Probability density for the hydrogen atom

[wakaranai]

Homework Statement

Consider a superposition state |Ψ⟩ = (|2, 0, 0⟩ + |2, 1, 0⟩) of the hydrogen atom.
a) Calculate the probability density at t = 0 and sketch the function in the (xy) plane.
b) Calculate ⟨Ψ|r|Ψ⟩ and mark on the graph the expected value of the position obtained in part (a). Interpret your result.

Relevant Equations

I'm not certain about the equations

I did this and I don't know if it's correct:

 

[PeroK]

Please see the guidelines for asking for homework help:

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.635513/

Here are some common mistakes to avoid:

• Don't simply say "I have no clue," "I have no idea where to start," or "I'm completely lost."
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[wakaranai]

Please see the guidelines for asking for homework help:

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.635513/

Here are some common mistakes to avoid:

• Don't simply say "I have no clue," "I have no idea where to start," or "I'm completely lost."
  These don't qualify as attempts. Instead, it suggests you haven't put much effort into reading and understanding your textbook and lecture notes, going over similar examples, etc. The helpers aren't here to answer your questions so that you don't have to read your book. We get that some people say this even though they've put in the effort. If this is you, please heed the next bit of advice.

So sorry, now i tried to include my attempt.
(English is not my first language)

 

[PeroK]

So sorry, now i tried to include my attempt.
(English is not my first language)

If ##\psi = \psi_1 + \psi_2##. Which of these is correct?
$$|\psi|^2 = |\psi_1|^2 + |\psi_2|^2$$$$|\psi|^2 = (\psi_1 + \psi_2)(\psi_1^* + \psi_2^*) = |\psi_1|^2 + |\psi_2|^2 + \psi_1\psi_2^* + \psi_2 \psi_1^* $$

 

[wakaranai]

If ##\psi = \psi_1 + \psi_2##. Which of these is correct?
$$|\psi|^2 = |\psi_1|^2 + |\psi_2|^2$$$$|\psi|^2 = (\psi_1 + \psi_2)(\psi_1^* + \psi_2^*) = |\psi_1|^2 + |\psi_2|^2 + \psi_1\psi_2^* + \psi_2 \psi_1^* $$

the second one...
in this case ψ_1=ψ*_1 (and the same for ψ_2)?

 

[PeroK]

the second one...
in this case ψ_1=ψ*_1 (and the same for ψ_2)?

Yes, those particular wave functions are real-valued.

Note also that the wave-function you have been given is not normalised:

Consider a superposition state |Ψ⟩ = (|2, 0, 0⟩ + |2, 1, 0⟩) of the hydrogen atom.

That wavefunction has a modulus squared of ##2##. I would have expected:
$$\ket \psi = \frac 1 {\sqrt 2}\big (\ket {2,0,0} + \ket {2,1,0} \big )$$

 

[wakaranai]

Yes, those particular wave functions are real-valued.

Note also that the wave-function you have been given is not normalised:

That wavefunction has a modulus squared of ##2##. I would have expected:
$$\ket \psi = \frac 1 {\sqrt 2}\big (\ket {2,0,0} + \ket {2,1,0} \big )$$

I don't know how to relate the result with the plot in xy plane

 

[PeroK]

View attachment 330311

I don't know how to relate the result with the plot in xy plane

You seem to have got a lot of simplification for your final answer. Are you sure about that? I would expect the result to be messy.

Note that, in the x-y plane, we have ##\cos \theta = 0##. That should simplify things. Even so, sketching that function wouldn't be easy!

 

[wakaranai]

You seem to have got a lot of simplification for your final answer. Are you sure about that? I would expect the result to be messy.

Note that, in the x-y plane, we have ##\cos \theta = 0##. That should simplify things. Even so, sketching that function wouldn't be easy!

So |Ψ|^2 = |Ψ_1|^2

 

[PeroK]

So |Ψ|^2 = |Ψ_1|^2

Looks like it.