Undergrad Given a Hamiltonian, finding the energy levels

Click For Summary
To find energy levels from a given Hamiltonian, one must solve the eigenvalue problem associated with that Hamiltonian, which can be done analytically for simple cases or numerically for more complex systems. When dealing with Hamiltonians that include angular momentum operators, it is crucial to verify whether the quantum numbers n, l, and m are appropriate for the system, as not all may be good quantum numbers. If the Hamiltonian can be transformed into terms that commute with Lz, it may be acceptable to use the |n, l, m> representation, but one must ensure that the number of quantum numbers corresponds to the degrees of freedom in the system. Additionally, operators commuting with the Hamiltonian must be identified to determine which quantum numbers can uniquely specify eigenstates. Overall, careful analysis of the Hamiltonian's structure and its commutation relations is essential for accurately determining energy levels.
Buggy Virus
Messages
2
Reaction score
0
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?
 
Physics news on Phys.org
Buggy Virus said:
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?
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.
Buggy Virus said:
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?
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.
 
  • Like
Likes BvU
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?
 
Buggy Virus said:
I am given an equation for a specific Hamiltonian [...]
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?
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.
 
Buggy Virus said:
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?
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##.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
996
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K