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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,862 answers , 20,637 comments
1,470 users with positive rep
502 active unimported users
More ...

Fermionic anti-commutation relations

+ 3 like - 0 dislike
38 views

For Pauli's exclusion principle to be followed by fermions, we need these anti-commutators $$[a_{\lambda},a_{\lambda}]_+=0 $$ and $$[a_{\lambda}^{\dagger},a_{\lambda}^{\dagger}]_+=0 $$ Then $$n_{\lambda}^{2}=a_{\lambda}^{\dagger}a_{\lambda}a_{\lambda}^{\dagger}a_{\lambda}=a_{\lambda}^{\dagger}\left(1-a_{\lambda}^{\dagger}a_{\lambda}\right)a_{\lambda}=a_{\lambda}^{\dagger}a_{\lambda}=n_{\lambda}.$$ which gives $ n_{\lambda}=0,1 $. Here we used the anti-commutator $$[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_+= \delta_{\lambda,\lambda^{\prime}} $$ But we could have used even a commutator instead of the anti-commutator and still got the same result i.e. if we choose $[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime}} $ then $n_{\lambda}^{2}=a_{\lambda}^{\dagger}a_{\lambda}a_{\lambda}^{\dagger}a_{\lambda}=a_{\lambda}^{\dagger}\left(1+a_{\lambda}^{\dagger}a_{\lambda}\right)a_{\lambda}=a_{\lambda}^{\dagger}a_{\lambda}=n_{\lambda} $ which also gives $n_{\lambda}=0,1 $
What conditions make us impose the last anti-commutation relation $$[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_+= \delta_{\lambda,\lambda^{\prime}} $$ instead of $[a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime}} $ ?

I mean, we do not need all relations to be anti-commuting. I can take 2 of them to be anti-commuting but the third one i.e. relation between creation and annihilation operator to be commuting and still maintain the Pauli's exclusion

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user cleanplay
asked Sep 15, 2013 in Theoretical Physics by cleanplay (80 points) [ no revision ]
Perhaps my eyes are deceiving me, but it seems that you actually used the anti-commutator you were wondering about in your second equality after "because."

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user joshphysics
@joshphysics made the change. I mean, I could have used even a commutator instead the anti-commutator and still got the same result

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user cleanplay
Related: physics.stackexchange.com/q/17893/2451

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Qmechanic
@Qmechanic : thanks, though it addresses the question but it is a bit vague for me.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user cleanplay

3 Answers

+ 3 like - 0 dislike

We need it because we want the "occupied" state to have $n_\lambda=1$; I will omit the $\lambda$ argument everywhere. In other words, we need $$ n a^\dagger |0\rangle \equiv a^\dagger a a^\dagger |0\rangle = 1\cdot a^\dagger $$ But the left hand side has the operator that is $$a^\dagger a a^\dagger |0\rangle = a^\dagger ([a,a^\dagger]_+ - a^\dagger a) = a^\dagger [a,a^\dagger]_+ $$ where the last term was dropped in the last term because $(a^\dagger)^2=0$. So we demand $$a^\dagger[a,a^\dagger]_+ |0\rangle = a^\dagger|0\rangle.$$ In combination with your other conditions, this is only possible if the anticommutator is one – we may "cancel" the $|0\rangle$ ket vector because a similar condition may be derived for $|1\rangle$ as the ket vector.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Luboš Motl
answered Sep 15, 2013 by Luboš Motl (10,248 points) [ no revision ]
@Motl I mean we could also use the commutator $ [a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime‌​}} $ and still get the same result. Why should I use only the anti-commutator ? I can use the anti-commutators $$[a_{\lambda},a_{\lambda}]_+=0 $$ and $$[a_{\lambda}^{\dagger},a_{\lambda}^{\dagger}]_+=0, $$ along with $$ [a_{\lambda},a_{\lambda^{\prime}}^{\dagger}]_{-}=\delta_{\lambda,\lambda^{\prime‌​}} $$

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user cleanplay
+ 3 like - 0 dislike

Suppose $a$ and $a^{+}$ operators satisfy $$ \left\{ a,a\right\} =0\mbox{ and }\left[a,a^{+}\right]=1 $$ We have basically $a^{2}=0$ and $aa^{+}=a^{+}a+1$.

Now consider $aaa^{+}$.

$$ 0=aaa^{+}=a\left(a^{+}a+1\right)=aa^{+}a+a=a^{+}aa+2a=2a. $$

So we get $a=0$.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Louis Yang
answered Mar 7, 2014 by Louis Yang (90 points) [ no revision ]
+ 1 like - 0 dislike

You are obviously wrong.

When you have operators with anti-commutation relations, you have :

$${a^+}_\lambda {a^+}_{\lambda'} + {a^+}_{\lambda'} {a^+}_{\lambda} = 0 \tag{1}$$

Taking $\lambda = \lambda'$, you get :

$${a^+}_\lambda {a^+}_{\lambda} = 0 \tag{2}$$

If you have operators with commutation relations, you have :

$${a^+}_\lambda {a^+}_{\lambda'} - {a^+}_{\lambda'} {a^+}_{\lambda} = 0 \tag{3}$$

Taking $\lambda = \lambda'$, you get :

$${a^+}_\lambda {a^+}_{\lambda} - {a^+}_{\lambda} {a^+}_{\lambda} = 0 \tag{4}$$ which is a trivial equation ($x=x$)

So, it is not true, with commutation relations, that you have : ${a^+}_\lambda {a^+}_{\lambda} = 0$, so your equation $n_\lambda ^2=n_\lambda$ is obviously false for operators with commutations relations.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Trimok
answered Sep 15, 2013 by Trimok (950 points) [ no revision ]
My question is only about the last anti-commutation relation which you did not use in your proof. I understand that you need the two anti-commutation relations that you have used, in order to prove the Pauli's exclusion principle. My question is on what basis we choose the third relation i.e the relation of the creation and annihilation operator to be of anti-commutator type and not commutator type. Read my comment in response to Lubos Motl's answer.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user cleanplay
@cleanplay : This is an other question... In short, this has to do with the spin-statistics theorem. For instance, in quantum field theory, if you write the hamiltonian for a fermionic field (for instance a Dirac field), you will find something like $H = \sum_k (b^+_kb_k - d_kd^+_k)$ ($b$ concerns particles and $d$ concerns anti-particles). But this hamiltonian has to be bounded below, and you have to choose anti-commutation relations, to have $H = \sum_k (b^+_kb_k + d^+_k d_k)$, up to a (infinite) constant.

This post imported from StackExchange Physics at 2014-05-04 11:38 (UCT), posted by SE-user Trimok

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$\hbar$ysics$\varnothing$verflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...