My understanding is that if we are dealing with a system of two electrons, the total wavefunction needs to be antisymmetric only when the two electrons have same value of n and l ( i.e. they are equivalent). In the case of non-equivalent electrons, the total wavefunction doesn't necessarily have to be antisymmetric.

Is it correct? If not, can you explain the requirement of antisymmetry in case of equivalent and non-equivalent electrons?

( I was reading about the term symbols and terms like ${^3S}$ had me question my understanding of Pauli exclusion principle, as both spin and space parts are symmetric here.)

  • $\begingroup$ I'm not sure what you mean by non-equivalent electrons. All electrons are indistinguishable. An n-electron wave function must always be fully antisymmetric. $\endgroup$
    – march
    Nov 22, 2015 at 16:51
  • $\begingroup$ @march By non-equivalent electrons, I mean electrons not having same values of both n and l quantum numbers. The equivalent electrons have both n and l values same. It has nothing to do with indistinguishability of electrons. $\endgroup$ Nov 22, 2015 at 17:08
  • $\begingroup$ In a state like $^3S$, the two paired electrons must be in different $s$ orbitals, e.g. the excited state of the Helium atom where one electron is in the $1s$ and the other in the $2s$ orbital. In that case, the space part is antisymmetric. $\endgroup$
    – LLang
    Nov 22, 2015 at 17:14
  • $\begingroup$ But you can't talk about this electron having this quantum number and that electron having that quantum number. That's almost the whole point of making the wave function antisymmetric. $\endgroup$
    – march
    Nov 22, 2015 at 17:25

1 Answer 1


The wavefunction is always antisymmetric. It does not matter if they have similar, different, or even identical numbers for $n$ and $l$.

But let's be clear. The wavefunction is a function from the configuration space into the joint spin state. So for a spin 1/2 particle the wavefunction is a function from $\mathbb R^3$ to $\mathbb C^2$ whereas for two spin 1/2 particles the wavefunction is a function from $\mathbb R^6$ to $\mathbb C^2\otimes\mathbb C^2$ and it has to be antisymmetric if they are indistinguishable fermions, symmetric if they are indistinguishable bosons, and no restrictions (could be symmetric, antisymmetric, or neither) if they are distinguishable.

And by the symmetries we mean things like $\Psi(\vec a,\vec b)=-\Psi(\vec b,\vec a).$

So let's do an example. If $\psi_{nlm}:\vec r\mapsto\psi_{nlm}(\vec r)$ is a solution with a particular value of $n,$ $l,$ and $m$ for the scalar (spin zero) Schrödinger equation for a 1/r potential then you could make a spin 1/2 solution by having $$\Psi:\mathbb R^3\rightarrow\mathbb C^2:\vec r\mapsto\psi_{nlm}(\vec r)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right].$$

For a pair of indistinguishable spin 1/2 particles (e.g. two electrons, two muons, etcetera) a possible wavefunction is

$$\Psi:\mathbb R^6\rightarrow\mathbb C^2\otimes\mathbb C^2,$$ $$\Psi:(\vec a,\vec b)\mapsto\psi_{nlm}(\vec a)\psi_{NLM}(\vec b)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right] \otimes\left[\begin{matrix}A\\ B\end{matrix}\right]-\psi_{nlm}(\vec b)\psi_{NLM}(\vec a)\left[\begin{matrix}\alpha\\ \beta\end{matrix}\right] \otimes\left[\begin{matrix}A\\ B\end{matrix}\right].$$

In fact you can build more complicated states out of these antisymmetric states. But the note complicated states will remain antisymmetric.

  • $\begingroup$ By non-equivalent electrons, I mean electrons not having same values of both n and l quantum numbers. The equivalent electrons have both n and l values same. It has nothing to do with indistinguishability of electrons. It is the terminology used in the book by Bransden and Joachain. $\endgroup$ Nov 22, 2015 at 17:09
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    $\begingroup$ @YogeshYadav I perfectly understood exactly what you meant and answered your question by telling you correct facts that work in all situations. All electrons are indistinguishable from each other because they are both electrons. $\endgroup$
    – Timaeus
    Nov 22, 2015 at 17:11
  • $\begingroup$ @YogeshYadav The antisymmetry has everything to do with the fact that you can't tell two electrons apart and nothing to so with anything else. I've edited to try to make it more clear, but please try to read the answer as telling you fundamental facts from which you can derive other facts. For instance, the Pauli Exclusion Principle is a consequence of this more general fact. $\endgroup$
    – Timaeus
    Nov 22, 2015 at 17:26
  • $\begingroup$ To be clear, in some cases (i.e. when the two electrons are well-separated relative to the sizes of their wavefunctions) one can get away with not antisymmetrizing the wavefunction, because it has only a tiny observable difference from the product state. This can be a subtle point. However, as Timaeus says, formally the wavefunction of several identical fermions is always antisymmetrized, so when in doubt you should just always do that (or use 2nd quantized notation). $\endgroup$
    – Rococo
    Nov 22, 2015 at 18:08
  • $\begingroup$ Shankar's QM book has a nice discussion on this (heading: "When can we ignore symmetrization and anti-symmetrization?") $\endgroup$
    – Rococo
    Nov 22, 2015 at 18:10

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