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Spectrum of Momentum operator in the Hilbert Space L^2([-L,L])

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the spectrum of the Momentum operator in the Hilbert Space defined by L^2([-L,L]), consisting of all square integrable functions ψ(x) in the range -L, to L

    2. Relevant equations

    We can get the resolvent set containting all λ in ℂ such that
    you can always find a vector ψ(x) in the Domain of [itex]\hat{p}[/itex] that satisfies the following equation:
    (λI - [itex]\hat{p}[/itex]) ψ(x) = [itex]\Phi[/itex](x)
    where [itex]\Phi[/itex](x) is any state in the Hilbert Space.

    The Spectrum of the operator is the complement of the resolvent set with respect to the complex plane

    The momentum operator, in position representation, is given by [itex]\hat{p}[/itex] =[itex]\frac{\hbar}{i}\frac{d}{dx}[/itex] with domain consisting of all functions in the Hilbert Space with square integrable derivatives and whose values at -L and L are equal.

    3. The attempt at a solution

    I apply the operators on the state ψ(x) to get the following expression for the LHS:
    λψ(x) - [itex]\frac{\hbar}{i}[/itex]ψ'(x)

    I then move the derivative term to the RHS and divide by lamda to get an expression for ψ(x):
    [itex]\frac{\hbar}{iλ}[/itex]ψ'(x) + ([itex]\Phi[/itex](x))/λ.

    Now here's the problem. If I take the absolute square of both sides and integrate, lamda just becomes a constant outside the integral. Now both integrals will converge, the second term due to it being part of the Hilbert Space and the first term being a prerequisite for being in the Hilbert Space. So the only value of λ that is not in the resolvent set (and hence in the spectrum) is 0. And even that is suspect, as that would mean we divided by zero in the initial steps. Even if I differentiate both sides before integrating (to test the second condition for being in the domain of p_hat), lamda still falls out, which instead leads to the entire complex plane being in the spectrum, as the derivative of [itex]\Phi[/itex] is not necessarily square integrable.

    I can't make sense of both answers, so surely I've done something terribly wrong. What is it?
     
  2. jcsd
  3. Oct 8, 2013 #2

    vanhees71

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    Is the "momentum operator" you use on the considered Hilbert space essentially self-adjoint? This is neither a trick nor a trivial question!
     
  4. Oct 8, 2013 #3
    It's symmetric within the defined domain but it's not self adjoint
     
  5. Oct 8, 2013 #4

    dextercioby

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    Where did you use that the Hilbert space is L^2[-L,L] ? This should normally put constraints on your 'wavefunction'* which would help you determine all possible things: the domain of hbar/i d/dx, the domain of its adjoint (if any), the deficiency indices and the spectrum of hbar/i d/dx.

    *such as psi(-L)=psi(L) =0.
     
  6. Oct 8, 2013 #5

    dextercioby

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    Let's turn to your method of finding the spectrum, i.e. the resolvent method.

    λψ(x) - ℏiψ'(x) = Φ(x). Now there's a trick: You're not allowed to divide by λ, because λ=0 can be a valid value in the resolvent set. So you get

    λψ(x) = Φ(x) + ℏiψ'(x).

    The conditions you set on functions ψ(x) in the domain of the operator p =ℏi d/dx at the end points -L and L single out that they are no longer valid for the derivatives of those functions. This tiny thing proves that the operator is symmetric, but not self-adjoint. So you can indeed conclude that there's no complex λ sending an arbitrary ψ (in particular the psi's at the 2 endpoints) into a sum of an arbitrary function and the derivative of ψ. If the resolvent set is empty, the spectrum of the derivative operator is the entire C.

    One shows that the ℏi d/dx on L^[-L,L] with ψ(-L)=ψ(L)=0 is closed, symmetric but not selfadjoint and has deficiency indices 1,1, hence admits a uniparameter set of self-adjoint extensions with real spectrum.
     
  7. Oct 13, 2013 #6

    dextercioby

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