# I Technical question about modern Double Slit w/ one photon

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1. Apr 4, 2016

### DavidReishi

My question is about the experiment in which detectors are used behind the slits to determine which slit the photon goes through. Specifically, it's about the detectors themselves. What I found is that these detectors are some kind of photoelectric detectors. My question is this. During the single-photon experiment, is the detector thought to interact with the light (for example, as in the photoelectric effect), and if not, how is it that the detector actually detects the passage of the photon, that is, without interacting with it?

2. Apr 4, 2016

### DrChinese

There are a couple of ways this is done. The simplest is to block one side completely of course.

But the way that is more interesting manner is to place polarizers in front of the slits. If they are oriented parallel, there is interference. If they are oriented perpendicular, there is no interference. Obviously if a photon goes through only 1 slit, the orientation of the other slit should not matter. The idea is described below.

http://arxiv.org/abs/1110.4309

Unfortunately, I do not have a reference for execution of this experimental setup. Does anyone else?

3. Apr 4, 2016

### Strilanc

Even in the ideal case, the detectors create an entangled copy of information defining the light's position. This counts as an interaction.

(The detector also amplifies the entangled copy of the information into a full-blown thermodynamically-irreversible measurement.)

Here's a diagram of a circuit showing the effects of the operations on the state:

The top wire represents the position of the photon (which slit it went through). The bottom wire represents the detector state. Both can be in two states, which we call Off and On or equivalently $|0\rangle$ and $|1\rangle$.

Going through the slit is emulated by a Hadamard gate, which changes the top wire's state from $|0\rangle$ to $\frac 1 {\sqrt 2} |0\rangle + \frac 1 {\sqrt 2} |1\rangle$. The position state is in superposition.

The detector is emulated by a Controlled-Not gate. It copies the position information onto the second wire, but the copy is entangled not independent. The system ends up in the state $\frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |11\rangle$. This is observationally distinguishable from the previous state, using only the first qubit, because the density matrix you get when you trace out the second qubit has gone from $\frac 1 2 \begin{bmatrix} 1&1\\1&1 \end{bmatrix}$ to $\frac 1 2 \begin{bmatrix} 1&0\\0&1 \end{bmatrix}$.

The result is that the qubit no longer self-interferes when you rotate it; it's stuck in the $\frac 1 2 \begin{bmatrix} 1&0\\0&1 \end{bmatrix}$ state at the center of the Bloch sphere. In practice that shows up as the interference pattern disappearing.

Last edited: Apr 4, 2016
4. Apr 4, 2016

### DavidReishi

Thanks for the link. I found the article indecipherable, but I dowloaded the pdf and'll give it another attempt.

Do you mind clarifying this? Are you saying that the simplest way to find out which slit the photon went through is to block one slit, and then see if it went through the other?

5. Apr 4, 2016

### DavidReishi

I really appreciate your writing everything out the way you did. Finding it very difficult, I read it four or five times. I hope you'll indulge a couple questions I have even if those questions end up demonstrating nothing other than my remarkable ignorance of the matter.

Would it be fair to say that implied in these words is the observation that, whether speaking of the photon or the entangled copy, the measurement involved a transfer of energy, i.e. the way energy is transfered when a photon is absorbed?

This observationally distinguishable change in the state of the system, does it correspond to a change in the state of an electron having taken place? If so, where? Somewhere in the detector? Or does it have nothing to do with an electron changing state?

6. Apr 4, 2016

### DavidReishi

Another question... I'm unclear, is the detector actually in the path of the photon? I figured it was kind of off to the side, like an audience member watching a tennis match. But now I wonder if that's wrong. Is the photon directed at the detector which reflects the light to the back plate at the same time as detecting it?

7. Apr 4, 2016

### Strilanc

No, that's not implied.

In quantum mechanics, energy is a complicated concept related to the eigenvalues of a matrix that defines how the system as a whole changes over time. It may not be associated with individual parts of the system. I recommend focusing on the information instead of the energy.

Quantum information can be embodied in many different ways, but it follows the same rules regardless. Whether it's the spin of an electron or the polarization of a photon isn't really relevant. What's relevant is the fact that there's a copy; that the system as a whole went from $|00\rangle + |01\rangle$ to $|00\rangle + |11\rangle$.

When you ignore the system as a whole, and focus on just one of the parts, some of the picture can be missing. Before the photon has hit the detector, you wouldn't be missing any of the picture by ignoring the detector. Afterwards, you are. That's the difference.

Yes, it's actually in the path of the photon. But the detector doesn't have to reflect the photon. For example, the detector could work by passing the photon through an SPDC crystal and directing one of the resulting two entangled photons into a photomultiplier. That will lower the frequency of the photon, and make its path deviate a bit, but otherwise leaves it unmolested.

In the ideal case there's no interaction at all except for the creation of an entangled copy.

8. Apr 5, 2016

### DavidReishi

I assure you I didn't ignore the system as a whole. I took what you wrote, read it four or five times, and then, to the small degree that I understood it, accepted it as true. But also, realize that I came to the problem with a particular question in mind...though I'm aware now that I asked it in a wrong way. What I wanted to know is whether, once through the slit, the photon undergoes absorption and a new photon is emitted, or whether it's the originally emitted photon that hits the back plate.

And I believe you led me to the answer. The SPDC. Thank you. One more question if you don't mind. Is it true that the SPDC crystal's efficiency at causing an entangled pair is extremely low, like one pair for every hundred-gazillion incoming photons, and if so, how exactly is it used with the one-photon-at-a-time experiment?

9. Apr 5, 2016

### StevieTNZ

Most photon detectors absorb the photon, so nothing hits the back plate. But I do believe there are photon detectors out there that note the presence of a photon but not absorb it

10. Apr 5, 2016

### vanhees71

This way to imprint the which-way information on the photon itself via polarizers in the slits has been famously used by Walborn et all in the quantum-erasure experiment:

http://journals.aps.org/pra/abstract/10.1103/PhysRevA.65.033818