Simultaneous diagonalization and replacement of operators with eigenvalues ?

[AxiomOfChoice]

Apparently, if I have a Hamiltonian that contains an operator, and that operator commutes with the Hamiltonian, not only can we "simultaneously diagonalize" the Hamiltonian and the operator, but I can go through the Hamiltonian and replace the operator with its eigenvalue everywhere I see it show up. Can someone please explain why this is? Thanks!

 

[kanato]

The proof that you can simultaneously diagonalize commuting operators should be available in any elementary quantum mechanics text. IIRC it goes something like this:

1. We examine an eigenvector of operator A with eigenvalue a: $$A|a> = a|a>$$
2. Now act with B (remember numbers commute with operators: $$BA|a> = a B |a>$$
3. We assume B commutes with A: $$BA |a> = AB |a>$$
4. Use 2 to replace in 3: $$A(B |a> )= a (B |a>)$$
5. Now we see that B|a> is an eigenvector of A with eigenvalue a. Thus, B can only scale |a> by a constant which means |a> is an eigenvector of B. I've assumed A has no degenerate eigenvectors; the situation gets a bit trickier under conditions of degeneracy, but you can still prove that commuting operators can be simultaneously diagonalized.

You can only replace the operator with its eigenvalue when the operator is operating on one of its eigenvectors, and then you have to use the eigenvalue associated with that eigenvector. And that's because of the definition of eigenvectors/eigenvalues.

 

[Entropia]

This phenomenon is known as the commutation of observables in quantum mechanics. When two operators commute, it means that their corresponding physical quantities can be measured simultaneously without affecting each other's outcome. This is because the operators share a common set of eigenstates, which are the states in which the operators have definite values.

When we diagonalize the Hamiltonian and the operator simultaneously, we are essentially finding a set of basis states in which both operators are diagonal, meaning they have definite values. This allows us to easily replace the operator with its corresponding eigenvalue in the Hamiltonian because the eigenvalue represents the measurement outcome of that operator in that particular basis state.

In other words, by diagonalizing both the Hamiltonian and the operator, we are able to simplify the mathematical representation of the system and make it easier to analyze and interpret. This concept is fundamental in quantum mechanics and is crucial in understanding the behavior of physical systems at the microscopic level.

 

[Entropia]

Simultaneous diagonalization and replacement of operators with eigenvalues is a powerful mathematical tool in quantum mechanics. It allows for a simplified representation of a system's dynamics, making it easier to analyze and understand.

When a Hamiltonian contains an operator that commutes with it, it means that the two operators have a common set of eigenstates. This means that the two operators can be simultaneously diagonalized, resulting in a set of eigenstates that are common to both operators.

This simultaneous diagonalization allows us to simplify the Hamiltonian by replacing the operator with its corresponding eigenvalue. This is possible because the eigenvalue represents the value of the operator when acting on its corresponding eigenstate.

This simplification is useful because it allows us to focus on the dynamics of the system in terms of its eigenstates and eigenvalues, rather than the more complicated operator. It also allows us to make predictions about the behavior of the system based on the eigenvalues, which are often easier to interpret physically.

In summary, simultaneous diagonalization and replacement of operators with eigenvalues is a useful technique in quantum mechanics that simplifies the representation of a system's dynamics and allows for easier analysis and interpretation.