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QFT1995

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In summary, the time ordering operator ##\mathcal{T}## and the normal ordering operator ##\hat{N}## commute with each other and can be applied in either order without changing the outcome. However, the notation of placing the time ordering operator inside the expectation value is unconventional and it is meant to be interpreted as acting within the expectation value. Additionally, it is important to note that when normal-ordering a string of operators, they should all be defined at the same spacetime point. Finally, if two expressions are equal under the time ordering operator, it does not necessarily mean that they are equal without the operator.

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QFT1995

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king vitamin

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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 [itex]t < t'[/itex], these orderings are identical so they commute, while if [itex]t > t'[/itex] they result in differing orders and don't commute.

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QFT1995

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$$\mathcal{T}\hat{N}= \mathcal{T}$$

and

$$\hat{N}\mathcal{T}= \hat{N}$$

correct?

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king vitamin

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Yes, that's right.

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QFT1995

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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.

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king vitamin

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QFT1995 said:

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 [itex]\langle 0 | (\mathrm{stuff})| 0 \rangle[/itex] is a c-number rather than an operator, so the normal/time ordering operators leave it unchanged.

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Since time in quantum theory (particularly also of course in QFT) is a real parameter it commutes with all operators.QFT1995 said:

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.

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king vitamin

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QFT1995 said:

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 [itex]\mathcal{O}_i(t_i)[/itex] defined at particular spacetime points, and then you order the resulting string of normal-ordered operators in time.

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QFT1995

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Ah thank you. This is the point I was missing. One last question. If I have two expressions say $$ \mathcal{T} A(\phi(x_1),\phi(x_2)\dots)=\mathcal{T}B(\phi(x_1),\phi(x_2)\dots),$$king vitamin said: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

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

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.

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king vitamin

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QFT1995 said: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?

In mathematics and physics, two operators are said to commute if their order of operation does not affect the end result. In the context of time and normal ordering operators, this means that the order in which these operators are applied does not change the final outcome.

Knowing if these operators commute is important in quantum mechanics and quantum field theory, as it affects the accuracy and validity of calculations. If the operators do not commute, it can lead to incorrect results and interpretations.

The commutativity of two operators can be determined by applying them in different orders to a given function or equation. If the end result is the same regardless of the order, then the operators commute. In the case of time and normal ordering operators, this can also be determined through mathematical proofs and equations.

If the operators do not commute, it means that the order of operation does affect the final outcome. This can lead to inconsistencies and errors in calculations, and may require more complex mathematical techniques to accurately solve problems.

Yes, the concept of commuting operators has applications in various fields including quantum mechanics, signal processing, and computer science. For example, in quantum computing, the commutativity of operators is important for designing efficient algorithms and reducing computational errors.

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