# Boundary value problem

1. Jan 24, 2007

### John O' Meara

Schodinger's equation for one-dimensional motion of a particle whose potential energy is zero is
$$\frac{d^2}{dx^2}\psi +(2mE/h^2)^\frac{1}{2}\psi = 0$$
where $$\psi$$ is the wave function, m the mass of the particle, E its kinetic energy and h is Planck's constant. Show that

$$\psi = Asin(kx) + Bcos(kx)$$ ( where A and B are constants) and $$k =(2mE/h^2)^\frac{1}{2}$$ is a solution of the equation.
Using the boundary conditions $$\psi=0$$ when x=0 and when x=a, show that
(i) the kinetic energy $$E=h^2n^2/8ma^2$$
(ii) the wave function $$\psi = A sin(n\pi\times x/a)$$ where n is any integer. (Note if $$sin(\theta) = 0 then \theta=n\pi$$)

My attempt:

A*sin(0) + B*cos(0) = 0, => 0 +B =0 => B = 0.
Therefore
A*sin(k*a)=0, Therefore $$(2mE/h^2)^\frac{1}{2}a = n\pi => E=n^2\pi^2h^2/2ma^2$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Jan 24, 2007
2. Jan 24, 2007

### John O' Meara

(ii) should read A*sin(n*pi*x/a), I'm just notgood at boundary value problems.

3. Jan 24, 2007

### HallsofIvy

Staff Emeritus
Well, first, you haven't shown that that $\psi$ does, in fact, satisfy the differential equation. After that, yes, B= 0. Now, assuming A is not 0, that is that $\psi$ is not itself identically 0, then yes, we must have sin(ka)= 0 so that $ka= (2mE/h^2)^{\frac{1}{2}}= n\pi$. E follows eactly as you say.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution[/QUOTE]

4. Jan 25, 2007

### John O' Meara

I was able to show that $$\psi$$ is a solution of the equation, it is (i) and (ii) that I had trouble doing espically (ii) i.e., $$\psi = A sin(n\pi\times x/a)$$. Where did he get the argument "$$n\pi\times x/a$$", or more importantly how does he expect me to get that argument of the sine.
Also remember in (i) the answer he has for E $$=h^2n^2/8ma^2$$, not what I got for E. Thanks for the help.

5. Jan 26, 2007

### John O' Meara

I hope someone can tell me how $$\psi$$ can go from $$Asin((2mE/h^2)^\frac{1}{2}x) to Asin(n\pi\times x/a)$$. Thanks for the help.

6. Jan 26, 2007

### HallsofIvy

Staff Emeritus
Well, when x= a, the argument is $n\pi$ what is $sin(n\pi)[/tex]? Remember that you were told that [itex]\psi(0)= 0$ and $\psi(a)= 0$. Knowing that cos(0)= 1 tells us that the second constant, B, must be 0. That leaves Asin(kx). We must have Asin(ka)= 0 and we don't want A= 0 (that would mean our function is always 0) so we must have sin(ka)= 0. For what x is sin(x)= 0? Multiples of $\pi$ of course:
$ka= n\pi$. For that to be true, k must be equal to $n\pi/a$

You were also told that $k= \sqrt{2ME/h^2}$ and you now know $k= n\pi/a[\itex] so [itex]n\pi/a= \sqrt{2ME/h^2}$. Solve that for E.