Wave function homework Problem 2.1 in Griffiths' book

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1. Nov 4, 2016

Tspirit

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. Nov 4, 2016

BvU

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 . perhaps you want to rephrase ?

3. Nov 4, 2016

Tspirit

Thank you for pointing out my improper way to describe my question and answering my question.

4. Nov 4, 2016

BvU

All in good spirit. You think you have it figured out now ?

5. Nov 4, 2016

Tspirit

No, I am thinking. Maybe tomorrow.

6. Nov 4, 2016

7. Nov 6, 2016

Jamison Lahman

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).