I do not understand why the extra 2 factor is there (energy of 1D electron gas at zero temperature)

[bluepilotg-2_07]

Homework Statement

For the 1D electron gas at zero temperature derive the average energy and show that it can be written as E_avg = d/d+2 where d is the dimension of the system.

Relevant Equations

E_avg = E/N
E_F = ((h_bar)^2 * (k_F)^2)/2m

Fermi energy is given by $$\epsilon_F = \frac{\hbar ^2 k_F ^2}{2m}$$
$N = \frac{2k_F L}{2\pi} \Rightarrow \frac{k_F L}{\pi}$ the factor of two in the numerator comes from the electrons having two spins.
$$E=\frac{2L}{2\pi} \int_{0}^{k_F} \frac{\hbar^2 k^2}{2m}\, 2\,dk$$ The two in front of the integral comes from the spin and the $2\, dk$ comes from the dimension of the system.
$$E = \frac{L\hbar^2 k_F^3}{3\pi m}$$
$$E/N = \frac{\hbar^2 k_F^2}{3m} \Rightarrow \frac{2\epsilon_F}{3}$$
The answer should be $\epsilon_F/3$

 

[TSny]

$N = \frac{2k_F L}{2\pi}$ ... the factor of two in the numerator comes from the electrons having two spins.

Can you show how you got this result? What are the possible values of $k$ for the electron states? Are you assuming periodic boundary conditions for the states?

 

[bluepilotg-2_07]

Hello, thanks for responding
I assumed periodic boundary conditions when solving the Schrodinger equation for the 1D system. At zero temperature the number of occupied states is equal to the number of electrons N. The electrons exist along some line of length L in the 1D system. If I have period $\pi$, I suppose N would actually be $\frac{L}{\pi} \cdot 2k_F$, 2 arising from the two possible electron spins and $k_F$ being the wavevector. This seems to solve my issue and does give me $\frac{1}{3} \epsilon_F$ for the average energy, but I still do not think I understand why this way of calculating gives N.
My initial result for N came from looking at my professor's notes for the 3D system and trying to reason out the dimensions of the 1D system.

 

[TSny]

I assumed periodic boundary conditions when solving the Schrodinger equation for the 1D system.
At zero temperature the number of occupied states is equal to the number of electrons N. The electrons exist along some line of length L in the 1D system.

Ok.

If I have period $\pi$, I suppose N would actually be $\frac{L}{\pi} \cdot 2k_F$, 2 arising from the two possible electron spins and $k_F$ being the wavevector. This seems to solve my issue and does give me $\frac{1}{3} \epsilon_F$ for the average energy, but I still do not think I understand why this way of calculating gives N.

When using periodic boundary conditions, the wavefunction for an energy eigenstate has the form $\psi(x) =\frac 1 {\sqrt{L}} e^{ikx}$. What are the allowed values of $k$ such that $\psi(x)$ satisfies the boundary condition?

 

[bluepilotg-2_07]

If my boundary conditions are $\psi(x+L)=\psi(x) \Rightarrow \frac{1}{\sqrt{L}}e^{ikL}=\frac{1}{\sqrt{L}}$, then allowed values of k should be $k=\frac{2\pi}{L}n$.

 

[TSny]

If my boundary conditions are $\psi(x+L)=\psi(x) \Rightarrow \frac{1}{\sqrt{L}}e^{ikL}=\frac{1}{\sqrt{L}}$, then allowed values of k should be $k=\frac{2\pi}{L}n$.

Ok. But you need to specify the allowed values of $n$. For periodic boundary conditions, an electron can travel in the positive x-direction or the negative x-direction. That is, there are states of positive x-component of momentum and also states of negative x-component of momentum.