Four momentum of a photon squared, trouble with combining wave-particle duality and energy-frequency relation

Since $p^2=E^2-\vec{p}^2=m^2$ and

$E=h\nu=\frac{hc}{\lambda}$ and

$|\vec{p}|=\frac{h}{\lambda}$

we have that

$p^2=\frac{h^2c^2}{\lambda^2}-\frac{h^2}{\lambda ^2}$

If I go to Planck-units ($c=1,h=1$), this becomes zero. Is this a correct thing to do? It doesn't feel right because it depends on the units (SI or Planck) I use.

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You're right that, if the result of a calculation depends on the units you use, there's something wrong with it. In fact, even more than that, you can't even get any result out of $p^2 = \frac{h^2c^2}{\lambda^2} - \frac{h^2}{\lambda^2}$ in SI units, because you'd be taking a difference between two quantities with different units.

In this case, if you're using SI units, the energy-momentum relation is $p^2 = \frac{E^2}{c^2} - |\vec{p}|^2$; that extra factor of $c^2$ is necessary to keep the units consistent.

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Ahh yes of course, it's confusing sometimes because my syllabus jumps from one to the other... –  PatronBernard May 28 '12 at 21:04