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If Klein Gordon equation is for spin-0 particles, I write massless fields as $\square A=0$, how can I say $A_\mu=\epsilon^\mu e^{-ikx}$ as a wave function of polarized photon (spin-1) ?

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  • $\begingroup$ By solving the equation? $\endgroup$ – Prahar Jan 16 '15 at 0:28
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    $\begingroup$ Yes. $A_\mu$ is just a wave equation of polarized photon when you solve the Klein-Gordon $\endgroup$ – aQuestion Jan 16 '15 at 0:32
  • $\begingroup$ I don't understand your question. Are you asking how you get $A_\mu = \epsilon_\mu e^{- i k x}$? This follows trivially from $\Box A_\mu = 0$ $\endgroup$ – Prahar Jan 16 '15 at 0:33
  • $\begingroup$ No. Suppose i solved the equation, do not worry about it. I just wonder how can one get photon wave function as a solution of KG? Photon is spin-1 particle, but KG is for spin-0 right? $\endgroup$ – aQuestion Jan 16 '15 at 0:36
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    $\begingroup$ Yes and no. The Klein-Gordon equation is satisfied by all fields, irrespective of their spin. The KG equation simply gives us the relativistic dispersion relation $p^2 - m^2 = 0$. Thus, a scalar, spinor or even a spin-1 field satisfies it. However, the spin-1 field satisfies an additional condition $\partial^\mu A_\mu = 0$ which projects out the scalar part of $A_\mu$ and leaves behind only the 3 spin-1 d.o.f. $\endgroup$ – Prahar Jan 16 '15 at 0:40
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As in the comments, fulfilment of the Klein Gordon equation is only a necessary condition for a field and it is fulfilled by all fields. For example, the Dirac equation for an electron implies the Klein Gordon equation, but not conversely. If you've never seen this, try working out the following. Begin with the Dirac equation $(i{\partial\!\!\!\big /} - m) \psi = 0$ and then impart the operator on the left hand side again to both sides of the equation: you'll find $(i{\partial\!\!\!\big /} - m)^2 \psi = 0$ is equivalent to the Klein Gordon equation for all four components of the Dirac spinor independently. $(i{\partial\!\!\!\big /} - m) \psi = 0$ is therefore a strictly tighter constraint than $(i{\partial\!\!\!\big /} - m)^2 \psi = 0$. Likewise for a massless spin one field: you of course get D'Alembert's equation. But you also need a Lorenz gauge condition, leading to the tighter constraint.

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    $\begingroup$ You don't square the left-hand side, you apply the conjugate: $(i\gamma^\nu \partial_\nu + m)(i\gamma^\mu \partial_\mu-m) \psi = -(\gamma^\mu\gamma^\nu \partial_\mu \partial_\nu+m^2)\psi$, and then use the Clifford algebra to show $\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu = \partial_\mu \partial^\mu$ from which it follows, $(\partial_\mu \partial^\mu + m^2)\psi = 0$. $\endgroup$ – JamalS Jan 16 '15 at 15:21
  • $\begingroup$ evaluation KG from Dirac field as $(-i \partial - m) (i \partial - m)\Psi= 0$ so it becomes $(\square+m^2)\Psi=0$ so if I have such a solution $\Psi$, it should satisfy both KG and Dirac. I think there is a typo in your answer. But still good explanation. $\endgroup$ – aQuestion Jan 16 '15 at 16:12
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$A^\mu$ is a quantum field, i.e. an operator-valued distribution, and not a wavefunction.

In QFT, there are no wavefunctions, since we have no naive position basis for our theory (the theory of the relativistic position representation is called Newton-Wigner theory). The analogue to the QM wavefunction is a wavefunctional that is a functional of time and the fields themselves.

Also, $A^\mu = \epsilon^\mu\mathrm{e}^{\mathrm{i}kx}$ is not a proper solution for the wave equation for the quantum field (though it is a solution of the equation in general). Since it is an operator-valued object, the proper way of writing this would be $A^\mu = \epsilon^\mu(k) a^\dagger_k \mathrm{e}^{\mathrm{i}kx} + \epsilon^\mu(k) a_k \mathrm{e}^{-\mathrm{i}kx}$, where the $a^{(\dagger)}_k$ are annihilation (/creation) operators for a photon with momentum $k$. In general, the free EM quantum field is mode decomposed as

$$ A^\mu = \int \frac{\mathrm{d}^3p}{(2\pi)^3}\frac{1}{\sqrt{2\lvert p\rvert}}\sum_\lambda\epsilon^\mu_\lambda(p) (a_{\lambda,p}\mathrm{e}^{\mathrm{i}px} + a^\dagger_{\lambda,p}\mathrm{e}^{-\mathrm{i}px})$$

where the $\epsilon_\lambda(p)$ are a basis for the space of polarisation vectors labelled by $\lambda$, and the $a_{\lambda,p}$ the ladder operators for a photon of momentum $p$ and polarisation $\epsilon_\lambda(p)$.

tl;dr: You cannot say that $A^\mu$ is the wavefunction of anything.

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  • $\begingroup$ I'll think about it for a while. $\endgroup$ – aQuestion Jan 16 '15 at 16:17
  • $\begingroup$ Not to be pedantic, but wouldn't the polarization vector pick up a conjugation? $\endgroup$ – ClassicStyle Sep 27 '16 at 7:03
  • $\begingroup$ @TylerHG The polarization vectors aren't complex vectors, they are real vectors in Minkowski space. Conjugating them has no effect. $\endgroup$ – ACuriousMind Sep 27 '16 at 11:58
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The photon wave function is a solution of a quantized Maxwell's equation , in the potential form. There the variables exist which will allow to build up the classical electromagnetic wave from an enormous number of photons, as shown in this link :

The photons also have polarizations so the wave function has many components, too. I don't want to scare you by the indices but the wave function of a single photon mathematically looks like the (complexified) classical electromagnetic potential →A(x,y,z), with some extra subtleties. (But its interpretation is different!)

As the other answer says the plane wave is a solution that emerges from all the differential equations used in quantum mechanical problems. The specific variables entering will depend on the specific appropriate equation for the particles under question.

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