Is the Secular Matrix of Perturbation Diagonal for Hydrogen's n=2 Level?

[Fraktal]

Homework Statement

Show for the n=2 level of hydrogen, that the secular matrix of the perturbation $\hat{V}$ is diagonal in the basis of states $\psi_{n,l,m}$.

Homework Equations

1. The n-th energy level splitting is found from solving the eigenvalue problem for the secular matrix:

$$H_{\alpha,\beta}=\langle n,\alpha |\left(\hat{H}_{K}+\hat{H_{S}}\right) |n,\beta \rangle$$

2. The perturbation is given by:

$$\hat{V}=-\frac{\hbar^{4}}{8m^{3}c^{2}}\Delta^{2}$$

3. Wave function in n-th level:

$$\psi_{n,l,m}(r)\chi_{S}(\sigma)=|n, \alpha \rangle$$

where $\alpha = {l,m,s}$

The Attempt at a Solution

I think I need to show that the $\hat{H_{K}}$ term in the 'relevant equation 1' is diagonal, but not sure how to do this.

.. or it could be some completely different method. I don't know how to start with this.

 

[mark726]

Thank you for your question. To show that the secular matrix of the perturbation \hat{V} is diagonal in the basis of states \psi_{n,l,m}, we need to first understand the structure of the secular matrix and how it is affected by the perturbation. The secular matrix is a matrix representation of the Hamiltonian operator in a specific basis, in this case the basis of states \psi_{n,l,m}. The elements of the matrix correspond to the expectation values of the Hamiltonian operator between different states in the basis.

In the absence of the perturbation \hat{V}, the Hamiltonian operator is diagonal in the basis of states \psi_{n,l,m} since it only contains the kinetic energy term \hat{H_{K}} and the potential energy term \hat{H_{S}}. This means that the expectation values of the Hamiltonian operator between different states in the basis will be zero unless they are the same state, in which case the expectation value will correspond to the energy of that state.

Now, when we introduce the perturbation \hat{V}, it will add an additional term to the Hamiltonian operator. However, since the perturbation only depends on the position of the electron, it will not affect the angular momentum or spin states of the electron. This means that the perturbation will only affect the energy levels, resulting in a splitting of the n-th energy level into sub-levels.

In the basis of states \psi_{n,l,m}, this means that the only non-zero elements of the secular matrix will be along the diagonal, since the perturbation only affects the energy levels of the states and not the states themselves. This is why the secular matrix of the perturbation \hat{V} is diagonal in the basis of states \psi_{n,l,m}.

To show this mathematically, we can write out the elements of the secular matrix as:

H_{\alpha,\beta}=\langle n,\alpha |\left(\hat{H}_{K}+\hat{H_{S}}+\hat{V}\right) |n,\beta \rangle

Since the perturbation only affects the energy levels, we can rewrite the above equation as:

H_{\alpha,\beta}=\langle n,\alpha |\left(\hat{H}_{K}+\hat{H_{S}}\right) |n,\beta \rangle + \langle n,\alpha |\hat