Show that it is a energy eigenstate and find the corresponding energy

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SUMMARY

The discussion centers on demonstrating that the wavefunction |ψ⟩ = (1/√2)(|210⟩ + |211⟩) is an energy eigenstate of the hydrogen atom. The key conclusion is that |ψ⟩ qualifies as an energy eigenstate if it is an eigenvector of the Hamiltonian operator for the hydrogen potential. The corresponding energy can be derived from the quantum numbers associated with the state vectors, specifically using the hydrogen atom's Hamiltonian.

PREREQUISITES
  • Understanding of quantum mechanics, specifically the hydrogen atom model.
  • Familiarity with wavefunctions and their normalization.
  • Knowledge of the Hamiltonian operator in quantum mechanics.
  • Basic concepts of angular momentum operators, including L² and Lz.
NEXT STEPS
  • Study the Hamiltonian operator for the hydrogen atom and its implications on energy levels.
  • Learn about the quantum numbers n, l, and m and their significance in determining energy eigenstates.
  • Explore the derivation of energy eigenvalues for hydrogen using the Schrödinger equation.
  • Investigate the role of angular momentum operators in quantum mechanics, particularly L² and Lz.
USEFUL FOR

Students and professionals in quantum mechanics, particularly those studying atomic physics and energy eigenstates in systems like the hydrogen atom.

Denver Dang
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Homework Statement


Hi.

I'm looking at a hydrogen atom, which normalized stationary states is defined as |nlm>

The hydrogen atom is described by the normalized wavefunction:
[tex]\left| \psi \right\rangle =\frac{1}{\sqrt{2}}\left( \left| 210 \right\rangle +\left| 211 \right\rangle \right)[/tex]
Now, show that [itex]\left| \psi \right\rangle[/itex] is an energy eigenstate, and find the corresponding energy.


Homework Equations



I'm told that:

[tex]{{L}^{2}}\left| nlm \right\rangle =l\left( l+1 \right){{\hbar }^{2}}\left| nlm \right\rangle[/tex]
[tex]{{L}_{z}}\left| nlm \right\rangle =m\hbar \left| nlm \right\rangle[/tex]
[tex]{{L}_{+}}={{L}_{x}}+i{{L}_{y}}[/tex]
[tex]{{L}_{-}}={{L}_{x}}-i{{L}_{y}}[/tex]


The Attempt at a Solution



In my mind, it seems so easy, but I don't have my book at my side, so I can't even check how it is done. Does it has something to do with:
[tex]H\left| \psi \right\rangle =E\left| \psi \right\rangle[/tex]

If so, what hamiltonian am I suppose to use ?

Well, I'm kinda lost right now, so I was hoping to get a push in the right direction.

Thanks in advance.


Regards
 
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Yes - it is an energy eigenstate if it is an eigenvector of the Hamiltonian.
You will need the Hamiltonian for the hydrogen potential.

Note: out of the quantum numbers shown in the state vectors, which refer to the energy eigenstates?
 
Ahh yes, thank you very much :D
 

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