Average energy of the electrons at T = 0

[Paolis]

According to the quantum mechanical free electron model the average energy is E=3E_F/5 for the 3D case. Nevertheless I saw in a specialised physics book that for the 1D model the average energy at T=0 is 0 and wanted to know if it is the same for the 3D case.

 

[mfb]

You know that the answer is not 0 for the 3D case.

It does not matter, as absolute energies do not play a role in quantum mechanics. The global energy scale is arbitrary anyway.

 

[Henryk]

Paolis,
Let me first derive the formula for the 3 D case. For a free electron model, electron energy is given as $E = \frac {\hbar^2 k^2}{2m}$. We can invert the formula and write $k = \sqrt {\frac {2mE}{\hbar^2}}$ or $k \sim \sqrt E$.
In 3D, the constant energy surface is a surface of a sphere. The number of electrons of energy not grater than at the surface is the reciprocal space volume of a sphere, that is $N \sim \frac 4 3 \pi k^3 \sim E^ {\frac 3 2}$
that gives the density of states as $g(E) = \frac {dN}{dE} \sim E^{\frac 1 2}$

Now, the average kinetic energy is $$\left< E \right> = \frac {\int^{E_F}_0 E\cdot g(E) \, dE} {\int_0^{E_F} g(E)\, dE} = \frac {\int^{E_F}_0 E\cdot E^{\frac 1 2} \, dE} {\int_0^{E_F} E^{\frac 1 2}\, dE} = \frac 3 5 \cdot E_F$$

That's how you got the formula for the average kinetic energy in the 3-D case.
In one dimension, the number of states from zero to $k_F$ is proportional to $k_F$, that is proportional to $\sqrt{(E)}$. Differentiating wrt to E, we get the density of states as $g(E) \sim E^{\frac 1 2}$
The average kinetic energy can be calculated as before but using a 1-D density of states function, that is
$$\left< E \right> = \frac {\int^{E_F}_0 E\cdot g(E) \, dE} {\int_0^{E_F} g(E)\, dE} = \frac {\int^{E_F}_0 E\cdot E^{-\frac 1 2} \, dE} {\int_0^{E_F} E^{-\frac 1 2}\, dE} = \frac 1 3\cdot E_F$$

So, the average kinetic energy in 1-D measure relative to the bottom of the band is $\frac 1 3 E_F$