Klein-Gordon Causality calculation

In summary, the homework statement is 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).
  • #1
furdun
3
0
[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
[tex]
<0|a_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0> \ \ (1).
[/tex]
I am unsure of why the term
[tex]
<0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0>
[/tex]
would vanish.

Homework Equations


The expansion of the field is given by
[tex]
\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})
[/tex]
and the normalization condition for states is
[tex]
<\textbf{p}|\textbf{q}> = (2\pi)^{3}\delta^{3}(\textbf{p}-\textbf{q}).
[/tex]


The Attempt at a Solution


Looking at the normalization condition given above I got,
[tex]
<0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0> = <0|\textbf{p}+\textbf{q}> = (2\pi)^{3}\delta^{3}(\textbf{p}+\textbf{q}).
[/tex]
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.
 
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  • #2
The quick answer is that ap kills the vacuum, so since:

[tex]<0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q} }|0>^* = <0|a_{\textbf{q}}a_{\textbf{p} }|0>=0[/tex]

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

Also, note that [itex] a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q} }|0>[/itex] is not the 1-particle state [itex]|\textbf{p}+\textbf{q}> [/itex], 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.
 
  • #3
Thank you very much that does help. Could you possibly point me in the direction as to why
[tex]
<0|\textbf{p};\textbf{q}> = 0
[/tex]
where [tex]|\textbf{p}, \textbf{q}>[/tex] is a two particle state. I remember this but can't recall why. Also how do I mark this as answered?
 
  • #4
furdun said:
Thank you very much that does help. Could you possibly point me in the direction as to why
[tex]
<0|\textbf{p};\textbf{q}> = 0
[/tex]
where [tex]|\textbf{p}, \textbf{q}>[/tex] 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 [itex]|0>[/itex] 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
[tex]
a_p |0>=0
[/tex]
for all p

and thus
[tex]
<0|a_p^\dagger = 0
[/tex]
for all p.

and thus
[tex]
<0|a_p^\dagger a_q^\dagger = 0
[/tex]

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

adam
 
  • #5
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.
 
  • #6
That makes complete sense, thank you for your help.
 

What is the Klein-Gordon causality calculation?

The Klein-Gordon causality calculation is a mathematical method used in theoretical physics to study the causal relationships between different events in a system. It is based on the Klein-Gordon equation, which describes the behavior of particles with spin 0, such as the Higgs boson.

Why is the Klein-Gordon causality calculation important?

The Klein-Gordon causality calculation is important because it allows researchers to understand the causal relationships between different events in a system. This is crucial for predicting and understanding the behavior of physical systems, particularly at the quantum level.

How is the Klein-Gordon causality calculation performed?

The Klein-Gordon causality calculation involves solving the Klein-Gordon equation for a particular system. This typically involves using mathematical techniques such as Fourier transforms and Green's functions to find the solutions. Once the solutions are found, the causal relationships between different events can be determined.

What are some applications of the Klein-Gordon causality calculation?

The Klein-Gordon causality calculation has many applications in theoretical physics, including in the study of particle physics, quantum field theory, and cosmology. It is also used in the development of new theories and models in physics, and in the prediction and interpretation of experimental results.

Are there any limitations to the Klein-Gordon causality calculation?

Like any mathematical model, the Klein-Gordon causality calculation has its limitations. It is based on certain assumptions and simplifications, and may not accurately describe all physical systems. Additionally, it is most applicable to systems at the quantum level, and may not be as relevant for larger-scale systems.

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