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Expectation value for momentum operator using Dirac Notation

  1. Mar 2, 2014 #1
    Question and symbols:

    Consider a state|ε> that is in a quantum superposition of two particle-in-a-box energy eigenstates corresponding to n=2,3, i.e.: |ε> = ,[1/(2^.5)][|2> + |3>], or equivalently:
    ε(x) = [1/(2^.5)][ψ2(x) + ψ3. Compute the expectation value of momentum: <p> = <ε|[itex]\widehat{}p[/itex]|ε>.

    Relavent equations:

    [itex]\widehat{}p[/itex] = -i[STRIKE]h[/STRIKE](∂ψ/∂x)

    for n=2,...; ψ = sqrt(2/L) sin(2πx/L)

    Attempt at solution:

    I've computed <ε|[itex]\widehat{}p[/itex]|ε>
    = .5[ <2| + <3| ] [itex]\widehat{}p[/itex] [ |2> + |3> ]
    = .5 [<2|[itex]\widehat{}p[/itex]|2> + <3|[itex]\widehat{}p[/itex]|2> + <2|[itex]\widehat{}p[/itex]|3> + <3[itex]\widehat{}p[/itex]|3>]

    = -i[STRIKE]h[/STRIKE]/L [ ∫L0 sin(2πx/L)(∂ψ/∂x)sin(2πx/L) + sin(3πx/L)(∂ψ/∂x)sin(2πx/L) + sin(2πx/L)(∂ψ/∂x)sin(3πx/L) +sin(3πx/L)(∂ψ/∂x)sin(3πx/L)

    = -i[STRIKE]h[/STRIKE]/L [ ∫L0 (L/2π)sin(2πx/L)cos(2πx/L) + (L/2π)sin(3πx/L)cos(2πx/L) + (L/3π)sin(2πx/L)cos(3πx/L) + (L/3π)sin(3πx/L)cos(3πx/L)

    I've tried to:

    1) convert the trigs to exponentials and work through exponentials
    2) directly integrate all of these
    3) convert the trigs into different trig identities

    I've been told that the answer = 0 but the closest I got when I converted to exponentials was 1.

    Sorry if the text doesn't look right, this is my first time. the "h" with the cross through it is plancks constant divided by 2π
     
    Last edited: Mar 2, 2014
  2. jcsd
  3. Mar 2, 2014 #2

    Simon Bridge

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    Have you tried exapnding the epsilon states in terms of momentum eigenstates?
    You will also benefit from learning LaTeX.

    i.e. $$\hat p = -i\hbar \frac{\partial}{\partial x}$$
     
  4. Mar 2, 2014 #3

    hilbert2

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    The expectation value of momentum is zero for any real-valued wavefunction. To see why this is true, note that

    ##\left<\hat{p}\right>=-i\hbar\int_{-\infty}^{\infty}\psi^{*}(x)\frac{d}{dx}\psi(x)dx##,
    which is obviously pure imaginary if the function ##\psi## is purely real. But on the other hand, the expectation value of ##\hat{p}## must be real because ##\hat{p}## is a hermitian operator. Therefore the only possibility is that ##\left<\hat{p}\right>=0##.
     
  5. Mar 3, 2014 #4

    Simon Bridge

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  6. Mar 4, 2014 #5

    vanhees71

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    Sigh :-(. Why are they asking to calculate things that don't exist? One cannot state it often enough: There is no momentum operator for a particle in a finite box!
     
  7. Mar 4, 2014 #6

    dextercioby

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    Oh, Hendrik, but of course there is: the momentum operator defined as the derivative wrt to position (drop -ihbar, it's irrelevant) exists as a linear operator on the subset of L^2 [0,L] of (eq. classes of) wavefunctions with the property that psi(0)=psi(L)= 0 (the so-called 'physical boundary conditions'). It can be shown to be linear and symmetric, but not self-adjoint. Its maximal domain of definition contains the basis set of Hamiltonian eigenfunctions, henceforth you can compute its matrix elements with any vector in their linear span. The OP's problem really makes physical sense.

    N.B. I took the liberty to alter the appearance of the quoted post.

    The mathematical theory of the finite 1D-box of 0 potential inside the box and (physically non-tractable) infinite potential outsde the 1D-box is fully presented in Akhiezer and Glazman, Vol.1, pp. 106 to 110.
     
    Last edited: Mar 4, 2014
  8. Mar 4, 2014 #7

    strangerep

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    I'm reasonably sure Hendrik meant "...there is no self-adjoint momentum operator for a particle in a finite box", but... I should let him speak for himself.

    Certainly, one can compute matrix elements. But one might object that since this operator is not self-adjoint, it doesn't qualify as a good quantum observable, hence is not "physical".
     
  9. Mar 4, 2014 #8

    strangerep

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    Getting back to what OP actually asked...

    Show how you're doing one of the integrals, e.g., the one involving ##\sin(3\pi x/L) \cos(2\pi x/L)## .
     
  10. Mar 5, 2014 #9

    vanhees71

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    Of course you can differentiate an [itex]\mathrm{L}^2([0,L])[/itex] function, but the operator [itex]-\mathrm{i} \partial_x[/itex] is not self-adjoint and thus doesn't represent an observable. So there is no momentum operator for the particle in a finite box with rigid boundary conditions. I've shown why, in my previous posting. It seems to be a subtlety, but this kind of inaccuracies give rise to a lot of misunderstandings in quantum theory. A very nice didactical paper on this issue is

    F. Gieres, Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys. 63 (2000) 1893
    http://arxiv.org/abs/quant-ph/9907069
     
  11. Mar 5, 2014 #10

    Simon Bridge

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    I agree with strangerep - we need to hear back from sddang (the OP) before we can do any more on this question.

    I don't think that an discussion of the existence or otherwise of the momentum operator or whatever is going to help with the question at hand.
     
  12. Mar 5, 2014 #11

    dextercioby

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    Well, it's expected that due to the unphysical V=infinity outside the box, psi =0 outside the box. But why would it be necessarily 0 at the 2 points which form the edges of the box ? That's why the operator is not self-adjoint, because with this constraint its domain is not maximal, it will allow for non-trivial symmetric extensions (actually self-adjoint), so that the so called 'physical boundary conditions' (1) are too restrictive. If you 'relax'

    [tex] \psi (0) = \psi (L) = 0 [/tex] (1)

    to

    [tex] \psi (0) = e^{-i\alpha} \psi (L) \neq 0 ~, \alpha\in\mathbb{R}[/tex] (2)

    the momentum operator (-ihbar derivative) becomes self-adjoint for any value of the real parameter, hence it will be a 'good observable'.
     
  13. Mar 5, 2014 #12

    vanhees71

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    Fine! That's a way out, although I don't know what's the physical interpretation of this situation now. Admittedly, the infinite potential is as unphysical as this. Note, however, that now the position operator becomes problematic. At least multiplication of the wave function with [itex]x[/itex] is not a self-adjoint operation anymore (but still hermitean as is the case with the momentum operator for rigid boundary conditions).

    A nice (well-known) example, how this model can be misleading is given by the similar case of the rigid rotator model. This describes a particle restricted to a circle of radius [itex]R[/itex]. It can be described by the square-integrable functions [itex]\psi(\varphi)[/itex], [itex]\varphi \in [0,2 \pi[[/itex] with periodic boundary conditions [itex]\psi(2 \pi)=\psi(0)[/itex]. The angular-momentum operator is given by [itex]\hat{J}=-\mathrm{i} \partial_\varphi[/itex]. Obviously you have the commutation relation
    [tex][\varphi,\hat{J}]=\mathrm{i}.[/tex]
    Now a naive conclusion is as for usual position and momentum that
    [tex]\Delta \varphi \Delta J \geq 1/2.[/tex]
    On the other hand there are proper eigenstates of [itex]\hat{J}[/itex] with eigenvalues [itex]0,1,\ldots[/itex] with eigenstates
    [itex]u_n(\varphi)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} n \varphi) \; n \in \mathbb{N}_0.[/tex]
    These form a complete set on the above defined Hilbert space.

    Now there's obviously a contradiction with the angle-angular-momentum uncertainty relation, because in any of these eigenstates the variance of angular momentum is 0.

    The resolution of the puzzle is again to carefully think about the question, whether the angle is a proper observable in this case!
     
  14. Mar 5, 2014 #13

    Simon Bridge

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    <puzzled> how does this help sddang?
     
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