# Self-adjointness of the real scalar field

1. Aug 17, 2015

### soviet1100

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 $\phi(\textbf{x}) = \phi^{\dagger}(\textbf{x})$ with the commutation relations

$[\phi(\textbf{x}), \Pi(\textbf{y})] = i \delta^{(3)}(\textbf{x} - \textbf{y})$

Fourier transforming the fields as

$\phi(\textbf{x}) = \int{\frac{d^{3}p}{(2\pi)^{3}} \tilde{\phi}(\textbf{p}) e^{i\textbf{p.x}}}$

where $\phi(\textbf{p}) = \phi^{\dagger}(-\textbf{p})$ ensures that $\phi(\textbf{x})$ is self-adjoint.
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I have verified that $\phi(\textbf{p}) = \phi^{\dagger}(-\textbf{p})$ does indeed make $\phi(\textbf{x})$ self-adjoint, but is this defined in order to make $\phi(\textbf{x})$ 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: Aug 17, 2015
2. Aug 17, 2015

### ChrisVer

No, the relation is taken in order to keep the field self-adjoint (it's proven from considering the field self-adjoint).

You also had the same thing in QM... it appears when your operators stopped having countable & discrete eigenstates/values. In that case it's not interesting to represent them with matrices.

3. Aug 17, 2015

### soviet1100

Thanks for the help, ChrisVer.

Just one more question. The fourier transformed field $\tilde{\phi}(\textbf{p})$ is also self-adjoint with $\tilde{\phi}(\textbf{p}) = \tilde{ \phi}^{\dagger}(\textbf{p})$, isn't it? But if this is the case, then along with $\tilde{\phi}(\textbf{p}) = \tilde{\phi}^{\dagger}(-\textbf{p})$, this implies $\tilde{\phi}(\textbf{p}) = \tilde{\phi}(-\textbf{p})$, right?