Solving Schrodinger equation in two dimensions

[6021023]

How is it that X = Ae^(ikx) + Be^(-ikx) is the same as X(x)=Asin(kx)+BCos(kx)? Is there a name for that rule?

If C = 0, I don't see why that doesn't turn

X = Ae^(sqrt(C)x) + Be^(-sqrt(C)x)

into

X = Ae^0 + Be^0

Just plug in right?

 

[gabbagabbahey]

How is it that X = Ae^(ikx) + Be^(-ikx) is the same as X(x)=Asin(kx)+BCos(kx)? Is there a name for that rule?

Just use Euler's formula and rename the constants to A and B when you are done...you should be familiar with this form of the solution, by now...it is VERY VERY VERY common in physics, and you can verify that it satisfies the ODE just by differentiating it twice.

If C = 0, I don't see why that doesn't turn

X = Ae^(sqrt(C)x) + Be^(-sqrt(C)x)

into

X = Ae^0 + Be^0

Just plug in right?

If C=0, your characteristic equation has a single eigenvalue of r=0, with multiplicity of 2, so the form of your general solution will be different (Remember, your ODE is second order, so you expect to have two linearly independent solutions, a constant is only one solution)...

 

[6021023]

So if C = 0, then

r1 = r2 = 0

so the equation is
X = Ae^(sqrt(C)x)+Bxe^(sqrt(C)x)
X = Ae^0 + B(0)e^0
X = Ae^0
X = A

Is that right?

 

[gabbagabbahey]

No, re-read the section of your calc book that deals with repeated roots to the characteristic equation...

 

[6021023]

Can I just skip over to X''(x)=0, and then integrate twice to solve for X?

Then that becomes X'(x) = X + C1
X(x) = XC2 + C1

 

[gabbagabbahey]

Yes, X(x)=Ax+B.

Now what can you say about $\Psi(\frac{\pm L_x}{2},y)$ and $\frac{\partial}{\partial x}\Psi(\frac{\pm L_x}{2},y)$ and why?

 

[6021023]

I thought that in the equation there are two Ys. One is Y(x) and the other is Y(y). So Y is actually Y(x,y)?

I don't know what to say about those equations. Any hints?

 

[gabbagabbahey]

$\Psi$ is the greek letter 'psi', not Y

 

[6021023]

I know that psi is not Y, I just don't know how to type it in here so I just use Y instead.

 

[gabbagabbahey]

Okay, then continue on...what are the answers to my questions in post #36?

 

[6021023]

Are you sure there isn't a typo? The first equation says Y(+Lx/2,y) and the second says dY(+Lx/2,y)/dx

 

[gabbagabbahey]

I'm sure. And stop using Y to represent the wavefunction, just type Psi...what is the value of the wavefunction at the boundaries? How about its partial derivatives?

 

[6021023]

I'm not even sure what the boundaries are.

 

[gabbagabbahey]

They are the boundaries of your potential well; $x=\pm\frac{L_x}{2}$ and $y=\pm\frac{L_y}{2}$...what else would they be?

 

[6021023]

What is the equation for the wave function?

 

[gabbagabbahey]

The wavefunction is $\Psi(x,y)$ and it is what you are trying to determine from Schroedinger's equation.

Get some sleep and answer these questions in the morning when your mind is fresh.:zzz:

What does Schroedinger's equation tell you about $\Psi(x,y)$ outsode the potential well (where $V=\infty$)?

What do you know about the continuity and differentiability of a physical wavefunction? What does that tell you about $\Psi(x,y)$ on the boundaries of the well? (Those will be your boundary conditions)

 

[6021023]

V is not in the denominator so it will not make anything go to 0. It will just create an infinitely large number.

A wavefunction is continuous and therefore differentiable everywhere. So at the boundaries it will also be differentiable.

 

[6021023]

How about this. Since the probability of finding the particle outside the box with dimensions Lx and Ly is 0, then the wavefunction must go to 0 outside of these limits.

 

[gabbagabbahey]

How about this. Since the probability of finding the particle outside the box with dimensions Lx and Ly is 0, then the wavefunction must go to 0 outside of these limits.

Not very sound logic; the reason that the probability is zero is because the wavefunction is zero; not the other way around. The reason the wavefunction is zero is because it is the only finite wavefunction (infinite wavefunctions are unphysical) )that will satisfy the Schroedinger equation with an infinite potential (You really should already know this; you need to spend more time studying your text!):

$$\left(\nabla^2+\frac{2mE}{\hbar^2}\right)\Psi(\textbf{r})=V(\textbf{r})\Psi(\textbf{r})$$

The RHS of the equation is always finite ($E$ is finite, $\Psi(\textbf{r})$ is finite and twice differentiable wherever [itex]V(\textbf{r})[itex]is continous); if [itex]V(\textbf{r})[itex]is infinite, $\Psi(\textbf{r})$ must then be zero in order to produce a finite product with it.<br /> <br /> Make sense?[/itex][/itex][/itex][/itex]