QM - Eigenfunction / Eigenvalue Problem

[danny271828]

Homework Statement

Find the eigenfunctions and eigenvalues for the operator:

a = x + $$\frac{d}{dx}$$

2. The attempt at a solution

a = x + $$\frac{d}{dx}$$

a$$\Psi$$ = $$\lambda$$$$\Psi$$

x$$\Psi$$ + $$\frac{d\Psi}{dx}$$ = $$\lambda$$$$\Psi$$

x + $$\frac{1}{\Psi}$$$$\frac{d\Psi}{dx}$$ = $$\lambda$$

x + $$\frac{d}{dx}$$ ln($$\Psi$$)= $$\lambda$$

$$\frac{1}{2}$$x$$^{2}$$ + ln($$\Psi$$) = $$\lambda$$x +c

$$\Psi$$ = e^(-$$\frac{1}{2}$$x$$^{2}$$+$$\lambda$$x+c)

$$\Psi$$ = e^(-$$\frac{1}{2}$$x$$^{2}$$+$$\lambda$$)$$\Psi$$(0)

Not sure from here... I think I plug into initial equation?

a$$\Psi$$ = $$\lambda$$$$\Psi$$

substituting and using chain rule I obtain...

x$$\Psi$$ + ($$\lambda$$-x)$$\Psi$$ = $$\lambda$$$$\Psi$$

so x + ($$\lambda$$ - x) = $$\lambda$$

so $$\lambda$$ = $$\lambda$$ nope I guess I don't do that... good check though, so I guess this is the correct way to solve for eigenfunction... Can someone help me with finding $$\lambda$$?

 

[meopemuk]

Normally, the spectrum of allowed eigenvalues $\lambda$ is obtained from the normalization condition. I.e., there should be a non-zero normalization constant (which you denoted by $\Psi(0)$) such that the integral of the square of the modulus of the wavefunction over entire space (the total probability of finding the particle) is equal to 1. In your case, this doesn't matter, because the wavefunction is normalizable for all real $\lambda$. So, all values of $\lambda$ are allowed. However, in more complex cases such as a particle in a potential well - the normalization condition would allow you to find the discrete spectrum of energy eigenvalues.

Eugene.

P.S. Perhaps, it is not a good idea to ask the same question in three different threads.

https://www.physicsforums.com/showthread.php?t=182374
https://www.physicsforums.com/showthread.php?t=182369

 

[danny271828]

Thanks Eugene...

Sorry about that... Kind of new to this... I havn't seen a way to delete your own posts... but there probably is one...