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Wave function homework Problem 2.1 in Griffiths' book

  1. Nov 4, 2016 #1
    upload_2016-11-4_21-29-25.png
    In the (b),I have some questions:
    (1) Does it mean ψ can be real or not real?
    (2) Why do the solutions of linear combination must have the same energy? As I know, these solutions are often different, as long as they are eigenvalues of time-independent Schrodinger equation.
    (3) In the sentence "......as well stick to ψ's that are real", what does "that" denote?
     
  2. jcsd
  3. Nov 4, 2016 #2

    BvU

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    Hi,

    Your question falls out of the blue for someone who doesn't have Griffiths at hand. Provide some more context (not by attaching pages upon pages of pictures, but in a few words -- that helps you to understand the question as well).

    This section is an introduction to the time independent Schroedinger equation (separation of variables ##\bf x## and ##t## in math lingo)

    Apparently (a) is no problem for you.
    (1) For (b) you ask what he says in the note ? When you type ψ no one knows if you mean his ##\Psi## or his ##\it \psi##. What is it specifically that isn't clear ?
    (2) that's not what it says. But if you express a solution with a given energy as a linear combination of other solutions, those better have the same energy !
    (3) you can also read "stick to real ##\it \psi##"
    so there is no reason to ask what "that" stands for any more :smile: . perhaps you want to rephrase ?
     
  4. Nov 4, 2016 #3
    Thank you for pointing out my improper way to describe my question and answering my question.
     
  5. Nov 4, 2016 #4

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    All in good spirit. You think you have it figured out now ?
     
  6. Nov 4, 2016 #5
    No, I am thinking. Maybe tomorrow.
     
  7. Nov 4, 2016 #6
  8. Nov 6, 2016 #7
    It is important to remember the two main purposes of ## \Psi ## and ## \psi ##: ## \Psi ## is defined to be complex by equation 2.14 as you have posted. I am sure Griffiths at some point has explained the concept of an expectation value where, when normalized, $$\int_{-\infty}^{\infty} |\Psi(x,t)|^{2}dx = 1$$
    The absolute value eliminates the complex nature of the wave function. Additionally, ## \psi ## is an eigenfunction which, when an operator is applied, produces an eigenvalue, in this case the energy. In his hint, he states that ## \psi ## and ## \psi^{*} ## both result in the same eigenvalue (energy), so he is saying you might as well make ## \psi ## real. Griffiths has a very comprehensive appendix on linear algebra that may prove helpful to read (if you continue into chapter 3, I would highly recommend reading the appendix).
     
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