Do the time and normal ordering operators commute?

[QFT1995]

Does the time ordering operator $\mathcal{T}$ commute with the normal ordering operator $\hat{N}$? i.e. is $$[ \mathcal{T},\hat{N}] =0$$ correct?

 

[king vitamin]

Be careful, as neither time-ordering nor normal-ordering are linear operators (in fact they don't satisfy the usual definition of an "operator" in quantum mechanics). When you time-order a set of time-dependent operators, it outputs the order of operators in ascending (right-to-left) chronological order, while when you normal-order a set of operators, you place them in some "canonical" ordering. So a given string of operators is always in the order of the last of either \mathcal{T} or \hat{N} which you applied to it.

Let's just look at a simple example. In the simple (Heisenberg picture) quantum harmonic oscillator, consider either
$$
\mathcal{T}\left\{ a(t) a^{\dagger}(t') \right\}
$$
or
$$
\hat{N}\left\{ a(t) a^{\dagger}(t') \right\}.
$$
If t < t', these orderings are identical so they commute, while if t > t' they result in differing orders and don't commute.

 

[QFT1995]

Okay thank you. If I apply normal ordering followed by time ordering, is that identical to just applying time ordering? In the same manner, if I apply time ordering followed by normal ordering, it that identical to just normal ordering? i.e. are

$$\mathcal{T}\hat{N}= \mathcal{T}$$
and
$$\hat{N}\mathcal{T}= \hat{N}$$
correct?

 

[king vitamin]

Yes, that's right.

 

[QFT1995]

Okay thank you. Also is $$ \mathcal{T} \langle 0 |\bigg\{ \phi(x_1) \phi(x_2)\dots \bigg\}|0\rangle $$
the same as
$$ \langle 0 |\mathcal{T} \bigg\{ \phi(x_1) \phi(x_2)\dots \bigg\}|0\rangle $$
where $\mathcal{T}$ is now on the inside.

 

[king vitamin]

Okay thank you. Also is $$ \mathcal{T} \langle 0 |\bigg\{ \phi(x_1) \phi(x_2)\dots \bigg\}|0\rangle $$
the same as
$$ \langle 0 |\mathcal{T} \bigg\{ \phi(x_1) \phi(x_2)\dots \bigg\}|0\rangle $$
where $\mathcal{T}$ is now on the inside.

I have never seen the former notation before, but I would say they are definitely different, since the object \langle 0 | (\mathrm{stuff})| 0 \rangle is a c-number rather than an operator, so the normal/time ordering operators leave it unchanged.

 

[QFT1995]

If you look on page 5 eq 4.2 of the paper http://www.sbfisica.org.br/~evjaspc/xviii/images/Nunez/Jan26/AdditionalMaterial/Coleman_paper/Coleman_paper.pdf (I have linked the paper here), the former notation is written.

 

[vanhees71]

Does the time ordering operator $\mathcal{T}$ commute with the normal ordering operator $\hat{N}$? i.e. is $$[ \mathcal{T},\hat{N}] =0$$ correct?

Since time in quantum theory (particularly also of course in QFT) is a real parameter it commutes with all operators.

Note that in relativistic QFT both time and position are not a priori observables but just real parameters. As in non-relativistic QM time is just a parameter introducing an ordering of causal effects (thus defining "future, present, and past" via the causal time arrow). In relativistic QFT position is just a label for the infinitely many degrees of freedom described by fields.

Concrete relativistic QFT models are built using the unitary representations of the Poincare group. The only type of QFT used for practical purposes (aka the Standard Model) realizes the Poincare group with local fields, being quantized as bosons or fermions such that the Hamiltonian is bounded from below (i.e., existence of a stable ground state is built in by construction). All observables are then defined as functions (or functionals) of fields.

A position operator can be constructed a posteriori only for massive particles or massless particles with spin $s \leq 1/2$.

Time never becomes an observable, because otherwise the Hamiltonian would be its "canonical conjugate" implying an energy spectrum that is not bounded from below. This argument by Pauli (1930) holds in both non-relativistic QM and relativistic QFT.

 

[king vitamin]

If you look on page 5 eq 4.2 of the paper http://www.sbfisica.org.br/~evjaspc/xviii/images/Nunez/Jan26/AdditionalMaterial/Coleman_paper/Coleman_paper.pdf (I have linked the paper here), the former notation is written.

I see. I believe this is just unusual notation, and you are meant to interpret the time-ordering symbol as acting inside the expectation values. Obviously, if you evaluated the expectation value before time-ordering, the time-ordering would not do anything! Notice that he does put T inside the expectation value in Eq 4.18.

As an aside, you'll notice that Coleman never normal-orders a string of operators which are defined at different spacetime points. The expectation values take the form
$$
\langle \mathcal{T} \left\{ \hat{N}[\mathcal{O}_1(t_1)] \hat{N}[\mathcal{O}_2(t_2)] \cdots \right\} \rangle
$$
That is, you normal-order some composite operators \mathcal{O}_i(t_i) defined at particular spacetime points, and then you order the resulting string of normal-ordered operators in time.

 

[QFT1995]

As an aside, you'll notice that Coleman never normal-orders a string of operators which are defined at different spacetime points. The expectation values take the form

​

$$\langle \mathcal{T} \left\{ \hat{N}[\mathcal{O}_1(t_1)] \hat{N}[\mathcal{O}_2(t_2)] \cdots \right\} \rangle$$

Ah thank you. This is the point I was missing. One last question. If I have two expressions say $$ \mathcal{T} A(\phi(x_1),\phi(x_2)\dots)=\mathcal{T}B(\phi(x_1),\phi(x_2)\dots),$$
does that mean $A(\phi(x_1),\phi(x_2)\dots)=B(\phi(x_1),\phi(x_2)\dots)$? Here $A$ and $B$ are functions of the fields.

 

[Demystifier]

Consider the product of operators $a(t_1)a^{\dagger}(t_2)$. I think it's quite clear that normal ordering and time ordering in this case do not commute.

 

[king vitamin]

Ah thank you. This is the point I was missing. One last question. If I have two expressions say $$ \mathcal{T} A(\phi(x_1),\phi(x_2)\dots)=\mathcal{T}B(\phi(x_1),\phi(x_2)\dots),$$
does that mean $A(\phi(x_1),\phi(x_2)\dots)=B(\phi(x_1),\phi(x_2)\dots)$? Here $A$ and $B$ are functions of the fields.

Have you tried playing with some simple expressions to check if this is a reasonable relation?