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This experiment combines the previous two: a double-slit setup with polarizers.
More precisely, we cover slit A with a horizontal polarizer and slit B with a vertical polarizer, the two polarizers are perpendicular. A third polarizer, which can be rotated, is placed after the slits so it affects photons from both paths.
Here is what we observe:
- When the third polarizer is set to 0° or 90° (aligned with one of the slit polarizers): no interference, the pattern looks like single-slit diffraction.
- When the third polarizer is set to 45° or -45° (diagonal): interference returns, we see the characteristic fringes of a double-slit pattern.
Why does this happen? The perpendicular polarizers on the slits "tag" each path: a horizontally polarized photon must have come from slit A, a vertically polarized one from slit B. This is called which-way information. When which-way information is available, even if we do not actually look at it, interference disappears.
The diagonal polarizer erases this tagging. It projects both paths onto the same polarization state, making them indistinguishable again. With no way to tell which slit the photon passed through, interference returns. This is the quantum eraser effect.
Why use polarizers instead of, say, placing a detector at one slit? A detector would absorb the photon entirely. Polarizers let most photons through, they interact with the photon just enough to tag it, without destroying it.Without spoiling the next experiment, polarizers are not the only way to interact with a photon without absorbing it. We will discover a second method in Experiment 4.
Here is an animation of Experiment 3. The light coming out of the laser is diagonally polarized. The polarizers can be rotated using the pink handles, and the slit parameters can be adjusted in the control panel. The dark curve shows the maximum diffraction envelope, and the purple curve shows the result of the current setup.
Curious of what our setup looks like? Here is a few pictures of what we have:
Here is a representation of the setup shown in the picture. All of the equipment is displayed, but the scaling is not proportional.
The raw data can be downloaded here:
To recreate a pattern, an actuator was moved along the beam. All data files have four columns. The first two are the angles of the polarizers, but only P1 is used. The third is the distance the actuator has travelled in $\mu m$%\unit{\micro\meter}.
The last column is the number of counts on D1, the detector. All acquisitions are for a two-second window.
Let us analyze these graphs. First, at 0 degrees and 90 degrees, there is no interference. By placing polarizer P1 at the same angle as one of the slit polarizers, we know for certain through which slit the photon passed. It behaves like a single-slit experiment. The patterns at 0 and 90 degrees show a slight shift relative to each other. This shift represents the spacing between the slits. Since each slit acts as a single slit, each produces a diffraction pattern centered in front of it.
For the graphs at 45 degrees and -45 degrees, interference is clearly visible, but the two patterns differ. The 45-degree graph has the same shape as the one from Experiment 2, with three large peaks (the pattern is not perfect — there is actually a phase offset in our polarization setup, which will be corrected in the next experiment). Upon closer inspection, the pattern at -45 degrees is the inverse of that at 45 degrees: all maxima become minima and vice versa. Summing both plots recovers the full diffraction envelope. (The same phenomenon occurs with the 0 and 90 degree graphs — adding them yields the same diffraction pattern, though less striking since they already showed diffraction individually.)
Why are the two diagonal plots inverted? There are two explanations. First, polarizers block a portion of the photons. Perpendicular polarizers block complementary portions, so summing them recovers all the photons.
In the first experiment, we learned how to describe a photon's polarization and how polarizers change it. In the second experiment, we learned how to describe a photon's position, like being at the source, at slit A, or at slit B, and how interference arises from superpositions of paths.
In this experiment, we put those two aspects together. A photon has both:
- A polarization
- A position in the setup
To write this compactly, we simply list them side by side. For example, the initial state of a diagonal photon located at source $S$ is denoted as follows:
That is all we need — every state will now be described by polarization and position.
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Step 0. Initial state. We start with the photon prepared in a diagonal state at the source:
Step 1. Double slit. At the double slit, the photon can go through slit $\ket{A}$ or $\ket{B}$. We ignore photons that hit the wall ($\ket{\text{lost}}$) and keep only the ones that pass. The distance from the source to each slit is $l$, so:
In the above, all we are doing is distributing $\ket{\diag}$ to $\ket{A}$ and $\ket{B}$.
Step 2. Polarization. Now comes the crucial step, as each slit has a polarizer. Path $\ket{A}$ has a polarizer $P_{\leftrightarrow}$, and path $\ket B$ has a polarizer $P_{\updownarrow}$:
In the above algebra, on the second line, we simply express $\ket{\diag}$ in terms of $\ket{\updownarrow}$ and $\ket{\leftrightarrow}$. On the third line we then apply $P_{\leftrightarrow}$ and $P_{\updownarrow}$ to the terms representing $\ket{\diag}$. Due to the action of polarizers, we lose~$\ket{\updownarrow}$ for $\ket{A}$ and $\ket{\leftrightarrow}$ for $\ket{B}$ respectively. On lines four and five we simplify and combine the terms to get a cleaner equation.
So after the slits, the surviving photons are either horizontal from $\ket{A}$ or vertical from $\ket{B}$. The terms $\ket{\text{lost}_{P_A}}$ and $\ket{\text{lost}_{P_B}}$ are just there to account for photons that were lost due to polarization. $\ket{\text{lost}_{P_A}}$, for example, describes the photons that tried to go through slit $\ket{A}$ but were blocked by its horizontal polarizer (because their polarization was not aligned). The same applies to $\ket{B}$ in a vertical manner with $\ket{\text{lost}_{P_B}}$.
Step 3. Reaching the detector. After the photons pass the slits, they propagate toward the detector. In reality, they can arrive at many points along the vertical axis, but let us focus on one arbitrary position, denoted by $\ket{y}$. The distance from slit $\ket{A}$ to this point is $a$ so the phase factor is $e^{ika}$, and from slit $\ket{B}$ it is $b$, so the phase factor is $e^{ikb}$. In the algebra below, we show how this appears in practice, and factor out $\ket{y}$, which will be useful later on. The "$\cdots$'' before and after the terms in the definition of $\ket{\psi_3}$ indicates that we are only showing one contribution, as the photons can hit many other places than $\ket{y}$, so there is a whole collection of similar terms. But given we only focus on an arbitrary position $\ket{y}$ here as an example, that is the only one we show.
Step 4. Calculating the intensity. To predict what the detector sees, we calculate the \textit{intensity} at $\ket{y}$. Intensity is the brightness at a spot on the detector, or more technically the likelihood of finding a photon there. Intensity comes from taking the squared magnitude of the total amplitude.Here we have two terms in superposition:
To find the intensity, we compute the squared magnitude of each amplitude. But here is the key point: because $\ket{\leftrightarrow}\ket{y}$ and $\ket{\updownarrow}\ket{y}$ are orthogonal states, they cannot add up or cancel each other — regardless of their phases. Their contribution to the intensity will invariably be the sum of each term's squared amplitude:
This is the mathematical reason why marking the paths with perpendicular polarizers destroys interference: the two terms live in orthogonal states, so they can never combine. We simply get the sum of two independent contributions.
So the resulting pattern is simply a flat constant at half the brightness of the ordinary double slit. The reason is that the two polarizations are orthogonal: the kets can no longer factorize in a way that lets the complex exponentials add up or cancel each other.
In equation form, the observed distribution is just the single-slit diffraction envelope scaled down by a factor of $1/2$:
where $\alpha = \frac{\pi w \sin\theta}{\lambda}$, with $w$ being the slit width and $\theta$ the angle from the center.
This is consistent with the intuition: each slit still produces its own diffraction pattern, but because their polarizations are perpendicular, the detector cannot "add'' them coherently. Instead, we only see the sum of two independent single-slit patterns, resulting in half the overall intensity and no interference.
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We now add a third polarizer P$_1$ after the slits. The result depends entirely on which basis we use to express the photon's polarization.
The state before P$_1$
Just before the third polarizer, the photon is in the state:
The two polarization states $\ket{\leftrightarrow}$ and $\ket{\updownarrow}$ are orthogonal. This is why there is no interference: the terms cannot combine.
Choosing a basis
Here is the key insight: the same state can be written in different bases.
In the H/V basis (horizontal/vertical):
The two terms have different polarizations $\rightarrow$ orthogonal $\rightarrow$ no interference.
In the diagonal basis (using $\ket{\leftrightarrow} = \frac{1}{\sqrt{2}}(\ket{\diag} + \ket{\adiag})$ and $\ket{\updownarrow} = \frac{1}{\sqrt{2}}(\ket{\diag} - \ket{\adiag})$):
Now each polarization state has both phase terms $\rightarrow$ interference is possible.
What P$_1$ does
The third polarizer selects one term from this superposition:
where $\Delta = a - b$ is the path difference.
Why does 45° restore interference?
At $0°$ or $90°$, the polarizer selects photons from one slit only. We know which path was taken $\rightarrow$ no interference.
At $45°$, the polarizer projects both paths onto the same final polarization $\ket{\diag}$. A photon that survives could have come from either slit — we cannot tell. The which-way information has been erased.
This is the quantum eraser effect.
Complementary pairs
The $45°$ and $-45°$ patterns are opposite: maxima $\leftrightarrow$ minima. Adding them:
This recovers exactly the pattern from before adding P$_1$. Same for $0°$ + $90°$:
The third polarizer does not create or destroy photons. It picks a decomposition: a way of splitting the photons into two complementary groups that always sum to the same total.
If you want to check out papers about this experiment, here are the two best ones we have found:
If you want a more interactive ressource, here are some Youtube videos we found to be accurate and insightful.
