# Relationship between Fermi Energy and Average Energy

Electrons are in a one dimensional box of length 2L. What is the relationship between Ef, the fermi energy and E, the average energy?

I looked up the formula for the relationship: all im getting is

     Eav = E0 + (3/5)*Ef


Where do I put the 2L?

This is for a board exam review for mechanical engineering; in our country, we're required to pass an exam before we can get our licence to operate. Fermi levels, semiconductor physics and other quantum mechanical topics were not taught in my regular curriculum and I therefore lack the most of the background to understand what is really happening.

• I feel compelled to say this - mechanical engineering and quantum mechanical engineering are not the same thing, and I marvel at an exam board that slips this kind of question in to a mechanical engineering exam. What country is this? – Floris Jun 28 '15 at 1:03
• @Floris Philippines – james Jun 28 '15 at 3:51
• Quite demanding - and good luck! I hope that they don't teach the uncertainty principle in mechanical engineering... "Will this bridge be strong enough? Well - it is and it isn't. We won't know until we observe it." :-) – Floris Jun 28 '15 at 3:54

You might find the wiki article on this topic helpful. Summarizing:

When you have a 1-D box, the energy states of an electron can be given by

$$E_n = E_0 + \frac{\hbar^2 \pi ^2}{2 m L^2} n^2$$

Now the things to note are this:

• Two electrons (with opposite spin) can occupy the same level
• The Fermi level is the energy of the last electron
• After each pair of electrons is added, $n$ increases by 1

With that given, if you have $2N$ electrons (so $N$ levels are occupied) you can find the mean energy of all electrons:

$$E_{av} = E_0 + \frac{2\sum_1^{N} E_n}{2N}$$

A sum of the first $n$ squares is of course (see here for derivation):

$$\sum_1^N n^2 = \frac{N^3}{3} + \frac{N^2}{2} + \frac{N}{6}$$

For sufficiently large values of $N$, we can ignore the second and third terms and end up with something that looks exactly like the answer you would get if you were integrating $n^2$. That is not a coincidence.

Assuming there are lots of electrons, we get

$$E_{av} = E_0 + \frac{\hbar^2 \pi ^2}{2 m L^2} \frac{N^3}{3N}$$

But since the Fermi level is

$$E_f = E_0 + \frac{\hbar^2 \pi ^2}{2 m L^2} N^2$$

We see that if we put $E_0 = 0$ (or we ignore it because it is so much smaller than the Fermi level for sufficiently large values of $N$), the average energy is

$$E_{av} = \frac{E_f}{3}$$

QVOD ERAT DEMONSTRANDVM

Single particle energy eigenstates for a system of particles in a box are given by $$E_n=\frac{\hbar^2 \pi^2}{2mL^2}\,n^2 + E_0.$$ The Fermi energy for a single particle is, by definition, the value of its energy that exhausts all the possible states given by $N$ indistinguishable particles; in the case at hand, for fermions (electrons), this is given by $$E_f=\frac{\hbar^2 \pi^2}{2mL^2}\,n_F^2|_{n_F=N/2} = \frac{\hbar^2 \pi^2}{2mL^2}\, \frac{N^2}{4}$$ because you can place at most $2$ fermions with different spins in the same state. The overall total energy is the sum over all electrons of all the maximum possible energy up to each one Fermi level for each particle, namely: $$E_{tot} = \int_0^N\,dN'\left(E_0 + E_F(N')\right)=NE_0 + \frac{\hbar^2 \pi^2}{2mL^2}\frac{1}{4} \frac{N^3}{3};$$ defining now $$\overline{E}=\frac{E_{tot}}{N}=E_0 + \frac{E_F}{3}$$ leads to the expected result. Notice that, if in higher dimensions, you may have different factors due to integrations in higher dimensional domains.

• We got to the same answer at almost exactly the same time... – Floris Jun 28 '15 at 1:22
• Ahah, happy hour, two at the price of one ;-)! – gented Jun 28 '15 at 1:26
• I had not heard it called that before... but it's a good word. Welcome on the site! – Floris Jun 28 '15 at 1:27