Degeneracy

Two or more linear independent quantum states (Eigenstate) share the same eigenvalue. An energy level EnE_n is degenerate if there exist multiple linearly independent Eigenstate ∣ψ1⟩,∣ψ2⟩,…,∣ψn⟩\left\lvert \psi_1 \right\rangle, \left\lvert \psi_2 \right\rangle, \ldots, \left\lvert \psi_n \right\rangle which satisfy

H^∣ψi⟩=En∣ψi⟩∀i=1,2,…,g\hat{H}\left\lvert \psi_i \right\rangle=E_n\left\lvert \psi_i \right\rangle\quad \forall i=1,2,\ldots,g

Where if g=1g=1, the energy level is non-degenerate. If g>1g>1, the energy level is degenerate. If something is degenerate, that implies there is no unique basis. What you want to write down should be a projection onto the Subspace.

H^≜(E00E10E2)∣0⟩≜(100)∣1a⟩≜(010)∣1b⟩≜(001)\hat{H}\triangleq\begin{pmatrix} E_0 & & 0 \\ & E_1 & \\ 0 & & E_2 \end{pmatrix} \quad \left\lvert 0 \right\rangle\triangleq\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} \quad \left\lvert 1a \right\rangle\triangleq\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} \quad \left\lvert 1b \right\rangle\triangleq\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} ⇒H^=E0∣0⟩⟨0∣+E1∣1a⟩⟨1a∣+E2∣1b⟩⟨1b∣\Rightarrow\hat{H}=E_0\left\lvert 0 \right\rangle\left\langle 0 \right\rvert+E_1\left\lvert 1a \right\rangle\left\langle 1a \right\rvert+E_2\left\lvert 1b \right\rangle\left\langle 1b \right\rvert

The Ket (State) ∣1a⟩,∣1b⟩\left\lvert 1a \right\rangle,\left\lvert 1b \right\rangle are degenerate if and only if E1=E2E_1=E_2. If E0=E1,E1≠E2E_0=E_1, E_1\neq E_2 then ∣0⟩\left\lvert 0 \right\rangle is degenerate with ∣1a⟩\left\lvert 1a \right\rangle.

This just means that you can’t detect the difference between states given only the energy/eigenvalue.

Remember the eigenvalue is observed as energy if the operator is the Hamiltonian.

This means the Hamiltonian alone is not enough to determine the state.

This is because if E is degenerate then

H∣ψ1⟩=E∣ψ1⟩H\left\lvert \psi_1 \right\rangle= E\left\lvert \psi_1 \right\rangle H∣ψ2⟩=E∣ψ2⟩H\left\lvert \psi_2 \right\rangle= E\left\lvert \psi_2 \right\rangle

Now just finding EE doesn’t mean we can distinguish the states.

We need another operator A^\hat{A} which commutes with H^\hat{H} to distinguish the states. Note that just because A^\hat{A} commutes with H^\hat{H} doesn’t mean we can distinguish the states.

A∣ψ1⟩=a1∣ψ1⟩A\left\lvert \psi_1 \right\rangle= a_1\left\lvert \psi_1 \right\rangle A∣ψ2⟩=a2∣ψ2⟩A\left\lvert \psi_2 \right\rangle= a_2\left\lvert \psi_2 \right\rangle

And if a1≠a2a_1\neq a_2 then we can distinguish the states.

BUT if [A^,H^]≠0[\hat{A},\hat{H}]\neq0 then this doesn’t always hold true for the entire Hilbert Space.

This is because if

H^∣ψ⟩=E∣ψ⟩\hat{H}\left\lvert \psi \right\rangle= E\left\lvert \psi \right\rangle ⇒A^H^∣ψ⟩=A^E∣ψ⟩\Rightarrow\hat{A}\hat{H}\left\lvert \psi \right\rangle= \hat{A}E\left\lvert \psi \right\rangle ⇒(H^A^−[H^,A^])∣ψ⟩=EA^∣ψ⟩\Rightarrow(\hat{H}\hat{A}-[\hat{H},\hat{A}])\left\lvert \psi \right\rangle = E\hat{A}\left\lvert \psi \right\rangle ⇒H^A^∣ψ⟩−[H^,A^]∣ψ⟩=EA^∣ψ⟩\Rightarrow\hat{H}\hat{A}\left\lvert \psi \right\rangle - [\hat{H},\hat{A}]\left\lvert \psi \right\rangle= E\hat{A}\left\lvert \psi \right\rangle

If [H^,A^]=0[\hat{H},\hat{A}]=0 (i.e., H and A commute) then

⇒H^A^∣ψ⟩=EA^∣ψ⟩\Rightarrow\hat{H}\hat{A}\left\lvert \psi \right\rangle = E\hat{A}\left\lvert \psi \right\rangle

This means that A^∣ψ⟩\hat{A}\left\lvert \psi \right\rangle is in the EE-eigenspace of H^\hat{H}. Another way to say this is that A^∣ψ⟩\hat{A}\left\lvert \psi \right\rangle is in the Set of eigenvectors for H that share eigenvalue E.

This means that A^\hat{A} maps a vector in the EE-eigenspace of H^\hat{H} to another vector in the EE-eigenspace of H^\hat{H}.