Hydrogen atom wave function Help

[Shafikae]

Consider a hydrogen atom whose wave function is at t=0 is the following superposition of energy eigenfunctions nlm(r)
(r, t=0) = *[2100(r) -3200(r) +322(r)

What is the probability of finding the system in the ground state (100? in the state (200)? in the state (322)? In another energy eigenstate?
For this part i found each eigen state and put it into an integral. Should there be limits of integration for r? If so, from where to where? I did the integration for (100) and (200) but for (322) i got something crazy.

What is the expectation value of the energy: of the operator L2, of the operator Lz
I have no clue what to do here.

 

[Shafikae]

Consider a hydrogen atom whose wave function is at t=0 is the following superposition of energy eigenfunctions [tex]\psi[/tex]_nlm(r)
[tex]\Psi[/tex](r, t=0) = [tex]\frac{1}{\sqrt{14}}[/tex]*[2[tex]\psi[/tex]₁₀₀(r) -3[tex]\psi[/tex]₂₀₀(r) +[tex]\psi[/tex]₃₂₂(r)

What is the probability of finding the system in the ground state (100? in the state (200)? in the state (322)? In another energy eigenstate?
For this part i found each eigen state and put it into an integral. Should there be limits of integration for r? If so, from where to where? I did the integration for (100) and (200) but for (322) i got something crazy.

What is the expectation value of the energy: of the operator L², of the operator L_z
I have no clue what to do here.

 

[grey_earl]

First, it helps if you don't douple-post. Or even triple-post, there is an Edit button underneath the posts you made.

For your problem: a) the wave functions (eigen functions of the Hamiltonian) also depend on the angular variables [tex]\phi[/tex] and [tex]\theta[/tex], not only on [tex]r[/tex], assuming you use standard spherical coordinates. In those, [tex]r[/tex] ranges from 0 to infinity.
b) you need the expression for [tex]L^2[/tex] and [tex]L_z[/tex] in spherical coordinates. Then you put them "in an integral". If you need more detailed explanations, show your integral and your results for the first to wave functions.

 

[berkeman]

(Looks like multiple posts have been merged -- please do not multiple post.)

 

[Shafikae]

Sorry I didnt know how to edit my post... and I only reposted because I realize it was suppose to be in the homework section. Yes I know they depend on all three variables.

[tex]\psi[/tex]₁₀₀ (r,[tex]\theta[/tex], [tex]\phi[/tex]) = [tex]\frac{exp(-r/a)}{\sqrt{pi*a³}}[/tex]

then i took the square of that and put it in an integral in terms of spherical coordinates...
my answer was -[tex]\frac{exp(-2r/a)}{a²}[/tex] (2r² + 2ar + a²)

I did this each for all 3 states and obtained l[tex]\Psi[/tex](r, t=0) l²

 

[dextercioby]

Well, before worrying about triple integrals and spherical angles, I would search the book/the notes/the internet for the orthonormality relation of the H-atom's Hamiltonian eignefunctions. This would help you see which integrals need to be performed (if any) and which not.

 

[Shafikae]

another part of this question is if the eigen function a parity operator...
I know that under parity operation r -> -r
(r,[tex]\theta[/tex], [tex]\phi[/tex]) -> (r, pi - [tex]\theta[/tex], [tex]\phi[/tex] + pi)

I think it is a parity operator unless I'm not really understanding it. Am I wrong?

 

[vela]

Fixed your LaTex. You'll find it easier to write the whole expression in LaTeX rather than using it just for special characters. You can click on the expressions to see how they were coded.

[tex]\psi_{100}(r,\theta, \phi) = \frac{\exp(-r/a)}{\sqrt{\pi a^3}}[/tex]

then i took the square of that and put it in an integral in terms of spherical coordinates...
my answer was [tex]-\frac{\exp(-2r/a)}{a^2}(2r^2 + 2ar + a^2)[/tex].

I did this each for all 3 states and obtained [tex]|\Psi(r, t=0)|^2[/tex].

From what you've written, it's not really clear what you're calculating or why.

In general, to find the probability that the atom is in a state ϕ, you would first calculate the amplitude

[tex]A_\phi = \langle \phi \vert \Psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} \phi^*(r,\theta,\phi)\Psi(r,\theta,\phi) r^2 \sin\theta \,d\phi\,d\theta\,dr[/tex]

and then square its modulus to get the probability P_ϕ=|A_ϕ|².

In this problem, the state ϕ happens to be an eigenstate of the hydrogen atom and Ψ is expressed as a linear combination of hydrogen atom eigenstates; therefore, you can take advantage the orthonormality of the eigenstates and avoid grinding through the integrals. (Or you could grind out the integrals, and if you do them correctly, you should see the property we're all referring to.)

another part of this question is if the eigen function a parity operator...
I know that under parity operation r -> -r
(r,[tex]\theta[/tex], [tex]\phi[/tex]) -> (r, pi - [tex]\theta[/tex], [tex]\phi[/tex] + pi)

I think it is a parity operator unless I'm not really understanding it. Am I wrong?

That's the effect of the parity operator P on the wave function ψ:

[tex]\hat{P}\psi(r,\theta,\phi) = \psi(r,\pi-\theta,\phi+2\pi)[/tex]

The question is asking you to determine whether

[tex]\hat{P}\psi(r,\theta,\phi) = \lambda \psi(r,\theta,\phi)[/tex]

holds for some constant λ.