Second quantization

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Hi

I have a question regarding second quantization. In the following link: http://books.google.dk/books?id=v5v...q="The introduction of quantum field"&f=false (equation 1.71) they write the kinetic energy operator in real space representation, i.e. using quantum field operators. Here they rigourously transform the creation/annihilation operators using equation 1.69 the page before.

Now, if we take a look at equation 1.101 on page 24, they write the potential energy operator V in momentum space, but notice how they do not make the same rigourous transformation of the creation/annihilation operators, but merely substitute ν with k in equation (1.62) in their indices.

My question is, why they do not make that rigourous transformation in the case with k, but only change the indices of the operators. Can you shed some light on this?

I will appreciate any help you can give me.

Best,
Niles.
 

Answers and Replies

  • #2
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I don't mean to be rude, but I thought this was a pretty general question. No one knows this?
 
  • #3
Hi

I have a question regarding second quantization. In the following link: http://books.google.dk/books?id=v5v...q="The introduction of quantum field"&f=false (equation 1.71) they write the kinetic energy operator in real space representation, i.e. using quantum field operators. Here they rigourously transform the creation/annihilation operators using equation 1.69 the page before.

Now, if we take a look at equation 1.101 on page 24, they write the potential energy operator V in momentum space, but notice how they do not make the same rigourous transformation of the creation/annihilation operators, but merely substitute ν with k in equation (1.62) in their indices.

My question is, why they do not make that rigourous transformation in the case with k, but only change the indices of the operators. Can you shed some light on this?

I will appreciate any help you can give me.

Best,
Niles.

the equation 1.62 you are referring to is missing in the google book version of that text
 
  • #4
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I'm afraid I don't really understand your question. They're basically showing what an operator looks like in second quantization. You can choose a basis for the creation/annihilation operators with respect to the momentum basis or the position basis. Any basis will do -- they're always related through some unitary transformation.

So what rigorous treatment are you referring to?
 
  • #5
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Hmm, let me refraise my question then: The operators ak1 and ak2 and their Hermitian conjugates in equation (1.103), are they given by (from equation (1.69))

[tex]
a_k = \sum\limits_\upsilon {\left\langle {k|\psi _\upsilon } \right\rangle ^* a_\upsilon }
[/tex]

?
 
  • #6
fzero
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Act on the vacuum:


[tex]

a_k |0\rangle = \sum\limits_\upsilon {\left\langle {k|\psi _\upsilon } \right\rangle ^* a_\upsilon } |0\rangle = \sum\limits_\upsilon \langle {k|\psi _\upsilon } \rangle^* |\psi _\upsilon \rangle = | k \rangle.

[/tex]
 
  • #7
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Act on the vacuum:


[tex]

a_k |0\rangle = \sum\limits_\upsilon {\left\langle {k|\psi _\upsilon } \right\rangle ^* a_\upsilon } |0\rangle = \sum\limits_\upsilon \langle {k|\psi _\upsilon } \rangle^* |\psi _\upsilon \rangle = | k \rangle.

[/tex]
I think you are missing a dagger. But in my post right before I wasn't doubting the validity of the general expression

[tex]

a_k = \sum\limits_\upsilon {\left\langle {k|\psi _\upsilon } \right\rangle ^* a_\upsilon }

[/tex]

but if the authors of the book mean the above when they simply write ak.
 
  • #8
fzero
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I think you are missing a dagger.
You're correct, I am.

But in my post right before I wasn't doubting the validity of the general expression

[tex]

a_k = \sum\limits_\upsilon {\left\langle {k|\psi _\upsilon } \right\rangle ^* a_\upsilon }

[/tex]

but if the authors of the book mean the above when they simply write ak.
No, you can define creation and annihilation operators for any state. [tex]a^\dagger_\upsilon[/tex] creates a particle in the state [tex]|\psi_\upsilon\rangle[/tex]. Completely independently we can define [tex]a^\dagger_{\mathbf{k}}[/tex] to create a particle in the state [tex]|\mathbf{k}\rangle[/tex]. However we can use an appropriate change of basis states to relate the creation operators and the relationship will be precisely of the form of the equation that you're asking about.
 
  • #9
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Great, thanks to all of you.

Best,
Niles.
 

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