ +
@r r
The
Z new LHS is now Z Z
3 @ a @ a
d r ( i~) ( + ) = d dr r2 (i~) +
Z Z @r r @r r
@ 2 2 a
= d dr ( i~) @r r + (i~) r
r
,vi Contents
Z Z
a
= d dr ( i~) r2 @@r + 2r + (i~) r2
Z Z r
h i
2@
= d dr ( i~) r @r + ( i~) [2r ra ]
Z
@ (2 a)
= d3 r ( i~) ( + )
@r r
Therefore, in order for A to be Hermitian a = 1:Thus
@ 1
A = i~ +
@r r
6
r r
D=p + p
r r
Now
r
p
r
@ x @ y @ z 1 x 1 y 1 z
= i~ + + = i~ +x + +y + +z
@x r @y r @z r r r3 r r3 r r3
x@ y@ z@
i~ + +
r @x r @y r @z
r x@ y@ z@
p = i~ + +
r r @x r @y r @z
r r
D=p + p
r r
2
r 3 x@ y@ z@
= i~ +2 + +
r3 r r @x r @y r @z
2 @
= ( i~) +2
r @r
1 @
= 2 ( i~) +
r @r
Thus 12 D is the same as the operator in the previous problem. It is, of
course, Hermitian
7.
We must take D in this problem
X as
D = 12 [p r + r p] = 12 (pj xj + xj pj )
(i) [D; xi ] = Dxi xi D
Consider
pj xj xi xi pj xj = pj xi xj xi pj xj = (xi pj i~ ij ) xj xi pj xj = i~xi
Similarly, one can show that
xj pj xi xi xj pj = i~xi
Therefore,
[D; xi ] = i~xi
(ii) [D; pi ] = Dpi pi D
, 0.1 Solutions Ch. 2 vii
pj xj pi pi pj xj = pj (pi xj + i~ ij ) pj pi xj = i~pi
Following the same steps as in (i), we get
[D; pi ] = i~pi
(iii)
X [D; Li ] = DLi Li D
ibc [Dxb pc xb pc D]
Consider
pj xj xb pc xb pc pj xj = pj xb xj pc xb pc pj xj
= (xb pj i~ bj ) xj pc xb pc pj xj = xb pj xj pc i~xb pc xb pc pj xj
= xb pj xj pc i~xb pc xb pc pj xj
= xb pj ( jc + pc xj ) i~xb pc xb pc pj xj = 0
Therefore,
[D; Li ] = 0
To prove the relation
ei D=~ xi e i D=~ = e xi (1)
We use the relation
1
eA Be A = B + [A; B] + 2! [A; [A; B]] + :::::(2)
Thus
2
ei D=~ xi e i D=~ = xi + i~ [D; xi ] + 2!
1 i
~ [D; [D; xi ]] + ::::::(3)
We have already found that :::
[D; xi ] = i~xi
Thus the right hand side of (3) is
2
xi + i~ [D; xi ] + i~ [D; [D; xi ]] + ::::::
2 2
= xi + i~ ( i~xi ) + 2!1 i
~ ( i~) xi + :::::
2
= xi + xi + 2! xi + :::::
2
= (1 + + 2! + :::::)xi
= e xi
This proves relation (1)
8.
Start with [x; p] = i~
(i) x; p2 = [x; p] p + p [x; p] = 2i~p
(ii) x2 ; p = x [x; p] + [x; p] x = 2i~x
(iii) x2 ; p2 = x2 ; p p + p x2 ; p = 4i~xp
9.
[x; F (d)] j i = xF (d) j i F (d)x j i = x [F (d) j i] F (d) [x j i] = x [F (d) j i]
(x d) [F (d) j i] = dF (d) j i
Hence
[x; F (d)] = dF (d) (1)
h d j x j d i = h j F y xF j i (2)
However, from the above relation
xF F x = dF = F d
Therefore,
F y xF x = d i.e. F y xF = x + d
, viii Contents
and (2) gives
h dj x j di = h j x + d j i = h j x j i + d
10. The double commutator [[x; H0 ] ; x] can be written as
[[x; H0 ] ; x] = 2xH0 x H0 x2 x2 H0 (1)
The diagonal elements are given by
hi j[[x; H0 ] ; x]j ii = 2 hi jxH0 xj ii 2Ei i x2 i (2)
where we have used the fact that x is Hermitian. To determine the …rst
term on the right hand side of (2), we insert the complete set of jni
P P 2
hi jxH0 xj ii = hi jxj ni hn jH0 xj ii = jhn jxj iij En (3)
n n
where we have used the relation hnj H0 = En hnj :The second term on the
right hand side of (2), is similarly given by
P P 2
i x2 i = hi jxj ni hn jxj ii = jhn jxj iij (4)
n n
From (2), (3) and (4) we obtain
P 2
hi j[[x; H0 ] ; x]j ii = 2 (En Ei ) jhn jxj iij (5)
n
Let us calculate the left hand side of (1) in a di¤erent way. First we note
that
px
[x; H0 ] = i~ (6)
m
and, similarly,
i~ ~2
hi j[[x; H0 ] ; x]j ii = hi j[px ; x]j ii = (7)
m m
Equating (5) and (7) we have
P 2 ~2
(En Ei ) jhn jxj iij =
n 2m
10. consider the matrix element
hij [[H; eik r ]; e ik r
X] jii = hij He e
ik r ik r
eik r He ik r e ik r Heik r +
ik r ik r ik r ik r
e e H jii = fhij He jni hnj e jii hij eik r H jni hnj e ik r jii
ik r
hij
Xe H jni hnj eik r jii + hij e ik r jni hnj eik r H jiig
= fEi hij eik r jni hnj e ik r jii En hij eik r jni hnj e ik r jii En hij e ik r jni hnj eik r jii+
Ei hijXe ik r jni hnj eik r jiig
2 2
= (Ei En ) hij eik r jni + (Ei En ) hij e ik r jni (1)
However,
2 2 2
hij e ik r jni = jhij [cos (k r) i sin (k r)] jnij = jhij cos (k r) jnij +
2 2
jhij sin (k r) jnij = hij eik r jni
Therefore, (1) can be written X as 2
hij [[H; eik r ]; e ik r ] jii = 2 (Ei En ) hij eik r jni (2)
One can also write the operator on the left hand side of (2) as
[[H; eik r ]; e ik r ] = 2H eik r He ik r e ik r Heik r (3)
We use for the second two terms on the right hand side of (3)
1
eA Be A = B + [A; B] + 2! [A; [A; B]] + :::::(4)
