Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Zero Energy States in 2D Systems

+ 1 like - 0 dislike
236 views

Since we are on a planar system (2D system) the massless Dirac equation reads $$\vec{\alpha}\cdot(\vec{p}-e\vec{A})\psi_E=E\psi_E$$

Here Dirac matrices are Pauli matrices ($\alpha^1=-\sigma^2$ , $\alpha^2=\sigma^1$)

To find zero-energy modes we write the wave function as $\psi_0= \begin{pmatrix} u\\ v \end{pmatrix}$, and choose Coulomb gauge for $\vec{A}$ assumed to be single valued and well behaved at the origin,

$$A^i=\epsilon^{ij}\partial_ja$$ $$B=-\nabla^2a$$

Then the Dirac equation reduces to pair,

(1) $$(\partial_x + i\partial_y)u - e(\partial_x+i\partial_y)au=0$$ $$(\partial_x - i\partial_y)v + e(\partial_x-i\partial_y)av=0$$

with solutions

$$u=\exp(ea)f(x+iy)$$ $$v=\exp(-ea)g(x-iy)$$

For example if we apply a constant magnetic field with 2 possible Coulomb gauges $$a_0^I=-{1\over 4} r^2B$$ $$a_0^{II}=-{1\over 2}x^2B$$

For $a_0^I$ case zero-energy states may be written as $$\psi_{0(n)}^I=e^{-eBr^2/4}(x+iy)^n$$

For $a_0^{II}$ case zero-energy states may be written as $$\psi_{0(k)}^{II}=e^{-eBx^2/2}e^{k(x+iy)}$$

My questions are:

  1. How can I solve the differential equations given in (1) to obtain $u$ and $v$ solutions given above ? Where do $f(x+iy)$ and $g(x-iy)$ arise from ?

  2. The author sets $f(x+iy) = (x+iy)^n$ and $g(x-iy) = e^{k(x+iy)}$. How can I determine these functions ?

Reference

R. Jackiw, Phys. Rev. D 29, 2375 (1984).

This post imported from StackExchange Physics at 2014-04-13 14:46 (UCT), posted by SE-user ucoskun
asked Apr 5, 2014 in Theoretical Physics by ucoskun (5 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...