Categories: Physics, Quantum mechanics.

Time-ordered product

In quantum mechanics, especially quantum field theory, a time-ordered product is a product of explicitly time-dependent operators, subject to certain ordering constraints.

Let us start with an unusual motivation. Suppose that some time-dependent operator $$\hat{A}(t)$$ is defined like so, as a product of $$N$$ time-dependent sub-operators $$\hat{a}_n(t)$$:

\begin{aligned} \hat{A}(t) \equiv \int_0^{t} \hat{a}_1(t_1) \bigg( \int_0^{t_1} \hat{a}_2(t_2) \bigg( \int_0^{t_2} \hat{a}_3(t_3) \bigg( \cdots \bigg) \dd{t_3} \bigg) \dd{t_2} \bigg) \dd{t_1} \end{aligned}

Crucially, the upper limits of the inner integrals depend on the surrounding variables, meaning that these integrals cannot simply be reordered.

An interpretation is that the rightmost $$\hat{a}_N(t_N)$$ is applied first, and then $$\hat{a}_{N-1}(t_{N-1})$$ secondly with $$t_{N-1} > t_N$$, and so on. This suggests there is a form of “time-ordering” here: the integrals sweep across all relative timings of $$\hat{a}_n$$, but preserve the ordering. Indeed, this could be rewritten as a time-ordered product (see the interaction picture for an example).

A more general and intuitive motivation goes as follows. Suppose we have a product of $$N$$ time-dependent operators $$\hat{a}_n(t)$$, each representing a certain event. Clearly, we would want to apply them in chronological order:

\begin{aligned} \hat{a}_N(t_N) \: \hat{a}_{N-1}(t_{N-1}) \: \cdots \: \hat{a}_2(t_2) \: \hat{a}_1(t_1) \qquad \mathrm{where} \qquad t_N > t_{N-1} > ... > \: t_2 > t_1 \end{aligned}

But what if the ordering of the arguments $$t_N, ..., t_1$$ is not known in advance? We thus define the time-ordering meta-operator $$\mathcal{T}$$, which reorders the operators based on the $$t$$-values such that they are always in chronological order. For example:

\begin{aligned} \mathcal{T} \big\{ \hat{a}_1(t_1) \: \hat{a}_2(t_2) \big\} \equiv \begin{cases} \hat{a}_1(t_1) \: \hat{a}_2(t_2) & \mathrm{if} \; t_2 < t_1 \\ \hat{a}_2(t_2) \: \hat{a}_1(t_1) & \mathrm{if} \; t_1 < t_2 \end{cases} \end{aligned}

This example suggests a general algorithm for $$\mathcal{T}$$: we need to consider every permutation of the operators $$\hat{a}_n(t_n)$$, and leave only the single one that satisfies our demands.

Mathematically, we do this by summing up all permutations, and multiplying each term with a product of Heaviside step functions $$\Theta$$, which remove the term if the ordering is wrong:

\begin{aligned} \mathcal{T} \big\{ \hat{a}_1 \cdots \hat{a}_N \big\} \equiv \sum_{p \in P_N}^{} \Theta\big(t_{p_1} \!\!-\! t_{p_2}\big) \cdots \Theta\big(t_{p_{N-1}} \!\!-\! t_{p_N}\big) \: \hat{a}_{p_1}(t_{p_1}) \: \cdots \: \hat{a}_{p_N}(t_{p_N}) \end{aligned}

With this, our earlier example for two operators $$\hat{a}_1$$ and $$\hat{a}_2$$ takes the following form:

\begin{aligned} \mathcal{T} \big\{ \hat{a}_1(t_1) \: \hat{a}_2(t_2) \big\} = \Theta(t_1 - t_2) \: \hat{a}_1(t_1) \: \hat{a}_2(t_2) + \Theta(t_2 - t_1) \: \hat{a}_2(t_2) \: \hat{a}_1(t_1) \end{aligned}

However, we are still missing an important detail: so far, we have quietly been assuming that the operators are bosonic (see second quantization). To include fermionic operators, we must allow the sign of each term to change, based on whether the permutation is even or odd:

\begin{aligned} \mathcal{T} \big\{ \hat{a}_1(t_1) \: \hat{a}_2(t_2) \big\} = \Theta(t_1 - t_2) \: \hat{a}_1(t_1) \: \hat{a}_2(t_2) \pm \Theta(t_2 - t_1) \: \hat{a}_2(t_2) \: \hat{a}_1(t_1) \end{aligned}

Where $$\pm$$ is $$+$$ for bosons, and $$-$$ for fermions in this case. The general definition of $$\mathcal{T}$$ is:

\begin{aligned} \boxed{ \mathcal{T} \big\{ \hat{a}_1 \cdots \hat{a}_N \big\} \equiv \sum_{p \in P_N}^{} (\pm 1)^p \bigg( \prod_{j = 1}^{N-1} \Theta\big(t_{p_j} \!-\! t_{p_{j+1}}\big) \bigg) \bigg( \prod_{k = 1}^N \hat{a}_{p_k}(t_{p_k}) \bigg) } \end{aligned}

1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.