# Mean energy of free electrons

My textbook states that the mean energy of free electrons within a solid state is given by $$E_{av} = \frac{U}{N} = \frac{\int_0^{k_f}D(k)E(k)dk}{\int_0^{k_f}D(k)dk}$$ where $D(k)$ is the density of states and $k_f$ the fermi-wave-vector.

With $D(k) = \frac{m}{\pi^2\hbar^2}k$ and $E(k) = \hbar^2 k^2 /2m$ I receive $$U=\frac{k_f^4}{8\pi^2} \, \,\, \, , \,\,\, N=\frac{m}{2\pi^2\hbar^2}k_f^2$$ thus $U/N = \hbar^2 k_f^2/4m = E_f/2$ which is obviously wrong.

I don't see my mistake.

If I convert $D(k)$ in $D(E)$ with $E=\hbar^2k^2/2m$ and calculate $$U=\int_0^{E_f}D(E) E\, dE$$ and $$N=\int_0^{E_f}D(E)dE$$ I get the correct solution of $E_{av} = 3/5 E_f$.

What am I doing wrong? I am quite sure that I didn't do the calculations wrong, so the error must be caused by assumptions.

• Are you perhaps missing a jacobi an factor in the integrals. What dimensionality are the integrals in? Jul 10, 2016 at 21:30
• this is just copied from my textbook. It doesn't mention any more factors Jul 11, 2016 at 5:43
• Sorry, missed that the intergrals where already in polar form with cutoff $k_f$. The point i was making, which has been pointed pout in one of the answers is that depending on dimnentionallity the density of states is different. In 1D it's constance for instance. Jul 11, 2016 at 9:06

The expression that you wrote for the density of states is for a free electron gas in 2D. Your answer is right, if you have a 2D free electron gas. However, for a 3D gas, the density of states per volume is given by $$D(k)=\frac{k^2 }{ \pi^2}$$

You get the above expression as follows:

$$D(k)\ dk\times Volume= \underbrace{2}_{\mbox{spin degeneracy}}\times \underbrace{4\pi k^2 dk}_{\mbox{3D volume element in k-space}} \div \underbrace{\left(\frac{2 \pi}{L}\right)^3}_\mbox{Volume occupied by 1 state in k-space}$$

where L is the length of the container.

Some brief explanation of the origins of the terms:

We multiply by 2 for spins because electrons are spin-1/2 particles and therefore each k-state will have a degeneracy of $2(1/2)+1=2$.

The $4 \pi k^2$ factor is obtained following the same reasoning as you would to find the volume occupied between $r$ and $(r+dr)$, by a sphere.

From periodic boundary conditions, we find that states are uniformly spaced by $\frac{2 \pi}{L}$ in k-space. Therefore the volume occupied by each k-state is simply the cube of that.

If you use this expression, you will indeed get the mean energy to be $\frac{3E_F}{5}$.

• How come? The density in my question is from my textbook as well and I get the same result if I calculate $D(k) = 1/V \frac{dN}{dk} \frac{dk}{dE}$ Jul 11, 2016 at 4:25
• I have edited my answer and included a bit more detail. Let me know if it makes sense. If there is still some confusion, can you please post the question from your textbook? Jul 11, 2016 at 21:21

I think that finally understood why you are confused.

The density of states which your question gave is for a 2D electron gas, and not a 3D gas. Therefore, your first answer of $E_F/2$ is correct!

You should get the same result even if you integrate with respect to E rather than k. The probably got a different answer because of an algebraic mistake.

Here is the calculation: $$\frac{dN}{dk}=\frac{dN}{dE} \frac{dE}{dk}$$ $$\frac{dE}{dk}=\frac{\hbar^2k}{m}$$ $$\frac{dN}{dk}=\frac{mk}{\pi^2 \hbar^2}$$ $$\implies \frac{dN}{dE}=\frac{m^2}{\hbar^4 \pi^2}=constant$$

$$U=constant \times \frac{E^2}{2}$$ $$N=constant \times E$$

Therefore, $$<E>=\frac{U}{N}=\frac{E_F}{2}$$

The density of states for a 3D gas is $\underbrace{2}_{\text{spin degeneracy}} \times \underbrace{4\pi k^2}_{\text{surface of sphere in momentum space}} \underbrace{\frac{V}{\pi^3}}_{\text{volume of single k-state}}$.

• How come? The density in my question is from my textbook as well and I get the same result if I calculate $D(k) = 1/V \frac{dN}{dk} \frac{dk}{dE}$ Jul 11, 2016 at 4:25