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QFT1995
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Does the time ordering operator ##\mathcal{T}## commute with the normal ordering operator ##\hat{N}##? i.e. is $$[ \mathcal{T},\hat{N}] =0$$ correct?
QFT1995 said: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.
Since time in quantum theory (particularly also of course in QFT) is a real parameter it commutes with all operators.QFT1995 said:Does the time ordering operator ##\mathcal{T}## commute with the normal ordering operator ##\hat{N}##? i.e. is $$[ \mathcal{T},\hat{N}] =0$$ correct?
QFT1995 said: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.
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
$$\langle \mathcal{T} \left\{ \hat{N}[\mathcal{O}_1(t_1)] \hat{N}[\mathcal{O}_2(t_2)] \cdots \right\} \rangle$$
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.
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.