Dirac Hydrogen Atom: Parity and Odd-Operator

In summary, the conversation discusses the properties of the odd-operator and how it relates to the parity operator. It is explained that if a state has a definite parity, then the diagonal elements of the odd-operator will be zero. This is due to the relationship between the parity operator and the odd-operator, where they must have opposite signs for the equation to hold.
  • #1
ChrisVer
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Hey I was reading through a text and came across:
"[ Having extracted the Dirac version of Schrodinger's equation of the H atom...] Since the states [itex] | j j_z l >[/itex] have definite parity, the odd-operator [itex] \vec{S} \cdot \hat{r}[/itex] will have vanishing diagonal elements. Also since [itex]\big(\vec{S} \cdot \hat{r} \big)^2 =1[/itex] then its offdiagonal elements will be [itex] \frac{1}{2} e^{\pm i \phi} [/itex] (we can choose the phase [itex]\phi=0[/itex])[...]"

I can understand the second statement from the Pauli matrices... However I think that I don't understand the 1st statement as it is... why would the diagonal elements of an odd-operator be zero if parity is definite?
 
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  • #2
I think what they mean as "fixed" is actually "definite".
 
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  • #3
Yup understand that... (a flaw in my translation :biggrin:)...I don't understand why the diagonal elements are zero.
 
  • #4
If ##A## is an odd operator and ##\pi## is a parity operator, then we have ##\pi^\dagger A \pi = -A##. If furthermore a state ##|\psi\rangle## which has a definite parity sandwhiches the expression ##\pi^\dagger A \pi = -A## from left and right in each side of the equation, you will get ##\langle \psi| A |\psi \rangle = -\langle \psi|A|\psi \rangle## because ##\langle \psi|\pi^\dagger A \pi|\psi \rangle = \langle \psi| A |\psi \rangle##. You see then we must require ##\langle \psi| A |\psi \rangle = 0## so that the previous relation can hold.
 
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1. What is the Dirac hydrogen atom?

The Dirac hydrogen atom is a model used in quantum mechanics to describe the behavior of a single electron orbiting around a nucleus. It takes into account both the electron's relativistic effects and its spin, making it a more accurate model than the classical hydrogen atom.

2. What is parity in the context of the Dirac hydrogen atom?

Parity is a quantum number that describes the symmetry of a physical system. In the context of the Dirac hydrogen atom, it refers to the behavior of the system under spatial inversion, or flipping the coordinates of all particles. Parity can be either even (+1) or odd (-1) for a given system.

3. What is the odd-operator in the Dirac hydrogen atom?

The odd-operator refers to a mathematical operator in the Dirac equation that determines the parity of a quantum state. It is a product of the parity operator and the spin operator, and it can help determine whether a state is even or odd under spatial inversion.

4. How does the odd-operator affect the energy levels of the Dirac hydrogen atom?

The odd-operator affects the energy levels of the Dirac hydrogen atom by introducing an additional term in the energy equation. This term is known as the spin-orbit coupling term and it arises due to the interaction between the electron's spin and its orbital angular momentum. The odd-operator determines whether this term is present or not, thus affecting the energy levels of the system.

5. What is the significance of the odd-operator in the study of the Dirac hydrogen atom?

The odd-operator is significant in the study of the Dirac hydrogen atom because it allows us to better understand the behavior of the system under spatial inversion. It also helps us to accurately predict the energy levels of the system and provides further evidence for the validity of the Dirac equation in describing relativistic effects in quantum systems.

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