Say you have a photon source that produces two kinds of photon,

  • a photon in a superposition with respect to polarization, |H> + |V>; and
  • a photon whose polarization is not in superposition, |H> OR |V>

The photon source randomly decides which type to output. Also, when outputting a photon which is not in superposition, the polarization is random. Is it possible to detect whether a photon's polarization has been measured?

Thanks for the help so far. I've refined the question and, in doing so, it may present "a moving target" - apologies for that!

I gather (from what I've read) that it's not considered possible. If I understand correctly, "the test" would require many photons to interfere with themselves before a statistical analysis would reveal whether they were in a superposition or not; so "testing" a single photon doesn't provide enough data for a statistical conclusion(?)

At this point I've tried to improve the question so that it's less ambiguous to those who have a better understand of the subject. Actually I'm afraid of introducing ambiguity through use of symbols and/or terms which I dont really understand. It's also quite possible that I'm incapable of understanding the answer to my question! (Thanks Emilio, you may be correctly anticipating the reason for my question , though you're loosing me with "reference frames" and "basis states").

  • $\begingroup$ I think you're confused about some details that make it hard to know what you mean. I think you should rephrase your question to simply: "Is it possible to detect whether a photon in a superposition state of two different polarization has been measured in one of the two states?" (if that's what you mean) $\endgroup$ Dec 30, 2017 at 6:11
  • 1
    $\begingroup$ "Being in a superposition state" is not a meaningful statement: you might claim that $|H⟩+|V⟩$ is a superposition state and $|H⟩$ isn't, but for someone in a different frame of reference (equally valid to yours) with basis states $|D⟩$ and $|A⟩$ ("diagonal" and "antidiagonal"), those states are better phrased as $|D⟩$ and $|D⟩+|A⟩$, i.e. the latter is a superposition and the former isn't. That means that "being in a superposition" is only ever meaningful when in explicit reference to a given basis. $\endgroup$ Dec 30, 2017 at 20:46
  • $\begingroup$ That's not to say that there isn't a valid question there (such as e.g. distinguishing between a superposition and a probabilistic mixture, with equal population weights, of two pre-specified states) but the phrasing needs to be a good bit more careful than the current version. $\endgroup$ Dec 30, 2017 at 20:47
  • $\begingroup$ ^To satifsy you, I think OP can add to the beginning that he's measuring in the basis of H & V and is interested in superpositions of H and V $\endgroup$ Dec 30, 2017 at 22:33
  • $\begingroup$ Isn’t it possible that a photon maintains single polarization until interfering with something like a slit? Then if the photons polarization is within a certain degree of aligning it would then rotate and continue on but now aligned with the slit. If the polarization is not close enough it will not make it through. $\endgroup$ Dec 31, 2017 at 0:26

2 Answers 2


Is it possible to detect whether a photon's polarization has been measured?

Yes, and this is how Quantum Key Distribution works to safely transmit information.

A quantum state that is prepared in a diagonal polarization has an equal probability of being in H or V, while having a 100% probability of being in D.

$|D\rangle = (|H\rangle + |V\rangle)$

(If you're confused here, read about the Stern-Girlac experiment and learn how having certainty about some measurements causes uncertainty in others. )

So if you always measure in the basis of in the A and D basis (which is just the horizontal and vertical basis shifted by 45%), you will always obtain either 100% or 0% since your state is either in A or D, or it isn't. BUT if an easedropper tries to measure in the H or V basis, then now your state becomes H or V (and is now in a superposition of A or D, while before it was in either one or the other!). This means that when you measure your easedropped state there's a 50% chance you DON'T obtain what you sent. So if you know what information you sent out, and you see that it's changing, then you know that someone is easedropping on you (and that, as you asked, we can detect that the polarization is being measured!) .

EDIT: If the question is, "Can an experimentalist create a single photon that is in a superposition state with 100% certainty without destroying it by measuring the state," the answer is yes. Experimentalists can create pairs of entangled photons, by measuring one of them they know the timing information about where the other photon is. Then if the photon's polarization is rotated such that it is in D, then it is in a superposition of H and V at the same time. Basically, we know that states that are in D are in superpositions of H and V, so it's not required for us to measure it.

  • $\begingroup$ Just as a caveat, this distinction is only possible statistically. It doesn’t work if you only have one. $\endgroup$
    – knzhou
    Dec 30, 2017 at 22:08

The question of the OP reveals issues with some of the basic concepts in quantum physics. Two particular concepts are those related to superposition and mixing, although the latter is not specifically stated.

Superposition implies a linear coherent summation of the terms $$ |\psi_1\rangle = \sum_n |n\rangle \alpha_n , $$ where $|n\rangle$ represents the basis and $\alpha_n$ denotes the coefficients. Normalization ($\langle\psi_1|\psi_1\rangle=1$) requires that $$ \sum_n |\alpha_n|^2=1 . $$ However, the basis is not unique. One can always define a different basis with the aid of a unitary transformation $$ |n\rangle \to |n'\rangle=U|n\rangle . $$ This implies that one can find a unitrary transformation that would give $|\psi_1\rangle$ as an element of the new basis. In that case $|\psi_1\rangle$ is not a superposition anymore.

Mixing is what happens when there is some ignorance about what is going on. A typical scenario is the one you provide in terms of the source. The source can either produce $|D\rangle = (|H\rangle + |V\rangle)/\sqrt{2} $, or $|H\rangle$ or $|V\rangle$, each with a certain propbability. The result is that the source produces a mixed state. The way that this is represented in quantum physics is in terms of a density operator $$ \hat{\rho} = |D\rangle P_D \langle D| + |H\rangle P_H \langle H| + |V\rangle P_V \langle V| , $$ where $P_D$, $P_H$ and $P_V$ are probabilities such that $$ P_D+P_H+P_V=1 . $$ Using this formalism, one can now proceed to make calculations to make predictions of what one would see in experiments.


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