Suppose in our double slit experimental setup with the usual notations $d,D$, we have a beam of light of known frequency $(\nu)$ and wavelength $(\lambda)$ – so we can describe it as $$ξ_0 = A\sin(kx-\omega t). \tag{1}$$ It passes through the two holes and moves ahead doing the usual interference stuff, so the final form of the wave will be $$ξ = ξ_1 + ξ_2 = 2A\cos(u/2)\sin(kx-\omega t+0.5*u) \tag{2}$$ where $u$ is the phase difference.
We can convert phase difference $u$ to path difference $q$. Now we choose the point of interest on the screen $(s)$ ,(which depends on path difference q and hence phase difference u). The amplitude at $s$ will be $$ξ = 2A\cos(as)\sin(kx-wt+as), \tag{3}$$ where $a$ is constant.
Now this amplitude is a set of waves which interfere with different phases, and is function of the variables $s,x,t$. Since I placed the screen at some fix distance $x=D$ from the wall with slits, $ξ$ reduces to a function of two variables $s,t$. Rewriting $$ξ_D=2A\cos(as)\sin(as -wt +kD), \tag{4}$$ this is also a wave description (but with different meaning).
The screen is along our $x$-axis (or to be precise $s$-axis). The intensity obtained on the screen is proportional to to absolute square of the wave amplitude written above, which in turn depends on $s$ (and t as well).
But the intensity is also proportional to number of photons. So we postulate that the probability that a photon hit a certain $s$ is proportional to the $$\text {intensity} = |\text {amplitude}|^2. \tag{5}$$
Now, the function $ξ$ I have written above is the wave function ($\psi$) from the quantum mechanics with $s$ acting as $x$ (in $\psi$)? If not, then what is the relation between them? (I will have some additional things to ask depending upon your response.) Thank you!
Best Answer
The functions you write down are solutions of Maxwell's equations (if you think of them as lone, Cartesian components) and, as such, have an exact relationship the one-photon quantum state of the quantum photon field.
Now, whether this is a photon wave function depends on your definitions. If you want to write down the quantum state of a one-photon, so called Fock state of the quantum photon field in position co-ordinates then you are doomed (the position co-ordinate components are what most people understand by "wavefunction"). There is no description of the photon whose squared modulus tells you the probability to find a photon, as there is with the nonrelativistic Schrödinger equation for the electron. This lack has to do with the fact that there is no nonrelativistic description of the photon: Maxwell's equations are already fully relativistic and indeed can be written in a form that shows them to be analogous to the Dirac equation for a massless particle.
However, what you can do is describe the probability amplitude for a photon to be absorbed by an ideal detector at a given point in space and time. This probability amplitude is what you have written down in your question.
This absorption probability amplitude is related to a one-photon Fock state $\psi$ of the quantum light field as follows:
$$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ \vec{\phi}_B(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{B}}^+(\vec{r},t)\left|\left.\psi\right>\right. \end{array}\tag{1}$$
where $\psi$ is the (Heisenberg picture) light field quantum state, $\mathbf{\hat{B}}^+,\,\mathbf{\hat{E}}^+$ are the positive frequency parts of the (vector valued) electric and magnetic field observables and, of course, $\left<\left.0\right.\right|$ is the unique ground state of the quantum light field.
This relationship is invertible, i.e., given the vector valued $\vec{\phi}_E,\,\vec{\phi}_B$, one can uniquely reconstruct the one-photon light field quantum state, so you can think of it as being a particular representation of the one-photon state.
For one photon states, $\vec{\phi}_E,\,\vec{\phi}_B$ fulfill Maxwell's equations; conversely, every classical solution to the Maxwell equations also defines a corresponding one-photon state through the inversion of (1).
The probability density to destructively detect the photon, when the state is properly normalised, is the analogue of the classical energy density (normalisation makes the classical energy density into a probability denstity), i.e.
$$p(\vec{r},\,t) = \frac{1}{2}\,\epsilon_0\,|\vec{\phi}_E|^2 + \frac{1}{2\,\mu_0}\,|\vec{\phi}_B|^2$$
See my answer here for more information and references.