## Saturday, November 22, 2014

### 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 =  ħ/ 2m   [3p2 N/ v] 2/3

Then:

e =    6.81  x 10 -23  J

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

v F  =   [2 e /  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 =  kB t

Then:  T F =    (e F)/ kB   =

(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 kB 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 kB T)   =    (4 kB T) -1

3. Let emd,  show that: f(d  ) = 1 – f( -d )

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

Solution:

If emd,    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 )