Expectation value of the vector potential operator and the classical limit

Question

I'm very much a beginner in QFT and am aiming for intuition and basic understanding.

Let's consider a 1D universe and forget about photon polarisation.

Imagine the following quantum state:

$$|\psi\rangle=|n_{k}\rangle$$

an energy eigenstate with $n$ photons with wavenumber $k$.

The expectation value of the vector potential field operator:

$$\hat{A}=\int \frac{dk}{\sqrt{2\omega}}\left(\hat{a}^{-}_{k}e^{i(kz-\omega t)}+\hat{a}^{+}_{k}e^{-i(kz-\omega t)} \right )\tag{1}$$

is

$$\langle\psi|\hat{A}|\psi\rangle=\langle n_{k}|\int \frac{dk}{\sqrt{2\omega}}\left(\hat{a}^{-}_{k}e^{i(kz-\omega t)}+\hat{a}^{+}_{k}e^{-i(kz-\omega t)} \right )|n_{k}\rangle\tag{2}$$

This will be zero because the destruction operators will either create or destroy a photon in the $|n_{k}\rangle$ state so that the projection of the remaining states onto the $\langle n_{k}|$ state is zero.

Question 1: Why does this even for very high numbers of photons not retrieve the classical limit where one finds a sinusoidal wave?

My best attempt at question 1:

I think this might be a manifestation of $\Delta E\Delta t \geq \hbar$ and the fact that we have here a quantum state for which $\Delta E=0$ so that $\Delta t$ must be infinite and we can infer nothing about the temporal behaviour of the state.

Indeed if we introduce some $\Delta E$ and consider instead the state $|\psi\rangle=\frac{1}{\sqrt2}(|n_{k}\rangle+|m_{k})\rangle$ we find:

$$\langle\psi|\hat{A}|\psi\rangle=\frac{1}{2}\left(\langle n_{k}|+\langle m_{k}|\right)\int \frac{dk}{\sqrt{2\omega}}\left(\hat{a}^{-}_{k}e^{i(kz-\omega t)}+\hat{a}^{+}_{k}e^{-i(kz-\omega t)} \right )\left(|n_{k}\rangle+|m_{k}\rangle\right)\tag{3}$$

Is not zero if $n=m\pm 1$

And will actually give back a real sinusoid since the imaginary part cancels.

Question 2:

Why does this not work with $n=m\pm 2$ which would still give zero?

My poor best attempt at question 2:

Is the state $|\psi\rangle=\frac{1}{\sqrt2}(|n_{k}\rangle+|(n\pm2)_{k}\rangle)$ unphysical maybe? Will an $|(n\pm1)_{k}\rangle$ always creep in?

I would love to read any insights!

Answer 1

Question 1: Why does this even for very high numbers of photons not retrieve the classical limit where one finds a sinusoidal wave?

Having a high number of photons is not enough to make a classical sinusoidal electromagnetic wave. You need a state which is a superposition of many different numbers of photons.

This is because there is an uncertainty relation between the number of photons and the phase (just search for "number phase uncertainty" to find more info). $$\Delta n\ \Delta \phi \ge 1$$

A sinusoidal electromagnetic wave has a very precisely defined phase (i.e. $\Delta\phi \ll 1$). Therefore the number of photons in this wave is highly uncertain (i.e. $\Delta n \gg 1$). So, for building a quasi-classical state $|\psi\rangle$ you need a sum of many different $|n_k\rangle$ basis states (e.g. with all states from $|n_k=900\rangle$ to $|n_k=1100\rangle$).

Answer 2

The states which have a classical limit are coherent states which, for each mode $k$, are of the form $|z\rangle =e^{-|z|^2/2}e^{z \hat a_k^\dagger}|0\rangle$. These have $$ \hat a_k|z\rangle= z|z\rangle, $$ and hence $$ \langle z |\hat a_k|z\rangle = z. $$