# Understanding the quantum eraser from a quantum information stand point

Is there a way to understand the quantum eraser experiment in terms of quantum information? In particular, is there a quantum circuit that would function as a quantum eraser experiment?

The issue I'm having is that the erasure (and the detection for that matter) does not appear to be a Unitary operator?

• – PyRulez Jan 16 '16 at 15:43
• – PyRulez Jan 16 '16 at 15:44

The answer is structured as follows:

1. I will first give the quantum circuit corresponding to a normal double slit (or interferometer),
2. then the circuit where the which-way information has been recorded,
3. a circuit where the which-way information is first recorded and then erased in a unitary way,
4. and finally a circuit where the which-way information is erased through a coherent measurement (postselected upon the measurement outcome), which can be done before or after ("delayed choice") the photon is detected; this is commonly referred to as the "quantum eraser".

## I. Interference without which-way information

An interferometer corresponds to the following circuit: Here, the first $H$ puts the qubit in the superposition of both paths/slits -- the states $\vert0\rangle$ and $\vert1\rangle$ correspond to the two paths/slits --, the matrix $$\Lambda = e^{i\phi/2}\vert0\rangle\langle0\vert + e^{-i\phi/2} \vert1\rangle\langle1\vert$$ introduces a phase shift between the two paths (this can e.g. be a function of the position on the screen, or the relative length of the interferometer arms), and the second $H$ makes the two paths interfere. If the output qubit is measured in state $\vert0\rangle$, the interference is constructive, otherwise desctructive.

You can easily check that this yields an interference pattern which varies like $\mathrm{prob}(0)=\cos(\phi)^2$.

## II. Which-way information destroys interference

Now imagine that we want to copy the which-way information: Then, what we do is this is, we copy the which-way information before traversing the interferometer onto qubit $c$. It is easy to see that this destroys the interference pattern, i.e., $\mathrm{prob}(0)=1/2$.

## III. Unitary erasure of which-way information

Finally, if we want to erase the which-way information again after traversing the interferometer in a unitary fashion, what we do is It is again easy to see that (since $\Lambda$ is diagonal) this undoes the effect of the first CNOT (i.e., erases the which-way information), and thus, we will observe the interference pattern again.

## IV. Quantum eraser and delayed choice experiment

The scheme of part III is not what is commonly denoted as a quantum eraser (thanks to Emilio Pisanty for the clarification.) In a quantum eraser, we want to erase the which-way information by measuring it in a basis which reveals no information about the path. The advantage is that this measurement thus commutes with the detection of the photon, and thus can be carried out much after the detection of the photon ("delayed choice"). The disadvantage is that we only recover our interference pattern if we condition on the successful erasure of the which-way information.

The procedure is described by the following circuit: Here, we copy the which-way information onto the qubit $c$ before traversing the interferometer. After the interferometer, the $H$ makes the two paths interfere, and we measure. Note that at this point this is nothing but the procedure of part II and thus, $\mathrm{prob}(0)=1/2$. (In particular, there is no interference whatsoever observed.)

Now we measure the which-way qubit. There are two ways of measuring it:

### 1. Learning the which-way information

Measurement in the $\vert0\rangle$, $\vert1\rangle$ basis reveals the which-way information. Going through the math, you can easily see that in this case, the distribution for $q$ is $\mathrm{prob}(0)=1/2$ for both outcomes $c=0$ and $c=1$.

(Note that the state before the measurement on $q$ does depend on the outcome of $c$ -- you can understand this by either changing the order of the two measurements, or by saying that the state before the measurement is entangled.)

### 2. Erasing the which-way information

Let us now instead measure $c$ in the $\vert+\rangle$, $\vert-\rangle$ basis. If we obtain outcome $\vert+\rangle$, we have effectively erased the which-way information. This can be understood by moving the measurement all the way to the CNOT, and noting that a CNOT in this setup, followed by a projection onto the $\vert+\rangle$ state on the target qubit, corresponds to the identity on the control qubit.

If we obtain $\vert+\rangle$, the (conditional!) probability distribution on $q$ will be thus $\mathrm{prob}(0)=\cos(\phi)^2$.

What happens if we obtain $\vert-\rangle$? In this case, one can easily check that the CNOT + projection yields a Pauli $Z$, i.e., there is an additional phase shift of $\pi$ introduced between the two paths of the interferometer. This is, in this case, we will still have interference, but the interference pattern will be shifted by half a spacing, i.e., $\mathrm{prob}(0)=\sin(\phi)^2=1-\cos(\phi)^2$.

So what does this teach us? As long as we don't know the measurement outcome on the $c$ qubit, we will see the average of both interference pattern, which is nothing but $$\tfrac{1}{2}\Big[\sin(\phi)^2+\cos(\phi)^2\Big]=\frac{1}{2}\ ,$$ and thus just the same as if we would not have measured $c$, or measured in the which-way basis. This, of course, makes a lot of sense -- observations on $q$ should not depend on whether (and how) $c$ has been measured.

Thus, in order to reveal the interference pattern when erasing the which-way information, we have to carry out the experiment repeatedly (a single run will clearly not allow to determine $\mathrm{prob}(0)$ or to see interference fringes). Once we have done so, we need to take all outcomes (clicks) on $q$ where the erasure works (i.e., where we obtained $\vert+\rangle$ on $c$) and look only at those, i.e., the conditional probability distribution $\mathrm{prob}(q=0|c=+)$. Once we do so, but only then, we will see an interference pattern. Nothing magic going on here.

I will refrain from interpretations here, but depending on one's personal interpretation, this might have some implications what the photon "does" when passing the double slit.

(Circuits created with qasm2circ.)

(Circuits may be simulated here.)

• you could use quirk to provide interactive versions of the quantum circuits. For example, your first circuit is this, while your second is this. It shows nicely how the probabilities change in the different cases, and encourages experimentation. – glS Dec 12 '17 at 21:02
• @glS I'm not a bit fan of quirk. Too much moving, and I usually don't understand what it is telling me. But feel free to write another answer with animated circuits. – Norbert Schuch Dec 12 '17 at 23:58