# Symmetry and two electron wave function

• I
• hokhani
In summary, the conversation discusses the left-right symmetry of a system with two identical orbitals and the notation used to indicate the number of spin-up electrons in each orbital. The question is whether the left-right symmetry of the system requires the eigenstates of the Hamiltonian to be in a specific form, such as a superposition of left-right symmetric and anti-symmetric states. The response is that there is no strict requirement for the eigenstates to follow this symmetry, but it may be necessary for certain operators, such as the Hamiltonian. However, for a general state, there is no such requirement.
hokhani
TL;DR Summary
Understanding the relation between system symmetry and wave-function symmetry
In the picture below we have two identical orbitals A and B and the system has left-right symmetry. I use the notation ##|n_{A \uparrow}, n_{A \downarrow},n_{B \uparrow},n_{B \downarrow}>## which for example ##n_{A \uparrow}## indicates the number of spin-up electrons in the orbital A. I would like to know is it possible to have an eigenstate as ##|1,1,0,0>## in this left-right symmetric system or, because of the symmetry of system, we must only have symmetric wave-functions as ##\frac{|1,1,0,0>\pm|0,0,1,1>}{\sqrt(2)}##?
Any help is appreciated.

The only symmetry that states must follow is the one related to interchange of two identical particles.

(That said, if you need states to be eigenstates of some operators, like the Hamiltonian, then certain symmetries may need to be respected. But for a general state, there is no such requirement.)

Lord Jestocost and hokhani
Thanks very much. In the second quantized form, the interchange of particles are included in the commutation of fermionic operators.
However, I would like to know whether the left-right symmetry of this system demands the eigenstates of the Hamiltonian be in the form ## \frac{1}{\sqrt(2)}(|1,1,0,0>\pm|0,0,1,1>)## or they can also have the form like ##|1,1,0,0>##?

What is the Hamiltonian?

I don't mean a specific Hamiltonian. I mean each left-right symmetric Hamiltonian for this system.

Sorry if I didn't convey myself. To explain in a better way; for a left-right symmetric system, can we have the eigenstates which doesn't have this symmetry?

hokhani said:
Sorry if I didn't convey myself. To explain in a better way; for a left-right symmetric system, can we have the eigenstates which doesn't have this symmetry?
If you have a degenerated eigenvalue with a left-right symmetric and a left-right anti-symmetric eigenstate, then take a superposition of them to get an eigenstate which is not left-right symmetric.

DrClaude and hokhani
gentzen said:
If you have a degenerated eigenvalue with a left-right symmetric and a left-right anti-symmetric eigenstate, then take a superposition of them to get an eigenstate which is not left-right symmetric.
Thanks, I got it. Like the parity in one-particle system, provided that there is no degeneracy, the eigenstates must have even or odd symmetry as ##\frac{1}{\sqrt(2)}(|1,1,0,0>\pm|0,0,1,1>)##. Otherwise, there is no demand for symmetry of the eigenstates.

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