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!

  • $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – lemon
    Commented Feb 14, 2015 at 10:10
  • $\begingroup$ You made many mistakes, you'd better see the article in Wikipedia with the double slit $\endgroup$
    – Sofia
    Commented Feb 14, 2015 at 11:56
  • $\begingroup$ Thanks for pointing out. In later stage I have interchangeably use a.s as u/2. Anyways I have corrected the same. Is that the mistake you were pointing at? $\endgroup$ Commented Feb 14, 2015 at 12:07

2 Answers 2


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.

  • $\begingroup$ Thanks for sharing this wonderful insight although I'm not sure how well for i understood this. Just as a take home message: suppose I do double slit experiment with a light and obtained some interference and hence some lambda 1. Now if I somehow do the experiment with electrons such that I obtain the very same interference and parallely calculate the see Broglie wavelength lambda 2 of these electrons. Will these lambda 1 and 2 be equal ( assuming the relativity property of light doesn't exist). $\endgroup$ Commented Feb 14, 2015 at 20:49
  • $\begingroup$ From my interpretation of your above answer, it seems to me that we simple cannot extend the concept of wave function as we know in Schrödinger equation to photon. So if the modified relevistic equation implied some new wave equation, then can then we say that these waves are same (related) to their EM waves. $\endgroup$ Commented Feb 14, 2015 at 20:56
  • $\begingroup$ @ManishKumarSingh Your last comment is pretty much spot on. The take home message is that a solution to the classical Maxwell equations uniquely defines a one photon state. The "energy density" calculated from this solution is proportional to the probability of absorbing the photon at a given point with an ideal detector. This last notion is subtly different from the notion of "finding a particle" that one sees used with the nonrelativistic Schrödinger equation to define "shapes" of atomic orbitals, for example. Interestingly, this "finding" notion is flawed in the relativistic Dirac .... $\endgroup$ Commented Feb 14, 2015 at 22:33
  • $\begingroup$ @ManishKumarSingh equation too. But for the experimental situation described, if you swapped photons for electrons and used the electron wavefunction to foretell detection densities for the electron and used your functions to foretell the same thing for photons in the analogous experiments, the results would be the same. I should have made that clearer. It is for this reason that some people call what you have written the "photon wave function". $\endgroup$ Commented Feb 14, 2015 at 22:39
  • $\begingroup$ @user929304 A ray is actually a definition of a plane wave, or at least a locally plane wave, i.e. a field with wavefronts that can be considered flat over several wavelengths normal to the ray. The disturbance is spread out over space: it therefore can still interfere with itself if the copies travel over different distances. But this is where the notion of coherence length comes in: the source must be coherent enough that the disturbance delayed by distance $x$ is highly correlated with that delayed by distance $x+\Delta$, where $\Delta$ is the path difference. If the light has ... $\endgroup$ Commented Jul 4, 2016 at 12:03

I refer to your main questions,

  • 1) 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?

If you would have written the function $ξ$ correctly as in the article about the 2slit experiment, yes, the amplitude on the screen were given by the function $ξ$ .

  • 2) 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. $

Yes and no. There are two issues here and you have to distinguish between them. A) The clarity of the pattern depends on the number of photons. B) But the positions of the minima and maxima don't, they are the same, no matter how many photons you send.

Let me explain: if you send only one photon through the double-slit, you'll see no pattern. The standard QM says that the pattern exists, but you cannot see it because the photon is absorbed in one point of the screen, s.t. you'll not see the pattern, only the point. But if you send one more photon, and one more, and so on, there will appear more points on the screen, and you'll see that they appear according to the pattern indicated by your eq. (5). The more photons you send, you'll get more of the same, one and the same pattern, more clear.

  • $\begingroup$ which means $ \vert\psi(x,t)\vert^2=\vert U(x)\vert^2 $ and we go back to his original question. $\endgroup$
    – E.phy
    Commented May 27, 2016 at 23:41

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