# Solving Schrodinger equation in two dimensions

1. Oct 14, 2009

### 6021023

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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.

2. Oct 14, 2009

### gabbagabbahey

Is this for a finite well or an infinite well?

3. Oct 14, 2009

### 6021023

Infinite potential well.

4. Oct 14, 2009

### 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?

5. Oct 14, 2009

### 6021023

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

6. Oct 14, 2009

### 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?

7. Oct 14, 2009

### 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?

8. Oct 14, 2009

### gabbagabbahey

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

9. Oct 14, 2009

### 6021023

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

10. Oct 14, 2009

### gabbagabbahey

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

11. Oct 14, 2009

### 6021023

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

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

12. Oct 14, 2009

### gabbagabbahey

Yes.

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

13. Oct 14, 2009

### 6021023

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

14. Oct 14, 2009

### 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 $\frac{2mE}{\hbar^2}$?

15. Oct 14, 2009

### 6021023

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

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).

16. Oct 14, 2009

### gabbagabbahey

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

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?

17. Oct 14, 2009

### 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.

18. Oct 14, 2009

### gabbagabbahey

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

What if both functions are constants?

19. Oct 14, 2009

### 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.

20. Oct 14, 2009

### 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?