0
$\begingroup$

I am looking into Landau Diamagnetism and am reading Dupre's paper. I am slightly confused at where he has got a term in his value of $E$ from.

He states that: $$ E=(n+1/2)\hbar\omega+\hbar^2k_z^2/2m $$ and I can easily get to the first part, but am confused where the second term $\hbar^2k_z^2...$ comes from as it always seems to disappear.

The Hamiltonian I am using is $$ H=(-\hbar^2/2m)*d^2/dy^2+\hbar^2k_z^2/2m+m\omega_c/2(y-\hbar k_x/m\omega_c) $$ however I may have the wrong indices (it could be $k_x$ or $k_z$ switched - I'm not certain)

Improve this question
$\endgroup$
2
  • $\begingroup$ It could be helpful to add a link to the "Dupre's paper" $\endgroup$
    – Roger V.
    Commented Apr 25, 2022 at 7:27
  • $\begingroup$ What do you mean when you say that term "disappear(s)"? Can you show us at exactly which step in your derivative it disappears? $\endgroup$
    – d_b
    Commented May 27, 2022 at 19:10

2 Answers 2

Reset to default
1
$\begingroup$

You can think about it classically. If you have a charged particle moving in a straight line and you put a magnetic field $\mathbf{B}$ in the $z$-direction, its motion will change but not the velocity in the $z$-direction (because Lorentz force is zero in that direction $F\propto\mathbf{v}\times\mathbf{B}$).

The same happens with the electrons in your Fermi gas (free electron model if you wish), the magnetic field is confining the motion in a plane perpendicular to $\mathbf{B}$ but not in the $z$-direction. So basically you have some quantized motion in the $xy$-plane but you have the equivalent to a 1D free particle in the $z$-direction. The energy of a free particle you can write it as you wish $E_z=\frac{p_z^2}{2m}=\frac{\hbar^2k_z^2}{2m}$ but $(E_z,p_z,k_z)$ are just free parameters that can go from $0$ to $\infty$.

Improve this answer
$\endgroup$
0
$\begingroup$

I trust the energy in the z-direction is negative as no way for the electron to go in that direction. This implies that k_z is imaginary as the wave in the z-direction must be decaying.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.