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I was studying with the book "Modern Quantum Chemistry" of Sazbo and Oustland. In the chapter 2, they talk about unrestrict (UR) determiants. It's explaneid that UR determinants are not eigenfunctions of $S^{2}$, but we can expand them as linear combinations of singlet $|1 \rangle$, doublet $|2 \rangle$, triplet $|3 \rangle$ states and so on. For exemple, UR determinants used as approximatons to a system that is a singlet (with $N^{\alpha} = N^{\beta}$) is

$$ |\Psi_{S} \rangle = a_{1} |1 \rangle + a_{2} | 3 \rangle + a_{3} |5 \rangle + ... $$

To UR determinants that represents a system with $N^{\alpha} = N^{\beta} + 1$ (a doublet) and $N^{\alpha} = N^{\beta} + 2$ (a triplet), we have

$$ |\Psi_{D} \rangle = b_{1} |2 \rangle + b_{2} | 4 \rangle + a_{3} |6 \rangle + ... $$

$$ |\Psi_{T} \rangle = c_{1} |3 \rangle + c_{2} | 5 \rangle + c_{3} |7 \rangle + ... $$

and so on. That is, in the expansion we use either even or odd eigenstates and the series begins in the eigenstate with the same spin of the system. I would like to know why this is so. I understand that, since $H, S_{z}$ and $S^{2}$ commute, we can construct states that are eigenstates of both of them, and that this set of states is complete, so any function can be write as a linear combination of these states. But I can't see why in the above expansions we exclude some of the eigenstates.

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  • $\begingroup$ I hope you didn't read my answer too early, because the bonus section was a little dodgy. Sorry. I replaced it with a more technically correct version. $\endgroup$
    – orthocresol
    Sep 16 at 23:57
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Forgive me, as my specialty is not in electronic structure theory. But the root cause of this is angular momentum coupling and that's a topic that I feel qualified enough to comment on.

In this answer, small letters ($s$, $m$) refer to quantum numbers and big letters are the corresponding quantum mechanical operators.

When you have two particles with spin quantum numbers $s_1$ and $s_2$, the spin quantum number $s$ of the resulting two-particle system has the possible values

$$s = |s_1 - s_2|, |s_1 - s_2| + 1, \ldots, s_1 + s_2 \tag{1}$$

(the absolute values can be left out if you assume, without loss of generality, that $s_1 \geq s_2$.) This fact is stated in pretty much any textbook that covers quantum mechanical angular momentum. Since the allowed values for $s$ increase by $1$ each 'step', the multiplicity, $2s + 1$, increases by $2$ each 'step'.

The simplest example is two spin-1/2 particles, e.g. electrons. In this case $|s_1 - s_2| = 0$ and $s_1 + s_2 = 1$, so the resulting coupled states are triplet ($s = 1$) and singlet ($s = 0$) states. Note that $m$ here refers to the projection quantum number (i.e. the $m$ in $\hat{S}_z|\psi\rangle = m\hbar|\psi\rangle$), not the multiplicity $2s + 1$.

$$\begin{align} |s = 1, m = 1 \rangle &= |\alpha\alpha\rangle \\ |s = 1, m = 0 \rangle &= 2^{-1/2}(|\alpha\beta\rangle + |\beta\alpha\rangle) \\ |s = 1, m = -1\rangle &= |\beta\beta\rangle \\ |s = 0, m = 0 \rangle &= 2^{-1/2}(|\alpha\beta\rangle - |\beta\alpha\rangle) \end{align}$$

In particular, note that it is not possible to construct a doublet state from two spin-1/2 particles. In general, it is not possible to construct a system with half-integer spin from two half-integer spin subsystems. We can generalise this quite easily:

Spin of system 1 ($s_1$) Spin of system 2 ($s_2$) Spin of overall system
Half-integer Half-integer Integer
Half-integer Integer Half-integer
Integer Half-integer Half-integer
Integer Integer Integer

Larger systems are constructed by repeatedly coupling together smaller systems. However, this table applies at every stage: so if you couple together system A which consists of two electrons (integer spin), with system B which consists of a third electron (half-integer spin), you get a three-electron system which must have half-integer spin.

From this it is not too hard to see that if you have an odd number of electrons, the resulting overall spin must be half-integer (i.e. multiplicities of $2, 4, 6, \ldots$); and if you have an even number of electrons, the resulting overall spin must be integer (i.e. multiplicities of $1, 3, 5, \ldots$).


Bonus

An extra mathematical proof of the table above is reproduced from Sakurai's Modern Quantum Mechanics, 3rd ed. It follows from the relationship between the 'uncoupled' and 'coupled' representation, which differ in the choice of the basis sets. In the 'uncoupled' form, we have simultaneous eigenstates of $\hat{S}_1^2$, $\hat{S}_{1z}$, $\hat{S}_2^2$, and $\hat{S}_{2z}$, i.e. quantum numbers of $s_1, m_1, s_2, m_2$. In the 'coupled' form we have simultaneous eigenstates of $\hat{S}^2$, $\hat{S}_1^2$, $\hat{S}_2^2$, and $\hat{S}_z$, i.e. quantum numbers of $s, s_1, s_2, m$. (In both cases, these are the maximal number of operators that mutually commute, so it's not possible to add any more quantum numbers to either set.)

Notice that $s_1$ and $s_2$ are 'shared' between both representations. So in order to cut down on the notation, we denote the 'uncoupled' basis states by $|m_1;m_2\rangle$; and the 'coupled' basis states by $|s;m\rangle$. Essentially, we cut out the quantum numbers that aren't unique.

The transformation between the two bases is given by

$$|s;m\rangle = \sum_{m_1,m_2} |m_1;m_2\rangle \langle m_1;m_2 | s;m\rangle,$$

which is an expansion of the coupled state $|s;m\rangle$ as a linear combination of uncoupled states $|m_1;m_2\rangle$. The scalar quantity $\langle m_1;m_2 | s;m\rangle$ is a Clebsch–Gordan coefficient. It can be shown that these coefficients vanish unless $m_1 + m_2 = m$; and this relies on the fact that the corresponding quantum mechanical operators add up:

$$\hat{S}_z = \hat{S}_{1z} + \hat{S}_{2z}.$$

Thus, for any arbitrary coupled state $|s;m\rangle$,

$$(\hat{S}_z - \hat{S}_{1z} - \hat{S}_{2z}) |s;m\rangle = 0.$$

Note that this is purely because the operators cancel each other out. It does not depend on the state $|s;m\rangle$ being an eigenstate of all three operators: in fact, in general it will not be, because this is an eigenstate of $\hat{S}^2$ which does not commute with $\hat{S}_{1z}$ or $\hat{S}_{2z}$. Left-multiplying by some arbitrary uncoupled basis bra $\langle m_1m_2|$, we get

$$\langle m_1m_2|(\hat{S}_z - \hat{S}_{1z} - \hat{S}_{2z}) |s;m\rangle = 0.$$

The trick here is now to apply the relevant operator to the relevant side of the braket. In particular, we want to apply $\hat{S}_z$ to the ket, and apply $\hat{S}_{1z}$ and $\hat{S}_{2z}$ to the bra, because these components are eigenstates of the respective operators. We don't need to worry about complex conjugates because the operators are Hermitian and thus have real eigenvalues.

$$\begin{align} \langle m_1m_2|(\hat{S}_z - \hat{S}_{1z} - \hat{S}_{2z}) |s;m\rangle &= \langle m_1m_2|\hat{S}_z|s;m\rangle - \langle m_1m_2|\hat{S}_{1z}|s;m\rangle- \langle m_1m_2|\hat{S}_{1z}|s;m\rangle \\ &= m\hbar \langle m_1m_2|s;m\rangle - m_1\hbar \langle m_1m_2|s;m\rangle - m_2\hbar \langle m_1m_2|s;m\rangle \\ &= (m - m_1 - m_2)\hbar \langle m_1m_2|s;m\rangle = 0 \\ \end{align}$$

which means that either $m = m_1 + m_2$, or $\langle m_1m_2|s;m\rangle = 0$. So essentially, all the terms in the sum

$$|s;m\rangle = \sum_{m_1,m_2} |m_1;m_2\rangle \langle m_1;m_2 | s;m\rangle$$

vanish, unless $m = m_1 + m_2$. This means that it's not possible to get a coupled state $|s;m\rangle$ if $m$ cannot be expressed as a sum of some $m_1$ and some $m_2$ (because all the terms would go to zero, and we'd get a zero state, which is unphysical).

Now, consider two particles which have half-integer spin, i.e. half-integer $s_1$ and $s_2$. Because $m_1$ is allowed to take values $-s_1, -s_1 + 1, \ldots, s_1$, it follows that $m_1$ also has half-integer values. The same is true of $m_2$.

But we showed above that for the coupled states, the only valid states are those where $m$ can be expressed as a sum of $m_1$ and $m_2$: this means that $m$ must be an integer. Furthermore, the quantum number $m$ is allowed to take values $-s, -s+1, \ldots, s$. Consider the case where $s$ were to be a half-integer: this would mean that $m$ had to be a half-integer as well, a contradiction. So $s$ must be an integer.

Altogether, this completes a proof that if $s_1$ and $s_2$ are half-integer, then $s$ must be an integer (i.e. the first row of the table above). The other cases follow in a similar fashion.

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