# Schrodinger equation problem

1. Oct 25, 2015

### Abdul.119

1. The problem statement, all variables and given/known data
Consider the time-dependent Schrodinger equation for a free particle in two spatial dimensions

Using the method of separation of variables, determine the wave function ψ(x,y,t)

2. Relevant equations

3. The attempt at a solution
Not sure how to do the separation here since it is a function of 3 variables, I'm inclined to think that Ψ(x,y,t) = f(x) g(y) j(t) can be assumed, hence substitute it in the equation and separate. Is it correct to assume that?

2. Oct 25, 2015

### TSny

Yes. See what you get.

3. Oct 25, 2015

### Abdul.119

I would get this

Then what? I can't separate each variable to each side, since there are 3 variables
Correction: in the LHS the g(t) should be g(y)

4. Oct 25, 2015

### TSny

The next step is standard: divide the equation by Ψ(x, y, t) = f(x)g(y)j(t).

5. Oct 25, 2015

### Abdul.119

Ok ok. Then I get this

Then what do I do with it?

6. Oct 25, 2015

### TSny

7. Oct 25, 2015

### Abdul.119

Yes, but I'm new to it. So I believe I should set the whole equation equal to a variable lambda, then the LHS would be

And do the same for the RHS where I would get two other separate equations, one in terms of f and one in terms of g? or would it just give:
?

Last edited: Oct 25, 2015
8. Oct 25, 2015

### TSny

Yes, except that you should think of $\lambda$ as a constant rather than a variable. To see this, note that your equation in post #5 must be satisfied for all allowed values of t, x, and y. So, suppose you arbitrarily pick a value for x and y. The right hand side is then some number. The left hand side is a function of time that must equal that number for any value of t. So, in fact, the left side must be independent of t. That is, the left side must be a constant which you can call $-\lambda$.

See if you can construct similar arguments to show that the individual terms on the right side of the equation in post #5 must each be equal to a constant (although not necessarily the same constant.)

9. Oct 25, 2015

### Abdul.119

Ok, then the terms in the RHS would be
Where lambda is just an arbitrary constant
Is the problem finished like this?

Last edited: Oct 25, 2015
10. Oct 25, 2015

### TSny

OK. But, you should distinguish the three constants by not using the same symbol $-\lambda$ for each one. Also, there is no need to write the constants with a negative sign. In the Wikipedia link, they wrote the constant with a negative sign for later convenience for the particular partial differential equation there. In your case, you will discover whether the constants should be positive or negative as you solve the individual equations.

The three constants are not completely arbitrary at this point. The equation in post #5 gives you a relation that the three constants must satisfy.

11. Oct 25, 2015

### Abdul.119

ħ
Ok, then I can say

And so

And same for the LHS of the equation in #5

12. Oct 25, 2015

### TSny

OK. These are standard differential equations which you can solve. How is the constant $\lambda$ related to the constants $B$ and $C$? (Use the equation in post #5)

13. Oct 25, 2015

### Abdul.119

From the equation, λ is related to B and C as
λ = B + C

Last edited: Oct 25, 2015
14. Oct 25, 2015

### Abdul.119

Then where do I go from there?

15. Oct 25, 2015

### TSny

Yes. So, λ will be determined once the values of B and C are known.

16. Oct 25, 2015

### TSny

You have 3 ordinary differential equations: one for j(t), one for f(x), and one for g(y). Try to solve them.

17. Oct 25, 2015

### Abdul.119

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I get really complicated stuff. For example for f(x), I multiply the equation by dx^2 and divide by f(x), then integrate with respect to f, and the rest of the terms are constants, the integration gives f(ln(f)-1). Am I doing something wrong here?

Last edited: Oct 25, 2015
18. Oct 25, 2015

### TSny

You can't multiply by dx2. You can't "break apart" the second derivative that way.

Simplify the equation by lumping all the constants together. For example, suppose you let $k_x$ be a constant such that $k_x^2 = \frac{2mB}{\hbar^2}$. Write the differential equation in terms of $k_x$. You will see that you get a familiar type of ordinary differential equation.

19. Oct 25, 2015

### Abdul.119

I need to take the square root to get the of the second power on df^2/dx^2 ?

20. Oct 26, 2015

### TSny

No.

Write the differential equation as $f''(x) = -k_x^2f(x)$, where the constant $k_x$ was defined in post #18. You're looking for a function $f(x)$ such that when you take its second derivative you get back the function multiplied by a negative constant.

Are you familiar with the form of the wavefunction for a free particle moving in one dimension? You might expect a similar form for $f(x)$.