# Orthogonality of Stationary States

## Homework Statement

I'm going back through some homework as revision, and came across this problem. It was marked as correct, but now i'm thinking it's unconvincing...

For a particle in an infinite square well, with ##V = 0 , 0 \leq x \leq L##, prove that the stationary eigenstates are mutually orthogonal.

##\int^{\infty}_{\infty} dx \phi^*_m \phi_n = \delta_{m,n}##

## The Attempt at a Solution

The stationary eigenstates are given by ##\phi_n (x) = \sqrt{\frac{2}{L}} \sin \frac{n \pi}{L}x##, and since the complex conjugate of a real function is just itself,

##\int^{\infty}_{\infty} dx \phi^*_m \phi_n = \frac{2}{L} \int^L_0 \sin \frac{m \pi }{L}x \sin \frac{n \pi}{L}x##.

Using the identity that ##\sin u \sin v = \frac{1}{2} [\cos (u-v) - \cos (u+v)]##

##\frac{2}{L} \int^L_0 \sin \frac{m \pi }{L}x \sin \frac{n \pi}{L}x = \frac{1}{L} \int^L_0 dx [\cos (\frac{(m-n)\pi}{L}x) - \cos (\frac{(m+n)\pi}{L}x)]##

Evaluating this I find that

##\int^{\infty}_{\infty} dx \phi^*_m \phi_n = \frac{1}{\pi} [\frac{\sin ((m-n)\pi)}{m - n} - \frac{\sin ((m+n)\pi)}{m+n}]##

Here I just said that this is zero for ##m \neq n##

The arguments of the sine functions are zero, regardless of whether or not ##m = n##, so what's missing here?

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Samy_A
Homework Helper

## Homework Statement

I'm going back through some homework as revision, and came across this problem. It was marked as correct, but now i'm thinking it's unconvincing...

For a particle in an infinite square well, with ##V = 0 , 0 \leq x \leq L##, prove that the stationary eigenstates are mutually orthogonal.

##\int^{\infty}_{\infty} dx \phi^*_m \phi_n = \delta_{m,n}##

## The Attempt at a Solution

The stationary eigenstates are given by ##\phi_n (x) = \sqrt{\frac{2}{L}} \sin \frac{n \pi}{L}x##, and since the complex conjugate of a real function is just itself,

##\int^{\infty}_{\infty} dx \phi^*_m \phi_n = \frac{2}{L} \int^L_0 \sin \frac{m \pi }{L}x \sin \frac{n \pi}{L}x##.

Using the identity that ##\sin u \sin v = \frac{1}{2} [\cos (u-v) - \cos (u+v)]##

##\frac{2}{L} \int^L_0 \sin \frac{m \pi }{L}x \sin \frac{n \pi}{L}x = \frac{1}{L} \int^L_0 dx [\cos (\frac{(m-n)\pi}{L}x) - \cos (\frac{(m+n)\pi}{L}x)]##

Evaluating this I find that

##\int^{\infty}_{\infty} dx \phi^*_m \phi_n = \frac{1}{\pi} [\frac{\sin ((m-n)\pi)}{m - n} - \frac{\sin ((m+n)\pi)}{m+n}]##

Here I just said that this is zero for ##m \neq n##

The arguments of the sine functions are zero, regardless of whether or not ##m = n##, so what's missing here?
When ##m=n##, you can't divide by ##m-n##, as you do in your computation.

Do the integral for ##m=n## separately.

When ##m=n##, you can't divide by ##m-n##, as you do in your computation.

Do the integral for ##m=n## separately.
When ##m=n##, you can't divide by ##m-n##, as you do in your computation.

Do the integral for ##m=n## separately.
Ah, I see.

For the case ##m = n## i'll have an integral of ##sin^2 (\frac{n \pi}{L} x)##

Thank you