Sum of spins of electrons in a wavefunction

Question

I wanted to ask a question about spin in an electron.

We were presented earlier today with the following information:

Using my synoptic knowledge, I recall that spin up $\alpha$ had a value of $+\frac{1}{2}$ and spin down $\beta$ had a value of $-\frac{1}{2}$.

Also, using the information above, $\alpha(1)\alpha(2)$ would have a total Spin quantum number $S$ of $\frac{1}{2} + \frac{1}{2} = 1$, $\beta(1)\beta(2)$ however I thought would result to being $S$ of $-\frac{1}{2} - \frac{1}{2} = -1$ and even so, one spin up and one spin down $\alpha(1)\beta(2) + \alpha(2)\beta(1)$ would total $\frac{1}{2} - \frac{1}{2} + \frac{1}{2} - \frac{1}{2}= 0$. These calculations contradict the data given above for the triplets and singlets states.

I'm confused. How are the values of $S$ for $\beta(1)\beta(2)$ and $\alpha(1)\beta(2) + \alpha(2)\beta(1)$ exactly derived?

Answer 1

I think the reason you're getting confused is because when we talk about "spin", we actually are referring to two separate observable values, the z-component of the spin (properly designated $\hat{S}_z$) and the square of the magnitude of the three-component spin vector (properly designated as $\hat{S}^2$).

The commonly used values of $+\frac12$ and $-\frac12$ are shorthand for $+\frac12 \hbar$ and $-\frac12\hbar$, the only values possible to measure for $\hat{S}_z$.

Because of quantum uncertainty requirements, we can know the values of both the z-component and the square of the magnitude of spin at the same time, but we cannot know either the x- or y-component while we know the z-component. As a result, we can only say that the spin vector lies somewhere on a the surface of a cone centered on the z-axis that expands upwards from the origin ($+\frac12$) or downwards from the origin ($-\frac12$).

For two uncorrelated spins (like in a triplet state), the z-components either add (2 positive) or (2 negatives) or cancel (1 positive and 1 negative), but that only means that the net z-component is 1, 0, or -1. Even if $\hat{S}_z=0$, there is still a net spin from the x- and y-components. The value of the magnitude (or technically its square) comes from $\hat{S}^2$, which is conserved as 1 for all three possibilities. That's the S value given in your notes. For the $\hat{S}_z=0$ case, that means the spin vector lies in the x-y plane as one of the possible radial vectors of a circle with radius 1 centered on the origin in that plane.

For the correlated spins (singlet state), the magnitude squared is 0, which occurs because the two spin vectors exactly oppose each other, and all three components cancel out, not just $\hat{S}_z$, though we have no measure of the x- and y- components that are cancelling. We only know the resulting magnitude is 0.

Answer 2

This is a popular piece of confusion, because of a terseness in notation. Basically, we are secretly talking about two quantum numbers here: The total magnitude of the electron spin, $S$, and the relative orientation of the electron spins, $M_S$.

As you correctly identify, for the $\alpha\beta + \beta\alpha$ configuration you get some quantum number that is $0$. This quantum number is, in fact, $M_S$; commonly interpreted as the orientation of the overall electron spin along the z-axis. Note, however, that $M_S$ for the negative linear combination is also $0$. So something else must be going on here - why do we call one of those a triplet and the other one a singlet state?

At this point, it makes sense to recall another case where you have already learned the difference between the total magnitude of an angular momentum, and its orientation in space; and that is atomic orbitals. For an s orbital, the angular momentum quantum number is $l=0$, and the magnetic quantum number $m_l$ is also $0$. For a p orbital, $l=1$, and $m_l$ can take values of $-1,0,+1$. In other terms, for a state with a given angular momentum quantum number $l$, we have $2l+1$ different ways to orient that momentum in space, which is described by the magnetic quantum number $m_l$.

Now, let's go back to our electron spins. First, it is important to realize that we are not talking about single electron quantum numbers any more, but instead the overall multi-electron system as a whole. For this, we add the magnetic quantum numbers $m_s$ of all electrons to arrive at the total quantum number $M_S$ (note the uppercase letters). Since each electron can have an individual $m_s$ of $+\frac{1}{2}$ or $-\frac{1}{2}$ (i.e., $\alpha$ or $\beta$ spin), we can have different "configurations" with different values of $M_S$.

In your two-electron system, the simplest case are the configurations with $M_S=+1$ ($\alpha\alpha$) and $-1$ ($\beta\beta$), and with our knowledge about the orientation of angular momenta above in hand, we see that these must belong to an overall $S=1$ state. But to such a state must also belong a $M_S=0$ configuration. Luckily, we have two of those at hand with out linear combinations, and since the $\alpha\beta+\beta\alpha$ one obeys the same exchange symmetry as the $\alpha\alpha$ and $\beta\beta$ configurations, we group the three of them together and thus have completed the $S=1$ triplet state.

We are left with the $\alpha\beta - \beta\alpha$ configuration. Its $M_S$ value is $0$, and there are no other configurations available to go along with it, so we have identified an $S=0$ singlet state.