Translation operator and parity operator

Question

(This is taken from Introduction to Quantum Mechanics by D. Griffiths, 3rd edition, Problem 6.18 .)

If a system has inverse symmetry, we know that [$\hat{H},\hat{\Pi}] =0$ where $\hat{\Pi}$ is the parity operator.

This means that eigenstates of the parity operator are eigenstates of $\hat{H}$. Namely:

$f(x) = \frac{1}{\sqrt{\pi \hbar} }\cos(px/\hbar)$

$g(x) = \frac{1}{\sqrt{\pi \hbar} }\sin(px/\hbar)$

This is easily seen by doing $\hat{\Pi} f(x) =f(x) $ and $\hat{\Pi} g(x) = -g(x) $.

The problem says that translation operator mixes these two states together, meaning that they must be degenerate.

Question

Show that the translation operator mixes these two states together (f and g), meaning that they must be degenerate.

This is what I did:

Translation operator: $\hat{T}u(x) = u(x-a)$
$$\hat{T}f(x)= \frac{1}{\sqrt{\pi \hbar} }\cos(pa/\hbar)\cos(px/\hbar) - \frac{1}{\sqrt{\pi \hbar} }\sin(pa/\hbar)\sin(px/\hbar) \\ = \cos(pa/\hbar)f(x)- \sin(pa/\hbar)g(x). $$

I can see that the states are mixed. But it doesn't have the same energy as $f(x)$ and $g(x)$. If I do $\hat{H}\hat{T}f(x) = E_n( \cos(pa/\hbar)f(x)- \sin(pa/\hbar)g(x) )$

The eigenvalue $E_n$ is multiplied by a constant. The only way this is true if I say that $E_n( \cos(pa/\hbar)f(x)- \sin(pa/\hbar)g(x) ) = E_n w(x) $.

Couldn't I show that they are degenerate by using the simple fact that $\hat{H} f(x) = E_n f(x) $ and $\hat{H} g(x) = E_n g(x) $, because [$\hat{H},\hat{\Pi}] =0$ ?

Answer 1

I dont get the question but from what I get you have the following let

$P$ is parity operator $T$ is translation operator and let $$[T,H]=[P,H]=0$$ also let

so we have $$T|\alpha\rangle=|\beta\rangle$$ $$P|\alpha\rangle=|\gamma\rangle$$ also all those three have same energy since $T$ and $P$ commutes with hamiltonian. now we have $$[T,P]=C$$ c is some operator which is not important

now $$\langle\alpha| TP|\alpha\rangle\neq\langle\alpha| PT|\alpha\rangle=\langle\alpha| TP|\alpha\rangle+\langle\alpha| C|\alpha\rangle\neq1$$ thus $\gamma$ $\beta$ states are different states with same energy thus here is your degeneracy. here the idea is to show hamiltonian has different eigen states with same energy. and party with translation enforces it.

Answer 2

Couldn't I show that they are degenerate by using the simple fact that $\hat{H} f(x) = E_n f(x) $ and $\hat{H} g(x) = E_n f(x) $, because [$\hat{H},\hat{\Pi}] =0$ ?

Where did this come from? $\hat{\Pi}$ does not mix f with g; in fact, it keeps them visibly apart and separate.

The problem asks you to assume separate $E_f$ and $E_g$ for f and g, respectively, and use $$ \hat{H}(\hat{T}f(x)) = \hat{T} \hat{H} f(x) = E_f (\hat{T} f(x) ), $$ which reads explicitly as $$ \hat{H}( \cos(pa/\hbar)f(x)- \sin(pa/\hbar)g(x) )=E_f( \cos(pa/\hbar)f(x)- \sin(pa/\hbar)g(x) )~~~\leadsto \\ - \sin(pa/\hbar)\hat{H}g(x)= - \sin(pa/\hbar) E_f~g(x),~~~~ \leadsto ~~~~E_f=E_g. $$