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A standard result in QFT is the expression of the interacting theory correlation functions in terms of field operators in the interaction picture and the free theory vacuum:

$$\langle\Omega|\mathcal T\{\phi(x)\phi(y)\}|\Omega\rangle=\lim_{T\rightarrow \infty(1-i\epsilon)}\frac{\langle 0|\mathcal T\left\{\phi_I(x)\phi_I(y)\exp[-i\int^T_{-T} dt H_I(t)]\right\}|0\rangle}{\langle0|\mathcal T\left\{\exp[-i\int^T_{-T} dt H_I(t)]\right\}|0\rangle} \tag{4.31}.$$

(I have numbered the equation as this is in reference to Peskin & Schroeder) The utility of this expression is obvious, but what is strange to me is that in order to obtain the numerator on the RHS above we take the previous step in the derivation:

$$\langle\Omega|\mathcal \phi(x)\phi(y)|\Omega\rangle=\lim_{T\rightarrow \infty(1-i\epsilon)}\frac{\langle 0|U(T,x^0)\phi_I(x)U(x^0,y^0)\phi_I(y)U(y^0,-T)|0\rangle}{\langle0|\exp[-i\int^T_{-T} dt H_I(t)]|0\rangle} \tag{i}$$

and apply the time-ordering operator to both sides, inside of which the time evolution operators can be moved past each other in the numerator, which leaves us with the same exponential as in the denominator.

My problem is that we then take 4.31 and treat it as if it is equal to (i), so that we can use a Taylor expansion and Wick's theorem etc. to calculate the interacting correlation functions. However, we appear to have made explicit use of the fact that everything commutes under the $\mathcal T\{\}$ "operator" to get there. Which in reality they do not.

I have, while writing this thought of an idea. Since the $U$ operators in (i) are integrals over $H_I(t)$ which (at least in $\phi^4$ theory) is simply $\int d^3z\text{ }\phi_I^4(z)$, we could commute this operator through the field operators $\phi(x)$ and $\phi(y)$, however this seems to require they be spacelike separated, which they will in general not be. So I am again at a loss.

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    $\begingroup$ Everything that's "out of time order" under the $\mathcal{T}$ symbol will be put back in order anyway. So operators that don't commute can be treated as commuting in time ordered correlation functions. $\endgroup$ Jun 12, 2021 at 21:54

2 Answers 2

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The main point is that the time ordering procedure ${\cal T}[~]$ does not take operators to operators, but symbols/functions to operators. This is similar to what happens with the normal ordering procedure $:~:$, cf. e.g. this, this & this Phys.SE posts.

Example:

$$ \begin{align}{\cal T}[e^{A(t_1)+B(t_2)}]~=~&{\cal T}[e^{A(t_1)}e^{B(t_2)}]\cr ~=~&\theta(t_1\!-\!t_2)e^{\hat{A}(t_1)}e^{\hat{B}(t_2)}\cr ~+~&\theta(t_2\!-\!t_1)e^{\hat{B}(t_2)}e^{\hat{A}(t_1)}.\end{align} $$

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  • $\begingroup$ I've read through a lot of answers (many given by you) on the subtleties of time-ordering and I thought I had understood it unfortunately. My objection is that if we symbolically rewrite (i) as 4.31, with the assumption that 4.31 would be wrong without the time-ordering operator, we shouldn't be able do manipulations in 4.31 that require the operators to have been moved into a specific order (i.e. the Taylor expansion of the Dyson series) and assume that this evaluates to something equal to (i). $\endgroup$
    – Charlie
    Jun 12, 2021 at 21:27
  • $\begingroup$ As maybe a simpler example (which you have specifically referenced but I can't find the exact question), if we apply a similar strategy to a formula like $e^Ae^B\rightarrow \mathcal T[e^Ae^B]=\mathcal T[e^{A+B}]$, I wouldn't expect to be able to algebraically manipulate the final expression and obtain a result that is valid without $\mathcal T$, since we have specifically neglected commutators in order to join the exponentials with the BCH formula. $\endgroup$
    – Charlie
    Jun 12, 2021 at 21:30
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jun 12, 2021 at 21:55
  • $\begingroup$ I think I understand now, thank you! $\endgroup$
    – Charlie
    Jun 12, 2021 at 22:01
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You can't actually "apply the time-ordering operator" to both sides of eq. (i) - as Qmechanic points out, the time-ordering operator acts on symbols, while the objects in eq. (i) are operators.

The argument is supposed to work as follows: We want to compute $\langle \Omega \vert \mathcal{T}[\phi(x)\phi(y)]\vert \Omega\rangle$. So we first look at what happens for $x^0 \geq y^0$, and the result is your eq. (i). Then, we look at what happens for $y^0 > x^0$ - this result is rarely explicitly written out but let's call it eq. (ii) - and we have $$ \langle \Omega \vert \mathcal{T}[\phi(x)\phi(y)]\vert \Omega\rangle = \theta(x^0 - y^0)\text{(i)} + \theta(y^0-x^0)\text{(ii)},$$ where the (i)/(ii) stand for the r.h.s of the respective equations and if you look at what the r.h.s. is explicitly, then you'll find the result is exactly the r.h.s. of eq. (4.31).

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