We know that if Hamiltonian commutes with parity operator and energy eigen values are non-degenerate then the corresponding wave function has well defined parity. But my question is what about degenerate eigenvalues(I know then in that case eigen function do not have definite parity), Is there a way to prove it mathematically that eigenfunctions of degerate eigenvalue do NOT have definite parity even when Hamiltonian commutes with parity operator.
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This statement cannot be proved because it is incorrect.
TLDR: Although the degenerate eigenstates of a Hamiltonian that commutes with the parity operator need not be eigenstates of the parity operator, you can always make an appropriate linear combination to obtain a new set of eigenstates of the Hamiltonian that are also eigenstates of the parity operator, and therefore have definite parity.
Longer answer: Consider two operators $\hat{A}$ and $\hat{B}$ that commute, and let $$ \hat{A}|\psi\rangle=\lambda|\psi\rangle $$ be the eigenvalue equation for $\hat{A}$. Then $\hat{B}|\psi\rangle$ is also an eigenstate of $\hat{A}$ with the same eigenvalue. To prove this: $$ \hat{A}(\hat{B}|\psi\rangle)=\hat{B}(\hat{A}|\psi\rangle)=\hat{B}(\lambda|\psi\rangle)=\lambda(\hat{B}|\psi\rangle), $$ where in the first step we used the fact that the operators commute. If $\lambda$ is a non-degenerate eigenvalue, then $\hat{B}|\psi\rangle$ is necessarily proportional to $|\psi\rangle$, so we can write: $$ \hat{B}|\psi\rangle=\mu|\psi\rangle $$ to show that $|\psi\rangle$ is also an eigenstate of $\hat{B}$ (with eigenvalue $\mu$). In your case, if $\hat{A}$ is the Hamiltonian and $\hat{B}$ the parity operator, then the eigenstates of the Hamiltonian are also eigenstates of the parity operator, and therefore they have definite parity.
If $\lambda$ is degenerate, then you are correct that we can no longer say that $|\psi\rangle$ is also an eigenstate of $\hat{B}$, because all we can say now is that $\hat{B}|\psi\rangle$ belongs to the subspace spanned by the eigenstates $|\psi^i\rangle$ that have the same eigenvalue $\lambda$, where $i=1,\ldots,g$ labels the different degenerate eigenstates. This is because any linear combination within the degenerate subspace is also an eigenstate of $\hat{A}$ in that subspace with the same eigenvalue $\lambda$. To show this, consider an arbitrary linear combination within that subspace: $$ |\chi\rangle=\sum_{i=1}^gc_i|\psi^i\rangle $$ for arbitrary coefficients $c_i$. Then, $$ \hat{A}|\chi\rangle=\hat{A}\left(\sum_{i=1}^gc_i|\psi^i\rangle\right)=\sum_{i=1}^gc_i\hat{A}|\psi^i\rangle=\sum_{i=1}^gc_i\lambda|\psi^i\rangle=\lambda\left(\sum_{i=1}^gc_i|\psi^i\rangle\right)=\lambda|\chi\rangle. $$ This means that $\hat{B}|\psi\rangle$ could be proportional to the result of one of these linear combinations rather than one of the original eigenstates, so the original eigenstates need not be eigenstates of the parity operator.
However, as the action of $\hat{B}$ on $|\psi\rangle$ is restricted to the degenerate subspace, then you can diagonalize $\hat{B}$ within that subspace to find the eigenstates of $\hat{B}$. These eigenstates are also eigenstates of $\hat{A}$ because they are simply linear combinations of the degenerate eigenstates. This means that you can always choose an appropriate linear combination of the original eigenstates of $\hat{A}$ to create a new set of eigenstates that are also eigenstates of $\hat{B}$. In your case, these new eigenstates will be degenerate eigenstates of the Hamiltonian and eigenstates of the parity operator, and therefore will have definite parity.
