Homework Help: General Solution to time independent schrodinger equation

1. May 6, 2010

hhhmortal

1. The problem statement, all variables and given/known data

This is really a maths problem I'm having.

I need to get the general solution for the infinite square well in the form:

u = A cos(kx) + B sin(kx)

I found the general solution to be:

u = A exp(ikx) + B exp(-ikx)

Using Euler's formula:

exp(ikx) = cos(kx) + isin (kx)
exp(-ikx) = cos(kx) - isin(kx)

Hence, general solution becomes :

u = A cos(kx) + Bcos(kx)

What am I missing out?

2. May 6, 2010

Cyosis

Show how you got from the exponential form to the double cosine form. I have a feeling you think that A sin x- B sin x=0.

3. May 6, 2010

hhhmortal

Yes you're right! That was a mistake I've made:

u = (A + B) cos(kx) + (A - B) i sin(kx)

I still dont know how to get the correct form from this..

4. May 6, 2010

Cyosis

The problem you're having is that you labeled the integration constants for both types of solutions the same. This causes confusion.

Instead use $u=C e^{ikx}+De^{-ikx}$ and find a relation between C,D and A,B (just relabel A and B in post #3).

Last edited: May 6, 2010
5. May 6, 2010

hhhmortal

I'm not sure what you mean. How would changing A & B to C & D make a difference?

6. May 6, 2010

Cyosis

Because the A in your first equation is not necessarily the same as the A in your second equation (post #1). Using the same name for different constants gets you into trouble. Just use the u given in my post with constants C and D and find a relation between those and A and B.

7. May 6, 2010

hhhmortal

Forgive me for asking again. I seem to be confused over something so simple.

I did as you said..I think and got:

C exp(ikx) = A cos(kx) + iB sin(kx)

D exp(-ikx) = A cos(kx) + iB sin(kx)

8. May 6, 2010

vela

Staff Emeritus
What people are trying to tell you is that $Ae^{ikx}+Be^{-ikx} \ne A\cos(kx)+B\sin(kx)$. You can't get one from the other because they're not equal to each other. You can, however, show that $Ce^{ikx}+De^{-ikx} = A\cos(kx)+B\sin(kx)$ if you choose A and B correctly. Remember that C and D are arbitrary constants. Any combination of them will just be another arbitrary constant.

9. May 6, 2010

hhhmortal

My previous post was wrong, sorry, is it simply:

C exp(ikx) = C cos(kx) + iC sin(kx)

D exp(-ikx) = D cos(kx) - iD sin(kx)

So,

C+D = A

(iC - iD) = B

Therefore:

C exp(ikx) + D exp(-ikx) = A cos(kx) + B sin(kx)

10. May 6, 2010

vela

Staff Emeritus
Yup, you got it.

11. May 7, 2010

hhhmortal

Thanks for the patience everyone