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Basic Field Calculations

+ 1 like - 0 dislike
94 views

I'm trying to derive the relation:

$\phi(x)\phi(y)=:\phi(x)\phi(y):+\langle 0|\phi(x)\phi(y)|0 \rangle$

but struggling to see the first few steps I need to make. I've made the substitutions

$\phi(x)=\phi^+(x)+\phi^-(x)$

and the same for $\phi(y)$, but it hasn't got me very far.

I'm then unsure what the it means when it says to compute

$\langle 0|\phi(x)\phi(y)|\bf{k_1},\bf{k_2} \rangle$.

Any help would be great!

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user user13223423
asked May 4, 2014 in Theoretical Physics by user13223423 (45 points) [ no revision ]
retagged May 4, 2014

1 Answer

+ 1 like - 0 dislike

It might be a good idea to start with the Fourier expansion of the fields:

$$\phi(x)=\int \frac{d^3k}{\sqrt{2\omega}(2\pi)^{3/2}}a^\dagger e^{ikx}+ae^{-ikx}$$

Normal ordering then means putting every $a^ \dagger$ to the left of any $a$ by which it is multiplied. The identity is then fairly trivial because $\langle 0|a^\dagger=a|0\rangle=0$ so we see that $\langle 0|\phi(x)\phi(y)| 0\rangle$ is going to contain exactly everything but the normal-ordered terms.

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user Danu
answered May 4, 2014 by UnknownToSE (505 points) [ no revision ]
Most voted comments show all comments
Is there not a more elegant way just using $\phi^+$ and $\phi^-$ and the commutation relations? I understand what you're saying, but I didn't think I need to resort to explicitly writing that expression in my calculations

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user user13223423
@user13223423 It's really just two lines... Plus, at least for me, it is good to try and get into the habit of writing many things out explicitly.

This post imported from StackExchange Physics at 2014-05-04 11:11 (UCT), posted by SE-user Danu
Yes I'll still give the right answer so I'll go with that. But how can the LHS not have anything to with the vacuum but the RHS have terms acting on the vacuum?

This post imported from StackExchange Physics at 2014-05-04 11:11 (UCT), posted by SE-user user13223423
@user13223423 Taking the vacuum expectation value simply 'kills' some of the terms. By the way, if you found this answer useful, please consider accepting it.

This post imported from StackExchange Physics at 2014-05-04 11:11 (UCT), posted by SE-user Danu
It's very helpful; but doesn't address the second half of my question

This post imported from StackExchange Physics at 2014-05-04 11:11 (UCT), posted by SE-user user13223423
I'm not quite sure what you mean by that second question. Once again writing the fields in terms of their mode expansions can give you a clearer expression in terms of only the creation and annihilation operators, which are easier to interpret.

This post imported from StackExchange Physics at 2014-05-04 11:11 (UCT), posted by SE-user Danu
The question I'm trying to answer says, "calculate the matrix elements.." followed by the equation in my question. I simply don't know what it means by this and I'm not sure if writing out the fields in terms of creation and annihilation operators will help for this part/ isn't necessary

This post imported from StackExchange Physics at 2014-05-04 11:11 (UCT), posted by SE-user user13223423

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