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Consider a standard double-slit experiment using light, where $|\Psi\rangle = \frac{1}{\sqrt{2}}(|L\rangle +|R\rangle)$ may represent the superposition state of a single photon passing through 'both slits,' interfering with itself.

Since the above state should be an eigenstate of some Hermitian operators (e.g., $|\Psi\rangle\langle\Psi|$), the superposition state could, in theory, be observed. However, how could one go about measuring this superposition state?

When dealing with multiple photons, the superposition of $|L\rangle$ and $|R\rangle$ can be indirectly inferred by examining the interference pattern. (So at least in this indirect sense they can be 'observed'?) But if only a single photon is used, how could we possibly observe the superposition state?

If there is no possible way (even in principle) to observe it, then what should we say about the fact that $|\Psi\rangle = \frac{1}{\sqrt{2}}(|L\rangle +|R\rangle)$ is an eigenstate of some Hermitian operator?

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  • $\begingroup$ A single photon doesn't have a state. The quantum mechanical state is the state of the ensemble. A single photon is simply the amount of energy that the electromagnetic field exchanges with the "detector". If you want to deal with "multiple photons", then the resulting wave function is a direct product of the single quantum wave functions (and you still would need to symmetrize that and build a Fock space). None of that is of any importance in the double slit experiment. Technically this isn't even quantum mechanics. The result is simply the classical intensity according to Maxwell. $\endgroup$ Commented Apr 9, 2023 at 9:10
  • $\begingroup$ If 1000 scientists each did a double slit experiment with a single photon .... when they got together to consolidate and compare results .... they would indeed realize the interference pattern. The wave functions you describe above could be wave functions of the virtual field of the photon (i.e. forces not energy) as opposed to the real photon energy transfer. An excited atom/electron is already interacting with the EM field even before real photon creation. $\endgroup$ Commented Apr 9, 2023 at 13:35

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how could one go about measuring this superposition state?

To measure $|\Psi \rangle \langle \Psi|$ physically, my first guess is that one would apply a known operation $U$ to shift the state to an eigenstate of a traditional observable like $x$ and then do the measurement in that basis. The $U$ would function such that only the state $|\Psi \rangle$ gets mapped to an exact eigenstate. I've seen this approach suggested for measuring in weird bases in a quantum information course.

Note also that alternatively, one could find $|\Psi(x)|$ through measurements of position. One could also find information about the phase through momentum measurements. Whether the phase is fully determined by momentum measurements I do not know off hand, but I have been wondering this for a while.

But if only a single photon is used, how could we possibly observe the superposition state?

You can't. The state is a whole function, it has more information than you could get just from measuring a single number. The approaches listed above are consistent with this, in that they also need a large number of samples to infer the state. Even when measuring the observable $|\Psi \rangle \langle \Psi |$, you would need many measurements; from a single measurement, the result you get could have come from infinitely many possible pre-measurement states. Okay, technically you'd need infinitely many measurements to be 100% sure of what the state is, but you can bound it really nicely with finitely many measurements.

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