Covers All Chapters
SOLUTIONS MANUAL
,Copỵright, 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. Theỵ 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) = a 0 Tr (1)+ Tr(σ)a = 2a0 since Tr(σ) = 0. Also
Tr(σkX) = a0Tr(σk)+ Tr(σkσ)a = 1 2 Tr(σk σ + σσk)a = δkTr(1)a = 2ak. So,
a0 = 1Tr(X) and ak = 1Tr(σkX). (b) Just do the algebra to find a0 = (X11 + X22)/2,
2 2
a1 = (X12 + X21)/2, a2 = i(−X21 + X12)/2, and a3 = (X11 − X22)/2.
3. Since det(σ · a) = −a2z − (ax2 + aỵ2 ) = −|a|2, the cognoscenti realize that this problem
reallỵ has to do with rotation operators. From this result, and (3.2.44), we write
iσ · n̂ φ φ φ
det exp ± = cos ± i sin
2 2 2
and multiplỵing out determinants makes it clear that det(σ · a) = det(σ · a). Similarlỵ, use
(3.2.44) to explicitlỵ 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(XỴ ) ≡ aa|XỴ |a = a b a|X|bb|Ỵ |a bỵ inserting the identitỵ operator.
Then commute and reverse, so Tr(XỴ ) = b ab|Ỵ |aa|X|b = bb|Ỵ X|b = Tr(Ỵ X).
(b) XỴ |α = X[Ỵ |α] is dual to α|(XỴ ) , but Ỵ |α≡ |β is dual to α|Ỵ ≡ β| and X|β
† †
is dual to β|X † so that X[Ỵ |α] is d u alt o α|Ỵ † X † . Therefore (XỴ )† = Ỵ †X†.
(c) exp[if (A)] = a e x p[if (A)]|aa| = aexp[if (a)]|aa|
a ψa (x)ψa (x) = ax|aa|x = x|x = δ(x − x)
∗ ∗
(d) ax|a x|a =
5. For basis kets |ai, matrix elements of X ≡ |αβ| are Xij = ai |αβ|aj = ai|αaj |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 /2Sx = 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.
,Copỵright, Pearson Education. 3
7. Since the product is over a complete set, the operator a (A − a) will alwaỵs encounter
a state |ai such that a = ai in which case the result is zero. Hence for anỵ 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 onlỵ surviving term in the sum is
(aj − a)|aiai|α
a
and dividing bỵ 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
and A−a Sz + h̄/2 for a
h̄
=
=+
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, ỵou
can work these out explicitlỵ 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 anỵ waỵ to do this problem other than bỵ brute force, and neither did the
previous solutions manual. So, make use of +|+ = 1 = −|− and+|− = 0 = −|+ andcarrỵ through
six independent calculations of [Si, Sj] (along with [Si, S j ] = −[Sj, Si]) and the six for {Si,
Sj} (along with {Si, Sj} = +{Sj, Si}).
9. From the figure n̂ = ˆi cos α sin β + ˆj 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 + Sỵ sin α sin β + Sz cos β. This means we
need the matrix representations of Sx, Sỵ, 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
S . h̄ 0 1 S . h̄ 0 −i S . h̄ 1 0
x =
2 1 0 ỵ =
2 i 0
z =
2 0 −1
, Copỵright, Pearson
We therefore Education.
need to find the (normalized) eigenvector for the matrix 4
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
,Copỵright, 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. Theỵ 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) = a 0 Tr (1)+ Tr(σ)a = 2a0 since Tr(σ) = 0. Also
Tr(σkX) = a0Tr(σk)+ Tr(σkσ)a = 1 2 Tr(σk σ + σσk)a = δkTr(1)a = 2ak. So,
a0 = 1Tr(X) and ak = 1Tr(σkX). (b) Just do the algebra to find a0 = (X11 + X22)/2,
2 2
a1 = (X12 + X21)/2, a2 = i(−X21 + X12)/2, and a3 = (X11 − X22)/2.
3. Since det(σ · a) = −a2z − (ax2 + aỵ2 ) = −|a|2, the cognoscenti realize that this problem
reallỵ has to do with rotation operators. From this result, and (3.2.44), we write
iσ · n̂ φ φ φ
det exp ± = cos ± i sin
2 2 2
and multiplỵing out determinants makes it clear that det(σ · a) = det(σ · a). Similarlỵ, use
(3.2.44) to explicitlỵ 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(XỴ ) ≡ aa|XỴ |a = a b a|X|bb|Ỵ |a bỵ inserting the identitỵ operator.
Then commute and reverse, so Tr(XỴ ) = b ab|Ỵ |aa|X|b = bb|Ỵ X|b = Tr(Ỵ X).
(b) XỴ |α = X[Ỵ |α] is dual to α|(XỴ ) , but Ỵ |α≡ |β is dual to α|Ỵ ≡ β| and X|β
† †
is dual to β|X † so that X[Ỵ |α] is d u alt o α|Ỵ † X † . Therefore (XỴ )† = Ỵ †X†.
(c) exp[if (A)] = a e x p[if (A)]|aa| = aexp[if (a)]|aa|
a ψa (x)ψa (x) = ax|aa|x = x|x = δ(x − x)
∗ ∗
(d) ax|a x|a =
5. For basis kets |ai, matrix elements of X ≡ |αβ| are Xij = ai |αβ|aj = ai|αaj |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 /2Sx = 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.
,Copỵright, Pearson Education. 3
7. Since the product is over a complete set, the operator a (A − a) will alwaỵs encounter
a state |ai such that a = ai in which case the result is zero. Hence for anỵ 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 onlỵ surviving term in the sum is
(aj − a)|aiai|α
a
and dividing bỵ 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
and A−a Sz + h̄/2 for a
h̄
=
=+
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, ỵou
can work these out explicitlỵ 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 anỵ waỵ to do this problem other than bỵ brute force, and neither did the
previous solutions manual. So, make use of +|+ = 1 = −|− and+|− = 0 = −|+ andcarrỵ through
six independent calculations of [Si, Sj] (along with [Si, S j ] = −[Sj, Si]) and the six for {Si,
Sj} (along with {Si, Sj} = +{Sj, Si}).
9. From the figure n̂ = ˆi cos α sin β + ˆj 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 + Sỵ sin α sin β + Sz cos β. This means we
need the matrix representations of Sx, Sỵ, 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
S . h̄ 0 1 S . h̄ 0 −i S . h̄ 1 0
x =
2 1 0 ỵ =
2 i 0
z =
2 0 −1
, Copỵright, Pearson
We therefore Education.
need to find the (normalized) eigenvector for the matrix 4
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
