How to Solve d^2 phi(y) / dy^2 = y^2 phi(y)?

[bon]

Homework Statement

See attached please

Homework Equations

The Attempt at a Solution

OK so I've done part (a) ... I just don't see how to do part (b)

I'm trying to solve d^2 phi(y) / dy^2 = y^2 phi(y), right? why should the answer include y^m? confused!

Any help would be great! Thanks

 

[kuruman]

Have you tried putting the assumed solution (equation 10) in the differential equation (equation 8)? Once you do that, consider the leading term in the expression you get in the limit y → ∞. The m is needed because all functions in the form of equation 10 show the same asymptotic behavior as y becomes large.

 

[bon]

Have you tried putting the assumed solution (equation 10) in the differential equation (equation 8)? Once you do that, consider the leading term in the expression you get in the limit y → ∞. The m is needed because all functions in the form of equation 10 show the same asymptotic behavior as y becomes large.

Thanks but Do you mean substitute it into equation (9) - (not 8) - but ignoring the epsillon term? I still don't see why we need the y^m term. As y becomes large phi (y) tends to zero as e^(-y^2) tends to zero faster than y^m tends to infinity..

 

[bon]

When i subs psi = A y^m e^(-y^2 /2) into 9 ignoring the epsillon term i get:

psi '' - y^2 psi = Ae^(-y^2/2) [ m(m-1) y^(m-2) + (-2m-1)y^m]

now this obviously tends to zero as y tends to infinity, so it satisfies the equation..but why did we need to y^m bit? Also, surely a whole load of other functions would have satisfied the DE?

 

[vela]

Try assume a solution of the form ψ(y) = p(y) e^-y2/2 and find the differential equation p(y) satisfies. Solve for p(y) using a series method, and show that if you require a normalizable solution ψ(y), the series has to truncate after a finite number of terms.

 

[bon]

Try assume a solution of the form ψ(y) = p(y) e^-y2/2 and find the differential equation p(y) satisfies. Solve for p(y) using a series method, and show that if you require a normalizable solution ψ(y), the series has to truncate after a finite number of terms.

Thanks but I am still completely confused.. :(

is my expression for psi'' - y^2 psi assuming the solution psi = Ay^m e^-y^2 in the post above correct?

I just still don't see where the y^m is coming from.. ill try to explain why. Consider equation (9) in the limit y-> infinity. we have

psi' - y^2 psi = 0. Now imagine you are solving that equation on its own, without reference to any other parts of the questions. Where does the y^m factor come into it? The equation doesn't even include an m! confused:S

sorry!

 

[bon]

anyone?

 

[vela]

Your confusion is understandable. The point of the problem could have been made a bit more explicitly.

Suppose the solution is given by ψ(y)=p(y)e^-y2/2 for some function p(y). If p(y) is a polynomial, then as y→∞, p(y)≈y^m, where m is the degree of p(y), and ψ(y)≈y^me^-y2/2. If p(y) isn't a polynomial, you'll get different asymptotic behavior. So what the problem is asking you to do is show that p(y) is a polynomial, as opposed to a function like sin y or e^y, where its series representation has an infinite number of terms.

 

[bon]

Aha ok i think i see now thanks!

So is my above expression for subsituting psi = ... into the equation (in post 4) correct?

 

[bon]

anyone?