I Given a Hamiltonian, finding the energy levels

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1. Mar 8, 2017

Buggy Virus

Hey, I just had a quick question about using hamiltonians to determine energy levels.
I know that the eigenvalue of the hamiltonian applied to an eigenket is an energy level.

H |a> = E |a>

But my question is if I am given an equation for a specific Hamiltonian:

H = (something arbitrary)

And asked to find the energy levels of the object I am given that hamiltonian for (say a molecule or a particle) and no other information, what strategy do I use?
If the Hamiltonian is comprised of angular momentum operators, can I just say my object is an arbitrary eigenket = |n, l, m> and find the general eigenvalue of when my hamiltonian is applied to that?

2. Mar 8, 2017

blue_leaf77

Solve it using calculus, much like how you solve the eigenvalue equation for the famous infinite well potential. If the Hamiltonian is defined on a finite-dimensional Hilbert space, using matrix mechanics can be easier.
You shouldn't do that. You have to check first whether those three quantum numbers represent good quantum numbers for your system. Consider the case where the Hamiltonian contains term like $\mathbf L \cdot \mathbf S$, in this case at least $m$ will not be a good quantum number because $L_z$ does not commute with $\mathbf L \cdot \mathbf S$ and hence you can't use it to specify the eigenstates.

3. Mar 9, 2017

Buggy Virus

If I were able to transform the Hamiltonian into terms that all commute with Lz would it be fine just to consider the system as an arbitrary |n, l, m> system?

4. Mar 9, 2017

A. Neumaier

You solve the eigenvalue problem for that Hamiltonian. In very simple cases analytically, in most cases numerically. For molecules, a famous package that you can use is Gaussian.

5. Mar 9, 2017

blue_leaf77

You actually have to find all operators which commute with the Hamiltonian to determine which quantum numbers can be used to uniquely specify each eigenstates. However, usually such task is aleviated by the number of degree of freedom. Usually the number of quantum numbers needed for unique identification is equal to the number of degree of freedom. If you find more than the number of degree of freedom, this operator should be a function of the other operators commuting with the Hamiltonian that you have found. For example, in 1D 1 particle system there should only be one quantum number for each eigenstate. For 3D systems and 1 electron such as hydrogen atom neglecting spin, you need three quantum numbers.
Just because a Hamiltonian commutes with $L_z$ does not mean that it also commutes with $L^2$. Consider a term like $kz$ where $k$ is a constant, this term commutes with $L_z$ but does not with $L^2$.

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