0
$\begingroup$

We know that radiation from a point source vanishes at large distances. So when an atom emits a photon, the expectation value of the electric field must vanish at large distances. How we can explain this fact when we treat the photon as a quantum state? In QM the photon state is |1> and the electric field operator is a sum over annihilation and creation operators of all modes. Is it possible to show that the expectation value of the electric field depends on the distance r from the atom? and that it will vanish as r goes to infinity?

$\endgroup$
0
$\begingroup$

The expectation value of the electric field on a Fock's state is zero. It is a well known result which can be verified by taking into account that the electric field is a linear combination of a photon creation and a photon annihilation operators. But the expectation value of both operators is zero on any eigenstate of the number operator. In oder to compare quantum description of the field with the classical fields, one has to move to states which do not correspond to a fixed number of photons.

$\endgroup$
  • $\begingroup$ Ok, sorry, let me ask my question again slightly different. how we can explain the fact that the expectation value for absorbing photon vanishes at large distances. This expectation value in QM is not zero in Fock state. It is proportional to 〈E^+ E^- 〉 $\endgroup$ – Jacob Jul 25 '19 at 23:05
  • $\begingroup$ @Jacob are you not mixing virtual photons (used to describe the static electric field) with real photons, i.e.. four vector length zero? Virtual photons are not "absorbed" they are under an integral of the problem at hand.. This might interest you motls.blogspot.com/2011/11/… $\endgroup$ – anna v Jul 26 '19 at 8:47
0
$\begingroup$

The key is to work with multimode fields.

Basically, the reason that field vanishes is because you are integrating field (or intensity) from an atom over a fixed area, like the size of your detector. If instead you were looking at the intensity over a fixed solid angle, you would probably find the intensity does not change with distance away from the emitter.

To resolve all these intricacies one needs to work with many photon modes, i.e. creation/anihilation operators as a function of wave-vector, in Fourier space, or as a function of position, in real space. Furthermore, this will only suffice for time-harmonic picture, even more care is needed for non-harmonic fields.

As an illistration consider this. Lets say you have a single-photon state propagating in the direction $\mathbf{\hat{k}}$ and with the wavelength $\lambda=2\pi/k$, where $\mathbf{k}$ is the wavevector. So your state of light is $|\psi\rangle=\hat{a}\left(\mathbf{k}\right)^\dagger |0\rangle$ (ignoring the polarization of light). What is the probability of detecting something on detector with detection area $S$, it will probably end up being somethething like $\int_S d^2 r'\langle\psi|\hat{a}^\dagger\left(\mathbf{r'}\right)\hat{a}\left(\mathbf{r'}\right)|\psi\rangle$, where $\mathbf{r'}$ is the position on the surface $S$ and $\hat{a}\left(\mathbf{r}\right)=\frac{1}{\left(2\pi\right)^3}\int d^3 k\, \hat{a}\left(\mathbf{k}\right)\exp\left(-i\mathbf{k}.\mathbf{r}\right)$ is the Fourier transform of the wavevector-based destruction (or creation) operators. The $1/r$ scaling will most likely arise once you start unravelling all these integrals - you have a 2D integral in space, but 3D in wavevector space, the mismatch will give you the scaling (I expect). We have not even touched upon the necessary commutation relations. What I am getting at here is that full treatment is more complex than $|0\rangle$ and $|1\rangle$ you see in simple accounts.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.