# 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