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

In summary: 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?
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LagrangeEuler
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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?
 
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But in the quantum linear harmonic oscillator case, you have even and odd eigenstates.
 
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LagrangeEuler said:
But in the quantum linear harmonic oscillator case, you have even and odd eigenstates.
Yes, you're right, even or odd.
 
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LagrangeEuler said:
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##.
 
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Thank you. But just to understand. Ground state of hydrogen is
[tex]\psi_{100}(r)=C\mbox{e}^{-\frac{r}{a_0}}[/tex]
and this is not even neither odd function. Right?
 
  • #7
LagrangeEuler said:
Thank you. But just to understand. Ground state of hydrogen is
[tex]\psi_{100}(r)=C\mbox{e}^{-\frac{r}{a_0}}[/tex]
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)##.
 
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  • #8
LagrangeEuler said:
Thank you. But just to understand. Ground state of hydrogen is
[tex]\psi_{100}(r)=C\mbox{e}^{-\frac{r}{a_0}}[/tex]
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.
 

1. What is a Hamiltonian?

A Hamiltonian is a mathematical operator used in quantum mechanics to describe the total energy of a system. It is represented by the symbol H and is a key component in the Schrödinger equation, which describes the time evolution of a quantum system.

2. What is a parity operator?

A parity operator is a mathematical operator that describes the symmetry of a system under spatial inversion. It is represented by the symbol P and is used to determine whether a system is symmetric or asymmetric with respect to its spatial coordinates.

3. How does a Hamiltonian commute with a parity operator?

A Hamiltonian commutes with a parity operator if the two operators share the same set of eigenvectors, meaning that they have common eigenstates. This indicates that the Hamiltonian and the parity operator have compatible properties and can be used to describe the same physical system.

4. What does it mean for a Hamiltonian to commute with a parity operator?

When a Hamiltonian commutes with a parity operator, it means that the Hamiltonian and the parity operator have the same set of eigenvectors and can be used interchangeably to describe the same physical system. This is useful in quantum mechanics because it allows for the simplification of calculations and the identification of important symmetries in a system.

5. Why is it important for a Hamiltonian to commute with a parity operator?

It is important for a Hamiltonian to commute with a parity operator because it allows for the identification of important symmetries in a system, which can simplify calculations and provide insights into the behavior of the system. This is particularly useful in quantum mechanics, where understanding the symmetries of a system is crucial in predicting its behavior.

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