Find the eigenstates and eigenvalues of the Hamiltonian.

[gabbagabbahey]

The wave function is defined as exp(-ax^2), so for a=0 it's 1.

No it isn't, you are forgetting about the normalization constant...which depends on $a$

Now, $$<H(0)>=\int V(x)dx$$ Now if I assume that this integral converges then it mustn't be bigger than zero, if it were zero then the integrand would equal almost everywhere to zero, but V(x)<0 for every x, so this integral is negative.

Again, $\langle H(0) \rangle$ is undefined, but $\lim_{a\to 0^{+}}\langle H(a)\rangle$ is well defined.

 

[MathematicalPhysicist]

Ok, thanks.

 

[MathematicalPhysicist]

wait a minute, how do I calculate this limit?
$$<H(a)>=a\hbar^2/2m + \sqrt(2a/\pi)\int exp-2ax^2)dx$$
it looks to me it converges to zero.

 

[jdwood983]

wait a minute, how do I calculate this limit?
$$<H(a)>=a\hbar^2/2m + \sqrt(2a/\pi)\int exp-2ax^2)dx$$
it looks to me it converges to zero.

Shouldn't there be a V in the integrand?

 

[MathematicalPhysicist]

Yes, there should, and square root is over (2a/pi).

I believe this question is the easiest from the questions from my HW.
:-)