Conmutative Hermitian operator in degenerate perturbation theory

[carllacan]

Hi.

In 2-fold degenerate perturbation theory we can find appropiate "unperturbate" wavefunctions by looking for simultaneous eigenvectors (with different eigenvalues) of and H° and another Hermitian operator A that conmutes with H° and H'.

Suppose we have the eingenvalues of H° are $E_n = a(-n)^2$, with eivenvectors $\vert n \rangle$ and $\vert -n \rangle$ ( where a is some complex number). Since the squared minus sign is what is causing the degeneracy would it be valid to choose $A = \sqrt{H°}$ so that its eigenvalues would simply be $E_n = a(n)$ and $E_n = a(-n)$?

Thank you for your time.

 

[WannabeNewton]

The usual choice is a set of operators that correspond to obvious symmetries of the system. For example let's say I'm calculating the Stark effect. Then I have rotational symmetry about the axis along which the electric field is aligned, say $z$ axis, so I can choose $J_z$ as one of the operators to use in my commuting set of observables for the degenerate perturbation theory calculation. I also have odd parity in the Hamiltonian so another operator to use is the parity operator. With these two operators, corresponding to the obvious azimuthal and odd parity symmetries of the Hamiltonian, I can reduce the Stark effect calculation to a very simple matrix determinant problem.

And no, what you are trying to do by taking the square root of an operator is a very dangerous thing.

 

[stevendaryl]

The usual choice is a set of operators that correspond to obvious symmetries of the system. For example let's say I'm calculating the Stark effect. Then I have rotational symmetry about the axis along which the electric field is aligned, say $z$ axis, so I can choose $J_z$ as one of the operators to use in my commuting set of observables for the degenerate perturbation theory calculation. I also have odd parity in the Hamiltonian so another operator to use is the parity operator. With these two operators, corresponding to the obvious azimuthal and odd parity symmetries of the Hamiltonian, I can reduce the Stark effect calculation to a very simple matrix determinant problem.

And no, what you are trying to do by taking the square root of an operator is a very dangerous thing.

I think you can make sense of the square-root of a Hermitian operator in the following sense: If $A$ is some Hermitian operator with positive eigenvalues, then you can define an operator $\sqrt{A}$ via:

$\sqrt{A} |\Psi \rangle = \sum_\alpha \sqrt{\alpha} |\alpha \rangle \langle \alpha | \Psi \rangle$

where $|\alpha\rangle$ is the eigenstate of $A$ with eigenvalue $\alpha$ (assuming non-degeneracy of the eigenvalues).

$\sqrt{A}$ is a meaningful operator, but in general it will be nonlocal, even if $A$ is local. That is, in a position basis $\sqrt{A} \Psi(x)$ may depend on $\Psi(x')$ for $x'$ far from $x$.