# What is the Normalisation Constant for a Particle in an Infinite Square Well?

• hhhmortal
In summary, normalizing a wave function is the process of ensuring that the total probability of finding a particle in a given space is equal to 1, which is a fundamental principle in quantum mechanics. This is necessary for accurate predictions of particle behavior. A wave function is normalized by dividing it by its norm, and if it is not normalized, it can lead to inaccurate predictions and violate fundamental principles. Any wave function can be normalized, but some may require more complex mathematical techniques.
hhhmortal

## Homework Statement

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?

## 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 I am 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.

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

hhhmortal said:
What I am 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:
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.

hhhmortal said:

## Homework Statement

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?

## 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 I am 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.

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:

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

diazona said:
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.

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

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.

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

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?

Yes.

## What is normalizing a wave function?

Normalizing a wave function is the process of ensuring that the total probability of finding a particle in a given space is equal to 1. This is important in quantum mechanics as it allows for accurate predictions of the behavior of a particle.

## Why is normalizing a wave function necessary?

Normalizing a wave function is necessary because it ensures that the total probability of finding a particle in a given space is equal to 1. This is a fundamental principle in quantum mechanics and allows for accurate predictions of the behavior of a particle.

## How is a wave function normalized?

A wave function is normalized by dividing it by its norm, which is the square root of the integral of the absolute square of the wave function over all space. This ensures that the total probability of finding a particle in a given space is equal to 1.

## What happens if a wave function is not normalized?

If a wave function is not normalized, it means that the total probability of finding a particle in a given space is not equal to 1. This can lead to inaccurate predictions and violate the fundamental principles of quantum mechanics.

## Can any wave function be normalized?

Yes, any wave function can be normalized as long as it satisfies the normalization condition. However, some wave functions may be more difficult to normalize than others and may require more complex mathematical techniques.

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