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Physics 325 Solution to Problem Set # 3
- Problem 5.3 in your textbook.
(a) We need
|Ψ±(~r 1 , ~r 2 )|^2 d^3 r 1 d^3 r 2 = 1 when ψa and ψb are orthonormal.
=> 1 = |A|^2
d^3 r 1
d^3 r 2
∣ψa(~r 1 )ψb(~r 2 ) ± ψb(~r 1 )ψa(~r 2 )
∣^2
= |A|^2
d^3 r 1
d^3 r 2
[∣
∣ψa(~r 1 )ψb(~r 2 )
∣^2 +
∣ψb(~r 1 )ψa(~r 2 )
∣^2 ± ψ a∗(~r 1 )ψ b∗ (~r 2 )ψb(~r 1 )ψa(~r 2 ) ± ψ∗ b (~r 1 )ψ∗ a (~r 2 )ψa(~r 1 )ψb(~r 2 )
]
= |A|^2
[ ∫
|ψa(~r 1 )|^2 d^3 r 1
|ψb(~r 2 )|^2 d^3 r 2 +
|ψb(~r 1 )|^2 d^3 r 1
|ψa(~r 2 )|^2 d^3 r 2 ± ∫ ψ∗ a(~r 1 )ψb(~r 1 )d^3 r 1
ψ∗ b (~r 2 )ψa(~r 2 )|^2 d^3 r 2 ±
ψ∗ b (~r 1 )ψa(~r 1 )d^3 r 1
ψ∗ a(~r 2 )ψb(~r 2 )|^2 d^3 r 2
]
=> 1 = |A|^2
[
]
= 2 |A|^2 => A =
(b) If ψa = ψb, then Ψ+ = A
[
2 ψa(~r 1 )ψa(~r 2 )
]
and: ∫ |Ψ+(~r 1 , ~r 2 )|^2 = 1 = 4 |A|^2
|ψa(~r 1 )|^2 d^3 r 1
|ψa(~r 2 )|^2 d^3 r 2 = 4 |A|^2 => A =
- Problem 5.4 in your textbook.
(a) For two non-interacting identiacl particles in an infinite square well potential,
Hˆ = p
2 1 2 m
p^22 2 m
for 0 ≤ x ≤ a and Hˆ = ∞ for x < 0 and x > a
where p 1 is the momentum operator for particle 1 and p 2 for particle 2. From the example, the two fermion ground state wave functions is:
ΨA(x 1 , x 2 ) =
a
[
sin(πx 1 /a) sin(2πx 2 /a) − sin(2πx 1 /a) sin(πx 2 /a)
]
Note that :
p^2 i 2 m
sin(nπxi/a) =
−¯h^2 2 m
d^2 dx^2
sin(nπxi /a) =
¯h^2 π^2 n^2 2 ma^2
sin(nπxi/a) = n^2 E 1 sin(nπx 1 /a)
Therefore HˆΨA(x 1 , x 2 ) =
p^21 2 m
ΨA +
p^22 2 m
ΨA =
a
[
E 1 sin(πx 1 /a) sin(2πx 2 /a)− 4 E 1 sin(2πx 1 /a) sin(πx 2 /a)
]
a
[
4 E 1 sin(πx 1 /a) sin(2πx 2 /a) − E 1 sin(2πx 1 /a) sin(πx 2 /a)
]
a
[
5 E 1 sin(πx 1 /a) sin(2πx 2 /a) − 5 E 1 sin(2πx 1 /a) sin(πx 2 /a)
]
= 5 E 1 ψA(x 1 , x 2 )
ie ΨA is an energy eigenstate with energy eigenvalue 5E 1.
(b) Distinguishable: ψ 11 = ground state (energy = 2E 1 ), ψ 12 and ψ 21 = first excited (energy = 5E 1 ) state, ψ 22 = second excited state (energy = 8E 1 ), ψ 31 and ψ 13 = third excited state (energy = 10E 1 ).
Bosons: ψ 11 = ground state (energy = 2E 1 ), ψ 12 + ψ 21 = first excited state (energy = 5E 1 ), ψ 22 = second excited state (energy = 8E 1 ), ψ 13 + ψ 31 = third excited state (energy = 10E 1 ).
Fermions: ψ 12 − ψ 21 = ground state (energy = 5E 1 ), ψ 13 − ψ 31 = first excited state (energy = 10E 1 ), ψ 23 − ψ 32 = second excited state (energy = 13E 1 ), ψ 14 − ψ 41 = third excited state (energy = 17E 1 ).
- Problem 5.14 in your textbook.
For the free electron gas, Equations 5.46 and 5.45 in your test gives us:
P =
Etot V
2¯h^2 (3π^2 N q)^5 /^3 30 π^2 m
V −^5 /^3 =>
dP dV
2¯h^2 (3π^2 N q)^5 /^3 30 π^2 m
V −^8 /^3 = −
P
V
Therefore, B = −V
dP dV
P
From problem 5.13(d) (see solution to selected problems), we have P = 3. 8 × 1010 N/m^2. Therefore, we would estimate the bulk modulus of copper to be (5/3) × 3. 8 × 1010 = 6. 3 × 1010 N/m^2.
- Problem 5.29 in your textbook.
(a) For distinguishable particles, there are 3 × 3 × 3 = 27 distiguishable states, ψi(x 1 )ψj (x 2 )ψk (x 3 ) where i, j, k = a, b, or c.
(b) For identical bosons, the wave function must be symmetric under the exchange of any pair of bo- son coordinates. There are 3 possible states where all 3 bosons are in the same single particle state: ψj (x 1 )ψj (x 2 )ψj (x 2 ) where j = a, b, c. These states are clearly symmetric under xi < − > xj.
Next, we consider when the cases when 2 particles are in the same single particle state, such as ψa(x 1 )ψa(x 2 )ψk (x 3 ) where k = b, c. This state is already symmetric under the exchange of x 1 and x 2. We must make it sym- metric under the exchange of x 1 and x 3 , and x 2 and x 3. We do so by adding the same state with the labels exchanged:
[
ψa(x 1 )ψa(x 2 )ψk (x 3 ) + ψa(x 3 )ψa(x 2 )ψk (x 1 ) + ψa(x 1 )ψa(x 3 )ψk (x 2 )
]
. There are 6 such states, the 2 above for k = b, c and 2 each when the 2 particles are in the states b and c.
Finally, we consider the case when the particles are all in different states, such as ψa(x 1 )ψb(x 2 )ψc(x 3 ). We make this symmetric by adding all permutations of the 3 labels: ψa(x 1 )ψb(x 2 )ψc(x 3 )+ψa(x 1 )ψc(x 2 )ψb(x 3 )+ ψb(x 1 )ψa(x 2 )ψc(x 3 ) + ψb(x 1 )ψc(x 2 )ψa(x 3 ) + ψc(x 1 )ψa(x 2 )ψb(x 3 ) + ψc(x 1 )ψb(x 2 )ψa(x 3 ). There is only one such state.
The total number of states for identical bosons is therefore 3 + 6 + 1 = 10.
(c) For identical fermions, the wave function must be antisymmetric under the exchange of any pair of particle coordinates. In particular, this means that no 2 particles can be in the same single particle state. This leaves only 1 possible state: all 3 particles in different single particle states, with odd permutations of particle pairs with a minus sign and even permutations with a + sign: ψa(x 1 )ψb(x 2 )ψc(x 3 ) − ψa(x 1 )ψc(x 2 )ψb(x 3 ) − ψb(x 1 )ψa(x 2 )ψc(x 3 ) + ψb(x 1 )ψc(x 2 )ψa(x 3 ) − ψc(x 1 )ψa(x 2 )ψb(x 3 ) + ψc(x 1 )ψb(x 2 )ψa(x 3 ).
