Computation of propagation amplitudes for KG field

Click For Summary

Discussion Overview

The discussion revolves around the computation of propagation amplitudes in the context of the Klein-Gordon field within quantum field theory. Participants explore the mathematical expressions involved in calculating the vacuum expectation value of the field operators and clarify the implications of annihilation and creation operators in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the equation for the Klein-Gordon field and the expression for the propagation amplitude, questioning why certain terms vanish in the computation.
  • Another participant clarifies that the annihilation operator \( a(p) \) does not annihilate the vacuum bra \( \langle 0| \), which leads to the non-vanishing of the term \( \langle 0| a(p) a^{\dagger}(q) |0\rangle \).
  • A further response emphasizes the importance of the bra-ket notation in understanding why terms with annihilation operators acting on the vacuum yield zero, while the term involving \( a(p) a^{\dagger}(q) \) is the only non-vanishing term.
  • Another participant reiterates the confusion surrounding the bra-ket notation and suggests a clearer representation of the equation to avoid misunderstandings.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of the annihilation and creation operators in the context of the vacuum state. While some clarify the reasoning behind the vanishing of certain terms, there remains a lack of consensus on the interpretation of the mathematical expressions involved.

Contextual Notes

The discussion highlights the complexities of operator algebra in quantum field theory, particularly regarding the treatment of vacuum states and the implications of bra-ket notation. There are unresolved aspects related to the assumptions made about the operators and their actions on the vacuum state.

Theage
Messages
11
Reaction score
1
Note: I'm posting this in the Quantum Physics forum since it doesn't really apply to HEP or particle physics (just scalar QFT). Hopefully this is the right forum.

In Peskin and Schroeder, one reaches the following equation for the spacetime Klein-Gordon field:
$$\phi(x,t)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\Big(a(p)e^{-ip\cdot x}+a^\dagger(p)e^{ip\cdot x}\Big)$$
Then they say that the propagation amplitude for a particle to go from a spacetime point x to a spacetime point y is \langle 0\vert\phi(x)\phi(y)\vert 0\rangle where the ket |0> is the vacuum. I understand this up to here. But then they start computing it as follows (equation 2.50):
$$\langle 0\vert\phi(x)\phi(y)\vert 0\rangle = \int\frac{d^3p\,d^3q}{(2\pi)^6}\frac{1}{2\sqrt{E_{p}E_{p'}}}\Big(a(p)e^{-ipx}+a^\dagger(p)e^{ipx}\Big)\Big(a(q)e^{-iqy}+a^\dagger(q)e^{iqy}\Big)$$

Clearly there will now be four terms when you expand the parentheses, and the book claims that all of these vanish except for the term with \langle 0\vert a(p)a^\dagger(q)\vert 0\rangle. Two questions:

a) Wouldn't this term vanish also since a(p) kills the vacuum bra, producing a zero?

b) Why doesn't the term with a^\dagger(p)a^\dagger(q) stay? In that term there are no annihilation operators to kill the vacuum, so surely the term should vanish.
 
Physics news on Phys.org
Theage said:
a) Wouldn't this term vanish also since a(p) kills the vacuum bra, producing a zero?

No. ##a_p## annihilates ##|0\rangle## which clearly means ##a^{\dagger}_p## annihilates ##\langle 0|##, not ##a_p##.

Theage said:
b) Why doesn't the term with a^\dagger(p)a^\dagger(q) stay? In that term there are no annihilation operators to kill the vacuum, so surely the term should vanish.

See above.
 
  • Like
Likes   Reactions: Theage
You forgot to "sandwich" the right-hand side of your equation between ##\langle 0|## and ##|0 \rangle##. Then it becomes immediately clear that all expressions with a annihilation operator acting to the right and a creation operator acting to the left (it becomes an annihilation operator when letting it act on the left argument of a scalar product) to the vacuum state, gives 0. The only non-vanishing term is thus indeed
$$\langle 0|a(p) a^{\dagger}(q) 0 \rangle =\langle a^{\dagger}(p) 0|a^{\dagger}(q)0 \rangle=(2 \pi)^3 \delta^{(3)}(\vec{q}-\vec{p}).$$
 
Last edited:
Again, it's all due to the damn bra-ket notation which is confusing. The equation should read:

## \langle 0, a (p) a^{\dagger} (q) 0\rangle = \langle a^{\dagger} (p) 0, a^{\dagger}(q)0\rangle ##
 

Similar threads

  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K