In a double-slit experiment with light, the observed peak intensity is 4 times higher than in the single-slit case.

Using classical physics, this can be explained as follows:

  • at the single-slit, the electric field vector has magnitude $E_0$ and the wave has intensity $I_0 \sim E_0^2$
  • at each slit of the double-slit, the electric field vector has the same magnitude $E_0$, and if they are in phase at the screen, the vector sum will be $2E_0$ and intensity $I \sim (2E_0)^2 = 4I_0$

Note that the field is $E_0$ at each of the 2 slits in the double-slit and at the single-slit, so the intensity $(\sim E_0^2)$ is the same at each of the 3 slits, which means that each slit passes the same amount of light. In other words, double-slit uses 2 times more light than single-slit. This is why in the quantum mechanical explanation that follows, I will compare double-slit/two-photons to single-slit/single-photon (not to single-slit/two-photons) and expect to get the same factor of 4.

Using quantum mechanics to explain the same thing, let’s consider the (position, momentum) coordinates $(x, p)$. The x axis is along the plate (that has the slits), and the slits are two points on this axis: $x_1$ and $x_2$.

$p$ is the momentum component along the x axis only, which determines the angle at which the photon will be detected on the screen and so the interference pattern. This means that in order to calculate the peak light intensity ratios between different cases (single slit, double slit), we just need to calculate the ratios between peak probability densities of $p$.

1) Single-slit/single-photon

The position-representation wave function is $$\psi(x) = \delta(x-x_1)$$ $\Rightarrow$ the momentum-representation wave function is its Fourier transform: $$\tilde \psi(p) = \frac 1 {\sqrt {2\pi}} \int \delta(x-x_1) e^\frac {-ipx} {\hbar} dx = \frac 1 {\sqrt {2\pi}} e^\frac {-ipx_1} {\hbar}$$ $\Rightarrow$ the probability density of $p$ is: $$P(p) = |\tilde \psi(p)|^2 = \frac 1 {2\pi}$$

2) Double-slit/single-photon $$\psi(x) = \frac 1 {\sqrt 2}[ \delta(x-x_1) + \delta(x-x_2)]$$ $$\Rightarrow \tilde \psi(p) = \frac 1 {2\sqrt {\pi}} \int [ \delta(x-x_1) + \delta(x-x_2)] e^\frac {-ipx} {\hbar} dx = \frac 1 {2\sqrt {\pi}} [e^\frac {-ipx_1} {\hbar} + e^\frac {-ipx_2} {\hbar}]$$ $$\Rightarrow P(p) = |\tilde \psi(p)|^2 = \frac 1 {2\pi} \left[1 + cos\left(\frac {p(x_1-x_2)} {\hbar}\right)\right]$$ The peak value of this is 2 times the one from single-slit (since $cos$ has max value 1). The usual factor of 4 (from classical physics) is because 2 slits also mean 2 times more photons in general, but now we are still using a single photon (as in the single-slit case).

3) Double-slit/two-photons

Now I would expect to get the usual factor of 4 (2 times previous case), but I actually get a factor of 8 (4 times previous case). The two photons have equal wave functions (calculated in previous case): $$\tilde \phi(p) = \tilde \psi(p) = \frac 1 {2\sqrt {\pi}} [e^\frac {-ipx_1} {\hbar} + e^\frac {-ipx_2} {\hbar}]$$ Each photon interferes with itself and with the other photon $\Rightarrow$ $$P(p) = |\tilde \phi(p) + \tilde \psi(p)|^2 = |2\tilde \psi(p)|^2 = 4 \frac 1 {2\pi} \left[1 + cos\left(\frac {p(x_1-x_2)} {\hbar}\right)\right]$$ The peak value of this is 8 times the one from single slit (instead of 4).

The only way we can get the expected factor of 4 is if $|\tilde \phi(p) + \tilde \psi(p)|^2 = |\tilde \phi(p)|^2 + |\tilde \psi(p)|^2$

But that would mean the wave functions of the two photons are "orthogonal", so they do not interfere. I guess this is why Dirac stated in his QM book that "each photon then interferes only with itself. Interference between two different photons never occurs".

Finally, suppose we replace the 2 slits with 2 antennas, and light with microwaves (lower frequency but still photons), a setup commonly used in wireless communications for beamforming. If we send the same signal on the 2 antennas, the resulting radiation pattern will have peaks (beams) and valleys depending on the angle, since it is an interference pattern just like in the double-slit experiment. So, in this case we have photon interference but I don’t think we can consider that each photon goes through both antennas and interferes only with itself.

So, are different photons actually interfering with each other?


2 Answers 2


"So, are different photons actually interfering with each other?"

There are some common confusions in the classical/historical explanations of "interference". First Dirac is correct that 2 photons never "interfere" especially if interference is defined as 2 photons cancelling .... that would be a violation of conservation of energy, unfortunately the cancel method is the most popular method taught in high school textbooks. His famous saying is that " a photon only interferes with itself" .... the modern interpretation is best represented by Feynman's quote, "every photon determines its own path".

Feynman proposed the path integral method to determine probabilities of photon paths ... the highest probabilities turn out to paths where the path length is an integer multiple of the wavelength. In the DSE there are no photons in the dark, the light bands have all the photons.

With microwaves/radio waves we have to be careful ... they are lower energy and created in large numbers. Let's take a radar example ... what can happen is that when 2 out of phase beams impinge on a metal surface half the electrons see one beam half the other beam ... net is that there's zero potential created in the antenna or metal surface. Where the beams are constructive a strong signal can be achieved.


So, in this case we have photon interference but I don’t think we can consider that each photon goes through both antennas and interferes only with itself.

Interference is a phenomenon that happens when solutions of wave equations are superimposed. The photon is a quantum mechanical entity a point particle with energy hν where ν is the frequency of the classical electromagnetic wave same energy photons build up and h Planck's constant.

The probability of finding the photon at (x,y,z,t) is described by a wavefunction and given by $Ψ^*Ψ$ . This means that one needs distribution of same boundary condition photons to see any interference.

So, are different photons actually interfering with each other?

No, In this single photon at a time experiment the build up of the interference pattern from seemingly random photon dots shows the cross section for "photon of given energy scattering off given double slits"


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