Normalization of a wavefunction

Click For Summary

Homework Help Overview

The problem involves normalizing a wavefunction for a particle of unit mass in an infinite square well, specifically the wavefunction u(x) = A sin(3πx/a). The task is to determine the normalization constant A.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of integrating the wavefunction squared over the appropriate limits, questioning the correct bounds for integration given the infinite square well constraints.
  • There is confusion regarding the behavior of the wavefunction outside the well, with some participants clarifying that it is zero outside the defined region.
  • Participants explore the implications of integrating from -∞ to ∞ and the need to consider the wavefunction's value in different sections of the domain.

Discussion Status

The discussion is ongoing, with participants actively clarifying the setup of the problem and the integration limits. Some guidance has been provided regarding the wavefunction's behavior outside the well and the need to integrate only where the wavefunction is non-zero.

Contextual Notes

Participants are grappling with the normalization condition and the implications of the infinite square well's boundaries on the integration process. There is an emphasis on correctly identifying the sections for integration based on the wavefunction's definition.

EmmaLemming
Messages
18
Reaction score
0

Homework Statement



This is a multi-choice question.

A particle of unit mass moving in an infinite square well,

V = 0 for lxl ≤ a
V = ∞ for lxl > a

is described by the wavefunction, u(x) = A sin (3∏x/a)

If the wavefunction is normalised, What is A?

a) 1/2a
b) 1/√2a
c) 1/√a


Homework Equations



I know that the integral of the wavefn squared is equal to 1 because it has to exist somewhere but when I tried integrating it, it either all went to 1 or ∞.

I know how to do this question, I just can't. An easy to follow mathematical proof would be most helpful.

The Attempt at a Solution



I am integrating between ∞ and -∞ is that correct?

so far I've got that

∫ A2 sin2 (3∏x/a) dx = 1

using the identity: cos (2x) = 1 - 2 sin2(x)

= A2/2 ∫ 1 - cos (6∏x/a) dx = 1


And now I'm stuck...
 
Physics news on Phys.org
EmmaLemming said:
is described by the wavefunction, u(x) = A sin (3∏x/a)

You need to be careful here. This is not true for all x. Its in an infinite square well, so what will the wave function be outside of that well?
 
Ah ok, I thought the wave function wouldn't exist outside of the square well so the wave function would be zero..?

I got the sin (3∏x/a) wavefunction given to me in the question so I just took it as true.
 
yes, you're right. The wave function outside the square well is zero. and inside the well, it is sin (3∏x/a). Also, you were right that the integral is from -∞ to ∞. But what is the integrand for lxl > a ?
 
I don't know. This is just a guess would it be,

between 0 and a

∫ A sin (3∏x/a) dx

This is where I get confused, because I thought to do this question all I have to do is square the wavefunction and integrate between -∞ and ∞.

Is there another step before hand?
 
the wave function is A sin (3∏x/a) for lxl ≤ a and it is zero for lxl > a

So you do need to 'square the wavefunction and integrate between -∞ and ∞', but the wave function will be zero for lxl > a

In other words, you need to identify the different 'sections' and integrate each section, to get the integration over all space.
 
Sorry if I'm being slow and thanks for your help.

but would one of the sections be an integral between a and -a ?

Do I do that and then integrate between -∞ and ∞?
 
no worries. Yeah, one section would be from -a to a. The total integral is from -∞ to ∞, so what would the other sections be?
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K