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Free particle in quantum mechanics, Dirac formalism

  1. Jun 30, 2011 #1
    The problem is very easy, maybe just something about eigenvectors that I'm missing. Go to the first two pages of the 5th chapter of ''Principles of Quantum Mechanics'', by Shankar, 2nd edition.

    1. The problem statement, all variables and given/known data

    Shankar wants to find the solution for a free particle in Quantum Mechanics. I've got a problem with what he calls ''trial solution'', and the change from an eigenvector to another.

    Hamilton operator: [itex]H=\frac{P^{2}}{2m}[/itex] for a free particle

    [itex]P[/itex], the momentum operator, [itex]m[/itex] the particle's mass

    State vector (ket): [itex]\left|\psi\right\rangle[/itex]


    2. Relevant equations

    The Schroedinger equation:

    [itex]i\hbar\\left|\dot{\psi}\right\rangle = H\left|\psi\right\rangle = \frac{P^{2}}{2m}\left|\psi\right\rangle[/itex]

    The time independent Schrodinger equation:

    [itex]H\left|E\right\rangle = \frac{P^{2}}{2m}\left|E\right\rangle = E\left|E\right\rangle[/itex]

    3. The attempt at a solution

    Shankar then says that ''any eigenstate of [itex]P[/itex] is also an eigenstate of [itex]P^{2}[/itex]'', which is very clear. Then he says: ''feeding the trial solution [itex]\left|p\right\rangle[/itex] '' into the time independent Schrodinger eq., he gets:

    [itex]\frac{P^{2}}{2m}\left|p\right\rangle = E\left|p\right\rangle[/itex]

    But I don't get it. Why did he change from the [itex]\left|E\right\rangle[/itex] eigenbasis to the [itex]\left|p\right\rangle[/itex] eigenbasis, and still maintained the eigenvalue E? Is there atheorem of eigenbasis I'm missing? I mean, [itex]\left|p\right\rangle[/itex] is an eigenvector of [itex]P[/itex], by definition, and then, of [itex]P^{2}[/itex]. Then, from the time independent Schr. eq. we see that [itex]\left|E\right\rangle[/itex] is an eigenvector of [itex]P^{2}[/itex] too. But that doesn't mean it's the same eigenbasis, does it? Or what does ''trial'' mean in this context (forgive me, my mother language is spanish).

    After that ''change of basis'' from [itex]\left|E\right\rangle[/itex] to [itex]\left|p\right\rangle[/itex], everything is fine. So please, explain it to me.

    Please, forgive me, I know this is a very stupid question, but I just don't get it.

    Thankyou very much for your patience.
     
    Last edited: Jun 30, 2011
  2. jcsd
  3. Jun 30, 2011 #2

    vela

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    The ket [itex]\vert p \rangle[/itex] is an eigenstate of both the momentum operator [itex]\hat{p}[/itex] and the Hamiltonian [itex]\hat{H}[/itex]. If you apply Hamiltonian to it, you will get the corresponding eigenvalue of the Hamiltonian, namely E.

    The [itex]\vert E \rangle[/itex] and [itex]\vert p \rangle[/itex] bases aren't the same. While a state [itex]\vert p \rangle[/itex] is an eigenstate of [itex]\hat{H}[/itex], a state [itex]\vert E \rangle[/itex] isn't necessarily an eigenstate of [itex]\hat{p}[/itex]. For example, you can show that a state with the wave function [itex]\psi(x) = e^{ikx} + e^{-ikx}[/itex] is an eigenstate of [itex]\hat{H}[/itex] with eigenvalue [itex]E=\hbar^2k^2/2m[/itex] but it's not an eigenstate of [itex]\hat{p}[/itex].
     
    Last edited: Jun 30, 2011
  4. Jun 30, 2011 #3
    Dear Lord, how could I be so blind??!! That's crystal clear!! Thank you so much, vela, for taking the time for answering such a silly question!! THANK YOU!!
     
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