# 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

In non-relativistic quantum mechanics the mass can, in principle, be considered an observable and thus described by a self-adjoint operator.

In this sense a quantum physical system may have several different values of the mass and a value is fixed as soon as one performs a measurement of the mass observable, exactly as it happens for the momentum for instance.

However, it is possible to prove that, as the physical system is invariant under Galileian group (or Galilean group as you prefer), a superselection rule arises, the well-known Bargmann mass superselection rule. It means that coherent superpositions of pure states with different values of the mass are forbidden.

Therefore the whole description of the system is always confined in a fixed eigenspace of the mass operator (in particular because all remaining observables,including the Hamiltonian one, commute with the mass operator).

In practice, the mass of the system behaves just like a non-quantum and given parameter. This is the reason, barring subtle technicalities (non separability of the Hilbert space if the spectrum of the mass operator is continuos), why the mass can be considered a given parameter rather than a self-adjoint operator in non-relativistic quantum mechanics.

In relativistic quantum mechanics the picture is quite different. First of all, one has to distinguish between elementary systems (elementary free particles in with Wigner's defintion) and compound (interacting) systems. The formers are defined as irreducible (strongly continuous) unitary representations of Poincaré group. Each such representation is identified by a set of numbers defining the eigenvalues of some observables which attains constant values in the representation in view if the irreducibility requirement. The nature of these numbers depend on the structure of the group one is considering.

Each such observable, in the irreducible Hilbert space of the system has the form $\lambda I$ where $\lambda$ is fixed a real number. Referring to Poincaré group, the mass operator turns out to be one of these elementary observables. Therefore, in relativistic quantum mechanics, the elementary systems must have trivial mass operator, which as before, can be considered as a given non-quantum parameter.

The picture changes dramatically if one focuses on compound systems: there the mass is simply the energy operator evaluated in the rest frame of the system. It generally shows a mixed spectrum made of a continuous part, due to the "relative" kinetic energy and, under that part, a point spectrum describing the possible masses of the overall system.

ADDENDUM. As Arnold Neumaier pointed out to me, neutrinos appear to have non-fixed values of the mass (i.e. the mass operator is not trivial) in view of the presence of weak interaction. In my view, it is disputable if they can be considered elementary particles since they include weak interaction in their description. Surely they are elementary from a pure physical viewpoint. Maybe Wigner's description is physically inappropriate.

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-1. In the relativistic case, mass is no longer a superselection rule, hence the mass operator is often nontrivial. Otherwise one couldn't have a nontrivial mass matrices of quarks and neutrinos, which are experimentally verified to show mass mixing. – Arnold Neumaier Aug 11 '14 at 11:12
It seem to me that I made a sharp distinction between the relativistic an non-relativistic case. In the relativistic case for elementary systems as pictured in Wigner's notion of elementary particle the mass is again a trivial operator as it is a Casimir of the representation, whereas it is not if considering compound systems. I have now stressed that "elementary" also means "free". – Valter Moretti Aug 11 '14 at 12:22
You made a sharp distinction but you still write ''Therefore, in relativistic quantum mechanics, the elementary systems must have trivial mass operator, which as before, can be considered as a given non-quantum parameter.'', which is wrong. – Arnold Neumaier Aug 11 '14 at 12:31
I'd just drop the corresponding paragraph. What you write about elementary systems in the sense of Wigner is technically correct, but misleading since elementary particles are not elementary systems in this sense. In today's theories, elementary particles are irreducible representations of the full symmetry group, which is not just the Poincare group. If the mass transforms nontrivially under the internal symmetry group, it is a matrix rather than a number. – Arnold Neumaier Aug 11 '14 at 12:38
Long discussions are here discouraged. Please ask instead at physicsoverflow.org. – Arnold Neumaier Aug 11 '14 at 12:48

Mass-squared is a Hermitian linear operator, it's a Casimir operator $\hat{C}_{1}=\hat{P}_{0}\hat{P}_{0}-\hat{P}_{i}\hat{P}_{i}$ for the Poincare group. It's Hermitian because the translation generators $\hat{P}_{\mu}$ are Hermitian. It commutes with all the generators of the Poincare group and so it's eigenvalues (mass-squared) are constant on each irreducible subspace.

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Of course, mass is an observable, although in simple models it is constant.

This is already the case classically. One cannot determine the path of as rocket that burns fuel (which forms a large fraction of its mass) without taking into account that the mass is variable.

The same holds in quantum mechanics, whenever the mass is not fixed by the modeling assumptions. Important case where the mass is given by a nontrivial operator M (then called the mass matrix) are

On the other hand, if the mass is a constant m, it is also represented by an operator, namely M:=m*1, where 1 is the identity matrix. This is a Hermitian operator, hence fits the standard description.

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I'll just supplement Prof. Kalitvianski's answer by adding that the mantra « observables as operators » only applies to the special kind of measurement called a « quantum measurement », which always involve amplification. Other kinds of measurement, such as measuring physical constants, are not modelled by observables either, like the speed of light, the mass of the electron, the charge on an electron, etc.

The Nobel prize winner Eugene Wigner wrote very careful articles about the theory of measurement, although it is considered superseded by many today, they are still an excellent foundation for understanding the more modern theories. They have been reprinted in the collection of very readable scientific essays of his, Symmetries and Reflections.

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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