Solving Schrodinger equation in two dimensions

[6021023]

Homework Statement

Solve the time independent Schrodinger equation for an electron in a 2-D potential well having dimensions Lx and Ly in the x and y directions respectively.

Homework Equations

d^2Y/dx^2 + d^2Y/dy^2 + 2m/h^2*EY = 0

The Attempt at a Solution

Y(x) = A exp(jkx) + B exp(-jkx)

I'm not even sure that is correct as a starting point, as that is what is given for the example in the book for the Schrodinger equation in one dimension, not two dimensions.

 

[gabbagabbahey]

Is this for a finite well or an infinite well?

 

[6021023]

Infinite potential well.

 

[gabbagabbahey]

Okay, so for |x|\leq L_x and |y|\leq L_y the schroedinger equation reduces to:

\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{2mE}{\hbar^2}\right)\Psi(x,y)=0

right?

Now use separation of variables; assume that \Psi(x,y)=X(x)Y(y)...what do you get when you plug that into Schroedinger's equation?

 

[6021023]

I forgot how to do separation of variables. How do you do that part?

 

[gabbagabbahey]

You assume the wavefunction is separable (i.e. assume \Psi(x,y)=X(x)Y(y))...substitute that into the Schrodinger equation and simplify...what do you get?

 

[6021023]

I get d^2Y/dx^2 + d^2Y/dy^2 + 2m/h^2*EX(x)Y(y) = 0

But doesn't the substitution also have to be done for the second derivatives? How would that be done?

 

[gabbagabbahey]

I get d^2Y/dx^2 + d^2Y/dy^2 + 2m/h^2*EX(x)Y(y) = 0

But doesn't the substitution also have to be done for the second derivatives? How would that be done?

No, you don't get that. And yes, the second derivatives act on \psi(x,y)=X(x)Y(y)

 

[6021023]

When you plug in for each of the second derivatives, do you treat the other variable as a constant?

 

[gabbagabbahey]

You tell me, how does one usually take the partial derivative of a multivariable function?

 

[6021023]

I googled it and found out that for multivariable partial derivatives the other variables are held constant.

So is this the answer?

Y(y)X(x)'' + X(x)Y(y)'' + 2m/h^2*E*X(x)Y(y) = 0

 

[gabbagabbahey]

Yes.

Now, what do you get when you divide both sides of the equation by X(x)Y(y)?

 

[6021023]

X(x)''/X(x) + Y(y)''/Y(y) + 2m/h^2*E = 0

 

[gabbagabbahey]

Right, now look carefully at that equation...what variable(s) is the quantity X''(x)/X(x) a function of ? What variable(s) is the quantity Y''(y)/Y(y) a function of? Under what conditions can the add together to get a constant value of \frac{2mE}{\hbar^2}?

 

[6021023]

X''(x)/X(x) is a function of x and Y''(y)/Y(y) is a function of y.

Are you asking when

X(x)''/X(x) + Y(y)''/Y(y) + 2m/h^2*E = 0

will be equal to 2m/h^2*E = 0 ?

If that's the case, then it will happen either when both X(x)''/X(x) and Y(y)''/Y(y) are equal to 0, or when X(x)''/X(x) = -Y(y)''/Y(y).

 

[gabbagabbahey]

X''(x)/X(x) is a function of x and Y''(y)/Y(y) is a function of y.

Right, and 'x' and 'y' are independent variables, right?

Are you asking when

X(x)''/X(x) + Y(y)''/Y(y) + 2m/h^2*E = 0

will be equal to 2m/h^2*E = 0 ?

No, I'm asking when X(x)''/X(x) + Y(y)''/Y(y) can equal -2m/h^2*E? How can a function of 'x' added to a function of 'y' result in a constant?

 

[6021023]

Yes, x and y are independent. A function of x and a function of y added together will result in a constant when the variables cancel each other out or become 0.

 

[gabbagabbahey]

Well, the variables can't cancel each other out, since they are independent.

What if both functions are constants?

 

[6021023]

If both functions are constant, then adding them together would result in a constant, but I don't see how they can be constants since they have variables in them.

 

[gabbagabbahey]

Are you telling me that you don't consider f(x)=1 to be a function of x? But you do consider f(x)=0 to be a function of x?

 

[6021023]

I just remembered that they can be constants since they are second derivatives. A variable can disappear when taking derivatives.

 

[gabbagabbahey]

Even if they weren't derivatives, a function of a variable can be equal to a constant...I'm not sure why you would think otherwise.

Anyways, yes they must be equal to some unknown constants...so, let's say X''(x)/X(x)=C, what does that make Y''(y)/Y(y)...do you know how to solve these two differential equations?

 

[6021023]

Yes, functions of a variable can be equal to a constant, I just suspected that wasn't the case here.

If X''(x)/X(x) = C, then X(x)''/X(x) + Y(y)''/Y(y) = -2m/h^2*E
C + Y(y)''/Y(y) = -2m/h^2*E
Y(y)''/Y(y) = -2m/h^2*E - C

If that is correct, then I think that the integral should be taken twice. However, I don't think that I know how to do that as the Y(y) on the bottom makes things more complicated.

 

[gabbagabbahey]

Yes, functions of a variable can be equal to a constant, I just suspected that wasn't the case here.

If X''(x)/X(x) = C, then


X(x)''/X(x) + Y(y)''/Y(y) = -2m/h^2*E
C + Y(y)''/Y(y) = -2m/h^2*E
Y(y)''/Y(y) = -2m/h^2*E - C

Right, now X''(x)/X(x)=C means X''(x)-CX(x)=0 right?...Surely you know how to solve a 2nd order linear, homogeneous ODE with constant coefficients? This is 1st year calc stuff...

 

[6021023]

It's been a while so I'm not too confident with solving the ODE. I think that I have to create a characteristic equation, and I'll find solutions s1 and s2, and plug that into an equation like Ae^(s1)+Be^(s2), then plugin initial values to find A and B? Is it something like that?

 

[gabbagabbahey]

Yes, exactly like that...Now just crack open your old calc textbook, review the section on constant coefficient ODEs and get to it...

 

[6021023]

X'' - CX = 0
r^2 - C = 0
r^2 = C
r1 = C
r2 = -C
X = Ae^(Cx) + Be^(-Cx)

Is that much correct? Now do I plugin initial and final conditions to solve for the constants, or do I first have to find the particular solution?

 

[gabbagabbahey]

Ermm... if r^2=C, doesn't that mean r=\pm\sqrt{C}?...And you'll want to look at 3 different cases: (1) C\equiv k^2>0; (2) C\equiv-k^2<0; and (3) C=0...what do the general solutions look like in each case?

 

[6021023]

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

if C = k^2 then the equation is
X = Ae^(kx) + Be^(-kx)

if C = -k^2 then it is
X = Ae^(ikx) + Be^(-ikx) (not sure about this one)

if C = 0 then it is
X = Ae^0 + Be^0 = A + B

 

[gabbagabbahey]

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

if C = k^2 then the equation is
X = Ae^(kx) + Be^(-kx)

Right.

if C = -k^2 then it is
X = Ae^(ikx) + Be^(-ikx) (not sure about this one)

Also correct, but it is more useful if you write it in the form X(x)=Asin(kx)+BCos(kx)

if C = 0 then it is
X = Ae^0 + Be^0 = A + B

No, if C=0 then your ODE becomes X''(x)=0...just integrate twice...what do you get?

 

[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 V(\textbf{r})is continous); if V(\textbf{r})is infinite, \Psi(\textbf{r}) must then be zero in order to produce a finite product with it.<br /> <br /> Make sense?