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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?

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1 Answer 1

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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).

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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 $\langle 0|:XX:|0\rangle = 0$). –  David Z Dec 10 '11 at 1:43
1  
all this sandwiching made me hungry –  lurscher Dec 10 '11 at 2:17
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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! –  David Z Dec 10 '11 at 2:56
    
Yes, I updated the answer with an explicit identity operator. –  Qmechanic Dec 10 '11 at 3:23
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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. –  Dilaton Dec 10 '11 at 10:40

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