My understanding is that in a double-slit experiment, quantum interference disappears if which-path information is available. How is available defined? Consider the following experiment:

SPDC is used to create an entangled pair of photons. The signal photon goes through a double-slit with a detector behind it. The idler photon hits the wall of the laboratory. Is which-path information available? After all, theoretically the information carried by the idler could be reconstructed from careful measurement of the wall's properties. In such a case is interference observed? How "available" must which-path information be?

  • $\begingroup$ You cannot extract information from the wall. In quantum mechanics you must always specify the complete arrangement you are interested in working with, so that you can calculate probabilities without ambiguity. If you wish to extract where the photon hit the wall, you are free to use a photographic plate. If you are interested in the momentum you can use something else, like a diffraction grating or a spectrometer. But Without specifying the full arrangement, you cannot proceed to make inferences. In general you cannot reconstruct. $\endgroup$
    – Prathyush
    Apr 2, 2013 at 14:13
  • $\begingroup$ If you specify exactly what you wish to do with the idler, we can analyse this setup carefully. Otherwise, This is just a normal interference experiment. $\endgroup$
    – Prathyush
    Apr 2, 2013 at 14:17
  • $\begingroup$ @Prathyush, Suppose you use a photographic plate to detect the idler photon's momentum vector. Then the signal photon's interference pattern disappears, correct? However the wall is a primitive form of photographic plate. You could in principle deduce information about the idler photon from the wall after the fact, no? It has correlated itself with the environment, no? $\endgroup$
    – user1247
    Apr 2, 2013 at 14:34
  • $\begingroup$ Ok lets use a photographic plate. The reason that insist is that different apparatus analyze/measure different pieces of information about the idler photon. I am not sure about the answer myself. How do you wish to infer which hole information from the position of the idler photon? $\endgroup$
    – Prathyush
    Apr 2, 2013 at 14:57
  • $\begingroup$ @Prathyush, Since the momentum of the idler+signal photon is conserved, knowing the idler photon's position gives position information about the signal photon (by extrapolation from the momentum vector). $\endgroup$
    – user1247
    Apr 2, 2013 at 15:32

4 Answers 4


This question actually has a very easy and rigorous answer. Having which-path information "available" is just a crude way of saying that the system is correlated with anything else. Usually, this is because the system has been decohered in whatever basis corresponds to the possible paths, which is usually position basis. In your case, the photon is actually never put into a coherent local superposition, and so interference will not be seen. Instead the SPDC process essentially creates a Bell state where one photon is thrown away. Skematically, the situation you describe is as follows. The splitting process is

$\vert S \rangle \otimes \vert I \rangle \to \frac{1}{\sqrt{2}} \Big [ \vert S_L \rangle \otimes \vert I_L \rangle + \vert S_R \rangle \otimes \vert I_R \rangle \Big ] = \vert \psi \rangle \qquad \qquad \qquad (1)$

where $S$ and $I$ stand for the signal and idle photons, respectively, and $L$ and $R$ stand for the left and right path. The reduced state of the signal photon is

$\rho^{(S)}=\mathrm{Tr}_I\Big[\vert \psi \rangle \langle \psi \vert \Big]$

(If you don't know what $\mathrm{Tr}$ means, or what a density matrix is, you absolutely must go learn about them. It doesn't take that long, and is crucial for understanding this question.) The measurement performed by the apparatus is essentially a measurement in the basis $\{ \vert \pm \rangle = \vert S_L \rangle \pm \vert S_R \rangle \} $. Here, getting a "plus" result in the laboratory means seeing the photon near a peak on the screen, and a "minus" result is seeing it in a trough.

You can check that measuring $\rho^{(S)}$ in the $\{ \vert \pm \rangle \}$ basis (or, in fact, any basis at all) gives equal probability of either outcome. This means no interference pattern, since photons are evenly spread over peaks and troughs. In particular, this is true no matter what happens to the idle photon; it could be carefully measured, or thrown away.

On the other hand, if you simply send the photon into a double slit experiment by sending it through a small hole and allowing the photon to enter either slit without being correlated with anything else, the evolution looks like

$\vert S \rangle \to \frac{1}{\sqrt{2}} \Big [ \vert S_L \rangle + \vert S_R \rangle \Big ] \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2)$

which doesn't involve a second photon that "knows" anything. In this case, a measurement in the $\{ \vert \pm \rangle \}$ basis gives "plus" with certainty (or near certainty), meaning we see an interference pattern because all (or most) of the photons only land at the peaks.

Finally, suppose we place a second particle like a spin-up electron in front of the right slit such that the electron's spin flips if and only if the photons brushes by it on the way through the right slit. In this case we'd get

$\frac{1}{\sqrt{2}} \Big [ \vert S_L \rangle + \vert S_R \rangle \Big ] \otimes \vert e_\uparrow \rangle \to \frac{1}{\sqrt{2}} \Big [ \vert S_L \rangle \otimes \vert e_\uparrow \rangle + \vert S_R \rangle \otimes \vert e_\downarrow \rangle \Big ] \qquad \qquad (3)$

Now, although nothing has really happened to the signal photon as it passed through the right slit--it doesn't, say, get slowed down or deflected--the electron now knows where the photon is. In fact, this state is identical to the first one we considered except with the electron in place of the idle photon. If we make a measurement on the signal photon, we now get either outcome with equal likelihood, meaning the interference pattern is lost.

The process of the electron getting entangled with photon is known as decoherence. (Note that we only use that word when the electron is lost, like it usually is. If the electron was still accessible and could potentially be brought back to interact again with the photon, we'd just say they had become entangled.) Decoherence is the key process, and plays a fundamental role in understanding how "classicality" arises in a fundamentally quantum world.


Make sure not to confuse two possible situations. The first is where the momenta of the idle and signal photon are correlated, and the slits are positioned to simply select for one of two possible outcomes, corresponding to equation (1) above:


The second is where the signal photon's spread over $L$ and $R$ is not caused by an initial event correlating it with idle photon, but simply by its own coherent spreading when it is restricted to pass through a small hole, corresponding to equation (2):


Note here that there is no violations of conservations of momentum, a subtle (for beginners) consequences of the infinite dimensional aspect of the photon's Hilbert space. (The fact that the two-slit experiment is the canonical example for introducing quantum wierdness is unfortunate because of these complications.) When the photon is confined to a small initial slit, it necessarily has a wide transverse momentum spread.

It might be helpful to concatenate these two cases:

enter image description here

Here, the idle photon is initially entangled with the signal photon, but the wall with the single slit destroys the signal photon for the $X/R_1$ outcome. When $Y/L_1$ happens, the signal photon can now be sent through 2 slits to produce and interference pattern. The idle photon's direction $X$ vs. $Y$ was correlated with the signal photon's $L_1$ vs. $R_1$, but it is never correlated with $L_2$ vs. $R_2$.

  • $\begingroup$ This looks like a really good answer, thanks! One big question that still bothers me is why, in the lone interfering photon example, it is not in actuality entangled in some way with the device used to create it. After all, the photon had to come from somewhere! Say you produce the photon from a laser -- then isn't at least one atom in the laser entangled with the photon, which would imply, by your logic, that the photon can never interfere!? $\endgroup$
    – user1247
    Apr 23, 2013 at 20:00
  • $\begingroup$ Good follow up question. The basic answer is that the natural, self-evolution of the photon will cause it to spread. In particular, it might start as a tightly-concentrated wavepacket (such as might be produced by a tiny hole in a wall at the top of this image: imgur.com/R8BFyUu ), but it will then spread out to have components near both L and R. This is a coherent superposition of different (position) eigenstate, and it is not entangled with anything else. $\endgroup$ Apr 23, 2013 at 20:22
  • $\begingroup$ Note that the photon story has subtleties. I've tried to talk about the two-dimensional subspace spanned by L and R, but of course the true photon Hilbert space is infinite dimensional. We ignore the many cases where the photon hits the wall between the slits and say that we've "renormalized" the wavefunction. But really the wall has just becoming entangled with photon, decohering it, and we are simply ignoring those cases where the photon is stopped and concentrate on the small component that passes through the slits (i.e. it lives in the L-R subspace). $\endgroup$ Apr 23, 2013 at 20:24
  • $\begingroup$ Wait, but I thought that was what I was talking about to begin with: the natural self-evolution of the photon causes it to spread and interfere. The idler photon is entangled with the signal photon, and so if the signal photon goes through the L slit, the idler photon must go through the corresponding R slit (by momentum conservation). Therefore the idler photon carries which-path information, and by your story there can be no interference even due to the natural self-evolution of the photon's wave function. So unless I am missing something your description is not consistent. $\endgroup$
    – user1247
    Apr 23, 2013 at 21:55
  • $\begingroup$ "The idler photon is entangled with the signal photon, and so if the signal photon goes through the L slit, the idler photon must go through the corresponding R slit (by momentum conservation)." You're confusing two different types of evolution. I'll explain in an edit to my original question. $\endgroup$ Apr 24, 2013 at 16:48

So, to be clear, my understanding of your setup is that you are doing SPDC in a noncollinear geometry, so you get photons entangled in transverse momentum, and you basically want to get the momentum of one photon from the other, by studying the wall.

To get interference, the momentum change must be indistinguishable in principle, not just practically. How could this happen? Well, the wall itself is also a quantum object, so if its two possible momenta from the photon are both within the uncertainty of its total momentum, it is not possible to distinguish the two cases.

In the case of this setup, really what you are suggesting is a quantum eraser experiment, in a way. While both photons exist, which-path information does too, but if the idler is absorbed in such a way that this information is no longer present in the object it interacted with, interference is re-established- at least in the formalism. Whether the wall preserves this information or not depends on its specific properties, but generically it would not. Especially once you consider the effects of finite temperature to redistribute the photon's energy and momentum across the atoms in the wall, such that each one only gets an incredibly tiny change to its state that is not distinguishable from its other interactions.

To compare this with a typical quantum eraser experiment, look here for example. Notice that they do their erasing with polarizers, but you can think of the polarizer itself as like a wall, when it is set at the correct angle to erase. After all, when light is changed by a polarizer it must also leave some tiny effect on the polarizer itself, but the recovery of interference fringes in their experiment (and many others) demonstrates that for a macroscopic object, any energy that's not deposited in a specially sensitive channel (like, say, a avalanche photodiode reaction) will in general erase quantum information.

(edit: this analysis is incorrect; see comments)


from the comments:

1) Every resource I can find (see wiki:en.wikipedia.org/wiki/Quantum_decoherence, for >example) disagrees with you, seeming to require interaction with the environment in order >to induce decoherence. And 2) If what you say were true then no photon would ever show >interference in a double slit experiment, since it is surely entangled in some way with a >particle in the past. It seems as though you are saying that the schrod. eq. does not apply >to entangled particles (wave funct diffusion -> interference)?

There is no inconsistency between what I am saying and what they are, but I have to be very careful to be clear about what I mean by 'decoherence,' and by 'environment.'

When you have two particles that are entangled, one can describe the system in terms of the possible measurements on the two objects together, or on one or the other individually. Looking at both objects together is more 'complete,' in the sense that it gives you all the information of individual measurements and the correlations as well, but on the other hand sometimes, like in the setup you've given, one of the particles is just being thrown away and you don't want to have to consider it.

Now, if you only have access to one of the objects, it turns out that object cannot be described by its own quantum state. Rather, you need to use the language of density matrices. So in this sense, you are right- the Schroedinger Equation is actually no longer true (but a slight generalization still holds). In the case you are describing, the density matrix for a single photon corresponds to a fully decohered mixture of traveling through the right and left slits.

To reconcile this with the other descriptions you've read, the key idea is to understand that decoherence is, in some sense, arbitrary. To get coherent effects, you need to have access to every part of the system that is entangled together, so if you can't do that you throw your hands up and say that it is decoherent. By doing this, you are saying that the system you are studying is entangled with the environment, with the environment simply defined as everything you aren't measuring. So effectively, when you throw away the second photon you have defined it as part of the environment, and you can still call it environmentally-based decoherence if you like.

So that brings us to your last, and very good, question- how is everything not entangled and decohered all the time? The short answer is that in the structure of quantum mechanics measuring something destroys all entanglement and acts as a sort of 'reset' on the state, after which you can prepare the object however you like. This is one of those issues that can be more and less obscure depending on how you interpret quantum mechanical measurement, but all this is really saying is that if you know an isolated object's initial conditions, of course you must be able to completely figure out what happens to it.

As far as references go, the most direct I've ever seen this point made is actually in the field of quantum computing. In that context the connection between decoherence and entanglement is called the 'principle of implicit measurement,' and it is stated as follows: if you throw away one part of your system, the effects are the same as if you had measured the properties of that part. Although it might not be obvious, this is identical to what I said above in terms of density matrices- and actually, in this wording it makes it extremely clear that you won't get interference in your second photon. You can find this in the Nielsen and Chuang book on Quantum Information, or restated in many different sets of lecture notes on google.

  • $\begingroup$ I'm not at all asking about an explanation for what happens when you try to detect which-path information on a single photon in the double-slit experiment. That is not relevant to this question. This question is about entangled pairs of photons. The wall in question is not the same wall that has the slits. The signal photon goes through the slits. Somewhere far away from the slits the entangled idler photon hits a wall. Different wall. $\endgroup$
    – user1247
    Apr 20, 2013 at 20:14
  • $\begingroup$ Oh, sorry for my misunderstanding. I've edited the above answer accordingly. $\endgroup$
    – Rococo
    Apr 21, 2013 at 0:57
  • $\begingroup$ Thanks for your edit. So, the question I want to really drive home here is: is the interference pattern destroyed if the idler photon's momentum is measurable in principle, even if the idler photon is totally isolated from the environment and the observer? In other words, the idler photon can be travelling through a vacuum, not entangled with the macroscoping environment or the observer. In 1000 years it will hit a wall which will destroy which-path information. Does the fact that the which-path information is measurable in principle in the intervening 1000 years relevant? $\endgroup$
    – user1247
    Apr 21, 2013 at 8:38
  • $\begingroup$ (continued)... in other words, I could measure which-path information, so in principle it is available, however if I choose not to measure it the which path information will be destroyed 1000 years from now. Is the interference pattern seen now or not? This is different from the quantum eraser experiment in that no information is needed from the idler photon in order to reconstruct the interference pattern. The interference pattern is either seen now or not. $\endgroup$
    – user1247
    Apr 21, 2013 at 8:40
  • $\begingroup$ My point is that your current answer "To get interference, the momentum change must be indistinguishable in principle" is ambiguous. What if the inteference is measurable in principle, but I choose not to, thus causing the information to eventually be lost, long after the interference pattern has been seen or not seen. Furthermore, your answer does not address the fact that (AFAIK) loss of interference requires the entanglement between the idler photon and the environment/observer, and the fact that the idler's momentum could be observable in principle yet not be entanlged with the environ $\endgroup$
    – user1247
    Apr 21, 2013 at 8:44

After all, theoretically the information carried by the idler could be reconstructed from careful measurement of the wall's properties.

This may not always be true if we take into account limit set by uncertainty principle. Such a comment would require knowledge about wall's properties.

If you can reconstruct the information to a degree of accuracy that tells you which slit the signal photon went through, then you will see no interference pattern.

  • $\begingroup$ I'm looking for an answer that goes into a bit more depth. For example, suppose that yes, you can in principle reconstruct the information from the wall. But what if the wall is completely isolated from the observer such that the observer is not entangled from the wall. Does the interference pattern still disappear? In this case there is no entanglement between the wall and the observer, yet at the same time there could be entanglement in the future which would allow the reconstruction of the information. $\endgroup$
    – user1247
    Apr 20, 2013 at 17:47
  • $\begingroup$ The fact that there is an entangled photon and that it hits the wall can not be neglected. This also brings into picture the interaction of idler photon with the wall. This does not allow the wall to be separated from the experiment's environment. $\endgroup$
    – Kamal
    Apr 20, 2013 at 18:22
  • $\begingroup$ My understanding is that decoherence would ultimately be due to the entanglement between the wall and the observer. If there is no entanglement between the wall and the observer, then there would be no decoherence and therefore no loss of interference. I don't know what you mean that the wall cannot be separated from the experiment. The wall can be in a vacuum in outer space, totally separated from the experimental observer who is somewhere else looking at the signal photon interference. $\endgroup$
    – user1247
    Apr 20, 2013 at 19:18

Yes, interference will be observed, (if you repeat with many photon pairs). For one single pair, you must remember that the photons are waves and spread out and take many paths (through all space and all time, if you believe Feynman). Also through both slits. Then "God throws the dice" and picks one tiny spot for each photon to land. So you really don't know which way the photon went for either photon, you only know where it ended up.

  • $\begingroup$ Yes, repetition of many photon pairs is assumed. From your response it is not clear that you appreciate the fact that given an entangled pair of photons, the detection of one can constrain information about the other. $\endgroup$
    – user1247
    Apr 2, 2013 at 10:37

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