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

AI Thread Summary
The discussion centers on understanding the factors involved in calculating the energy of a one-dimensional electron gas at zero temperature. The Fermi energy is derived from the equation involving the wavevector \( k_F \) and incorporates a factor of two due to electron spin. The average energy per electron is calculated to be \( \frac{1}{3} \epsilon_F \), which aligns with the number of occupied states being equal to the number of electrons \( N \) under periodic boundary conditions. The allowed values of \( k \) for the electron states are determined by the boundary conditions, leading to \( k = \frac{2\pi}{L}n \). The discussion emphasizes the importance of these factors and conditions in accurately determining the properties of the system.
bluepilotg-2_07
Messages
11
Reaction score
2
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##
 
Physics news on Phys.org
bluepilotg-2_07 said:
##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?
 
  • Like
Likes bluepilotg-2_07 and hutchphd
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.
 
bluepilotg-2_07 said:
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.

bluepilotg-2_07 said:
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?
 
  • Like
Likes 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##.
 
bluepilotg-2_07 said:
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top