How to Normalize a Wavefunction with Modulus x?

[Jimmy87]

Homework Statement

Normalise the wavefunction in the diagram which is given by:

psi(x) = A {x} < a

{x} is supposed to be mod x.

Homework Equations

None specifically

The Attempt at a Solution

I know that the square integral of the wavefunction needs to be set equal to 1. I am unsure exactly what I square as I haven't seen a wave-function written with a mod x. I know that mod x just means x with positive values. I know I need to square the wavefunction first so would that be:

psi(x) squared = A^2 a^2?

The answer for the square integrable of this wavefunction is A^2(2a) according to my textbook but they don't show you how. You would then have to choose A equal to 1/sqrt 2a and plug that into the original function to normalise it. I don't see how the square integral is A^2(2a) unless you take the area of the graph to be the integral of the function which is A(2a) but then if you square this then you get A^2(2a)^2?

 

[blue_leaf77]

The wavefunction should be written like
$$
\psi(x) = A \hspace{0.5cm} \textrm{for} \hspace{2mm} |x|<a
$$
and zero otherwise. A mathematical expression like $|x|<a$ is equivalent to $-a<x<a$.

The normalization reads as
$$
\int_{-\infty}^\infty |\psi(x)|^2 = 1
$$
What do you get if you calculate the left side using the given wavefunction?

 

[Jimmy87]

The wavefunction should be written like
$$
\psi(x) = A \hspace{0.5cm} \textrm{for} \hspace{2mm} |x|<a
$$
and zero otherwise. A mathematical expression like $|x|<a$ is equivalent to $-a<x<a$.

The normalization reads as
$$
\int_{-\infty}^\infty |\psi(x)|^2 = 1
$$
What do you get if you calculate the left side using the given wavefunction?

Yes sorry, what you wrote is exactly what is written in my textbook. This is what I am not sure about. I know I need to square the function first so I will get A^2 but I don't know what else to square is mod x part of the actual function or not? I can't actually get my teeth into this function. I am happy taking the square integral of complex exponentials but this simple one has thrown me!

 

[blue_leaf77]

is mod x part of the actual function or not?

I was aware that that's part of your confusion, that's why I purposely added "for" to the right of the wavefunction.

The wavefunction should be written like
$$
\psi(x) = A \hspace{0.5cm} \textrm{for} \hspace{2mm} |x|<a
$$
and zero otherwise.

The above lines describe the nature of $\psi(x)$. This just means that $\psi(x)$ has values of zero within $-a$ and $a$, and zero outside this range. Now you have to incorporate this behavior of $\psi(x)$ into the normalization integral.

 

[Jimmy87]

I was aware that that's part of your confusion, that's why I purposely added "for" to the right of the wavefunction.

The above lines describe the nature of $\psi(x)$. This just means that $\psi(x)$ has values of zero within $-a$ and $a$, and zero outside this range. Now you have to incorporate this behavior of $\psi(x)$ into the normalization integral.

Thanks. Could you give me some guidance on how to do this? Since the integral of any function is the area under the graph can I not say that the integral for $\psi(x)$ is A(2a)?

 

[blue_leaf77]

Since the integral of any function is the area under the graph can I not say that the integral for $\psi(x)$ is A(2a)?

That's right except for that the "graph" you are talking about should be $|\psi(x)|^2$, instead of $\psi(x)$.

 

[PeroK]

Yes sorry, what you wrote is exactly what is written in my textbook. This is what I am not sure about. I know I need to square the function first so I will get A^2 but I don't know what else to square is mod x part of the actual function or not? I can't actually get my teeth into this function. I am happy taking the square integral of complex exponentials but this simple one has thrown me!

I recall someone else had a similar problem recently when they couldn't accept a constant as a function, because it didn't have x in it. Try the equivalent

$\Psi(x) = A\frac{e^x}{e^x}$ for $|x| < a$

If you need x in your definition of a function.

 

[Jimmy87]

That's right except for that the "graph" you are talking about should be $|\psi(x)|^2$, instead of $\psi(x)$.

Thanks for your help. Ok so I think I get it. In my graph do we call the y-axis Ψ(x). So |ψ(x)|2 would be the square of that graph which means the amplitude A would become A^2. And the integral will be the area underneath the |ψ(x)|2 graph which would be A^2(2a), namely height multiplied by width?

 

[blue_leaf77]

Thanks for your help. Ok so I think I get it. In my graph do we call the y-axis Ψ(x). So |ψ(x)|2 would be the square of that graph which means the amplitude A would become A^2. And the integral will be the area underneath the |ψ(x)|2 graph which would be A^2(2a), namely height multiplied by width?

Yes.

 

[Jimmy87]

Yes.

Thanks. The solution to the normalisation is A is set equal to 1/sqrt2a. This means that when you take the square integral of the original function using limits of -a to +a using this expression for A you should get 1. Is there a way of showing this mathematically by actually plugging in the limits and doing the integration as you would normally. For example if you square your new wavefunction you get 1/2a. And you can't really integrate this the usual way as you would end upw ith the integral of 1/a which is ln(a). I can see how you get it by using the area of the graph as the integral because then you get 1/2a multiplied by 2a which gives 1.

 

[DrClaude]

Is there a way of showing this mathematically by actually plugging in the limits and doing the integration as you would normally. For example if you square your new wavefunction you get 1/2a. And you can't really integrate this the usual way as you would end upw ith the integral of 1/a which is ln(a). I can see how you get it by using the area of the graph as the integral because then you get 1/2a multiplied by 2a which gives 1.

When you have a function that is defined by parts, the integral looks like
$$
\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{-a} f(x) dx + \int_{-a}^{a} f(x) dx + \int_{a}^{\infty} f(x) dx
$$
In your case, on the right hand side, the first and last integrals are 0 because $f(x)=0$ over those ranges.

 

[Jimmy87]

When you have a function that is defined by parts, the integral looks like
$$
\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{-a} f(x) dx + \int_{-a}^{a} f(x) dx + \int_{a}^{\infty} f(x) dx
$$
In your case, on the right hand side, the first and last integrals are 0 because $f(x)=0$ over those ranges.

Thanks. But with the middle integral are you not taking the integral of 1/2a and then evaluating from -a to +a? How do you end up with one from that integral?

 

[DrClaude]

But with the middle integral are you not taking the integral of 1/2a and then evaluating from -a to +a?

I just wrote a generic function $f(x)$. In your case, $f(x) = \left| \psi(x) \right|^2$.

 

[PeroK]

Thanks. The solution to the normalisation is A is set equal to 1/sqrt2a. This means that when you take the square integral of the original function using limits of -a to +a using this expression for A you should get 1. Is there a way of showing this mathematically by actually plugging in the limits and doing the integration as you would normally. For example if you square your new wavefunction you get 1/2a. And you can't really integrate this the usual way as you would end upw ith the integral of 1/a which is ln(a). I can see how you get it by using the area of the graph as the integral because then you get 1/2a multiplied by 2a which gives 1.

You've really got a problem with the constant function. There's not much to say except that the integral of the constant function 1/a is most certainly not ln(a). It's x/a. And that's just about the simplest integral there is.

 

[Jimmy87]

You've really got a problem with the constant function. There's not much to say except that the integral of the constant function 1/a is most certainly not ln(a). It's x/a. And that's just about the simplest integral there is.

Yes it is really strange I can't quite grasp it as I find other complex integrations such as gaussian functions fine. So in this problem we are integrating 1/a with respect to what? x? So are you saying that the integral of a constant with respect to x is x times the constant?

 

[DrClaude]

So are you saying that the integral of a constant with respect to x is x times the constant?

Yes:
$$
\int A \, dx = A \int dx = A x
$$

 

[Jimmy87]

Yes:
$$
\int A \, dx = A \int dx = A x
$$

Ok, I see so I have done this to show what I think you mean for the problem in question. So is this what you mean I should get:

 

[DrClaude]

Ok, I see so I have done this to show what I think you mean for the problem in question. So is this what you mean I should get:

Correct!

 

[Jimmy87]

Correct!

Thank you!