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Why is inertial mass not an observable in QM?

  1. Nov 29, 2011 #1

    kith

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    In classical mechanics, I can measure the inertial mass of a particle by measuring force and acceleration: m=F/a. In QM, this equation only holds for expectation values <F> and <a>. Does this lead to the fact that inertial mass is not an observable?

    Is there a deeper underlying principle which determines if a classical observable has a quantum mechanical analogue? Like, it must be obtainable by some kind of "direct" measurement?

    For example, I'd guess that moment of inertia is not an observable in QM since mathematically, it is very similar to inertial mass. Also, time isn't an observable, but this is probably a different case (see the other thread).
     
    Last edited: Nov 29, 2011
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  3. Nov 29, 2011 #2
    I don't think there's any difficulty in defining a mass operator mathematically. And operationally we can, for example, measure the kinetic energy of a free particle with definite momentum, then use E=p^2/2m to get the mass.
     
  4. Nov 29, 2011 #3

    kith

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    Now that you mention it, I probably just assumed that there is no mass operator, because nobody talks about it. Which is not even true, because we talk of mass eigenstates in neutrino oscillations.
     
  5. Nov 29, 2011 #4
    There is no mass operator in (non-relativistic) quantum mechanics.

    You could define one, but it turns out that every physical state of a single particle system must be an eigenvector of mass, technically we say that the mass is a super-selected observable.

    The reason for this super-selection is that you want a galilean-invariant theory: try to prepare a system in a superposition of different mass states, if you change to another inertial frame (x=vt+x0) you will measure different transition probabilities, but this is incompatible with your assumption. So QM makes sense only on states with definite mass, this is equivalent to say that the mass is a classical observable: you don't need an operator for it!

    The neutrino mass matrix is not an operator: it is a matrix in the 3 dimensional space of flavours (electron,muon,tau), not on the Hilbert space of the states of neutrinos.
     
  6. Nov 29, 2011 #5
    I don't think that's quite true, Galilean-invariance could be compatible with a non-superselection theory. It's discussed in Weinberg Vol1 section 2.4,pg 62 and section 2.7. I didn't go through the math myself, but his point is we can enlarge Galilean group a bit by introducing a mass operator, and this enlargement will only make the superselected theory non-superselected, but makes no difference to other predictions of the theory at all.
     
  7. Dec 1, 2011 #6

    A. Neumaier

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    This is misleading.

    Compare it with saying that the Pauli matrices are not operators since they are matrices in the 2-dimensional space of spins (up, down), not on the Hilbert space of states of an electron. While this is true in a very literal interpretation, it is false in the slightly looser sense in which physicists interpret their terminology.

    When they talk about the Pauli matrices as operators on electron space, they mean, in strictly rigorous terms, the tensor product of the Pauli matrices on the spin part and the identity on the position part of the wave function.

    In the same sense, the neutrino mass matrix is a well-defined operator on the Hilbert space of neutrino
    states.
     
  8. Dec 3, 2011 #7
    I think that saying that mass is not an observable would be misleading.

    It is trivial to built a mass operator in quantum mechanics [itex]\hat{m} \equiv m \hat{1}[/itex]

    Now note that [itex]\langle m \rangle = \langle \Psi | \hat{m} | \Psi \rangle = m \langle \Psi | \hat{1} | \Psi \rangle = m [/itex].

    That is, mass in an entirely classical observable in QM.

    Your force expression verifies [itex]\langle ma \rangle = m \langle a \rangle [/itex]

    Any classical observable function of position and momentum has a quantum counterpart obtained from substituting momentum by momentum operator and position by position operator.

    The units of moment of inertia are {kg m2}. This means that the position operator enters in its definition.

    Time is evolution parameter in QM and is associated to a classical observable (reading of a classical clock).
     
  9. Dec 7, 2011 #8

    kith

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    Does this imply that neutrino oscillations can be pictured in the following way:
    A flavour eigenstate with well-defined momentum oscillates, because it is no eigenstate of H=P²/2m where mass is an operator. Or is this picture too naive, because it ignores QFT?
     
  10. Dec 7, 2011 #9

    A. Neumaier

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    One cannot use p^2/2m since this is the nonrelativistic kinetic energy, whereas neutrinos, being almost massless, must travel with almost the speed of light, hence cannot be described in a nonrelativistic approximation. The right formula H=gamma_0(gamma dot p +m) for the kinetic energy comes from the Dirac equation.
    This has little to do with QFT per se.

    All that matters is that in a basis of flavor eigenstates, the mass matrix m is not diagonal. Thus the dynamics with the Hamiltonian with operator mass changes the flavor state.
     
  11. Dec 7, 2011 #10

    kith

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    Ah, sure, my bad.

    Thanks for the reply!
     
  12. Dec 7, 2011 #11
    Even in Classical Mechanics, physical quantities like mass, charge, are not dynamical variables for particles, but, instead parameters that enter in the Lagrangian.

    In Quantum Mechanics, what is called an observable usually has a classical analog that is expressible as a function of coordinates and momenta - the canonical observables. There are other observables that have no Classical analog - most notably intrinsic spin.

    Quantum mechanics maintains that the fundamental observables are identified with the generators of a continuous group of some fundamental symmetry (spatial translations and rotations, for example).

    Among the observables in non-relativistic QM, position and Hamiltonian take a priviledged place. Thus, time in non-rel QM is treated as an independent parameter, very asymmetrically compared to position. Similarly, the Hamiltonian of a system is defined as the generator of temporal evolution. Mass in QM enters as a parameter that relates the momentum operator to the kinetic energy part of the Hamiltonian.

    As of now, no one is able to calculate what the masses of the elementary particles of the Standard Model should be. They are regarded as fitting parameters of the theory.
     
  13. Dec 7, 2011 #12

    A. Neumaier

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    The mass of a star or a rocket and the charge of a condensator depend dynamically on time. Thus, in general, mass is an observable like any other. Only if a system is such that it exchanges no mass with the surrounding, can mass be regarded as a parameter.
     
  14. Dec 7, 2011 #13
    I was referring to the Classical Mechanics of systems of particles. In classical mechanics, these corpuscles cannot be destroyed or created, not can their chemical composition change. Thus, whatever mass they had is fixed. The mass of a system of particles can change by simply changing the number of particles within the system. Thus, the law of conservation of mass in Classical Mechanics is simply a trivial restatement of the law of conservation of the corpuscles.

    In the Classical Mechanics of continuous media, the law of conservation of mass gives rise to a continuity equation. But, this equation is derived with the above implicit assumption. Namely, the mass flow current density at any point in the continuous medium is identified in the following manner. Consider a small volume [itex]\Delta V[/itex]. The corpuscles that are within this volume of the material have a flow velocity [itex]\vec{u}(\vec{x}, t)[/itex]. The total mass of these particles is given by the density at that point [itex]\rho(\vec{x}, t)[/itex] and is equal to:
    [tex]
    \Delta m = \rho \, \Delta V
    [/tex]
    Next, consider the volume to be an oblique cylinder with a base [itex]\Delta A[/itex] and a unit normal [itex]\hat{n}[/itex] and an axis along the particular flow velocity [itex]\vec{u}[/itex] and with a length [itex]u \, \Delta t[/itex]. Then, all of the corpuscles that are in this cylinder will flow through the considered area in a time [itex]\Delta t[/itex]. You should convince yourself that the volume of this cylinder is:
    [tex]
    \Delta V = (\vec{u} \cdot \hat{n}) \, \Delta A \, \Delta t
    [/tex]
    Therefore, the mass flow is:
    [tex]
    \frac{\Delta m}{\Delta t} = (\rho \, \vec{u} \cdot \hat{n}) \, \Delta A
    [/tex]
    But, the very definition of the mass flow density is:
    [tex]
    \frac{\Delta m}{\Delta t} = (\vec{J} \cdot \hat{n} ) \, \Delta A
    [/tex]
    Comparing the last two expressions, we get the following expression for the mass flow current density:
    [tex]
    \vec{J} = \rho \, \vec{u}
    [/tex]
    As we see, the assumptions that corpuscles are the carriers of the mass content of the medium is implicit in this derivation.

    The continuity equation (Law of conservation of mass in differential form) is:
    [tex]
    \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \, \vec{u}) = 0
    [/tex]
    This is one of the fundamental equations of fluid dynamics. If you interpret [itex]\rho[/itex] as charge/entropy density, then this is a continuity equation for those quantities. It means that the corpuscles are the carriers of the particular parameter of the liquid. This is the only way that a quantity can change for an open system in Classical Mechanics.

    The above continuity equation is within the Eulerian description of motion of a continuous medium. If we use the nabla rule:
    [tex]
    \nabla \cdot (\rho \, \vec{u}) = (\vec{u} \cdot \nabla) \rho + \rho \, (\nabla \cdot \vec{u})
    [/tex]
    and use the substantial time derivative:
    [tex]
    \frac{D \rho}{D t} \equiv \frac{\partial \rho}{\partial t} + (\vec{u} \cdot \nabla) \rho
    [/tex]
    then the continuity equation may be written as:
    [tex]
    \frac{D \rho}{D t} + \rho \, (\nabla \cdot \vec{u}) = 0
    [/tex]

    But, if we recall that the divergence of the flow velocity:
    [tex]
    (\nabla \cdot \vec{u}) = \frac{1}{\Delta V} \, \frac{D \Delta V}{D t}
    [/tex]
    is the relative change of the volume of a fluid element (containing the same corpuscles flowing with the liquid), then, we have:
    [tex]
    \frac{1}{\rho} \, \frac{D \rho}{D t} = -\frac{1}{\Delta V} \, \frac{D \Delta V}{D t}
    [/tex]
    [tex]
    \Delta V \, \frac{D \rho}{D t} + \rho \, \frac{D \Delta V}{D t} = 0
    [/tex]
    [tex]
    \frac{D}{D t} \left( \rho \, \Delta V \right) = 0
    [/tex]
    This means, that the total mass contained with a fluid element remains fixed. This is the Law of Conservation of mass in Lagrange's description of the motion of the continuous medium. If we regard the volume element as containing one such corpuscle, it means that the mass of this corpuscle is fixed.

    EDIT:
    Furthermore, dynamical variables are those that describe the state of a mechanical system in Hamiltonian's formalism, i.e. the canonical variables, and not some quantity that changes with time.
     
    Last edited: Dec 7, 2011
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