Wave function homework Problem 2.1 in Griffiths' book

[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?

 

[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 ?

 

[Tspirit]

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 ?

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

 

[BvU]

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

 

[Tspirit]

No, I am thinking. Maybe tomorrow.

 

[Tspirit]

 

[Jamison Lahman]

View attachment 108454
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?

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).