# Electric field of a point source in QM

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?

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.