Is the electric field the same as probability amplitude of a photon? I am a novice in Quantum Mechanics and have seen many authors interchangeably using the two terms in the introductory textbooks. But I have never seen it written explicitly anywhere 'The probability amplitude of a photon is its electric field' so i wonder if there is a subtlety here.
 A: Some lesser known property:
The probability amplitude for a single photon is actually the same math as the classical electric field. So if you for example split light in half on a half-reflecting mirror, half the light will go one way and half will go another way. For the photon the probability amplitudes will follow the same path as the light (since both are essentially waves), and the result will be equivalent.
So I think it is generally helpful to think of photons with this in mind. That said, the interpretation of what is going on is different. In the single photon case, the wave is the probability of finding a photon at a specific spot. In the classical case, it's finding a value of the electric field (which represnts how much of a force a charge is felt at that spot).
A photon also has an electric field, but performing a measurement of the electric field of a photon is actually different from measuring the location of a photon. Measuring the position of a photon is typically done by having it interact with a medium, which typically involves making a measurement of the energy of a photon (which is proportional to the energy of the electric field, which is $\propto E^2$. Measuring the electric field of a photon is a different quantum measurement which is $\propto E$. A single photon is not actually in an eigenstate of E, which means that a measurement will end up giving you a distribution of values of the electric field. The actual distribution of a single photon is a hermite-gaussian function, and is determined through "second quantization," which is the standard method of figuring out how the electric field is quantized.
A: Short answer: not quite. There are two important features here. First, the electric field of the photon is not the coordinate, it's related the momentum canonical conjugate to the coordinate. The coordinate is the solenoidal part of the vector potential (that's gauge invariant, believe it or not), and the solenoidal part of the electric field is the negative of the canonically conjugate momentum.
Second, the photons in the electric field are two infinite numbers of harmonic oscillators (one for each polarization at each momentum). Thus, the electric field is not the probability amplitude, itself. It's part of of the energy density. Recall that for a harmonic oscillator the energy is $\hbar\omega$ times the number (density), and that number density is the analogue to probability.
In detail, this is the mode space energy for photons (note: mode space is Fourier transform of space but not time, and I use the unitary convention for Fourier transforms):
\begin{align}
    H &= \int \mathrm{d}^3k\, \left(\frac{\epsilon_0}{2} \left[\mathbf{E}\cdot\mathbf{E} - \left(\hat{k}\cdot\mathbf{E}\right)^2\right] + \frac{k^2}{2\mu_0} \left[\mathbf{A}\cdot\mathbf{A} - \left(\hat{k}\cdot\mathbf{A}\right)^2\right] \right).
\end{align}
First, we rescale the fields $\mathbf{A}\rightarrow \sqrt{\mu_0}A$ and $\mathbf{E}\rightarrow \sqrt{\mu_0}A$. Second, we introduce a polar basis for the wave numbers with $\hat{k}$ the radial direction, $\hat{\varphi}$ the aziumuthal, and $\hat{\theta}$ the polar directions. The Hamiltonian becomes:
\begin{align}
   H &= \int \mathrm{d}^3k\, \left(\frac{1}{2c^2} \left[E_\theta^2 + E_\varphi^2\right] + \frac{k^2}{2} \left[A_\theta^2 + A_\varphi^2\right] \right).
\end{align}
Formally, to get the number operator, we need to divide the integrand by $\hbar c k$.
\begin{align}
   N &= \int \mathrm{d}^3k\, \left(\frac{1}{2\hbar c^3 k} \left[E_\theta^2 + E_\varphi^2\right] + \frac{k}{2\hbar c} \left[A_\theta^2 + A_\varphi^2\right] \right).
\end{align}
In reality, you would need to quantize (impose commutation relations $[A_i, E_j] \propto -i\hbar\delta(\mathbf{k}_A - \mathbf{k}_E)\delta_{ij}$, with $i,j$ drawn from $\{\theta,\varphi\}$, construct ladder operators, etc).
The integrand of $N$ is the density of photons per mode, up to a commutator. If you want to try to bring that back to real space, though, you're going to run into a problem: the expression won't be a local integral of a number density, unlike the energy which is a local integral of a density. It's going to have a form
$$\int \mathrm{d}^3x\,\mathrm{d}^3y\, \mathbf{E}(\mathbf{x})\cdot M_E(|\mathbf{x}-\mathbf{y}|) \mathbf{E}(\mathbf{y}) + \text{ similar for }\mathbf{A}.$$
Where $M_E$ is a matrix (and the vector potential will have a related, but different, matrix).
Thus, there isn't a simple expression for the number density of photons in real space.
A: It's not. The classical EM wave is composed of many many many photons. So the electric fields in it have nothing to do with probability of finding a photon.
In the Quantum theory, the electric field is an operator. It is itself probabilistic and its values are uncertain prior to measurement. As for the photon's wavefunction, a postion-space wavefunction-ish thing can be attributed to a photon, but there are many subtleties regarding the position basis in quantum field theory. It only exists in an approximate sense. In the momentum basis, the photon states evolve according to $|p\rangle (t)=|p\rangle e^{-ipct}$. So this is spiritually the same as the evolution of classical plane EM waves, which also have a $e^{-ipct}$ factor attached for the time evolution.
