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Can a photon be a superposition of multiple frequency states? Kind of similar to how an electron can be a superposition of multiple spin states.

randomafk
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1 Answers1

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Yes. Consider quantizing electromagnetic fields in a box. This corresponds to photons being trapped inside of said box since photons are just the mode quanta of the EM fields. The Hilbert space (called Fock space in this case) of the quantized radiation is found to be spanned by states $$ |\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad N=1,2,\dots $$ which represents a state with $N$ photons in the box with momenta $\mathbf p_i = \hbar \mathbf k_i$ and polarizations $\mu_i$ plus the vacuum state $|0\rangle$ containing no photons. Now suppose that at some point in time, the state of the system is $$ |\psi\rangle = \frac{1}{\sqrt 2}\left(|\mathbf k_1,\mu_1\rangle+|\mathbf k_2,\mu_2\rangle\right) $$ This represents a state in which there is a single photon in the box that is in a superposition of states; the vector $|\mathbf k_1, \mu_1\rangle$ represents a state with a single photon having momentum $\hbar \mathbf k_1$ and polarization $\mu_1$ while the vector $|\mathbf k_2, \mu_2\rangle$ represents a state with a single photon having momentum $\hbar \mathbf k_2$ and polarization $\mu_2$. In particular, recall that the frequency of a photon is related to $\mathbf k$; $E =\hbar\omega = \hbar c|\mathbf k|$ so that this state represents a photon in the box that is in a superposition of states corresponding to different frequencies.

joshphysics
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  • Of course, while it is pretty common to observe electrons while being agnostic regarding their spin, it's pretty hard in practice to observe a photon without collapsing it to a rather definite frequency, since they don't have much else defining them. –  Apr 30 '13 at 02:41
  • @ChrisWhite I have to admit, I don't quite understand what you mean when you say "it is pretty common to observe electrons while being agnostic regarding their spin." I'd like to though; would you mind elaborating a bit? – joshphysics Apr 30 '13 at 02:43
  • I think ChrisWhite just means that it's common to observe electrons in a superposition of spin states. (Obviously we can take any pure spin state and regard it as a superposition by choosing a basis where the state is not a basis element) – sjasonw Apr 30 '13 at 02:56
  • Why is the $|\psi\rangle$ state that of a single photon when $\mathbf{k}_2,\mu_2$ represents a two photon state? – DilithiumMatrix Apr 30 '13 at 03:04
  • @joshphysics I just meant you often have detectors that measure the energy, say, of an electron without knowing its spin, whereas I've never heard of a photon detector that doesn't tell you what the photon's frequency is. (But maybe that's my astro bias - we're not really interested in knowing about photons if we don't know their frequencies.) –  Apr 30 '13 at 05:26
  • @zhermes: It's a fault in notation. What was meant above is that a ket with N different $k_i$ is an N-particle state. The state $|\mathbf k_2, \mu_2\rangle$ is a one-photon state (it should probably be called $|\mathbf k', \mu'\rangle$ to avoid confusion) – Neuneck Apr 30 '13 at 06:22
  • @Zhermes Neuneck's got it right. I'll edit the post with a comment to make it more clear. – joshphysics Apr 30 '13 at 15:50
  • I don't see anything to disagree with in this answer, but it seems much, much more complicated than it needs to be. The wave equation is linear, so superpositions are valid states. –  Apr 30 '13 at 19:37
  • @BenCrowell I don't think my answer is all that complicated; I just wanted it to be concrete (regarding how photon states arise in quantum) and somewhat mathematically precise. You should post another response! – joshphysics Apr 30 '13 at 19:43