Second quantization

  • Thread starter Niles
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  • #1
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Main Question or Discussion Point

Hi

Say I have the following two fermionic creation/annihilation operators

[tex]
c^\dagger_ic_j
[/tex]

1) Yesterday, my lecturer said that the following is valid

[tex]
c^\dagger_ic_j = \delta_{i,j}c_jc^\dagger_i
[/tex]

Can you guys explain to me, where this formula comes from? I originally thought that it was one of the anti-commutator relations, but it cannot come from there.



2) Say I have en expression of the form

[tex]
c_{k+q}^\dagger c_{k-q'} c^\dagger_{k'-q'}c_{k'}
[/tex]

If the operators are fermionic, then if I want to have all dagger-operators on the LHS and non-dagger operators on the RHS, then do I have to use anti-commutator relatations in order to rewrite the expression?

Likewise, if they were bosonic operators, then I would have to use commutator relations in order to rewrite the expression?

Best,
Niles.
 
Last edited:

Answers and Replies

  • #2
Avodyne
Science Advisor
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1) Your lecturer is wrong.

2) Yes.
 
  • #3
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Thank you, it is kind of you to answer so quickly.
 
  • #4
A. Neumaier
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[tex]
c^\dagger_ic_j = \delta_{i,j}c_jc^\dagger_i
[/tex]
probably he meant c^*_ic_j = \delta_{i,j}+c_jc^*_i
which is the CCR for bosons.
 
  • #5
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Thanks. I have another question related to fermionic operators, so I'll just ask it here. It is regarding the relation

[tex]
\delta (t - t')\left\langle {\left\{ {c_i (t),c_i^\dag (t')} \right\}} \right\rangle = \delta (t - t')
[/tex]

The curly brackets denote an anti-commutator. Is there an easy way of showing this? The way I would show this is to look at the case t = t' and the case t != t'.
 
Last edited:
  • #6
A. Neumaier
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probably he meant c^*_ic_j = \delta_{i,j}+c_jc^*_i
which is the CCR for bosons.
Sorry, this is not quite true; [tex]c^*_ic_j = -\delta_{i,j}+c_jc^*_i[/tex]
 
  • #7
A. Neumaier
Science Advisor
Insights Author
2019 Award
7,211
3,093
Thanks. I have another question related to fermionic operators, so I'll just ask it here. It is regarding the relation

[tex]
\delta (t - t')\left\langle {\left\{ {c_i (t),c_i^\dag (t')} \right\}} \right\rangle = \delta (t - t')
[/tex]

The curly brackets denote an anti-commutator. Is there an easy way of showing this? The way I would show this is to look at the case t = t' and the case t != t'.
You cannot distinguish cases in this way since this is meant in the sense of distributions. Thus you need to multiply both sides by a function f(t,t'), integrate over t and t', and simplify before you can interpret the statement.
 

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