Orthogonality of Stationary States

[BOAS]

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.

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

Homework Equations

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?

 

[Samy_A]

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.

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

Homework Equations

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.

 

[BOAS]

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