General Solution to time independent schrodinger equation

[hhhmortal]

Homework Statement

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?

 

[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.

 

[hhhmortal]

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

Instead I get:

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

I still don't know how to get the correct form from this..

 

[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).

 

[hhhmortal]

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).

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

 

[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.

 

[hhhmortal]

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.

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)

 

[vela]

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.

 

[hhhmortal]

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.

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)

 

[vela]

Yup, you got it.

 

[hhhmortal]

Thanks for the patience everyone