Normalising Wavefunctions

1. Jan 18, 2016

Jimmy87

1. The problem statement, all variables and given/known data
Normalise the wavefunction in the diagram which is given by:

psi(x) = A {x} < a

{x} is supposed to be mod x.

2. Relevant equations
None specifically
3. 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 in to 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?

2. Jan 18, 2016

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 $-a<x<a$.

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

3. Jan 18, 2016

Jimmy87

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!

4. Jan 18, 2016

blue_leaf77

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.

5. Jan 18, 2016

Jimmy87

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)?

6. Jan 18, 2016

blue_leaf77

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

7. Jan 18, 2016

PeroK

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.

8. Jan 18, 2016

Jimmy87

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?

9. Jan 18, 2016

blue_leaf77

Yes.

10. Jan 18, 2016

Jimmy87

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.

11. Jan 18, 2016

Staff: Mentor

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.

12. Jan 18, 2016

Jimmy87

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?

13. Jan 18, 2016

Staff: Mentor

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

14. Jan 18, 2016

PeroK

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.

15. Jan 18, 2016

Jimmy87

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?

16. Jan 18, 2016

Staff: Mentor

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

17. Jan 18, 2016

Jimmy87

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:

18. Jan 18, 2016

Staff: Mentor

Correct!

19. Jan 18, 2016

Thank you!!