Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

Please welcome our new moderators!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

122 submissions , 103 unreviewed
3,497 questions , 1,172 unanswered
4,545 answers , 19,342 comments
1,470 users with positive rep
408 active unimported users
More ...

Generalized propagator

+ 3 like - 0 dislike
4 views

I don't understand how to calculate this generalized two-point function or propagator, used in some advanced topics in quantum field theory, a normal ordered product (denoted between ::) is subtracted from the usual time ordered product (denoted T):

$\langle X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau')\rangle ~=~ T ( X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau')) ~-~ : X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau'):$

My question is can the rhs of this propagator be derived or the meaning of the subtraction of the time ordered product explained and motivated in simple words?

asked Dec 9, 2011 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
Dear @Qmechanic thanks for improving my question (I am not a native English speaker and have to find out how LaTex works on this site too...). It looks better now :-)

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user Dilaton

1 Answer

+ 4 like - 0 dislike

If the operators $X_i$ can be written as a sum of an annihilation and a creation part

$$X_i~=~A_i + A_i^\dagger, \qquad A_i|0\rangle~=~0,$$

where

$$ [A_i(t),A_j(t^\prime)] ~=~ 0, $$

and

$$ [A_i(t),A_j^\dagger(t^\prime)] ~=~ (c~{\rm numbers}) \times {\bf 1}, $$

i.e. proportional to the identity operator ${\bf 1}$, then one may prove that

$$ T(X_i(t)X_j(t^\prime)) ~-~:X_i(t)X_j(t^\prime): ~=~\langle 0 | T(X_i(t)X_j(t^\prime))|0\rangle ~{\bf 1}. \qquad (1) $$

Proof: The time ordering is defined as

$$ T(X_i(t)X_j(t^\prime)) ~=~ \Theta(t-t^\prime) X_i(t)X_j(t^\prime) +\Theta(t^\prime-t) X_j(t^\prime)X_i(t)$$ $$~=~X_i(t)X_j(t^\prime) -\Theta(t^\prime-t) [X_i(t),X_j(t^\prime)]$$ $$~=~X_i(t)X_j(t^\prime) -\Theta(t^\prime-t) \left([A_i(t),A_j^\dagger(t^\prime)]+[A_i^\dagger(t),A_j(t^\prime)]\right). \qquad (2)$$

The normal ordering moves the creation part to the left of the annihilation part, so

$$:X_i(t)X_j(t^\prime):~=~ X_i(t)X_j(t^\prime) - [A_i(t),A_j^\dagger(t^\prime)].\qquad (3)$$

The difference of eqs. (2) and (3) is the lhs. of eq. (1) :

$$ T(X_i(t)X_j(t^\prime)) ~-~:X_i(t)X_j(t^\prime): $$ $$~=~ \Theta(t-t^\prime)[A_i(t),A_j^\dagger(t^\prime)] + \Theta(t^\prime-t)[A_j(t^\prime),A_i^\dagger(t)],\qquad (4)$$

which is proportional to the identity operator ${\bf 1}$ by assumption. Now sandwich eq. (4) between the bra $\langle 0 |$ and the ket $|0\rangle $. Since the rhs. is proportional to the identity operator ${\bf 1}$, the unsandwiched rhs. must be equal to the sandwiched rhs. times the identity operator ${\bf 1}$. Hence also the unsandwiched lhs. must be equal to the sandwiched lhs. times the identity operator ${\bf 1}$. This yields eq. (1).

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user Qmechanic
answered Dec 10, 2011 by Qmechanic (2,580 points) [ no revision ]
How do you get away without sandwiching on the left side of (4)? I would expect $\langle 0|T(X_i(t)X_j(t^\prime))|0\rangle ~-~\langle 0|:X_i(t)X_j(t^\prime):|0\rangle~=~\langle 0 | T(X_i(t)X_j(t^\prime))|0\rangle$ (which of course is a consequence of $\langle 0|:XX:|0\rangle = 0$).

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user David Z
all this sandwiching made me hungry

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user lurscher
Oh, I see now, so $\langle 0|T(XX)|0\rangle$ is implicitly multiplied by the identity operator. (right?) Should've noticed that in the first place... thanks for clarifying!

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user David Z
Yes, I updated the answer with an explicit identity operator.

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user Qmechanic
Thanks a lot @Qmechanic, this clear proof is exactly what I needed. The only thing left to do for me now is to check that the preconditions are valid in my specific case.

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user Dilaton

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...