Weyl Fermion in an infinite well

In summary, the conversation is about solving the Dirac equation with a mass function and finding the appropriate conditions for the wave function at the boundary. The article suggests using the Noether current to determine the condition, which leads to the conclusion that the wave function must have equal magnitudes at the boundary. However, the conversation reveals some discrepancies in the results and further discussion leads to the understanding that two different spinors must be used for the two boundaries, resulting in the correct solution.
  • #1
Paul159
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TL;DR Summary
I try to solve the Weyl equation in a infinite well with infinite mass condition.
Hello everyone,

I have a problem with bounds states of the 1D Weyl equation. I want to solve the Dirac equation

##−i\hbar \partial _x\Psi+m(x)\sigma _z \Psi=E\Psi## with the mass ##m(x)=0,0<x<a##, ##m(x)=\infty,x<0,x>a##. ##\Psi=(\Psi_1,\Psi_2)^T## is a two component spinor. Outside the well, ##\Psi=0##. Inside the well we have the plane wave equation
$$\Psi(x)=A e^{ikx} \begin{pmatrix} 1\\1 \end{pmatrix}+Be^{−ikx} \begin{pmatrix} -i\\i \end{pmatrix}$$. Of course the "wave function" is discontinuous at ##x=0,x=a##. I found this article where they talk about this problem. The condition they choose is that the Noether current is 0 at the well boundary. It is quite simple to find that we get from that ##|A|=|B|##. So we can write## B=Ae^{-i\phi}## where ##\phi## is real. After that I used eq. 33 of the article : at each boundary we must have ##\Psi_2/\Psi_1=ie^{i\alpha}## where ##\alpha=\pi## at ##x=0## and ##\alpha=0## at ##x=a##.

The condition at ##x=0## gives me that ##\cos \phi = \sin \phi -1##, so ##\phi = \pi/2##. The second condition gives me that ##e^{2ika} = -i##, so ##k_n = (n + 3/4) \pi /a##. The problem is that in the article they found ##k_n = (n + 1/2) \pi /a##.

If someone have already done this exercise, can you help me ?


Thanks !
 
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  • #2
Ok I get it. You have to take two different spinors for ##x = 0## and ##x = a##. The first condition at ##x = 0## will give you the trivial property ##-i = -i##. The condition at ##x = a## will give you ##e^{2ika} = -1##, such that ##k_n = (n + 1/2) \pi /a##.
You can lock this topic thanks.
 

FAQ: Weyl Fermion in an infinite well

What is a Weyl Fermion?

A Weyl Fermion is a type of fermion particle that was first theorized by physicist Hermann Weyl in 1929. It is a massless, relativistic particle that follows the laws of quantum mechanics and exhibits spin 1/2.

What is an infinite well?

An infinite well is a hypothetical system in which a particle is confined within a certain region by an infinitely high potential barrier. This means that the particle cannot escape from the well and is confined to a specific energy level.

What is the significance of studying Weyl Fermions in an infinite well?

Studying Weyl Fermions in an infinite well allows us to understand the behavior and properties of these particles in a controlled environment. It also provides insights into how these particles interact with confinement and how their energy levels are affected.

How are Weyl Fermions different from other fermions?

Weyl Fermions have unique properties that distinguish them from other fermions. They are massless, have a half-integer spin, and exhibit a chirality that determines their direction of motion. They also do not follow the Pauli exclusion principle, meaning multiple Weyl Fermions can occupy the same energy level.

What are the potential applications of Weyl Fermions in an infinite well?

Weyl Fermions in an infinite well have potential applications in quantum computing, as their unique properties can be harnessed for faster and more efficient data processing. They also have potential uses in developing new materials with unique electronic properties and in studying topological phases of matter.

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