Schrodinger Equation : Infinite Square Well

[MisterMan]

Hi, I am having trouble understanding an example from a textbook I am reading on the Schrödinger equation. The example deals with an infinite square well in one dimension. With the following properties:

V = 0\,where -a \leq x \leq a

V = \infty\,|x| \geq a

Where V is the potential.
The time-independent Schrödinger equation is given by this equation:

\frac{\hbar}{2m}\frac{d^2\psi}{dx^2} + E\psi = 0

where psi is:

\psi = Acos(kx) + Bsin(kx)

where k=\sqrt{\frac{2mE}{\hbar^2}}

and A and B are constants.

The boundary conditions in the book are said to be :

Acos(ka) + Bsin(ka) = 0
and
Acos(ka) - Bsin(ka) = 0

because we need to satisfy the Schrödinger equation when the potential is infinite when x is equal to plus or minus a.
The book then goes on to say:
Hence,
B = 0 and cos(ka) = 0 => k = \frac{n\pi}{2a} n=1, 3, 5,...
or
A = 0 and sin(ka) = 0 => k = \frac{n\pi}{2a} n=2, 4, 6,...

My question is this: Is psi equal to Aexp(ikx) in this example? In which case shouldn't the case of sin(ka) = 0 be irrelevant as A = 0? The solution to this equation is given in the book as:

\psi = \sqrt{\frac{1}{a}}cos(\frac{nx\pi}{2a}) for n odd

\psi = \sqrt{\frac{1}{a}}sin(\frac{nx\pi}{2a}) for n even

when |x| <= a and psi = 0 when |x| > a.

I hope I have made my problem clear, any help is appreciated.

Thanks

 

[Fredrik]

exp(ikx), exp(-ikx), sin(kx) and cos(kx) are all solutions of the time-independent equation (which can also be written as Hψ=Eψ). That set of four solutions isn't linearly independent, but any subset with exactly two members is. So to say that the general solution is of the form A sin(kx)+B cos(kx) is the same thing as saying that it's of the form C exp(ikx)+D exp(-ikx) or E sin(kx)+F exp(ikx). If you impose the boundary conditions on the second expression, you should get the same result as if you impose them on the first.

I'm not sure what you did to eliminate the sin solution. If you just add the two boundary conditions, you get the "B=0 and..." result, and if you subtract one of the boundary conditions from the other, you get the "A=0 and..." result.

 

[Dyson]

You may solve the problem like this:
The solution of the equation can be written as the superposition of e^{ikx} and e^{-ikx}, then,
\phi=Ae^{ikx}+Be^{-ikx}
In terms of the boundary condition,
0=Ae^{ika}+Be^{-ika}
0=Ae^{-ika}+Be^{ika}
These are two equations on two variables A and B. A and B mustn't be 0 at the same time because it will lead to \phi \equiv 0. And this means we can not find the particle in the Well. It evidently violates the \int \mid \phi \mid ^{2} dx=1!
In order to satisfy the condition, the determinant of the coefficents before A and B must be 0. So, we can get sin(2ka)=0. It means sin(ka)=0 orcos(ka)=0. Or we can rewrite ka=\frac{n\pi}{2}. Substituting it into the two equation, you can get the result.

 

[PianoDentist]

Hello, I am stuck on exactly the same problem so I thought rather than make a whole new thread Id ask here.

I understand (I think) the sines and cosines for different eigenfunctions with different n values...but I am not sure how to get the 1/SQRT (a) factor. Is this the normalisation factor or am I totally up the wrong path? it would be a great help if someone could show me how to do it

 

[dextercioby]

Yes, that is from the normalization factor. When you compute the integral between [-a,a] of sine squared, you get the <A> factor as 1/sqrt{a}.

 

[PianoDentist]

Hello, I am stuck on exactly the same problem so I thought rather than make a whole new thread Id ask here.

I understand (I think) the sines and cosines for different eigenfunctions with different n values...but I am not sure how to get the 1/SQRT (a) factor. Is this the normalisation factor or am I totally up the wrong path? it would be a great help if someone could show me how to do it

dont worry I got it. well I did it for the sin parts anyway. I am guessing doing it for the cos solutions gives same result