Hamiltonian commutes with a parity operator -- What does that mean?

[LagrangeEuler]

If Hamiltonian commutes with a parity operator $Px=-x$ are then all eigenstates even or odd? Is it true always or only in one-dimensional case?

 

[PeroK]

If Hamiltonian commutes with a parity operator $Px=-x$ are then all eigenstates even or odd?

Not necessarily. If the eigenstates are non-degenerate, then they must be even (edit or odd).

https://web.pa.msu.edu/people/mmoore/Lect21_Parity.pdf

This concept extends to 3D.

https://en.wikipedia.org/wiki/Even_and_odd_functions#Generalizations

 

[LagrangeEuler]

But in the quantum linear harmonic oscillator case, you have even and odd eigenstates.

 

[PeroK]

But in the quantum linear harmonic oscillator case, you have even and odd eigenstates.

Yes, you're right, even or odd.

 

[vanhees71]

But in the quantum linear harmonic oscillator case, you have even and odd eigenstates.

Sure, but since the eigenstates are non-degenerate they are also eigenstates of the parity operator and thus either even or odd.

For a system having degenerate energy eigenstates (i.e., if the Hamiltonian has an energy eigenvalue with more than one linearly independent eigenstate) then you can define an orthonormal set of common energy and parity eigenstates but you can build superpositions of energy eigenstates to the same eigenvalue (being then of course again an eigenstate of this eigenvalue) which is no parity eigenstate.

Take as an example the hydrogen atom (in the usual QM 1 non-relativistic solutions of the Coulomb problem). The only non-degenerate eigenstate is the ground state with $n=1$, which necessarily has $\ell=0$ (and consequently also $m=0$) and thus is parity-even (i.e., an eigenstate of parity with eigenvalue $+1$; the usual common eigenstates of the energy (quantum number $n$, $E_n=-1 \text{Ry}/n^2$) angular momentum $\vec{L}^2$ (quantum number $\ell$, eigenvalues $\hbar^2 \ell(\ell+1)$), and $L_z$ (quantum number $m$, eigenvalues $m \hbar$). The $n \in \mathbb{N}$ and $\ell \in \mathbb{N}_0$ and for any fixed $\ell$ the $m \in \{-\ell,-\ell+1,\ldots,\ell-1,\ell \}$, are all also parity eigenstates with parity eigenvalue $(-1)^{\ell}$. Take the first excited energy level with $n=2$, for which you can have $\ell=0$ or $\ell=1$ (in general for given $n$ the possible values for $\ell$ are $\{0,1,\ldots,n-1\}$). So you can build a new energy eigenstate to this same $n$ which is no parity eigenstate by simply superimposing the (unique) eigenstate with $n=2$ and $\ell=0$ (being parity eigenstates with eigenvalue $+1$) with any superposition of the three other eigenstates with $n=2$ and $\ell=1$ (being parity eigenstates with eigenvalue $-1$). Then of course the superposition is not a parity eigenstate but still an energy eigenstate to the eigenvalue $E_2$.

 

[LagrangeEuler]

Thank you. But just to understand. Ground state of hydrogen is
\psi_{100}(r)=C\mbox{e}^{-\frac{r}{a_0}}
and this is not even neither odd function. Right?

 

[PeroK]

Thank you. But just to understand. Ground state of hydrogen is
\psi_{100}(r)=C\mbox{e}^{-\frac{r}{a_0}}
and this is not even neither odd function. Right?

What definition of even are you using? It's spherically symmetric, so $\psi(\vec r) = \psi (-\vec r)$.

 

[vanhees71]

Thank you. But just to understand. Ground state of hydrogen is
\psi_{100}(r)=C\mbox{e}^{-\frac{r}{a_0}}
and this is not even neither odd function. Right?

This is even under parity. The parity operator acts on a wave function as $\hat{P} \psi(\vec{x})=\psi(-\vec{x})$. Since for $\ell=0$ the wavefunction depends only on $r=|\vec{x}|$ it's automatically even under parity.