2
$\begingroup$

It has been a long time I haven't done this and I am having a hard time writing things down in second quantization notation. Let us have a $n$-body system where the spin part and orbital parts are decoupled.

If want to write a two particle state, let's say a triplet state $\alpha$, with parallel spins, an electron with energy $\epsilon_1$ and another with $\epsilon_2$, I would write something like $$|\alpha\rangle=c^\dagger_{1\uparrow}c^\dagger_{2\uparrow}|\emptyset\rangle$$ where $|\emptyset\rangle$ is the void state, and $\uparrow,\downarrow$ are the spin states.

Same for a singlet $\beta$ with two electrons in the same energy $$|\beta\rangle=c^\dagger_{1\uparrow}c^\dagger_{1\downarrow}|\emptyset\rangle$$

But how do I distinguish a state $\gamma$ with $m_z=0$ (different energies, opposite spins)? What does this do? $$|\gamma\rangle=c^\dagger_{1\uparrow}c^\dagger_{2\downarrow}|\emptyset\rangle$$ This could be either a triplet (with opposite spins) or a singlet, what am I missing?

Edit: I now realize that there has to be a difference in the Fock space, between writing $|n_{1\uparrow},n_{1\downarrow},n_{2\uparrow},n_{2\downarrow}\rangle=|1001\rangle$ and $|0,1,1,0\rangle$ but I don't know how to interpret these states in terms of the singlet and the unparalleled triplet

$\endgroup$

1 Answer 1

2
$\begingroup$

The point is that the wavefunction is symmetric in spin for the triplet and antisymmetric for the singlet. So in your notation, the triplet is $$\frac1{\sqrt2}(c^\dagger_{1\uparrow}c^\dagger_{2\downarrow} + c^\dagger_{1\downarrow}c^\dagger_{2\uparrow})|\emptyset\rangle$$ and the singlet is $$\frac1{\sqrt2}(c^\dagger_{1\uparrow}c^\dagger_{2\downarrow} - c^\dagger_{1\downarrow}c^\dagger_{2\uparrow})|\emptyset\rangle.$$

$\endgroup$
6
  • $\begingroup$ Why is not the other way around? $\endgroup$
    – Mauricio
    Commented Apr 15, 2020 at 18:01
  • $\begingroup$ @Mauricio Why do you think it's the other way around? $\endgroup$
    – knzhou
    Commented Apr 15, 2020 at 18:02
  • $\begingroup$ You are telling me to check on the a/symmetry of the spin part and not the orbital part, why is that? $\endgroup$
    – Mauricio
    Commented Apr 15, 2020 at 18:04
  • $\begingroup$ @Mauricio Because you were asking about whether something is a spin singlet or spin triplet. That has nothing to do with the orbital behavior, you can infer it from the spins alone. $\endgroup$
    – knzhou
    Commented Apr 15, 2020 at 18:10
  • $\begingroup$ @Mauricio Meanwhile, the total wavefunction (spin plus orbital) is automatically antisymmetric -- that's the whole point of using second quantization notation, of course. $\endgroup$
    – knzhou
    Commented Apr 15, 2020 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.