# Relativistic genarization of Quantum Harmonic Oscillator

I am trying to find out relativistic description of a quantum harmonic oscillator. For a classical relativistic oscillator mass is a function of co-ordinates(http://arxiv.org/abs/1209.2876). $$m(x)=m-\frac{m\omega^{2}x^{2}}{c^{2}}$$ I want to get a quantum mechanical model for this particular hamiltonian $$H=\sqrt{p^{2}c^{2}+m(x)^{2}c^{4}}$$ First,I thought of re-writing this hamiltonian with the help of $\alpha$ and $\beta$ matrices.I got $$H=\vec{\alpha}.\vec{p}+\beta m(x)$$ But by adopting this prescription implies wave function should be $4$-component spinors. How can I get $4$-component spinors which would reduce to $1$-component wave function in non-relativistic quantum mechanics? If someone can suggest some other methods to solve for the hamiltonian that would also help.

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Just a terminological comment, not just questioning your wording but the wording in the aforementioned paper as well. "Harmonic" is something that is composed of sines and cosines (the solutions to the motion of the oscillator in this case). If the solutions aren't sines and cosines, e.g. because the energy isn't simply quadratic, one shouldn't call it "harmonic" oscillator. It's a relativistic generalization of the harmonic oscillator. – Luboš Motl Apr 22 '13 at 16:53
You are right.Changed it. – WInterfell Apr 22 '13 at 16:57
Your hamiltonian should have $m^2 c^4$. Also I would suggest that you can try to find (and I believe you will succeed) the eigenvalues and eigenstates of the operator under the square root. On the first thought, $H=\sqrt{Q}$ and $Q$ should have the same eigenvectors and correspondingly related eigenvalues. Note the reasoning behind the Klein-Gordon equation. You should also keep in mind that you should be carefull with physical interpretation of your results -- relativistic QM is not totally physically consitent and should be replaced with QFT. – Peter Kravchuk Apr 22 '13 at 19:33
Moreover, I think that your hamiltonian in $\alpha$ and $\beta$ matrices is not obviously the correct square root -- $m$ and $p$ do not commute. Finally, these matrices are some non-commuting operators, not initially present in the theory -- you are introducing new degrees of freedom, specifically spin. Just thought that it can be not exactly what you want. – Peter Kravchuk Apr 22 '13 at 19:42