Dirac said that a photon can only interfere with itself. This is consistent with the tensor product of two photon spaces representation. On the other hand, it is known that there is interference between distinct sources, between two photons, between two photons again, and between two electrons. This is related to this SE question, and this one.

Let $\frac 1 {\sqrt 2}\left(|1\rangle|2\rangle + |2\rangle|1\rangle\right)$ be a two-photon state.

Let's consider interference between the two photons. If one "rotates" one of the wavefunctions with a phase factor, one expects that this will affect the interference, and the result will depend on the phase factor. So, the interference between $|1\rangle$ and $|2\rangle$ should be different than the interference between $|1\rangle$ and $e^{i\varphi}|2\rangle$, and that between $e^{i\varphi}|1\rangle$ and $|2\rangle$ (where $\varphi$ is an arbitrary overall phase). This can be checked experimentally by using a phase shifter.

Suppose that there is an algorithm to calculate the interference between the two photons, and that this algorithm receives as input the two-photon state. The state $\frac 1 {\sqrt 2}\left(|1\rangle|2\rangle + |2\rangle|1\rangle\right)$ is equivalent to $\frac 1 {\sqrt 2}\left(|1\rangle|2'\rangle + |2'\rangle|1\rangle\right)$, where $|2'\rangle=e^{i\varphi}|2\rangle$. If the algorithm which gives us the interference depends only on the two-photon state, then it apparently gives the same result, even when we phase-shift the second photon.

Apparently, the interference between photons depends on the way we decompose the two-photon state in two photons: in terms of $|1\rangle,|2\rangle$ we obtain different interference than in terms of $|1\rangle,e^{i\varphi}|2\rangle$, respectively $e^{i\varphi}|1\rangle,|2\rangle$. So, either the Hilbert space does not contain all the needed information, or I am missing something (which is by far more likely). Could you please explain me where I am wrong, and what is the correct description of interference using the two-photon state?

P.S. Let me emphasize that I am fully aware that QM has been tested successfully by so many experiments, and that most likely I am missing something. I just don't know what.


2 Answers 2


Two photons do not interfere with each other. One photon interferes with itself.

You cannot add together the quantum states belonging to different Hilbert space.

Ok, two photons can have interference, because they are indistinguishable, and thus should always be described in whole. You can't use $|1\rangle+|2\rangle$ to describe the state of two photons, because that is not a symmetric state. It is not even a state in 2-photon space.

You should add some labels to clarify your thinking. For example, $|1\rangle_1+|2\rangle_2$.

  • $\begingroup$ Thank you. I updated the question with some references which seem to prove that there may be interference between different sources, different photons and different electrons. $\endgroup$ Sep 3, 2012 at 16:09
  • $\begingroup$ Thanks for the update. I can't see how labeling the photons allows me to avoid what I wrote in my second update. Namely, that the state obtained by shifting the phase of one of the photons is equivalent, in the two-state Fock space, with the original one. So I can't see where the information about that phase shift is in the vector state from the two-photon space. Yet, it has physical effects. $\endgroup$ Sep 4, 2012 at 4:42
  • $\begingroup$ @CristiStoica: I don't understand you. You have specify the state of the two photons. What next? $\endgroup$
    – Siyuan Ren
    Sep 4, 2012 at 6:12
  • $\begingroup$ The state of two photons remains unchanged if we shift the phase of one of them. But the interference pattern changes. So apparently the same two-photon state yields distinct interference patterns, depending on how we choose to write it in terms of two photon states. What I am missing? If I am wrong, then what is the correct algorithm to get the interference pattern solely from the two-photon state? $\endgroup$ Sep 4, 2012 at 6:24
  • $\begingroup$ @CristiStoica: I don't know where to begin. You should read those papers you cite, and understand what exactly two photon interference means. It does not in any way mean that you add the quantum states of two different photons together; it means you superimpose two different states in the two-photon system's space. $\endgroup$
    – Siyuan Ren
    Sep 4, 2012 at 8:41

$\newcommand{\ket}[1]{|#1\rangle}$The states $$\ket{\phi^+}=\frac{1}{\sqrt{2}}\left(\ket{1}\ket{2}+\ket{2}\ket{1}\right)\textrm{ and }\ket{+}=\frac{1}{\sqrt{2}}\left(\ket{1}+\ket{2}\right)$$ don't really have much to do with each other, and from their structure you can see that they are very different.

If you want to obtain the state $\ket{\varphi}=\frac{1}{\sqrt{2}}\left(\ket{1}+e^{i\varphi}\ket{2}\right)$ in port 1 of your system, starting from the common, entangled state $\ket{\phi^+}$, then you'd have to do a projective measurement on to $\ket{\varphi}$ on port 2, which leaves the desired state in port 1. (Check it!). Of course, this has the attendant risk that the measurement will yield $\ket{\varphi}$'s orthogonal complement, $\ket{\overline{\varphi}}=\frac{1}{\sqrt{2}}\left(\ket{1}- e^{i\varphi}\ket{2}\right)$, on port 2, thus collapsing port 1 onto $\ket{\overline{\varphi}}$. Thus measurements on port 1 done irrespectively of coincidence counts with port 2 will not yield any information; this reflects the fact that the reduced density matrix for either port in the state $\ket{\phi^+}$ is a completely mixed state.


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