I previously asked Proof that $L=S=0$ for filled electron subshells? which motivated me to look more deeply into the restrictions the Pauli exclusion principle places on multi-particle angular momentum states.

It's well known that 2 distinguishable spin-1/2 particles could occupy up to 4 different states: $$ |++\rangle, |--\rangle, |+-\rangle, |-+\rangle $$

Or, in the angular momentum coupled basis:

\begin{align} |1, 1\rangle =& |++\rangle\\ |1, 0\rangle =& \frac{1}{\sqrt{2}}\left(|+-\rangle + |-+\rangle\right)\\ |1, -1\rangle =& |--\rangle\\ |0, 0\rangle =& \frac{1}{\sqrt{2}}\left(|+-\rangle - |-+\rangle\right) \end{align}

Here the top three states represent the total spin-1 triplet and the bottom state represents the spin-0 singlet. Notably, the triplet states a symmetric with respect to particle exchange while the singlet states are anti-symmetric with respect to particle exchange. Again, for distinguishable particles all 4 states are allowed. However, Fermions must respect the Pauli exclusion principle which says that the multi-particle state must be anti-symmetric with respect to particle exchange.

Thinking about this some more. Suppose we have, now, $N$ fermionic particles, each with intrinsic spin 1/2 and total orbital angular momentum $l=1$. This will be to build up to 6 electrons filling the $p$ shell. The single particle Hilbert space is then $\mathcal{H}_i$ and the total Hilbert space is

\begin{align} \mathcal{H} = \bigotimes_{i=1}^{N} \mathcal{H}_i \end{align}

The dimension of the single particle Hilbert space is $3\times 2 = 6$. The dimension of the multiparticle space is $6^N$. For $N$ from 1 to 6 this gives $\text{dim} = \left\{6, 36, 216, 1296, 7776, 46656\right\}$

However, for identical fermions the multiparticle Hilbert space is now the alternating tensor product of the single particle Hilbert space. The states are slater determinants of single particle states. My understanding is that the dimension of this space is given by

$$ _6 C_N = \frac{6!}{N!(6-N)!} $$

Since you must choose $N$ unique states from the set of 6 available single particle sates. This leads to the dramatically reduced dimensionalities of $\text{dim}=\left\{6, 15, 20, 15, 6, 1 \right\}$ for $N$ from 1 to 6.

I know immediately for $N=1$ that this Hilbert space decomposes to a total spin 1/2 and total spin 3/4 subspace since it is composed of one spin-1 and one spin-1/2 component. However, for 2 spins it is already complicated for me to determine the total spin subspaces. The only way I would know how to do it is first decompose the 36 dimensional subspace from the distinguishable case into total spin subspaces (I could do this without too much trouble) then explicitly write down the states of those subspaces and determine which ones are anti-symmetric and remove all the others. This would be terribly time consuming and wouldn't easily generalize to larger $N$. Alternatively I could write down all of the anti-symmetric states available, but then it's not obvious to me how to assign spin subspaces to particular states.

My questions are as follows. They are ordered from major to minor questions

  • Is there a way to know, on symmetry or group theoretic ground, the angular momentum decomposition of the anti-symmetrized space based on the angular momentum of the constituent spaces or, even, the decomposition of the distinguishable Hilbert space?
  • Similar to above, is there a group-theoretic way to determine whether a particular subspace of the total distinguishable Hilbert space is anti-symmetric, symmetric, or mixed symmetry?
  • when you take the antisymmetric subspace of the distinguishable multi-particle Hilbert space are you guaranteed that your states always come in complete spin subspaces? Proof for this?
  • Am I correct about the dimensionality of the alternating Hilbert space?
  • How to notate the antisymmetric tensor product in Latex?

Even more succinctly: I know how to decompose a multi-particle Hilbert space into irreducible representations. What is a generic procedure to sort these irreducible representations based on their symmetrization properties?


1 Answer 1


So the goal is to take a tensor product space, and figure out its rotationally closed subspaces, which are also called irreducible representations (irreps). When applied to quantum angular momenta, the rotationally closed subspace are the combinations that are eigenstates of total angular momentum.

There is a deep mathematic theory (Schur-Weyl duality) that relates these subspaces to the representations of the symmetric group (aka: the permutation group). Moreover, these representations are related to Young tableaux via the Robinson-Schensted correspondence. The representations of the symmetric group are ultimately related to the partitions of an integer, that is, how many ways the integer can be expressed as a sum of smaller (or equal) positive integers.

A literature search on the aforementioned terms will land you in graduate level mathematics that are difficult to penetrate. Here, I'll try to present a physics oriented approach that will hopefully make some of the abstract concepts a little more concrete.

You start with the fundamental representation, as a single box in a Young diagram:

│  │

This represents the 2 dimensional spin up / spin down irrep.

Now take the tensor product with itself:

╭──┐   ╭──┐   ╭──┬──┐   ╭──┐
│  │ X │  │ = │  │  │ + │  │
└──┘   └──┘   └──┴──┘   ├──┼
                        │  │

The tensor product is formed by combining the two diagrams on the left in all possible ways that form a legitimate Young diagram.

Each diagram on the right represents an irrep of total angular momentum. You can find the dimension of the representation by using the remarkable Hook length formula:

$${\rm dim}\,W(n, r) = \Pi_{(i,j)\in Y(n)}\frac{r+j-i}{{\rm hook}(i, j)}$$

$n$ is the number of boxes in the diagram and $r$ is the dimension of the fundamental irrep ($r=2$). Here $(i, j)$ is an integer labeling the row and column number, the product runs over all boxes in the diagram. ${\rm hook}(i, j)$ is the hook length of the box given by:

$$ {\rm hook}(i,j)=1+{\rm arm}(i,j)+{\rm leg}(i,j)$$

The arm (leg) length of a box is the number of boxes to-the-right (below) the box.

Applying the hook length formula to the above equation gives:

$${\bf 2}\otimes{\bf 2}={\bf 3}_S\oplus{\bf 1}_A$$

which means combining two doublets yields a triplet and a singlet, which is what we learned in elementary quantum mechanics.

(Tensor side note: had we been using 3-vectors as our fundamental irrep, the hook length formula yields:

$${\bf 3}\otimes{\bf 3}={\bf 6}_S\oplus{\bf 3}_A$$

which tells us a cartesian tensors has a six dimensional part and a three dimensional part (that transforms like vector, aka the cross product).

In Minkowski space, it tells us 4-tensors look like:

$${\bf 4}\otimes{\bf 4}={\bf 10}_S\oplus{\bf 6}_A$$

where we recognize the 10 dimensions of the stress energy tensor $T_{\mu\nu}$, and the six of electromagnetism $F_{\mu\nu}$).)

Remarkable hook length formula, indeed.

Further consideration will show that the subspaces in this case are symmetric or antisymmetric.

Note: we combined 2 fundamental irreps. The partitions of 2 are:

$$ 2 = 2 $$ $$ 2 = 1 + 1 $$

Each partition corresponds to a Young diagram. Each diagram has a number of standard Young tableaux (that is, each box gets filled with a number from 1 to n such that numbers increase going to the left to right and top to bottom). Another hook length formula tells us how many standard tableaux exists for each diagram:

$${\rm dim}\pi_n = \Pi_{(i,j)\in Y(n)}\frac{n!}{{\rm hook}(i, j)}$$

This is the number of irreps of that dimension (and the number of irreps of the symmetric group of that dimension). Here it's both 1:

│1 │2 │
│1 │
│2 │

It will become clear why the horizontal (vertical) boxes are (anti)symmetric.

To show the power of the Young tableaux we need to combined 3 irreps. If it is spin 1/2, then the hook length formula tells us:

$${\bf 2}\otimes{\bf 2}\otimes{\bf 2}={\bf 4}_S\oplus{\bf 2}_M\oplus {\bf 2}_M$$

The dimension of the symmetric combination is 4:

$$|\frac 3 2,+\frac 3 2\rangle = |\uparrow\uparrow\uparrow\rangle $$ $$|\frac 3 2,+\frac 1 2\rangle = (|\downarrow\uparrow\uparrow\rangle+|\uparrow\downarrow\uparrow\rangle+|\uparrow\uparrow\downarrow\rangle)/\sqrt 3 $$ $$|\frac 3 2,-\frac 1 2\rangle = (|\downarrow\downarrow\uparrow\rangle+|\downarrow\uparrow\downarrow\rangle+|\uparrow\downarrow\downarrow\rangle)/\sqrt 3 $$ $$|\frac 3 2,-\frac 3 2\rangle = |\downarrow\downarrow\downarrow\rangle $$

while the antisymmetric combination has dimension 0: there is no singlet state. (Tensor side note: had we used $r=3$ and not $r=2$ in the hook length formula, we would have a one dimensional antisymmetric space, which is spanned by $\epsilon_{ijk}$...remarkable...how does it work?).

The question remains: how do the diagrams tell us how to combine permutation of indices or spin states? For that, we'll look at another $n=3$ diagram, with two standard fillings:

│1 │2 │
│3 │

│1 │3 │
│2 │

For specifics, we'll focus on the top one. From here we compute the Young symmetrizer and apply it to particle labels (or indices if we're doing rank-3 tensors).

First we need the symmetric group on 3 letters $(1,2,3)$:

$$ S_3 =\{e,e_{23},e_{12},e_{123},e_{132},e_{13}\}$$

where the permutation, e.g. $e_{123}$ means $(1,2,3)\rightarrow (2,3,1)$. Those are all six elements, with $e$ being the identity.

First: find all permutations that leave the tableaux "row equivalent". Tableaux are row equivalent if each row has the same numbers:


and similarly for column equivalence, with the addition that we include the parity of the permutation:


The Young symmetrizer is then the product of these two, as follows:

$$ S =R*C = \{e+e_{12}-e_{13}-e_{132} \}$$

You then apply these permutations to $|\uparrow\uparrow\downarrow\rangle$, and normalize, to get:

$$|\frac 1 2, +\frac 1 2\rangle= (|\uparrow\uparrow\downarrow\rangle - |\downarrow\uparrow\uparrow\rangle)/\sqrt 2 $$

Easy peasy.

  • $\begingroup$ Thank you very much for this great tutorial! I've seen in a few places the relationship between irreps of angular momentum spaces and young tableaux but I've never seen a clear explanation until this one! I'm still not totally clear on the Young symmetrizer. I understand how you constructed the Young symmetrizer $S$ for a particular Young Tableaux. Why, in this case, does it make sense to apply the symmetrizer to the state $|\uparrow \uparrow \downarrow \rangle$ and not some other state? $\endgroup$
    – Jagerber48
    Commented May 4, 2021 at 15:48
  • $\begingroup$ How would you do this when taking the tensor product between states from different representations? $\endgroup$ Commented Apr 11, 2023 at 17:21
  • $\begingroup$ like combing $l=1$ orbital angular momentum and $s=\frac 1 2$? I would make the $l=1$ from 2 $s=\frac 1 2$'s and proceed normally with 2 x 2 x 2 = 4 + 2 + 2, but keep only irreps that have the two spin half's symmetric, you'd be left with the 4 and one of the 2's, which makes sense: a vector boson plus a spinor has a $j=\frac 3 2$ and a $j=\frac 1 2$ total angular momentum eigenstate. I would write is "3 x 2 = 4 + 2", but I am not sure if you can get that from the hook length formula... $\endgroup$
    – JEB
    Commented Apr 12, 2023 at 2:57
  • $\begingroup$ I was thinking along the lines of something like this. Is it always true that only the symmetric sum can produce the highest allowed state? Also, can the hook length formula account for the "accidental" zeros of CG coefficients? $\endgroup$ Commented Apr 12, 2023 at 14:28

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