Klein-Gordon Causality calculation

[SOLVED] Klein-Gordon Causality calculation

Homework Statement

In Peskin and Schroeder on page 27 it is stated that when we compute the Klien-Gordon propagator in terms of creation and annihilation operators the only term that survived the expansion is
$$<0|a_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0> \ \ (1).$$
I am unsure of why the term
$$<0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0>$$
would vanish.

Homework Equations

The expansion of the field is given by
$$\phi (x) = \int \frac{d^{3}p}{(2 \pi)^{3}} \frac{1}{\sqrt{2E_{\textbf{p}}}}(a_{\textbf{p}}}e^{-ip\cdot x} + a^{\dagger}_{\textbf{p}}e^{ip\cdot x})$$
and the normalization condition for states is
$$<\textbf{p}|\textbf{q}> = (2\pi)^{3}\delta^{3}(\textbf{p}-\textbf{q}).$$

The Attempt at a Solution

Looking at the normalization condition given above I got,
$$<0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0> = <0|\textbf{p}+\textbf{q}> = (2\pi)^{3}\delta^{3}(\textbf{p}+\textbf{q}).$$
However this mean that (1) is not the only surviving term, and from my calculations this also gives a factor of 2 that should not be there. I am unsure of how this term vanishes.

StatusX
Homework Helper
The quick answer is that ap kills the vacuum, so since:

$$<0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q} }|0>^* = <0|a_{\textbf{q}}a_{\textbf{p} }|0>=0$$

the term vanishes (alternatively, let the operators act to the left rather than the right).

Also, note that $a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q} }|0>$ is not the 1-particle state $|\textbf{p}+\textbf{q}>$, but is the two particle state consisting of one particle with momentum p and another with momentum q (it is true that the total momentum is then p+q, and maybe that's what you mean, but if so it's confusing notation). Furthermore, the inner product of the vacuum with any n-particle state is zero, be it the 2-particle state you should have used or even the one-particle state |p+q>, so your second equality is also wrong.

Thank you very much that does help. Could you possibly point me in the direction as to why
$$<0|\textbf{p};\textbf{q}> = 0$$
where $$|\textbf{p}, \textbf{q}>$$ is a two particle state. I remember this but can't recall why. Also how do I mark this as answered?

olgranpappy
Homework Helper
Thank you very much that does help. Could you possibly point me in the direction as to why
$$<0|\textbf{p};\textbf{q}> = 0$$
where $$|\textbf{p}, \textbf{q}>$$ is a two particle state. I remember this but can't recall why. Also how do I mark this as answered?

the state you call $|0>$ is *not* a 2-particle state of zero momentum, it is the vacuum--it has no particles. It is orthogonal to any state that has particles. It is a basic property of the creation and annihilation operators that
$$a_p |0>=0$$
for all p

and thus
$$<0|a_p^\dagger = 0$$
for all p.

and thus
$$<0|a_p^\dagger a_q^\dagger = 0$$

and thus
$$<0|a_p^\dagger a_q^\dagger|0>=0$$
for all p and q. And the above is certainly not equal to the delta function expression you gave in your first post. cheers,

StatusX
Homework Helper
There are a few ways to see it:

1. ap annhilates the vacuum state, from which it follows by my last post.
2. The states are different eigenstates of the Hermitian operator N giving the number of particles in the system.
3. The inner product on a Fock space is defined so that only states in the same number N-particle subspace can have a non-vanishing inner product.

Of course, these are all related to each other. Also, things get a little less clear in an interacting theory, and often we have to redefine things (renormalize) so that these statements remain true.

That makes complete sense, thank you for your help.