• Support PF! Buy your school textbooks, materials and every day products Here!

Orthogonality of Stationary States

  • Thread starter BOAS
  • Start date
  • #1
555
19

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?
 

Answers and Replies

  • #2
Samy_A
Science Advisor
Homework Helper
1,241
510

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.
 
  • #3
555
19
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
 

Related Threads on Orthogonality of Stationary States

Replies
6
Views
2K
Replies
7
Views
1K
  • Last Post
Replies
2
Views
3K
Replies
1
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
921
  • Last Post
Replies
16
Views
1K
Top