Covers All 8 Chapters
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1 - Fundamental Concepts
2 - Quantum Dynamics
3 - Theory of Angular Momentum
4 - Symmetry in Quantum Mechanics
5 - Approximation Methods
6 - Scattering Theory
7 - Identical Particles
8 - Relativistic Quantum Mechanics
,Copyright, Pearson Education. 2
Chapter One
1. AC{D, B} = ACDB + ACBD, A{C, B}D = ACBD + ABCD, C{D, A}B = CDAB +
CADB, and {C, A}DB = CADB+ACDB. Therefore −AC{D, B}+A{C, B}D−C{D, A}B+
{C, A}DB = −ACDB + ABCD − CDAB + ACDB = ABCD − CDAB = [AB, CD]
In preparing this solution manual, I have realized that problems 2 and 3 in are misplaced
in this chapter. They belong in Chapter Three. The Pauli matrices are not even defined in
Chapter One, nor is the math used in previous solution manual. – Jim Napolitano
�
2. (a) Tr(X) = a0 Tr(1)�+ � Tr(σ� )a� = 2a�0 since Tr(σ� ) = 0. Also�
1
Tr(σk X) = a0 Tr(σk ) + � Tr(σk σ� )a� = 2 � Tr(σk σ� + σ� σk )a� = � δk� Tr(1)a� = 2ak . So,
a0 = 21 Tr(X) and ak = 12 Tr(σk X). (b) Just do the algebra to find a0 = (X11 + X22 )/2,
a1 = (X12 + X21 )/2, a2 = i(−X21 + X12 )/2, and a3 = (X11 − X22 )/2.
3. Since det(σ · a) = −a2z − (a2x + a2y ) = −|a|2 , the cognoscenti realize that this problem
really has to do with rotation operators. From this result, and (3.2.44), we write
� � �� � � � �
iσ · n̂φ φ φ
det exp ± = cos ± i sin
2 2 2
and multiplying out determinants makes it clear that det(σ · a� ) = det(σ · a). Similarly, use
(3.2.44) to explicitly write out the matrix σ · a� and equate the elements to those of σ · a.
With n̂ in the z-direction, it is clear that we have just performed a rotation (of the spin
vector) through the angle φ.
� � �
4. (a) Tr(XY ) ≡ a �a|XY |a� = a b �a|X|b��b|Y
� � |a� by inserting
� the identity operator.
Then commute and reverse, so Tr(XY ) = b a �b|Y |a��a|X|b� = b �b|Y X|b� = Tr(Y X).
(b) XY |α� = X[Y |α�] is dual to �α|(XY )† , but Y |α� ≡ |β� is dual to �α|Y † ≡ �β| and X|β�
is dual to �β|X † so
� that X[Y |α�] is dual�to �α|Y † X † . Therefore (XY )† = Y † X † .
(c) exp[if
� ∗(A)] = a exp[if
� (A)]|a��a| = a� exp[if (a)]|a��a|
(d) a ψa (x )ψa (x ) = a �x |a� �x |a� = a �x�� |a��a|x� � = �x�� |x� � = δ(x�� − x� )
� �� � ∗ ��
5. For basis kets |ai �, matrix elements of X ≡ |α��β| are Xij = �ai |α��β|aj � = �ai |α��aj |β�∗ .
For spin-1/2 in the√| ± z� basis, �+|Sz = h̄/2� = 1, �−|Sz = h̄/2� = 0, and, using (1.4.17a),
�±|Sx = h̄/2� = 1/ 2. Therefore
� �
. 1 1 1
|Sz = h̄/2��Sx = h̄/2| = √
2 0 0
6. A[|i� + |j�] = ai |i� + aj |j� =
� [|i� + |j�] so in general it is not an eigenvector, unless ai = aj .
That is, |i� + |j� is not an eigenvector of A unless the eigenvalues are degenerate.
, Copyright, Pearson Education. 3
7. Since the product is over a complete set, the operator a� (A − a� ) will always encounter
�
a state |ai � such that a� = ai in which case the result is zero. Hence for any state |α�
� � � �� �
(A − a� )|α� = (A − a� ) |ai ��ai |α� = (ai − a� )|ai ��ai |α� = 0=0
a� a� i i a� i
If the product instead is over all a� =
� aj then the only surviving term in the sum is
�
(aj − a� )|ai ��ai |α�
a�
and dividing by the factors (aj − a� ) just gives the projection of |α� on the direction |a� �. For
the operator A ≡ Sz and {|a� �} ≡ {|+�, |−�}, we have
� �� �
�
� h̄ h̄
(A − a ) = Sz − Sz +
a�
2 2
� A − a� Sz + h̄/2 h̄
and �� �
= for a�� = +
a� �=a��
a −a h̄ 2
Sz − h̄/2 h̄
or = for a�� = −
−h̄ 2
It is trivial to see that the first operator is the null operator. For the second and third, you
can work these out explicitly using (1.3.35) and (1.3.36), for example
� �
Sz + h̄/2 1 h̄ 1
= Sz + 1 = [(|+��+|) − (|−��−|) + (|+��+|) + (|−��−|)] = |+��+|
h̄ h̄ 2 2
which is just the projection operator for the state |+�.
8. I don’t see any way to do this problem other than by brute force, and neither did the
previous solutions manual. So, make use of �+|+� = 1 = �−|−� and�+|−� = 0 = �−|+� and
carry through six independent calculations of [Si , Sj ] (along with [Si , Sj ] = −[Sj , Si ]) and
the six for {Si , Sj } (along with {Si , Sj } = +{Sj , Si }).
9. From the figure n̂ = î cos α sin β + ĵ sin α sin β + k̂ cos β so we need to find the matrix
representation of the operator S · n̂ = Sx cos α sin β + Sy sin α sin β + Sz cos β. This means we
need the matrix representations of Sx , Sy , and Sz . Get these from the prescription (1.3.19)
and the operators represented as outer products in (1.4.18) and (1.3.36), along with the
association (1.3.39a) to define which element is which. Thus
� � � � � �
. h̄ 0 1 . h̄ 0 −i . h̄ 1 0
Sx = Sy = Sz =
2 1 0 2 i 0 2 0 −1
We therefore need to find the (normalized) eigenvector for the matrix
� � � �
cos β cos α sin β − i sin α sin β cos β e−iα sin β
=
cos α sin β + i sin α sin β − cos β eiα sin β − cos β
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1 - Fundamental Concepts
2 - Quantum Dynamics
3 - Theory of Angular Momentum
4 - Symmetry in Quantum Mechanics
5 - Approximation Methods
6 - Scattering Theory
7 - Identical Particles
8 - Relativistic Quantum Mechanics
,Copyright, Pearson Education. 2
Chapter One
1. AC{D, B} = ACDB + ACBD, A{C, B}D = ACBD + ABCD, C{D, A}B = CDAB +
CADB, and {C, A}DB = CADB+ACDB. Therefore −AC{D, B}+A{C, B}D−C{D, A}B+
{C, A}DB = −ACDB + ABCD − CDAB + ACDB = ABCD − CDAB = [AB, CD]
In preparing this solution manual, I have realized that problems 2 and 3 in are misplaced
in this chapter. They belong in Chapter Three. The Pauli matrices are not even defined in
Chapter One, nor is the math used in previous solution manual. – Jim Napolitano
�
2. (a) Tr(X) = a0 Tr(1)�+ � Tr(σ� )a� = 2a�0 since Tr(σ� ) = 0. Also�
1
Tr(σk X) = a0 Tr(σk ) + � Tr(σk σ� )a� = 2 � Tr(σk σ� + σ� σk )a� = � δk� Tr(1)a� = 2ak . So,
a0 = 21 Tr(X) and ak = 12 Tr(σk X). (b) Just do the algebra to find a0 = (X11 + X22 )/2,
a1 = (X12 + X21 )/2, a2 = i(−X21 + X12 )/2, and a3 = (X11 − X22 )/2.
3. Since det(σ · a) = −a2z − (a2x + a2y ) = −|a|2 , the cognoscenti realize that this problem
really has to do with rotation operators. From this result, and (3.2.44), we write
� � �� � � � �
iσ · n̂φ φ φ
det exp ± = cos ± i sin
2 2 2
and multiplying out determinants makes it clear that det(σ · a� ) = det(σ · a). Similarly, use
(3.2.44) to explicitly write out the matrix σ · a� and equate the elements to those of σ · a.
With n̂ in the z-direction, it is clear that we have just performed a rotation (of the spin
vector) through the angle φ.
� � �
4. (a) Tr(XY ) ≡ a �a|XY |a� = a b �a|X|b��b|Y
� � |a� by inserting
� the identity operator.
Then commute and reverse, so Tr(XY ) = b a �b|Y |a��a|X|b� = b �b|Y X|b� = Tr(Y X).
(b) XY |α� = X[Y |α�] is dual to �α|(XY )† , but Y |α� ≡ |β� is dual to �α|Y † ≡ �β| and X|β�
is dual to �β|X † so
� that X[Y |α�] is dual�to �α|Y † X † . Therefore (XY )† = Y † X † .
(c) exp[if
� ∗(A)] = a exp[if
� (A)]|a��a| = a� exp[if (a)]|a��a|
(d) a ψa (x )ψa (x ) = a �x |a� �x |a� = a �x�� |a��a|x� � = �x�� |x� � = δ(x�� − x� )
� �� � ∗ ��
5. For basis kets |ai �, matrix elements of X ≡ |α��β| are Xij = �ai |α��β|aj � = �ai |α��aj |β�∗ .
For spin-1/2 in the√| ± z� basis, �+|Sz = h̄/2� = 1, �−|Sz = h̄/2� = 0, and, using (1.4.17a),
�±|Sx = h̄/2� = 1/ 2. Therefore
� �
. 1 1 1
|Sz = h̄/2��Sx = h̄/2| = √
2 0 0
6. A[|i� + |j�] = ai |i� + aj |j� =
� [|i� + |j�] so in general it is not an eigenvector, unless ai = aj .
That is, |i� + |j� is not an eigenvector of A unless the eigenvalues are degenerate.
, Copyright, Pearson Education. 3
7. Since the product is over a complete set, the operator a� (A − a� ) will always encounter
�
a state |ai � such that a� = ai in which case the result is zero. Hence for any state |α�
� � � �� �
(A − a� )|α� = (A − a� ) |ai ��ai |α� = (ai − a� )|ai ��ai |α� = 0=0
a� a� i i a� i
If the product instead is over all a� =
� aj then the only surviving term in the sum is
�
(aj − a� )|ai ��ai |α�
a�
and dividing by the factors (aj − a� ) just gives the projection of |α� on the direction |a� �. For
the operator A ≡ Sz and {|a� �} ≡ {|+�, |−�}, we have
� �� �
�
� h̄ h̄
(A − a ) = Sz − Sz +
a�
2 2
� A − a� Sz + h̄/2 h̄
and �� �
= for a�� = +
a� �=a��
a −a h̄ 2
Sz − h̄/2 h̄
or = for a�� = −
−h̄ 2
It is trivial to see that the first operator is the null operator. For the second and third, you
can work these out explicitly using (1.3.35) and (1.3.36), for example
� �
Sz + h̄/2 1 h̄ 1
= Sz + 1 = [(|+��+|) − (|−��−|) + (|+��+|) + (|−��−|)] = |+��+|
h̄ h̄ 2 2
which is just the projection operator for the state |+�.
8. I don’t see any way to do this problem other than by brute force, and neither did the
previous solutions manual. So, make use of �+|+� = 1 = �−|−� and�+|−� = 0 = �−|+� and
carry through six independent calculations of [Si , Sj ] (along with [Si , Sj ] = −[Sj , Si ]) and
the six for {Si , Sj } (along with {Si , Sj } = +{Sj , Si }).
9. From the figure n̂ = î cos α sin β + ĵ sin α sin β + k̂ cos β so we need to find the matrix
representation of the operator S · n̂ = Sx cos α sin β + Sy sin α sin β + Sz cos β. This means we
need the matrix representations of Sx , Sy , and Sz . Get these from the prescription (1.3.19)
and the operators represented as outer products in (1.4.18) and (1.3.36), along with the
association (1.3.39a) to define which element is which. Thus
� � � � � �
. h̄ 0 1 . h̄ 0 −i . h̄ 1 0
Sx = Sy = Sz =
2 1 0 2 i 0 2 0 −1
We therefore need to find the (normalized) eigenvector for the matrix
� � � �
cos β cos α sin β − i sin α sin β cos β e−iα sin β
=
cos α sin β + i sin α sin β − cos β eiα sin β − cos β
