# Show that wave function in coordinates x,y is normalized

Tags:
1. Sep 27, 2015

### Crista

1. The problem statement, all variables and given/known data
A particle is described by the state of the following wave function.

wavefunction(x,y) = 30/[(a^5)(b^5)]^1/2 * x(a-x) * b(b-y)

2. Relevant equations
integral from 0 to i of x^n * (1-x)^m dx = (n!m!)/(n+m+1)!

3. The attempt at a solution
I know that normalizing means taking the integral from negative infinity to positive infinity of the probability density squared with respect to x, but I just don't know how to take that with respect to x and y. I tried looking at problems that involved a particle in a 2d box, but doing the math seems extremely difficult.

2. Sep 27, 2015

### andrewkirk

I presume when you say that is the wavefunction you mean that is the wavefunction's representation in the position (X) basis.
In that case your statement equates to saying that
$$\langle x,y | \psi\rangle = \frac{30}{\sqrt{a^5+b^5}}x(a-x)b(b-y)$$
and you are asked to prove that $\langle \psi|\psi\rangle = 1$.

What you need to do is expand $\langle \psi|\psi \rangle= 1$ in the X basis as follows:

$$\langle \psi|\psi\rangle = \int_{-\infty}^\infty \int_{-\infty}^\infty \langle \psi | x,y\rangle\langle x,y | \psi\rangle\, dy\,dx$$

Now use the fact that $\langle u|v\rangle = \langle v|u\rangle^*$ to simplify that and substitute in your function for $\langle x,y | \psi\rangle$. You should find that the integral factorises nice and easily.

ETA: But I don't think the integrals will exist. Are you sure you wrote the function correctly? It should go to zero as $x,y\to\pm\infty$, but what you wrote doesn't.

3. Sep 28, 2015

### Crista

It shows now boundaries, but people have suggested that I integrate from 0->a according to x and 0->b according to y.

4. Sep 28, 2015

### Crista

also, its (a^5)*(b^5) but the way you stated how to integrate it according to x and then y helped a lot. Not sure if the answer is correct, but I shall ask my teacher when I get it back. thank you!