1

I am trying to self study QED so I apologize if my question seems silly. As I realize, all physical processes should stem from some "hermitian" operator in the quantum language. As it is well known, both in the QM region (SHO) and the QED area, the creation and annihilation operators are non-hermitian (actually these pair are complex adjoints to eachother). Now, my question is how could photon creation, which is known to occur in multiple circumstances (e.g. spontaneous emission from an atoms excited state, or radiation due to a single electron having non-vanishing acceleration), correspond to a non-hermitian operator.

Seyed Mohsen Ayyoubzadeh
  • 246

1 Answers1

3

You are right that the photon creation operator $a^\dagger$ and the photon annihilation operator $a$ are non-hermitian. You are also right that all physical processes (like for example the electron-photon interaction) need to be hermitian operators.

The solution is that the physical processes always come as hermitian combinations of the creation and annihilation operators.

Example: The electron-photon interaction (taken from this answer to How do electrons and photons interact?) can be represented by $$ \hat{V}_\text{int}=e\hat{\mathbf{r}}\cdot\left(\mathbf{E}_0\hat a+\mathbf{E}_0^\ast\hat a^\dagger\right).$$

Although operators $\hat{a}$ and $\hat{a}^\dagger$ are non-hermitian, the combination $\hat{V}_\text{int}$ is hermitian.

Thomas Fritsch
  • 39,317
  • Fair enough. But the question remains that since any combination of the creation and annihilation operator should have equal weight of both to result in an hermitian operator, how could the net result of such an interaction be, for example as in the photon emission from an excited atom's case, the creation of a photon. – Seyed Mohsen Ayyoubzadeh Jun 17 '19 at 21:03
  • @SeyedMohsenAyyoubzadeh Remember that this operator always acts on a state (in this case of an atom and photons). Consider for example an initial state $|\Psi\rangle$ with no photons. Then, the term $\hat{a}|\Psi\rangle$ is zero, meaning that absorption doesn't occur. But the term $\hat{a}^\dagger|\Psi\rangle$ is always non-zero. – Thomas Fritsch Jun 17 '19 at 21:31
  • Thanks, however, $a\left|n\right\rangle=\sqrt n \left| {n - 1} \right\rangle$ and ${a^\dagger}\left| n \right\rangle = \sqrt {n + 1} \left| {n + 1} \right\rangle$, do these along with your answer (equal operators weight), imply that the relative probability of having ${n + 1}$ photons to ${n - 1}$ photons for an excited atom and ${n}$ photons (confined perhaps in a cavity) would always be $1 + \frac{1}{n}$, regardless of the details of the system such as the energy levels of the atom, etc? Also, what would be the mechanism (i.e. physical explanation) of photon absorption in such a system? – Seyed Mohsen Ayyoubzadeh Jun 18 '19 at 01:11