Is there any "singlet state" for 3 or more spin 1/2 particles?

1,024

Solution 1

For any even number of spin $1/2$ particles there is at least one state with zero total spin. However, this state is not attained by antisymmetrizing all the spin states, because as you said, this is simply impossible for $n > 2$ particles; the antisymmetrization gives exactly zero. You're getting yourself confused thinking about spatial wavefunctions, which have nothing to do with the problem.

Explicitly, consider $4$ particles. Suppose the first two are in an antisymmetrized singlet state, and the last two are as well. Both the first two and last two individually have no spin. Hence the combination of both of these pairs also has no spin. This is an example of a singlet state, and it is not constructed from antisymmetrizing all four spins, which would be impossible.

Solution 2

Group theoretically, the number of singlets contained in the composition of N=2m doublets is $$ \frac{(2m)!}{m!(m+1)!}~~, $$ so 1 for N=2, 2 for N=4, 5 for N=6, etc, and 0 for odd N, of course.

You see this from direct plug in of the general formula for the Kronecker product of N doublets, eqn (19) of Zachos 1992. You should be able to recognize the sequence as the diagonal of the Catalan triangle , so Catalan numbers.

The multiplicities of arbitrary products of arbitrary reps can be gotten by integrating representation characters over the SU(2) group invariant measure and possess interesting properties, e.g. Curtright et al 2017.

Share:
1,024

Related videos on Youtube

slaaidenn
Author by

slaaidenn

Updated on December 22, 2020

Comments

  • slaaidenn
    slaaidenn over 1 year

    Every system with $N$ or more electrons lies in a Hilbert space $H=H_{\text{space}} \otimes H_{\text{spin}}$, with $H_{\text{space}}=H_{\text{space}}^{1}\otimes\cdots\otimes H_{\text{space}}^{N}$ and $H_{\text{spin}}=H_{\text{spin}}^{1}\otimes\cdots\otimes H_{\text{spin}}^{N}$, $H^i$ being the $i$-th particle space. So the system has a state $|\Psi\rangle = |\Psi\rangle_{\text{space}} \otimes |\Psi\rangle_{\text{spin}} \in H$.

    What I couldn't come up with is an antisymmetric spin ket $|\Psi\rangle_{\text{spin}}$ when there was more than 3 electrons. This would mean that the only way of antisymmetrizing $|\Psi\rangle$, for $N\geq 3$, is by antisymmetrizing only the spatial part. I think it's strange, since for $N=2$ we do have an antisymmetric spin ket (the singlet state), so why wouldn't be such kets for $N\geq 3$?

    Ignoring the spatial part, and assuming $N\geq 3$, if we want to describe $N$ identical spins $\sigma_k = \pm$, we need to antisymmetrize the ket $|\sigma_1\rangle |\sigma_2\rangle \cdots |\sigma_N\rangle$ in the following way

    \begin{equation} |\Psi\rangle_{\text{spin}} = \frac{1}{\sqrt{N}}\sum_{p \in S_N}sg(p)|\sigma_{p(1)}\rangle |\sigma_{p(2)}\rangle \cdots |\sigma_{p(N)}\rangle \end{equation}

    Let's take, for example, the following ket (which we want to antisymmetrize)

    \begin{equation} |\phi\rangle=\underbrace{|+\rangle\cdots|+\rangle}_{n}\underbrace{|-\rangle\cdots|-\rangle}_{m} \quad (n+m=N) \end{equation}

    If we only look at the permutations $p$ which don't change $|\phi\rangle$, we end up with a subgroup $S_n S_m\subset S_N$, made up of:

    $S_n$ = permutations $\alpha\in S_N$ which don't change the "$\underbrace{|+\rangle\cdots|+\rangle}_{n}$" part and don't touch the "$\underbrace{|-\rangle\cdots|-\rangle}_{m}$" part

    and:

    $S_m$ = permutations $\beta\in S_N$ which don't touch the "$\underbrace{|+\rangle\cdots|+\rangle}_{n}$" part and don't change the "$\underbrace{|-\rangle\cdots|-\rangle}_{m}$" part

    With $S_n S_m$ being all the permutations of the form $\alpha \circ \beta$

    But the thing is that half of the elements of $S_n$ are even and the other half are odd, so the following sum is zero:

    \begin{align} A(|\phi\rangle) \stackrel{\text{def}}{=} &\sum_{p\in S_n S_m} sg(p) |\phi\rangle = \sum_{\alpha\beta\in S_n S_m} sg(\alpha\beta) |\phi\rangle = \sum_{\alpha\beta\in S_n S_m} sg(\alpha)sg(\beta) |\phi\rangle = \\ = &\sum_{\alpha\in S_n}\sum_{\beta\in S_m} sg(\alpha)sg(\beta) |\phi\rangle = \underbrace{\left(\sum_{\alpha\in S_n} sg(\alpha)\right)}_{0} \sum_{\beta\in S_m}sg(\beta) |\phi\rangle = 0 \end{align}

    And a similar computation could have been done to every permutation of $|\phi\rangle$, so, noticing that the original ket $|\Psi\rangle_{\text{spin}}$ is a sum of terms like $A(|\phi'\rangle)$, with $|\phi'\rangle$ being permutations of $|\phi\rangle$ which do change it (unlike before), it turns out that $|\Psi\rangle_{\text{spin}} = 0$ for every $N>2$ ! (with $|\Psi\rangle_{\text{spin}}$ antisymmetric)

  • slaaidenn
    slaaidenn about 4 years
    That 4-particle state you give is not totally antisymmetric. Why is it possible to have such state? Isn't it forbidden by spin-statistics theorem/symmetrization postulate?
  • knzhou
    knzhou about 4 years
    @adiselann No, only the total wavefunction must be antisymmetric, so presumably the position wavefunction compensates. However, since the position wavefunction can always take care of things, we can ignore the antisymmetrization postulate as long as we ignore the position wavefunction as well.
  • JEB
    JEB about 4 years
    Look at a beryllium atom. It has 4 electrons with zero spin and zero angular momentum: $[He]2S^2=1S^22S^2$. The pairs of electron each occupy different radial quantum numbers.