# What happens to the wave function after applying the D'Alembert operator?

Is the result of applying the D'Alembert operator on a wave function always zero? Or are there exceptions?

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First of all, zero would also be a result. Second of all, the function entering the massless Klein-Gordon equation (given by the D'Alembert operator) isn't a wave function but a field - classical or quantum field. If it were a wave function, the energy wouldn't be positively definite because there would be both solutions with $\exp(i\omega t)$ and $\exp(-i\omega t)$, positive and negative-frequency solutions, and energy would be unbounded from below. That's why "second quantization" is a must in relativistic theories and the function can't be an ordinary wave function. –  Luboš Motl Jul 7 '13 at 5:15

The Klein-Gordon equation for a free particle is as follows:

$$\left(\hat\Box+\mu^2\right)\Psi=0$$

Where $\mu=\pm \frac{mc_0}{\hbar}$ is the so-called scaled mass and $\hat\Box$ is the D' Alembert's operator in question. Solving for $\hat\Box\Psi$,

$$\hat\Box\Psi=-\mu^2\Psi=\left(i\mu\right)^2\Psi$$

So, yes, there are results. If the mass $m=0$, however, then $\mu=0$, and thus $i\mu=0$, and thus $\left(i\mu\right)^2=0$. And thus

$$\hat\Box\Psi=0$$

Conclusion: D'Alembert's operator kills the wavefunction (or more accurately, field) only for massless particles, i.e. it operates on the wavefunction (field) to yield $0$ iff the particle is massless. For non-zero masses, this will not be the case.

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Why the downvote? Speaking as a non-expert, dimension10's answer seems clear. If there is a problem with it that caused you to downvote I would be interested to know what the problem is. –  John Rennie Jul 7 '13 at 8:08
It's true what you say, but it is too formal--- you can just say it without the trivial manipulations, let the reader figure it out. –  Ron Maimon Aug 22 '13 at 22:06
@RonMaimon: I know, but I was pretty sure that would get the answer in the low quality queue which I'd not want it to go to. –  Dimensio1n0 Aug 23 '13 at 1:58