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Free particle - Eigenstates expansion

  1. Jul 16, 2011 #1
    Hi everybody!!
    Two question for you:

    1) Take a free particle, moving in the x direction.
    Its (time indipendent) wave function, in terms of the momentum is [itex]\psi(x)=\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}[/itex].

    Now, i know the momentum of the particle: p.
    So i should not know anything about its position, according to the uncertainty principle.

    Effectively i don't get any information on "where" the particle is looking at the wave function; but the fact that the [itex]\psi[/itex] depends on the position does mean something particular?

    2) Take an eigenstate of the hamiltonan operator [itex]|E\rangle[/itex].
    Now, they say that this wave function, in terms of the momentum is:[tex]\psi_p=\langle p|E\rangle[/tex] or, expressed with position eigenstates:[tex]\psi_x=\langle x|E\rangle[/tex]But those are just numbers, so it should't be like this instead:[tex]\psi_p=\int |p\rangle\langle p|E\rangle\;dp[/tex]and [tex]\psi_x=\int |x\rangle\langle x|E\rangle\;dx[/tex]??

    Thanks for your help!!
  2. jcsd
  3. Jul 16, 2011 #2


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    No, nothing particular except for the lack of a direct probabilistic interpretation, since the wavefunction is not normalized.

    psi(x) is an eigenfunction of the momentum operator in the coordinate representation. psi(p) as the Fourier transformation of psi(x) would be then the eigenfunction of the coordinate operator in the momentum representation.

    No, you have to pay attention,

    [tex]\psi_x=\int |x\rangle\langle x|E\rangle\;dx[/tex]

    is not correct since it's not an expansion of the same function wrt a basis of (generalized) eigenvectors of some self-adjoint operator (in this case x). So the correct version would be

    [tex] |E\rangle = \int_{x} dx ~ \langle x|E\rangle |x\rangle [/tex]
  4. Jul 16, 2011 #3


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    I'll answer (2), and you've pretty much got it straight already. ψ(p) = <p|ψ> is the wavefunction, a complex-valued function of p representing the probability amplitude for finding the particle with momentum p. On the other hand, |ψ> = ∫|p><p|ψ> is the state, a superposition of momentum eigenstates with coefficients <p|ψ>. They're usually interchangeable, and which you use depends on how comfortable you feel with the Dirac bra-ket notation that is used for states.
  5. Jul 16, 2011 #4


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    Numbers [itex]\psi(x)=<x|\psi>[/itex] instead of state vectors [itex]|\psi>[/itex] appear in Schrödinger's Equation, if you work in a chosen representation (like the position or momentum representation). To do this, you multiply
    [tex]i\hbar\partial_t |\psi> = H|\psi>[/tex]
    by [itex]<x|[/itex] respectivly [itex]<p|[/itex].
    Last edited: Jul 16, 2011
  6. Jul 17, 2011 #5
    Thanks for the attention!!
    But I still have some problem with the 2) part.

    Basically, when I have to solve the eigenvalue problem for a generic hamilonian [itex]H|\psi\rangle=E|\psi\rangle[/itex] what exactly is the [itex]\psi[/itex] i'm plugging in?

    I mean, I have to choose the representation (coordinate or momentum) to use before solving that, becaouse in the coordinate one the momentum is an operator [itex]-i\hbar\nabla p[/itex] while in the momentum representation the position operator is [itex]i\hbar\nabla x[/itex], so effectively I have to decide what to use first.

    So i can guess that there is a wave function of my particle, [itex]|\psi\rangle[/itex], but the one i'm actually using to solve the hamiltonian is [itex]\psi_x=\langle x|\psi\rangle[/itex] or [itex]\psi_p=\langle p|\psi\rangle[/itex].

    And so all the eigenfunction of the hamiltonian I get are function in the representation i choose, for example in the coordinate rapresentation the eigenfunction for the particle in the infinite well is like [itex]|\psi_n\rangle=\sqrt{\frac{\pi}{a}}{\sin(n\pi x)dx}[/itex]; and if I integrate that in a fixed interval it gives me the probability of finding the particle in that interval, so it certainly is in the coordinate representation.

    What confuses me is that I've concluded that the eigenfunction [itex]|\psi_n\rangle=\sqrt{\frac{\pi}{a}}{\sin(n\pi x)dx}[/itex] is a wave function in the coordinate representation; BUT also a ket. So it is a [itex]\psi_x=\langle x|\psi\rangle[/itex] and a [itex]|\psi\rangle[/itex] at the same time.


    However should not the momentum wave function be in the form [tex]\psi_p=\frac{1}{\sqrt{2\pi\hbar}}\int_{\mathbb{R}}\psi(x)e^{i\frac{p}{\hbar}x}\;dp[/tex]with [itex]\psi(x)=\langle x|p\rangle[/itex]??

    (I've already answered myself to this one, haven't I? :tongue2:)
    Last edited: Jul 17, 2011
  7. Jul 17, 2011 #6


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    tunafish, You probably have it straight now, but just in case let me summarize. When you write H|ψ> = E|ψ>, you're using Dirac notation and |ψ> is a state vector. Choosing a representation means for example that you insert |x>'s everywhere: ∫dx'<x|H|x'><x'|ψ> = E<x|ψ>. Now you can say p is a differential operator, p = -iħ∂/∂x, and put that in the Hamiltonian. The resulting equation can be solved for <x|ψ> = ψ(x), which is a function.
  8. Jul 17, 2011 #7


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    Your notation is still not right.
    Yes, this expression doesn't make sense. On the left-hand side, you have a state vector (ket, element of Hilbert space). On the right hand side, you have the component ψ(x) of this vector in the position representation (a wave function, if you take x as variable). Also, the dx is wrong here.

    How do you get this expression? The right way to proceed is this: start with the Schrödinger Equation for the state vector, apply the bra <x| or <p| to both sides and evaluate the scalar product for the Hamiltonian. If you write this out step by step, we are able to tell you what went wrong.
    No, <p|x> is the exponential. Do you know, how to get this expression? You start with ψ(p)=<p|ψ> and insert the completeness relation in respect to the position states in between <p| and |ψ>.
  9. Jul 17, 2011 #8
    Ok, i figured that out, my thanks to Bill_k and Kith!!

    Kith, please tell me if this is correct (but I think it is):

    1) It's [itex]\langle x|\psi\rangle=\sqrt{\frac{\pi}{a}}\sin(n\pi x)[/itex]
    yeah, i can see why now. And i don't know why on earth i've put that dx!!

    2) The momentum wave function is indeed wrong.
    In fact:[tex]\psi_p=\langle p|\psi\rangle=\int_{\mathbb{R}}{\langle p|x\rangle\langle x|\psi\rangle\;dx}[/tex]
    where [itex]\langle p|x\rangle=\frac{e^{-i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}[/itex] and [itex]\psi_x=\langle x|\psi\rangle [/itex]

    Thanks to both of you guys!!
    Last edited: Jul 17, 2011
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