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I'm looking at two-electron Helium atoms where one electron is in the ground state (due to if it were in other states, it's de-excitation would simply lead to the ionization of the electron). The other electron is in an arbitrary excited state. What does this notation mean in terms of the allowed wavefunctions?: enter image description here

Specifically, for the singlet state, I understand that the both electrons must have opposite spin to give a total overall spin of Zero (correct me if I'm wrong). But why does this mean suddenly the wavefunction, $\psi_{+}$ is allowed to be symmetric. I thought all fermion wavefunctions had to be anti-symmetric havinga $\psi_{-}$ state.

For the triplet state, as I understand it, the electrons have a total spin number of 1 meaning that they have to have the same spin as each other (+1/2 each or -1/2 each). Is this correct? Further to this, I don't specifically understand the notation in the equations above. Specfically, I don't understand the last bit added on the end: enter image description here What does this actually mean?

I understand this might be a silly question as it could just be me mixing up notation but any help will be appreciated. Thank You! :)

EDIT: See AV23 Comment below. Answer! :) Basically, find the $\hat{S}^{2}$ eigenvalues to deduce S using below:

enter image description here

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The $|00\rangle$ and $|1,M_S\rangle$ represent the spin singlet and triplet states. The overall wavefunction must contain both the 'space' part and the 'spin' part. We can schematically express this as follows: $$\psi \sim \psi(\mathbf{r}_1, \mathbf{r}_2) |s\rangle$$

Now, Pauli's exclusion principle demands the antisymmetry of the overall wavefunction. For the singlet state, as $|00\rangle = \frac{1}{\sqrt 2}\left(|+-\rangle - |-+\rangle\right)$ is already antisymmetric, $\psi_+$ can be antisymmetric only if the space part is symmetric.

The triplet states: $$|1, 1\rangle = |++\rangle$$ $$|1, 0\rangle = \frac{1}{\sqrt 2}\left(|+-\rangle + |-+\rangle \right)$$ $$|1, -1\rangle = |--\rangle$$ are all symmetric, and therefore, the space part corresponding to them must be antisymmetric.

Note that the triplet state $|1, 0\rangle$ does not have both individual spins in the same direction. However, it still has a total spin of $1$.

In summary, $\psi_+$ and $\psi_-$ are both antisymmetric; it is just that their space & spin parts are symmetric & antisymmetric and antisymmetric & symmetric respectively.

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  • $\begingroup$ For the |00> singlet state, why can't the minus sign in the spin state, be a plus sign making the spin part symmetric? (and then we can make $\psi_{+}$ anti-symmetric still by making the space part antisymmetric). This surely still corresponds to |00> = $| S M_{S} >$, no?, where S=0 and $M_{S}$ = 0? $\endgroup$ – SomePhysicsStudent Apr 23 '15 at 19:18
  • $\begingroup$ No - the $|00\rangle$ state is an eigenstate of the spin operator with eigenvalues 0, 0. Your proposed change would actually make it the triplet state $|1, 0\rangle$, which has a different measurable total spin. Remember that coefficients (in this case, $+$ or $-$) in the interference/superposition of base states can play a major role in determining physical quantities in quantum mechanics. $\endgroup$ – AV23 Apr 23 '15 at 19:22
  • $\begingroup$ You can read more about these states here: en.wikipedia.org/wiki/Triplet_state and here: en.wikipedia.org/wiki/Singlet_state $\endgroup$ – AV23 Apr 23 '15 at 19:23
  • $\begingroup$ Thank You Very Much! I think I just need to improve my fundamental understanding a little more! Thanks! :) $\endgroup$ – SomePhysicsStudent Apr 23 '15 at 19:30
  • $\begingroup$ @SomePhysicsStudent It might interest you to know that the requirement that a half-integer spin particle should follow the Fermionic statistics is due to special relativity. Thus, quantum theories that do not live in a geometric spacetime, for example, pure spin-states can have either Bosonic or Fermionic statistics (see, Schwinger Bosons). This is what is happening here with the spin-part of the wave-function. Both the Bosonic and Fermionic statistics are valid for them--but since the overall state still lives in a geometric spacetime, the overall wave-function follows Fermionic statistics. $\endgroup$ – Dvij Mankad Mar 6 at 16:54

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