- #1
soviet1100
- 50
- 16
Hello,
This problem is in reference to the QFT lecture notes (p.18-19) by Timo Weigand (Heidelberg University).
He writes:
For the real scalar fields, we get self-adjoint operators [itex] \phi(\textbf{x}) = \phi^{\dagger}(\textbf{x}) [/itex] with the commutation relations
[itex] [\phi(\textbf{x}), \Pi(\textbf{y})] = i \delta^{(3)}(\textbf{x} - \textbf{y}) [/itex]
Fourier transforming the fields as
[itex] \phi(\textbf{x}) = \int{\frac{d^{3}p}{(2\pi)^{3}} \tilde{\phi}(\textbf{p}) e^{i\textbf{p.x}}} [/itex]
where [itex] \phi(\textbf{p}) = \phi^{\dagger}(-\textbf{p}) [/itex] ensures that [itex] \phi(\textbf{x}) [/itex] is self-adjoint.
___________
I have verified that [itex] \phi(\textbf{p}) = \phi^{\dagger}(-\textbf{p}) [/itex] does indeed make [itex] \phi(\textbf{x}) [/itex] self-adjoint, but is this defined in order to make [itex] \phi(\textbf{x}) [/itex] self-adjoint, or is it an independent truth?
Also, in non-relativistic quantum mechanics, the hermitian adjoint (dagger) of an operator was represented by the transpose conjugate of the matrix representing the operator. In QFT, we have fields that are operators. Is there a matrix representation of these field operators too?
Thanks.
This problem is in reference to the QFT lecture notes (p.18-19) by Timo Weigand (Heidelberg University).
He writes:
For the real scalar fields, we get self-adjoint operators [itex] \phi(\textbf{x}) = \phi^{\dagger}(\textbf{x}) [/itex] with the commutation relations
[itex] [\phi(\textbf{x}), \Pi(\textbf{y})] = i \delta^{(3)}(\textbf{x} - \textbf{y}) [/itex]
Fourier transforming the fields as
[itex] \phi(\textbf{x}) = \int{\frac{d^{3}p}{(2\pi)^{3}} \tilde{\phi}(\textbf{p}) e^{i\textbf{p.x}}} [/itex]
where [itex] \phi(\textbf{p}) = \phi^{\dagger}(-\textbf{p}) [/itex] ensures that [itex] \phi(\textbf{x}) [/itex] is self-adjoint.
___________
I have verified that [itex] \phi(\textbf{p}) = \phi^{\dagger}(-\textbf{p}) [/itex] does indeed make [itex] \phi(\textbf{x}) [/itex] self-adjoint, but is this defined in order to make [itex] \phi(\textbf{x}) [/itex] self-adjoint, or is it an independent truth?
Also, in non-relativistic quantum mechanics, the hermitian adjoint (dagger) of an operator was represented by the transpose conjugate of the matrix representing the operator. In QFT, we have fields that are operators. Is there a matrix representation of these field operators too?
Thanks.
Last edited: