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Physics 325 Solution to Selected Problems #3 Winter 2004
- Prob. 5.5 in your textbook ψn =
2 /a sin(nπx/a), ψm =
2 /a sin(mπx/a).
We follow the notation of the text: 〈(x 1 − x 2 )^2 〉± = 〈(∆x)^2 〉d ∓ 2 |〈x〉ab|^2.
a.) For distinguishable particles:
〈(x 1 − x 2 )^2 〉 = 〈(∆x)^2 〉d = 〈x^2 〉a + 〈x^2 〉b − 2 〈x〉a〈x〉b
a
∫ (^) a
0
x^2 sin^2 (nπx/a)dx +
a
∫ (^) a
0
x^2 sin^2 (mπx/a)dx −
a^2
∫ (^) a
0
x sin^2 (nπx/a)dx
∫ (^) a
0
x sin^2 (mπx/a)dx
Letting y = nπx/a and z = mπx/a:
= 2 a
( (^) a nπ
) 3 ∫^ nπ 0
y^2 sin^2 (y)dy + 2 a
( (^) a mπ
) 3 ∫^ mπ 0
z^2 sin^2 (z)dz
a^2
( (^) a nπ
) 2 ( (^) a mπ
) 2 ∫^ nπ 0
y sin^2 (y)dy
∫ (^) mπ
0
z sin^2 (z)dz
a
( (^) a nπ
) 3 (^ y^3 6
y cos(2y) 4
nπ 0
a
( (^) a mπ
) 3 (^ z^3 6
z cos(2z) 4
mπ 0
−
a^2
( (^) a nπ
) 2 ( (^) a mπ
) 2 (^ y^2 4
cos(2y) 8
nπ 0
( (^) z 2 4
cos(2z) 8
mπ 0
a
( (^) a nπ
) 3 (^ (nπ)^3 6
nπ 4
a
( (^) a mπ
) 3 (^ (mπ)^3 6
mπ 4
a^2
( (^) a nπ
) 2 ( (^) a mπ
) 2 (^ (nπ)^2 4
)( (^) (nπ) 2 4
= a
2 3 − a
2 2 n^2 π^2
2 3 − a
2 2 m^2 π^2 − 8 a^2
= a
2 6 − a
2 2 π^2
n^2
m^2
b.) For identical bosons, 〈(x 1 − x 2 )^2 〉 = 〈(∆x)^2 〉d − 2 |〈x〉ab|^2.
〈x〉ab =
x ψ∗ n(x)ψm(x)dx = 2 a
∫ (^) a
0
x sin(nπx/a) sin(mπx/a)dx
a
∫ (^) a
0
x
cos
( (^) (n − m)πx a
− cos
( (^) (n + m)πx a
dx
a
( (^) a (n − m)π
) 2 ∫^ (n−m)π 0
y cos(y)dy −
a
( (^) a (n + m)π
) 2 ∫^ (n+m)π 0
y cos(y)dy
a
( (^) a (n − m)π
cos(y)
∣∣(n−m)π 0 −^
a
( (^) a (n + m)π
cos(y)
∣∣(n+m)π 0
= 0 for n − m even, and = 2 a π^2
(n + m)^2
(n − m)^2
for n − m odd.
Therefore, |〈x〉ab |^2 = 0 for n − m even, while for n − m odd:
|〈x〉ab|^2 = 4 a
2 π^4
(n + m)^2
(n − m)^2
= 64 a
(^2) n (^2) m 2 π^4 (n^2 − m^2 )^4
For n − m even : 〈(x 1 − x 2 )^2 〉 = a^2 6
a^2 2 π^2
n^2
m^2
For n − m odd : 〈(x 1 − x 2 )^2 〉 = a^2 6
a^2 2 π^2
n^2
m^2
128a^2 n^2 m^2 π^4 (n^2 − m^2 )^4
c.) For identical fermions: 〈(x 1 − x 2 )^2 〉 = 〈(∆x)^2 〉d + 2|〈x〉ab|^2 , and we use the results from part b.:
For n − m even : 〈(x 1 − x 2 )^2 〉 = a^2 6 −^
a^2 2 π^2
n^2 +^
m^2
For n − m odd : 〈(x 1 − x 2 )^2 〉 = a
2 6 − a
2 2 π^2
n^2
m^2
(^2) n (^2) m 2 π^4 (n^2 − m^2 )^4
- Prob. 5.6 in your textbook a. For distinguishable particles, the three particles states are:
ψ(x 1 , x 2 , x 3 ) = ψa(x 1 )ψb(x 2 )ψc(x 3 ) or ψb(x 1 )ψa(x 2 )ψc(x 3 ) or...
b. For identical bosons, the state must be symmetric under exchange of any pair of particles, so we add all possible permutations:
ψ(x 1 , x 2 , x 3 ) =
[
ψa(x 1 )ψb(x 2 )ψc(x 3 ) + ψb(x 1 )ψa(x 2 )ψc(x 3 ) + ψc(x 1 )ψb(x 2 )ψa(x 3 )
- ψa(x 1 )ψc(x 2 )ψb(x 3 ) + ψb(x 1 )ψc(x 2 )ψa(x 3 ) + ψc(x 1 )ψa(x 2 )ψb(x 3 )
]
c. For identical fermions, the state must be antisymmetric under exchange of any pair of particles. The Slater determinant gives us:
ψ(x 1 , x 2 , x 3 ) =
[
ψa(x 1 )
ψb(x 2 )ψc(x 3 ) − ψc(x 2 )ψb(x 3 )
− ψb(x 1 )
ψa(x 2 )ψc(x 3 ) − ψc(x 2 )ψa(x 3 )
ψa(x 2 )ψb(x 3 ) − ψb(x 2 )ψa(x 3 )
)]
