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"One electron is presumed to be in the ground state, the 1s state. An electron in an upper state can have spin antiparallel to the ground state electron ($S=0$, singlet state, parahelium) or parallel to the ground state electron ($S=1$, triplet state, orthohelium)." From HyperPhysics

When they say "parallel to the ground state electron" then is it assuming that they are both spin up, or both spin down? If so, isn't it then ignoring the $S=1$ state with spin up and spin down:

$$|1\rangle |0\rangle = \frac{|+\rangle|-\rangle + |-\rangle|+\rangle}{\sqrt 2}$$

Therefore, if one electron is presumed to be in the ground state, 1s, state, if the spins can be opposite, a second electron can also occupy the ground state in $S=1$ orthohelium. Is this correct?

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The definition of orthohelium as having two parallel electron spins is correct. In the state $$\frac1{\sqrt2}\big(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\big)$$ the two spins are also parallel. It may seem that they are not, but note that this state has $m_z=0$, so both spins are perpendicular to the quantization direction. Indeed, the operator $S_+$ will turn this state into $$\big(|\uparrow\uparrow\rangle\big)$$.

Also, this is an eigenstate of $S^2$ with eigenvalue 2$\hbar^2$, just like the other members of the triplet, while $S_1 \cdot S_2$ has eigenvalue $+\frac{\hbar^2}{4}$. The positive sign indicates that the spins are parallel. Indeed for the singlet this eigenvalue is $-\frac{3\hbar^2}{4}$, which indicates antiparallel alignment.

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  • $\begingroup$ Unmotivated down voting is best avoided. $\endgroup$
    – my2cts
    Dec 19, 2021 at 0:36
  • $\begingroup$ Hmm, this issue got me confused, and your assertion seemed wrong to me (therefore the downvote, which I now retracted) – I thought more about it and I was wrong. I still think your demonstration is not ideal: I think the best way to show that they are parallel would be to represent the state in a basis where this is evident. I'll try to add my own answer along those lines. (And I think when I downvoted the $S_1 \cdot S_2$ arugment was still missing). $\endgroup$ Dec 19, 2021 at 13:05
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The spins are not antiparallel in this state as @my2cts answer correctly states (as demonstrated by $\vec S_1 \cdot \vec S_2$ having a positive eigenvalue).

I was confused by this at first too, especially the argument, that the measurements of $S_{1z}$ and $S_{2z}$ will point to opposite directions due to entanglement seemed convincing to me.

But it just looks that way because the $S_z$ basis is the wrong basis to look at. They have to point in opposite directions in this basis, because the total $m_z = 0$. So you are just asking a spin state in the $x$-$y$-plane for its $z$-component. This way you can't find out whether the spins are parallel or antiparallel in that plane.

To make the parallel orientation of the spins manifest in the expression for the state, you'll have to rewrite the state in a basis in the plane.

We rewrite our state terms of the eigenstates $\lvert \pm \rangle$ of $S_x$: $$ S_x \lvert \pm \rangle = \pm \frac \hbar 2 \lvert \pm \rangle$$ In terms of those states we have $$ \lvert \uparrow / \downarrow \rangle = \frac{1}{\sqrt{2}} \big( \lvert + \rangle \pm \lvert - \rangle \big)$$ in terms of these states the two product states become $$ \lvert \uparrow \downarrow \rangle = \frac 1 2 \big( \lvert + \rangle + \lvert - \rangle \big) \otimes \big( \lvert + \rangle - \lvert - \rangle \big) = \frac 1 2 \big( \lvert++\rangle - \lvert+-\rangle + \lvert -+ \rangle - \lvert -- \rangle \big) $$ $$ \lvert \downarrow \uparrow \rangle = \frac 1 2 \big( \lvert++\rangle - \lvert-+\rangle + \lvert +- \rangle - \lvert -- \rangle \big) $$ So the triplet state with $m_z = 0$ is given by the following in this basis: $$ \frac{\lvert \uparrow \downarrow \rangle + \lvert \downarrow \uparrow \rangle}{\sqrt{2}} = \frac{\lvert ++ \rangle - \lvert -- \rangle}{\sqrt{2}}$$

And in this form we can see manifestly that the spins are in parallel. (Similarly, this could be done for the eigenstates of $S_y$.)

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With such a two-electron spin wave function, both electrons cannot be in the 1s state due to the electron parity. (The wave function should be antisymmetric relative to exchange of two particles.) When both electrons are in the 1s orbital, the only possible spin function reads $$\frac{1}{\sqrt{2}} \left( |\uparrow \downarrow \rangle-| \downarrow \uparrow\rangle \right)$$

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  • $\begingroup$ The overall wavefunction must be anti-symmetric but then why can't it be that the spatial is the anti-symmetric bit, not the spin part? $\endgroup$ Jun 9, 2015 at 16:59
  • $\begingroup$ For example? psi=1s(1)1s(2)-1s(2)1s(1)? Than it is zero $\endgroup$
    – freude
    Jun 9, 2015 at 20:15
  • $\begingroup$ I think this answer gives a correct statement, but ignores the question. The OP asks about whether a state with one electron in the 1s orbital and the other in the 2s orbital can have anti-aligned spins. To which I think yes, precisely the state suggested in the question. $\endgroup$
    – WillG
    Dec 18, 2021 at 3:43
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Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric.

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