What Eigenvalues Lead to Square-Integrable Eigenfunctions?

[Felicity]

Homework Statement

solve the eigenvalue problem

∫_(-∞)^x dx' (ψ(x' ) x' )=λψ(x)

what values of the eigenvalue λ lead to square-integrable eigenfunctions?

The Attempt at a Solution

∫_(-∞)^xdx' (ψ(x' ) x' )=λψ(x)

differentiate both sides to get

ψ(x)x=λ d/dx ψ(x)


ψ(x)x/λ= d/dx ψ(x)

2xe ^{x^2} =d/dx e ^{x^2}

so ψ(x) = e ^{x^2} and λ = 2

but this is not square integrable so either this is incorrect or there are other solutions I am not seeing

Can anyone help me find what I am missing?

Thank you

 

[Dick]

You want to solve the differential equation, ψ(x)x/λ= d/dx ψ(x) Separate the variables. You should get a solution with a lambda in it. Figure out what values of lambda make it square integrable.

 

[yaychemistry]

Hello,
I think there are other solutions.
Take your equation,
x\psi\left(x\right) = \lambda\frac{d}{dx}\psi\left(x\right)
and substitute \psi\left(x\right) = \exp\left(f\left(x\right)\right) and see if you don't get an equation for f\left(x\right) which has a solution that depends on \lambda.

Also, don't forget your solution has to be finite at the lower endpoint of the integral (-\infty) in the original problem statement.

 

[Felicity]

You want to solve the differential equation, ψ(x)x/λ= d/dx ψ(x) Separate the variables. You should get a solution with a lambda in it. Figure out what values of lambda make it square integrable.

ok i separated variables, integrated and got

ψ(x)=e^{1/(2λ) x2}

however I don't see how this can be square integrable since for any value of λ I can think of the integral of the square will equal infinity. I feel like I am missing something obvious here but I don't know what it is

 

[Dick]

ok i separated variables, integrated and got

ψ(x)=e^{1/(2λ) x2}

however I don't see how this can be square integrable since for any value of λ I can think of the integral of the square will equal infinity. I feel like I am missing something obvious here but I don't know what it is

Will it equal infinity even if lambda is negative?

 

[Felicity]

of course! I am so embarrassed, thank you