# Normalising wave functions.

1. Dec 20, 2009

### hhhmortal

1. The problem statement, all variables and given/known data

A particle of unit mass moving in an infinite square well, V=0 for |x|<= a , V=∞ for |x|>a , is described by a wave function u(x) = Asin(3πx/a).

(i) If I normalise the wave function, what is A?

(ii) And what is the energy of state described by this wave function?

3. The attempt at a solution

(i) Normalised the wave function by saying it is only valid from 0<x<=a and every where else is 0.

So I square the wave function, integrate it and I get:

A²a = 1 --> 1/√a

(ii) And the energy of state decribed by this wave function would be 9h²/8ma² , since we know its on n=3.

What Im puzzled about is if the width of the well is 2a or a? I know it is 'a' because the denominator of the wave function tells me it is. But when I use the general formula of:

A= √(2/L) I know L = a so I instead get:

A= √(2/a) which isn't true?

Thanks.

2. Dec 20, 2009

### diazona

Why only $0 < x \le a$? Take another close look at that question to figure out the range over which the potential is non-infinite.

The form of the wave function tells you nothing, in general, about the width of the well. You have to get that from other information in the problem - the width of the region in which the potential is zero.

3. Dec 20, 2009

### Cryxic

Your result is close but awkward on the first part. Go back and redo the integral. You should get the square root of 2 divided by a, not 1 divided by a. The bounds of integration are 0 to a. Given what you've presented here, the width of the well is a.

You have the right idea for the second part, but it should be:

$$E = 9\pi^2h^2/(2ma^2)$$

4. Dec 20, 2009

### hhhmortal

It's actually from -a< x <a which will give a well of width 2a.

If I integrate now this will become 2a(A²) = 1 --> 1/√2a

But If I use A= √(2/L) I know L = 2a so I instead get:

A = √(1/a)

Im clearly going wrong somewhere

5. Dec 21, 2009

### White

Probably posting your integration helps others to spot your mistakes.

if you really do not see it on your own.

Well you probably remember cos^2 + sin^2 = 1, so integrating over full peroids of a sine squared (or cosine squared) gives you half of the peroid length.

6. Dec 22, 2009

### brain90

For the first part the answer is A=1/√a

7. Dec 24, 2009

### hhhmortal

Yea it was a mistake I made with the integration, I finally got the correct answer. Is the energy state for this wave function simply:

9h²/8ma²

since we know n=3 from the wave function given?

8. Dec 24, 2009

### brain90

Yes.

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