# Is mass an observable in Quantum Mechanics?

One of the postulates of QM mechanics is that any observable is described mathematically by a hermitian linear operator.

I suppose that an observable means a quantity that can be measured. The mass of a particle is an observable because it can be measured. Why then the mass is not described by a linear hermitian operator in QM?

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see this –  Jack Jan 12 '12 at 14:20
What about $\hat{L}[\Psi\rangle = m^2 [\Psi\rangle$ (I mean if angular momentum of the particle depends linearly on it's mass squared)... ;-P ? –  Dilaton Jan 12 '12 at 17:04
@Nemo: the angular momentum is quantized and proportional to $\hbar$ solely. Besides, it is the quasi-particle angular momentum who is quantized and can take the eigenstates. A bound particle cannot have an eigenvalue of $L_z$ because the particle is always in a mixed state ;-) –  Vladimir Kalitvianski Jan 12 '12 at 17:20
@Nemo: huh? Where have you seen that? –  David Z Jan 13 '12 at 0:55
ohhhh, gotcha ;-) I've seen $J \sim m^2$ but I think the fact that you used $L$ threw me off. –  David Z Jan 13 '12 at 8:44
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On one hand, in non relativistic QM mass is a numerical parameter in a theoretical construction, so there is no need in an additional operator while transition form CM to QM.

On the other hand, in relativistic QM the energy is an operator reducing to mass (rest energy) in a particular case $\vec{p}=0$. Normally the energy of a free particle has a continuous spectrum (not quantized, nor calculated), so it is still an external parameter to the theory for an "elementary" free particle.

In case of a compound system, the rest mass can be (in principle) calculated from masses and interaction forces of its constituents.

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This is too vague –  Revo Jan 12 '12 at 16:40