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


1 Answer 1


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 $$\frac1{\sqrt2}(c^\dagger_{1\uparrow}c^\dagger_{2\downarrow} - c^\dagger_{1\downarrow}c^\dagger_{2\uparrow})|\emptyset\rangle.$$

  • $\begingroup$ Why is not the other way around? $\endgroup$
    – Mauricio
    Apr 15, 2020 at 18:01
  • $\begingroup$ @Mauricio Why do you think it's the other way around? $\endgroup$
    – knzhou
    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
    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
    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
    Apr 15, 2020 at 18:12

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