# Why doesn't a plane wave solution represent a single photon? [closed]

Why doesn't a plane wave solution represent a single photon? And what is meant by the quantum-mean field being zero?

EDIT:

This post is an extension to a previous post I made asking about the photon in QFT. I was asked by the person helping me to start a new question so he can respond in the form of an "answer" post and thereby keeping all "answers" on topic. The link is here Quantum State of Photon Question

• A plane wave solution to what? In what context is the mean of "the quantum field" (which quantum field?) said to be zero, and what do you not understand about it? C'mon, don't expect us to read your mind! Feb 9 '16 at 22:13
• Sorry I should have prefaced this by saying that this post is an extension to my previous post asking about the photon in QFT. I was asked to start a new question so as to keep all answers on topic as the person answering wanted to respond in the form of an "answer" post. Here is the link. physics.stackexchange.com/questions/234839/… Feb 9 '16 at 22:16
• Great. It will take some time to write it all down. In order for the question to be well received, as pointed out by ACuriousMind, you may want to slightly modify the text. You may write "... a plane wave classical solution to Maxwell's equation ..." and "Why is the quantum-mean of the electromagnetic field associated to a single photon zero?". Feb 9 '16 at 22:24

Ok so, let's start by defining the photon in the context of quantum theory. A photon is one of the quantum states of definite energy and momentum of the theory of Quantum Electrodynamics. In QED without fermions, a photon of momentum $k$ and spin $s$ is defined a the state

$$|\gamma, \vec{k},s\rangle=a^{\dagger}(\vec{k},s)\,|0\rangle$$ Where $a^{\dagger}(\vec{k},s)$ is the operator that appears in the Fourier-expansion of the electromagnetic four-potential operator $A^{\mu}(x)$:

$$A^{\mu}(x)=\int\frac{d^{3}k}{(2\pi)^{3}}\ \frac{1}{\sqrt{2k^{0}}}\ \sum_{s}\ \left\{e^{-ik_{\mu}x^{\mu}}\ a(\vec{k},s)\ e^{\mu}(\vec{k},s)+e^{ik_{\mu}x^{\mu}}\ a^{\dagger}(\vec{k},s)\ e^{\mu\ *}(\vec{k},s)\right\}$$ Depending on the conventions on the normalization of states, you may find different expressions for both the four-potential operator and the normalized state of the photon. In the conventions to which I'm used to, one usually writes $$|\gamma, \vec{k},s\rangle=\sqrt{2k^{0}}\ a^{\dagger}(\vec{k},s)\,|0\rangle$$ Why it is so depends on the commutation relations between the operators of the theory. The quantization of the electromagnetic field is not an easy subject, so in order to get useful results we shall work under assumptions which would be theoretically justified if one started from the very beginning. First of all, let's assume that the $0$ component of the electromagnetic four-potential can be set to zero: $A^{0}=0$ (this is not the case in QED with fermions, but this assumption can be made in our case, by setting $e^{0}(\vec{k},s)=e^{0\ *}(\vec{k},s)=0$). Second of all, let's state without derivation that $$[a(\vec{k},s),a^{\dagger}(\vec{k}',s')]=(2\pi)^{3}\ \delta^{(3)}(\vec{k}-\vec{k}')\ \delta_{ss'}$$ $$[a(\vec{k},s),a(\vec{k}',s')]=[a^{\dagger}(\vec{k},s),a^{\dagger}(\vec{k}',s')]=0$$ (again, the factor of $(2\pi)^{3}$ depends on the normalization conventions). This being said, you can measure the mean electromagnetic four-potential carried by a photon is the usual way: $$\langle A^{\mu}(x)\rangle_{\gamma,k,s}=\langle{\gamma,k,s}|A^{\mu}(x)|\gamma,k,s\rangle$$ So let's calculate $$A^{\mu}(x)|\gamma,k,s\rangle=\int\frac{d^{3}p}{(2\pi)^{3}}\ \frac{1}{\sqrt{2p^{0}}}\ \sum_{\sigma}\ \left\{e^{-ip_{\mu}x^{\mu}}\ a(\vec{p},\sigma)\ e^{\mu}(\vec{p},\sigma)+e^{ip_{\mu}x^{\mu}}\ a^{\dagger}(\vec{p},\sigma)\ e^{\mu\ *}(\vec{p},\sigma)\right\}\sqrt{2k^{0}}\ a^{\dagger}(\vec{k},s)\,|0\rangle=e^{-ik_{\mu}x^{\mu}}\ e^{\mu}(\vec{k},s)\ |0\rangle+\int\frac{d^{3}p}{(2\pi)^{3}}\ \frac{1}{2p^{0}}\ \sum_{\sigma}\ e^{ip_{\mu}x^{\mu}}\ e^{\mu\ *}(\vec{p},\sigma)\ |(\gamma,p,\sigma),(\gamma,k,s)\rangle$$ by using the commutation relations, where $$|(\gamma,p,\sigma),(\gamma,k,s)\rangle=\sqrt{2p^{0}}\sqrt{2k^{0}}\ a^{\dagger}(\vec{p},\sigma)\ a^{\dagger}(\vec{k},s)\ |0\rangle$$ is a state containing two photons. This way, we find that $$\langle A^{\mu}(x)\rangle_{\gamma,k,s}=e^{-ik_{\mu}x^{\mu}}\ e^{\mu}(\vec{k},s)\ \langle{\gamma,k,s}|0\rangle+\int\frac{d^{3}p}{(2\pi)^{3}}\ \frac{1}{2p^{0}}\ \sum_{\sigma}\ e^{ip_{\mu}x^{\mu}}\ e^{\mu\ *}(\vec{p},\sigma)\ \langle{\gamma,k,s}|(\gamma,p,\sigma),(\gamma,k,s)\rangle$$ Now of course the inner product between states containing different numbers of photons is zero. So $$\langle A^{\mu}(x)\rangle_{\gamma,k,s}=0$$ i.e. the mean four-potential carried by a photon is zero. As a consequence, the electric and magnetic field carried by a photon too must be zero (the derivatives can be brought out of and into the average sign). The direct calculation of the electromagnetic field-strength $F^{\mu\nu}$ is done as follows: $$\langle F^{\mu\nu}(x)\rangle_{\gamma,k,s}=\langle{\gamma,k,s}|\partial^{\mu}A^{\nu}(x)-\partial^{\nu}A^{\mu}(x)|\gamma,k,s\rangle$$ A direct computation shows that this indeed is zero.

So what does this mean? Does this mean that the photon carries zero electromagnetic field? Of course not. This only means that if you were able to measure the electromagnetic field carried by a photon many times, the point-by-point (i.e. separately in space and in time) average of the field would tend to zero in the limit of infinite measurements, and not because the photon is an oscillating wave: as I said, the average would be zero from point to point. On the other hand, a plane wave potential corresponds to a non-zero, fixed value of the electromagnetic field. Thus by no means you can identify it with a photon. On the other hand, complicated superpositions of states with different number of photons, from zero to infinity, can have non-zero average electromagnetic field. This article explains how one can get a specific classical field configuration from a superposition of photonic states.

• Wow thank you so much, this helps me tremendously! (Sorry for the late reply!) I very much appreciate you taking the time to write this up. Feb 10 '16 at 3:21
• I will take some time tomorrow to read through the article you linked as I find this subject extremely fascinating. Feb 10 '16 at 3:23
• Happy I was able to help :-D Feb 10 '16 at 4:02
• Hello again! This answer was so good and I would like to put it in my doctoral dissertation and I need a source to cite. Do you think you can provide me with a source that includes this derivation? Jul 1 '16 at 18:47
• Not at the moment. I'll have a look at a couple textbooks and let you know asap. Jul 1 '16 at 19:33