Solve a PDE with Separation of Variables

  • #1
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Homework Statement



Solve the 2-D time-independent Schrödinger equation with V (x,y) = 0:



Homework Equations





2/2m ( ∂2Ψ(x,y)/∂x2 + ∂2Ψ(x,y)/∂y2 ) = EΨ(x,y)​



The Attempt at a Solution



I started by getting -ћ2/2m to one side:


( ∂2Ψ(x,y)/∂x2 + ∂2Ψ(x,y)/∂y2 ) = -2mE/ћ2 Ψ(x,y)​


Here is my problem: I'm not sure what to do with everything on the right side of the = sign.

What follows is my misguided attempt at a solution:




I assumed a solution of the form


Ψ(x,y) = X(x)Y(y)​


Then I substituted this in for Ψ(x,y), giving:



Y(y) d2X(x)/dx2 + X(x) d2Y(y)/dy2 = -2mE/ћ2 X(x)Y(y)​



Then I divided both sides by X(x)Y(y):



1/X(x) d2X(x)/dx2 + 1/Y(y) d2Y(y)/dy2 = -2mE/ћ2


And here is where I hit the road block. If the two 2nd order ODEs are equal to a constant, then it is pretty obvious that since they add together to get -2mE/ћ2
and are equal to eachother, then the constant they are equal to has to be -mE/ћ2. I can't really see how to solve those ODEs. I mean, I can't just use the auxilery equation, because that would be m2 = 0, which would give me two solutions of zero. Obviously I am missing something pretty important here.

Any insight would be much appreciated. Please keep in mind that this is very new to me, and any explanations will have to be pretty elementary if I am to understand.

Thanks!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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You must have solved an equation like X''(x)=C*X(x) before. Because that's what you've got. I think the general solution can be expressed as the sum of exponential functions. Or if C is negative, sin's and cos's.
 
  • #3
Your are thinking about the Laplace equation:

[tex]\nabla^{2} \Psi= 0[/tex]

The time-independent Schrodinger is what is called the Helmoltz equation:

[tex](\nabla^{2} - k^{2})\Psi= 0[/tex]

where [tex]k^{2}= \frac{-2mE}{h^{2}}[/tex]

What you have is fine, so lets just define

[tex]\frac{X''}{X}=k^{2}_{x}[/tex]

and

[tex]\frac{Y''}{Y}=k^{2}_{y}[/tex]

So, [tex]k^{2}_{x} + k^{2}_{y}=k^{2}[/tex]

You solve like you normally do. At this point you cannot say what [tex]k^{2}_{x}[/tex] and [tex]k^{2}_{y}[/tex] actually are.
But the value of E will determine the sign of [tex]k^{2}[/tex]. This in turn gives the signs of gives the sign of [tex]k^{2}_{x}[/tex] and [tex]k^{2}_{y}[/tex] from our relationship.

One of the values of E is allowed, while that other is not. This depends on the physics of the problem: boundary conditions, initial conditions, unbounded variables, finiteness, etc. Also, [tex]k^{2}_{x}[/tex] and [tex][k^{2}_{y}][/tex] depend on the physics.

Like, is this a 2D Well or a Free Particle? The solutions are very similar and use the PDE you mentioned.

If you haven't notice yet, E is the separation constant between the spatial and time parts of the Schrodinger Eq.
 
Last edited:

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