2
SOLUTIONS MANUAL
, 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) = a 0 Tr (1 )+ Tr(σ)a = 2a0 since Tr(σ) = 0. Also
Tr(σkX) = a0Tr(σk) + Tr(σkσ)a = 1
2 Tr(σkσ + σσk)a = δkTr(1)a = 2ak. So,
2 2
a0 = 1 Tr(X) and ak = 1 Tr(σkX). (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) = −az2 −(a2x + a y2 ) = −|a|2, the cognoscenti realize that this problem
really has to do with rotation operators. From this result, and (3.2.44), we
iσ · n̂ φ φ φ
write 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 ) ≡ aa|XY |a= a b a|X|bb |Y |aby inserting the identity operator.
Then commute and reverse, so Tr(XY ) = b ab|Y |aa|X|b = bb|Y X|b = Tr(Y X).
is dual to β|X† so that X[Y |α] is du a lto α|Y †X†. Therefore (XY )† = 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
(c) exp[if (A)] = a exp[if (A)]|aa| = aexp[if (a)]|aa|
(d) a ψa∗ (x)ψa(x) = ∗
ax|a x|a = ax|aa|x = x|x = δ(x − x)
, 3
5. For basis kets |ai, matrix elements of X ≡|αβ| are Xij = ai|αβ|aj = ai|αaj|β∗ . For spin-1/2 in
the√| ± zbasis, +|Sz = h¯ /2 = 1, −|Sz = h¯ /2 = 0, and, using (1.4.17a),
±|Sx = h¯ /2 = 1/ 2.
Therefore . 1 1 1
|Sz = h̄/2Sx = h̄ /2| = √ 0 0
2
, 4
6. A[|i+ |j] = ai|i+ aj |j= [|i+ |j] so in general it is not an eigenvector, unless ai = aj.
That is, |i+ |jis not an eigenvector of A unless the eigenvalues are degenerate.
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) |aiai|α= (ai −a)|aiai|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)|aiai|α
a
and dividing by the factors (aj −a) just gives the projection of |αon the direction |a. For
the operator A ≡Sz and {|a} ≡{|+, |−}, we have
h̄
— S h
¯
(A −a) = Sz 2 z +
a 2
A −a Sz + h¯ / 2 h̄
and = for a
a−a h̄ = +2
a= a
Sz − for a =
or = — h
h̄ /2 ¯
− h¯
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 + h¯ / 2 1 1
= h̄
+ 1 = [(|++|) − (|−−|)+ (|++|) + (|−−|)] = |++|
S h̄ h̄ z 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, 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 + 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
S . S . S .
x = y = z =
2 1 2 i 2 0 −1
0 0
We therefore need to find the (normalized) eigenvector for the matrix
SOLUTIONS MANUAL
, 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) = a 0 Tr (1 )+ Tr(σ)a = 2a0 since Tr(σ) = 0. Also
Tr(σkX) = a0Tr(σk) + Tr(σkσ)a = 1
2 Tr(σkσ + σσk)a = δkTr(1)a = 2ak. So,
2 2
a0 = 1 Tr(X) and ak = 1 Tr(σkX). (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) = −az2 −(a2x + a y2 ) = −|a|2, the cognoscenti realize that this problem
really has to do with rotation operators. From this result, and (3.2.44), we
iσ · n̂ φ φ φ
write 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 ) ≡ aa|XY |a= a b a|X|bb |Y |aby inserting the identity operator.
Then commute and reverse, so Tr(XY ) = b ab|Y |aa|X|b = bb|Y X|b = Tr(Y X).
is dual to β|X† so that X[Y |α] is du a lto α|Y †X†. Therefore (XY )† = 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
(c) exp[if (A)] = a exp[if (A)]|aa| = aexp[if (a)]|aa|
(d) a ψa∗ (x)ψa(x) = ∗
ax|a x|a = ax|aa|x = x|x = δ(x − x)
, 3
5. For basis kets |ai, matrix elements of X ≡|αβ| are Xij = ai|αβ|aj = ai|αaj|β∗ . For spin-1/2 in
the√| ± zbasis, +|Sz = h¯ /2 = 1, −|Sz = h¯ /2 = 0, and, using (1.4.17a),
±|Sx = h¯ /2 = 1/ 2.
Therefore . 1 1 1
|Sz = h̄/2Sx = h̄ /2| = √ 0 0
2
, 4
6. A[|i+ |j] = ai|i+ aj |j= [|i+ |j] so in general it is not an eigenvector, unless ai = aj.
That is, |i+ |jis not an eigenvector of A unless the eigenvalues are degenerate.
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) |aiai|α= (ai −a)|aiai|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)|aiai|α
a
and dividing by the factors (aj −a) just gives the projection of |αon the direction |a. For
the operator A ≡Sz and {|a} ≡{|+, |−}, we have
h̄
— S h
¯
(A −a) = Sz 2 z +
a 2
A −a Sz + h¯ / 2 h̄
and = for a
a−a h̄ = +2
a= a
Sz − for a =
or = — h
h̄ /2 ¯
− h¯
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 + h¯ / 2 1 1
= h̄
+ 1 = [(|++|) − (|−−|)+ (|++|) + (|−−|)] = |++|
S h̄ h̄ z 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, 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 + 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
S . S . S .
x = y = z =
2 1 2 i 2 0 −1
0 0
We therefore need to find the (normalized) eigenvector for the matrix
