# Circular polarisation destroying interference pattern for double slit

I have been trying to mathematically explain the empirical result that putting orthogonal polarisers (quarter-wave plates) behind the two slits of a double-slit setup will erase the interference pattern. The trouble is, my analysis predicts an interference pattern. I must have made a silly mistake, but I can't spot it. I was hoping somebody could help me find and correct it.

Here goes:

Assume monochromatic, coherent, linearly polarised light with polarisation vector $\vec{x}$ is directed towards a double slit. By definition $\vec{x}$ is the direction of the vector of maximal electric (magnetic) field intensity associated with each photon. Let $\vec{y}$ be the direction of maximal magnetic field intensity of photon. Both $\vec{x}$ and $\vec{y}$ are perpendicular to the direction of propagation.

A quarter-wave plate is placed behind each slit, perpendicular to the direction of propagation. The fast axes of the plates behind slits 1 and 2 have directions $\vec{u}\equiv\frac{1}{\sqrt{2}}(\vec{x}+\vec{y})$ and $\vec{v}\equiv\frac{1}{\sqrt{2}}(\vec{x}-\vec{y})$ respectively, so that $\vec{x}=\frac{1}{\sqrt{2}}(\vec{u}+\vec{v})$ and $\vec{y}=\frac{1}{\sqrt{2}}(\vec{u}-\vec{v})$. The effect of the plates is to impose a half-period phase delay of on the component of the polarisation vector aligned with the slow axis. This circularly polarises photons passing through the two slits, in opposite directions.

Because the light is coherent, every photon that has been emitted but not yet entered the plates has a maximum electric field in its polarisation direction at the same time. Let the electric and magnetic field intensity of all photons, in the period between emission and striking a plate, be given by $A\sin kt$. That is, the electric and magnetic field vectors are $\vec{x}A\sin kt$ and $\vec{y}A\sin kt$ respectively.

Then, for a photon that has passed through the first slit, the electric field at time $t$ has components $A\sin kt$ and $A\sin(kt-\frac{\pi}{2})=-A\cos kt$ in directions $\vec{u}$ and $\vec{v}$ respectively. Hence the electric field after passing the plate has constant strength $A$ and the direction steadily rotates as $\vec{u}\sin kt-\vec{v}\cos kt$. The normalised polarisation vector is $\vec{p}_1(t)\equiv\frac{1}{\sqrt{2}}(\vec{u}\sin kt-\vec{v}\cos kt)$.

For a photon that has passed through the second slit,the electric field at time $t$ has components $-A\cos kt$ and $A\sin kt$ in directions $\vec{u}$ and $\vec{v}$ respectively. Again the electric field after passing the plate has constant strength $A$ and rotates with direction being $-\vec{u}\cos kt+\vec{v}\sin kt$. The normalised polarisation vector is $\vec{p}_2(t)\equiv\frac{1}{\sqrt{2}}(-\vec{u}\cos kt+\vec{v}\sin kt)$.

If the photon reaches the far field screen at time $t$, it has electrical field of strength $A$ pointing in direction $\vec{p}_1(t)$ or $\vec{p}_2(t)$ according to which slit it passed through.

The time taken for a photon to reach a position $\vec{r}$ on the screen will depend on the distance from the slit it has passed through to the point $\vec{r}$. For every point $\vec{r}$ there will be a difference $d(\vec{r})$ by which the distance travelled via the first slit exceeds the distance travelled via the second slit, to that point. The difference may be negative. Hence there will be a difference $\theta(\vec{r})\equiv\frac{k\,d(\vec{r})}{c}$ in the phases of (sine and cosine functions of) photons reaching that point.

Hence, at time $t$, the electrical field at point $\vec{r}$ on the screen will be (ignoring the diminution of intensity by distance): \begin{align*} \frac{A}{\sqrt{2}}(\vec{u}\sin kt&-\vec{v}\cos kt) + \frac{A}{\sqrt{2}}\big(-\vec{u}\cos (kt+\theta)+\vec{v}\sin (kt+\theta)\big)\\ &=\frac{A}{\sqrt{2}}\big( \vec{u}\sin kt-\vec{v}\cos kt\\ &\ \ \ \ \ \ \ \ -\vec{u}(\cos kt\cos\theta-\sin kt\sin\theta) +\vec{v}(\sin kt\cos\theta+\cos kt\sin\theta) \big)\\ &=\frac{A}{\sqrt{2}}\big( \vec{u} (\sin kt -\cos kt\cos\theta+\sin kt\sin\theta)\\ &\ \ \ \ \ \ \ \ +\vec{v}(-\cos kt +\sin kt\cos\theta+\cos kt\sin\theta) \big)\\ \end{align*} Since $\vec{u}$ and $\vec{v}$ are orthogonal, the magnitude of this is: \begin{align*} \frac{A^2}{2}\big( (&\sin kt -\cos kt\cos\theta+\sin kt\sin\theta)^2 +(-\cos kt +\sin kt\cos\theta+\cos kt\sin\theta)^2\big)\\ &\propto \sin^2 kt +\cos^2 kt\cos^2\theta+\sin^2 kt\sin^2\theta -2\sin kt\cos kt\cos\theta\\ &\ \ \ \ \ -2\cos kt\sin kt \cos\theta\sin\theta +2\sin^2 kt\sin\theta\\ &\ \ \ \ \ +\cos^2 kt+\sin^2 kt\cos^2\theta+\cos^2 kt\sin^2\theta -2\cos kt\sin kt\cos\theta\\ &\ \ \ \ \ +2\sin kt\cos kt\cos\theta\sin\theta -2\cos^2 kt\sin\theta\\ &=2 -4\cos kt\sin kt\cos\theta +2(\sin^2 kt-\cos^2 kt)\sin\theta\\ &=2 -2\sin 2kt\cos\theta -2\cos 2kt\sin\theta\\ &\propto 1-\sin(2kt+\theta) \end{align*} This will vary sinusoidally by $\vec{r}$ and hence will give an interference pattern.

What have I done wrong? There should be no interference pattern because the photons have been marked with which-way information by the differently oriented plates behind the two slits. Thank you in advance for any help.

You're missing the very last step: averaging the intensity over time. Since there's a $t$ inside the argument of the sinusoid, the contribution of that term is zero.
Averaging over time is always done in these kinds of problems, though often implicitly. For example, if you just have one slit, the electric field might be proportional to $\cos(\omega t)$, in which case the instantaneous intensity would be proportional to $\cos^2(\omega t)$. But nobody ever says that since we see on timescales much longer than $1/\omega$. We thus average over a full oscillation, getting a constant.