This question is related to: Mach-Zehnder interferometer and the Fresnel-Arago laws

Let us say we have unpolarised wave taking the form: $$\psi=\psi_0 e^{i(kx-\omega t)+i\phi(t)}$$ Where $\phi$ varies randomly with time. If I split this wave into two and send it through e.g. a double slit, one of the beams will experience a phase change due to an optical path length difference. When we combine these two waves one will take the form: $$\psi=\psi_0 e^{i(kx-\omega t)+i\phi(t)}$$ But what about the other?

Their are 3 possibilities: $$\psi=\psi_0 e^{i(k(x+x_0)-\omega t)+i\phi(t)}$$ $$\psi=\psi_0 e^{i(kx-\omega (t+t_0))+i\phi(t+t_0)}$$ $$\psi=\psi_0 e^{i(k(x+x_0)-\omega (t+t_0))+i\phi(t+t_0)}$$
Where $x_0$ and $t_0$ are constants. Which of these 3 is correct and why?

  • $\begingroup$ I don't understand how the wave you describe is unpolarized. And what physical phenomenon are you describing with this random phase ? $\endgroup$ – Dimitri May 12 '16 at 13:25
  • $\begingroup$ @Dimitri I may have overly complicated the question in order to make it not just 'optics' (in my experience (my) optics questions rarely get answered :) ). Your answer explains what I am after. $\endgroup$ – Quantum spaghettification May 12 '16 at 13:56
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    $\begingroup$ Okay :) I was just surprised that this phase factor is a global change of the shape of the wave that does not depend on the position. If for instance you're describing emission of a wave with random phase at point $x_0$, this phase factor would depend on the position. $\endgroup$ – Dimitri May 12 '16 at 14:05
  • $\begingroup$ @Dimitri Actually I was about to post a question along those lines myself. In the locations I have looked (i.e. lecture notes provided by my university, and the paper linked to in the linked question) indicate that for a unpolarised EM wave we have e.g. $E_x=E_0 \cos(kx-\omega t)$, $E_y=E_0 \sin(kx-\omega t+\phi(t))$ where $\phi(t)$ is random and depends only on $t$. I see no reason why it should not also depend on $x$. $\endgroup$ – Quantum spaghettification May 12 '16 at 15:08
  • $\begingroup$ I thought the same. If the source emits light with a random phase $\phi(t)$, the random phase factor at point $x$ and time $t$ should be $\phi(t-x/c)$ to account for propagation delay. $\endgroup$ – Dimitri May 12 '16 at 15:22

The two waves are interfering after having followed different paths, so $x$ must be different between the two. But you are observing them at the same time $t$ which must be the same for the two waves. So answer 1 is the good one.


If your first equation is correct at the point of splitting, then your second equation is incorrect. Both $x$ and $t$ will have changed.

The two amplitudes just prior to recombination are given by the third (assuming that the two arms are identical e.g., no phase element in one of the legs, and identical path lengths as in the "classic" free space Mach-Zehnder) with $x_0 = ct_0$ (assuming we are in air, not a fiber for example).

Imagine freezing time. Then the phase change is determined by $\exp(kx)$. But time does not stand still. The phase also changes by the passage of time: $\exp(\omega t)$.

If the path lengths are not identical, then you need a $x_1$ and a $t_1$ for one of the legs. If there is a phase element, you have to have to deal with that.

  • $\begingroup$ Are you sure I would need to add a $t_1$, as Dimitri suggests we are observing the two waves at the same time? $\endgroup$ – Quantum spaghettification May 12 '16 at 13:54
  • $\begingroup$ @garyp I am confused with your awnser. If the path lenghts are not identical, you just need a $x_1$ and a $x_2$ but one time $t_1$, as interference occur at the same time for the two waves. The wave is delocalized, I think the "delay" you're thinking of is included in the fact that the parts of the two interfering waves that were superposed (in phase) at the point of splitting do not superpose anymore at the point of interference. $\endgroup$ – Dimitri May 12 '16 at 14:09

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