Select Solutions to Quantum Statistics Problems

1.Find the Fermi sphere parameters: e _F ,  v _F  and   T _F for He3 at absolute zero, viewed as a gas of  non-interactive fermions. (The density of the liquid is 81 kg/ m ³).

Solution:

We first find the number of electrons per unit volume. (N/V)

N/ V =  

6.020  x 10 ²⁶  atoms/ kmol (81 kg/ m ³)/ 3 kg/ kmol

(Why 3 kg/ kmol in denominator? Because the atomic weight = 3)

Then: N/ V =  1.625   x 10 ²⁸  atoms

The Fermi Energy, e _F  =  ħ²  / 2m   [3p² N/ v] ^2/3

Then: 

e _F  =    6.81  x 10 ^-23  J

To get: v _F  , note: e _F  =   ½ m (v _F )²

v _F  =   [2 e _{F  /}  m]^1/2  =   165  m/s

Finally, the Fermi temperature is found based on the fact it is tied to the energy (as the energy increases, the temperature increases.  We have for the thermodynamic temperature: 

Since   T _F =  k_B t

Then:  T _F =    (e _F)/ k_B   =

(6.81       x 10 ^-23  J) /  (1.38 x 10 ^-23 JK^-1 ) = 4.93 K 

2. a) Show that (- ¶f / ¶ e)  evaluated at the Fermi level (e  = m) has the value (4 k_B T) ^-1. Thus, the lower the temperature, then the steeper the slope of the Fermi-Dirac function.

Hint: Use f(e)  =  1/ [exp (m - e)/ t + 1]

Solution:

We have:

f(e)  =  1/ {exp (m - e)/ t + 1} =   [exp (m - e)/ t + 1] ^-1

So:    - ¶f / ¶ e    = 

 - {- [exp (m - e)/ t + 1] ^-2 ·  1/ t  (exp (m - e)/ t)}

For e  = m:

- ¶f / ¶ e   =  exp (e - e)/ t + 1] ^-2 ·  1/ t  (exp (e - e)/ t)}

=  exp  (0) / t + 1] ^-2 ·  1/ t  (exp  (0)  / t)} = 1/ t  (2) ^-2

- ¶f / ¶ e   =   1/ 4 t   =    1/ (4 k_B T)   =    (4 k_B T) ^-1

3. Let e =  m +  d,  show that: f(d  ) = 1 – f( -d )

Hint: Let f(d  ) =  1/ [exp  (e- m )/ t + 1]

Solution:

If e =  m +  d,    then:    d   =  e  -    m

f(d  ) =  1/ [exp  (e- m )/ t + 1]

But:  f( -d  ) =  1/ [exp - (e- m)/ t + 1]

 =  1/ [exp   (m - e)/ t + 1]

And:

1 -    f( -d  ) =  1 -   1/ [exp   (m - e)/ t + 1]

=    [exp   (m - e)/ t]/  [exp   (m - e)/ t + 1]

=  l exp (- e/ t) /   l exp (- e/ t) +1

Since:  l =   exp (m / t)

Multiply numerator and denominator by using:  [exp (e/ t)]/ l

l exp (- e/ t)  (exp (e/ t)/ l) /[ l exp (- e/ t) + 1] (exp (e/ t)/ l)

=  1/   [1 +   1/ l  (exp (e/ t)]

Or:    1 -    f( -d  ) =    1/ [1 +  exp   -m (exp (e/ t)]

=   1/ [exp  (e- m)/ t + 1]

But:    f( -d  ) =   1/ [exp  (e- m)/ t + 1]

Therefore:      f(d ) = 1 – f( -d )