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Momentum and energy eigenvalues

  1. Aug 10, 2014 #1

    dyn

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    Hi. I will give you a question I have looked at and then tell you where I am confused.

    The wavefunction for a particle of mass m is ψ(x) = sin(kx)exp(-iωt) where k is a constant.

    (i) Is this particle in a state of defined momentum ? If so , determine its momentum.
    (ii) Is this particle in a state of defined energy ? If so , determine its energy.

    For (i) i thought if k is a constant then p=[itex]\hbar[/itex]k must also be a constant ie a definite momentum. The solution says to apply the momentum operator to ψ which shows that ψ is not an eigenfunction of momentum. Does p=[itex]\hbar[/itex]k not always apply ? If ψ is not an eigenfunction of momentum what happens when momentum is measured ?
    For (ii) i just assumed E=[itex]\hbar[/itex]ω which in this case turns out to be right although according to the solution i should have used the energy operator i[itex]\hbar[/itex]dψ/dt which gives an eigenvalue of E=[itex]\hbar[/itex]ω. Could i instead apply the Hamiltonian operator to give an energy eigenvalue of [itex]\hbar[/itex]^2k^2/2m ? Are these 2 methods equivalent ?
     
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  3. Aug 10, 2014 #2

    bhobba

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    Apply the momentum operator -i d/dx and see if it is ie it must be an eigenstate of it. From a quick calc I did it doesnt look like it.

    That cant be done until you know the Hamiltonian - it's not like the momentum operator.

    But once you do - same thing.

    Thanks
    Bill
     
    Last edited: Aug 11, 2014
  4. Aug 10, 2014 #3

    dyn

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    So I can always apply i[itex]\hbar[/itex]d/dt but only apply the Hamiltonian when I know its form ? But do both methods always give equivalent answers even though they will be in different form ?
    Applying the Hamiltonian for an infinite well gives E= [itex]\hbar[/itex]^2k^2/2m. Doesn't this answer imply the momentum is [itex]\hbar[/itex]k ?
     
  5. Aug 11, 2014 #4

    bhobba

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    This is basic QM. The momentum operator (basically - there is a subtle caveat associated with Lagrangian's and the concept of generalised momentum - the momentum operator applies to the generalised momentum - but no need to go into that here) is always the same.

    Mate you are intermixing concepts.

    The momentum operator is as I stated. Its exactly the same in classical mechanics, momentum is the derivative of mass times velocity - always (except for the caveat previously).

    The energy changes depending on the Hamiltonian.

    Thanks
    Bill
     
  6. Aug 11, 2014 #5

    dyn

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    Sorry , I think my last post was misleading. I meant ; if I know the Hamiltonian and am asked to calculate the energy eigenvalue can I use either i[itex]\hbar[/itex][itex]\partial[/itex]/[itex]\partial[/itex]t or apply the Hamiltonian. These 2 methods give 2 different answers but are they equivalent ?
    My other question is what happens if momentum is measured and the particle is not in a momentum eigenfunction ? Do we get no measurement or a superposition of measurements ?
     
  7. Aug 11, 2014 #6

    jtbell

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    If it is not in a momentum eigenstate, its state can be expressed as a superposition of momentum eigenstates. When you measure the momentum, you get one of the momentum eigenvalues that are contained in the superposition, randomly "chosen", with a probability equal to the complex square of the coefficient of that eigenstate in the superposition.
     
  8. Aug 11, 2014 #7

    dyn

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    Thanks. In the "real world" when you make a measurement, the particle doesn't know if is in a momentum eigenstate or a superposition of eigenstates , so how would you know what your result means ?
     
  9. Aug 11, 2014 #8

    bhobba

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    You need to understand the axioms of QM.

    First axiom is given an observation there is a Hermitian operator, O, called the observations observable, such that its possible outcomes are the eigenvalues of the observable.

    The second axiom is called the Born Rule. At your level it would be expressed as - if the system is in state |u> the expected outcome (ie it's average) is <u|O|u>.

    A little math shows (any book on QM will give the detail) only in the case if its an eigenstate can we know with 100% certainly what the outcome will be.

    Thanks
    Bill
     
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