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A free particle

  1. Jan 12, 2014 #1
    This is a question for those with Shankar in their disposal. When studying from this book, Exercise 5.1.1 puzzled me. Now, it is easy to solve the exercise from 5.1.9. However, how would we expand psi in terms of the energy eigenfunctions,* if we did not have 5.1.9*, but instead started fresh? It seems to me that the factor of m/(2mE)^(1/2) would then be absent from our integrand, instead of being the same as the integrand in Exercise 5.1.1. Can somebody shed some light on this confusion of mine? Thanks.
     
  2. jcsd
  3. Jan 12, 2014 #2

    Simon Bridge

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    So - if you started fresh, with no prior knowledge of 5.1.9, but using the lessons already in the same section and previous, how would you proceed?

    Bear in mind the goal of the calculation.
     
  4. Jan 12, 2014 #3
    See the attachment. Obviously there must be something I am missing. Thanks.
     

    Attached Files:

  5. Jan 12, 2014 #4

    Simon Bridge

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    ## % the following are definitions to make writing QM equations easier:
    \renewcommand{\ket}[1]{\left| #1 \right\rangle}
    \renewcommand{\bra}[1]{\left\langle #1 \right|}
    \renewcommand{\braket}[2]{\left\langle #1 \right|\left. #2\right\rangle}
    \renewcommand{\dE}{\;\text{d}E}
    ##
    You have to think carefully about what each term means - start further back.

    What are you trying to find out?
    What does the |E,a> signify compared with |ψ>
    What does the first equation at the top of the page mean?

    Stuff like that ... to understand what is going on you need to spell out your reasoning as you go.

    Note: if you rely on posting screenshots it will quickly get tedious
    instead: you can write out equations here. i.e. your first line is:
    $$\ket{\psi}=\int_0^\infty \sum_a \braket{E,a}{\psi}\ket{E,a}\dE$$

    ... to see how I did that, use the "quote" button at the bottom of this post.

    Your second line just asserts that ##\ket{\psi}## is a solution to the time dependent Schrödinger equation. If it is a free particle, then V=0. But are there other initial conditions?

    What you seem to be saying is that your approach is to expand the wavefunction in terms of energy eigenstates, which may include continuous and discrete eigenvalues ... so you sum over the discrete states and integrate over the continuous ones.

    What you are trying to do is work out the result without having an explicit form for the eigenstates.

    Is this correct?
     
  6. Jan 13, 2014 #5
    Yes, that is correct. All I am trying to do is to expand psi in terms of of the energy eigenstates. My derivation for ## \braket{E,a}{\psi}## is based on the one leading to 4.3.11a. The only initial condition is psi at t=0, which does not matter for U(t). I still cannot reconcile my result with that of the exercise. What also confuses me is that if I had started out with
    $$\ket{\psi}=\int_0^\infty \sum_a \braket{E,a}{\psi}\ket{E,a}\dE$$
    and then gone to p, I would get a term involving p in the integrand that is not present in 5.1.9
     
  7. Jan 13, 2014 #6

    Simon Bridge

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    Start with the end in mind.
    What was the exercise trying to show compared with what you are doing?
    You seem remarkably reluctant to commit to stating it yourself - I know there's a statement in the book, but writing it out yourself can focus the mind: it's a good discipline and makes sure that we are on the same page.

    At first I thought you were puzzling about how to evaluate ##\ket{E,a}\bra{E,a}## - but that no longer seems to be the case. It may help if you attempted to articulate what you are having trouble with.

    Note:
    A lot of the time the math is simpler in another representation - or it is better for showing the properties of interest.
     
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