# State vector after quantum eraser

Suppose I have a particle in a two-slit setup. At the slits, it's in state

$|\psi\rangle = |x_1\rangle + |x_2\rangle$

I entangle it with some detector:

$|\psi'\rangle = |x_1\rangle|d_1\rangle + |x_2\rangle|d_2\rangle$

Then, using a "quantum eraser," I remove the detection information. If I understand correctly, the state more or less goes back to

$|\psi\rangle = |x_1\rangle + |x_2\rangle$

Or, written in the joint state space:

$|\psi'\rangle = (|x_1\rangle + |x_2\rangle)\otimes|random\rangle$

Is this really all there is to it?

For example, I understand that in the two slit experiment we've erased information from detectors at the slits. We've also used polarizing filters with photons, and erased polarization information. These have different physical implementations.

What if I turned the latter experiment on its head? Suppose a particle encounters two slits. On the left there's a left-circular (|$L\rangle$) polarizer and on the right there's a right-circular one (|$R\rangle$). Call the which-path states $|l\rangle$ and $r\rangle$, so that we have:

$|\psi'\rangle = |L\rangle|l\rangle + |R\rangle|r\rangle$

If things are as simple as they seem, then after erasing the which-path info, I should get:

$|\psi'\rangle = (|L\rangle + |R\rangle)\otimes|random\rangle$

$|\psi\rangle = |L\rangle + |R\rangle = |H\rangle$

That is, it should come out horizontally polarized. Is this guaranteed? Or can the effect of "erasure" be more interesting somehow?

Erasure doesn't work the way you think it does. Basically it goes like this...

The most obvious way to 'erase' entangled information is to just undo the operation that created it:

Notice how the top qubit has been restored to a pure state by the end, despite that nasty entanglement in the middle.

But you can get a bit fancier than just repeating the same operation twice. Why not do a rotation (H) that swaps the X and Z axes first? Then we can use a controlled-Z instead of a controlled-X to unmake the entanglement:

This is convenient because a Z gate only applies to parts of the state where the qubit is On. That is to say, it acts like a control on itself. So you can freely swap it with its own controls:

And now that we have an operation that only depends on the second qubit, we can apply the deferred measurement principle. Measured controls are equivalent to quantum controls, as far as the other qubits are concerned anyways:

But really this whole 'controlled operation' business is kind of annoying, right? Why not just hope that the measurement result is Off, so that we don't have to apply the operation? If it happens to come out On, we'll just throw away that run and try again:

And finally, if we're going to be post-selecting anyways, why not just post-select directly on the X axis instead of swapping X and Z then post-selecting on the Z axis as a way to indirectly post-select on X?

And that last circuit is what people call quantum erasure.

In other words... Create a pair of entangled photons $|00\rangle + |11\rangle$, check if they're both in the $|0\rangle + |1\rangle$ state by running one of them through a diagonal polarizing filter (which measures the polarization and secretly throws out the $|0\rangle - |1\rangle$ outcomes), and... my gosh when the one photon makes it through the filter the other photon is in the $|0\rangle + |1\rangle$ state instead of an entangled state! And if we use a horizontal polarizer instead, they keep ending up in the $|0\rangle$ state! It's almost as if measurements on one predicts the other!...

The important thing to realize here is that erasure requires an interaction between the two entangled qubits. That interaction can be blatant, as in the first circuit, or kind of subtle, as in the last circuit where we throw away whole runs after the fact but don't really talk about how this requires coordination and ultimately is just an obfuscated conditioning of the first qubit on the second qubit.

Does that clear things up?

• Thanks! I think I follow. Perhaps my question now becomes: is there a way to physically realize your first circuit diagram in the case of my setup with opposite circular polarizing filters at slits? Or more broadly: in what circumstances can we set up an experiment where the erasure doesn't require postselection?
– A_P
Commented Jun 25, 2016 at 22:01
• @monk If you don't want to filter out runs (i.e. post-select), you should probably avoid using filters. If you want to do complicated quantum things with optics, review the Wikipedia article on linear optical quantum computing. Commented Jun 26, 2016 at 2:19
• Then perhaps put a QWP on the left and a 3/4 WP on the right (and start with all H-polarized photons)? Thanks for the reference. I read it, but understood very little. What I'm trying to do is come up with an experimental setup where a photon appears to be L-polarized in one "path" (for some definition of path) and R-polarized in another, but then always passes an H filter. BTW, maybe you know the answer to this question too? physics.stackexchange.com/questions/263821/…
– A_P
Commented Jun 26, 2016 at 22:18
• @monk If you want the photon to always pass an H filter, all paths need to end up H polarized. You can have deviations along the way, like temporarily rotating the top path's polarization to vertical but then back to horizontal, but in the end all cases will have to be horizontal or else some of them won't make it through the filter. It's also worth noting that, even if you've entangled the polarization with the path, just operating on the path won't be enough to get rid of the vertical polarization (not counting post-selection, of course). It would break reversibility. Commented Jun 26, 2016 at 22:26

The Mach-Zehnder interferometer version of the quantum eraser actually goes like this.

In step 1, we put a polarizer in the upper path:

The state just before the second beam-splitter is $$\frac{1}{\sqrt{2}}\begin{pmatrix}e^{i\theta}\\-1\end{pmatrix}$$. The beam-splitter action is given by $$\begin{bmatrix}1&-1\\1&1\end{bmatrix}$$, which results in $$\frac{1}{2}\begin{pmatrix}e^{i\theta}+1\\e^{i\theta}-1\end{pmatrix}$$. Notice how it "mixes" the two states. And since $$|\frac{e^{i\theta}+1}{2}|^2 = \frac{1+cos \theta}{2}$$, we get interference at the detectors.

In step 2, we introduce polarizers before the phase shifter:

There are two paths (upper/lower) and two polarizations (vertical/horizontal). Using the basis $$|UV\rangle, |UH\rangle, |LV\rangle, |LH\rangle$$, the state before $$BS_2$$ is now $$\frac{1}{2}\begin{pmatrix}0\\e^{i\theta}\\-1\\0\end{pmatrix}$$. Notice how the only component of the upper path belongs to the horizontal polarization, and vice versa for the lower. This means that $$BS_2$$ (now given by $$\begin{bmatrix}1&-1\\1&1\end{bmatrix} \otimes I$$) won't mix them: $$\frac{1}{2\sqrt{2}}\begin{pmatrix}1\\e^{i\theta}\\-1\\e^{i\theta}\end{pmatrix}$$. $$D_1$$ is as likely (1/8) to get a V or H photon.

In step 3, we introduce a 45$$^{\circ}$$ polarizer after $$BS_2$$:

It behaves as follows: $$\begin{bmatrix}\frac{1}{2}&\frac{1}{2}&0&0\\\frac{1}{2}&\frac{1}{2}&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$

It will "mix" the V and H components of the upper path, turning it into:

$$\frac{1}{2\sqrt{2}}\begin{pmatrix}\frac{e^{i\theta}+1}{2}\\\frac{e^{i\theta}+1}{2}\\-1\\e^{i\theta}\end{pmatrix}$$

Once again, we see interference. Note that we could have introduced a similar polarizer before the other detector, turning it into the unentangled state $$\frac{1}{4\sqrt{2}} \begin{pmatrix}e^{i\theta}+1\\e^{i\theta}-1\end{pmatrix} \otimes \begin{pmatrix}1\\1\end{pmatrix}$$.