I know that different representations of this algebra correspond to different spin. One can sort the representation according to the casimir. For any simple Lie algebra, the operator

$$ T^2 = T^{\alpha}T^{\alpha} $$

comutes with all the group generators such that

$$ [T^2,T^{\alpha}]=0 $$

So $T^2$ takes a constant value on each irreductible representation. Thanks to Shur's lemma, one knows that $T^2$ is proportional to the identity for all representation.

If one considers the rotation algebra $o(3)$, the Casimir takes the form

$$ C=\delta^{ij}J^iJ^j=s(s+1) $$

where $J^i$ are the generators of $o(3)$ and $s$ is a half integer that labels each representation of dimension

$$ d = 2s + 1 $$

For instance :

  • $s=0 \Rightarrow d=1$ and $S_i=0$
  • $s=1 \Rightarrow d=2$ and $S_i=\frac{1}{2}\sigma_i$
  • $s=2 \Rightarrow d=3$ and $S_i=\Sigma_i$
  • ...

where the $S_i$ are matrix representations of $J^i$ and $\sigma_i$ are the Pauli matrix. Moreover there is an isomorphism between $su(2)$ and $o(3)$.


Now if one looks at the Lorentz algebra, $o(1,3)$, one can do the same work but this time, there are two Casimirs that can be written

$$ C_{\pm}=s_\pm(s_\pm+1) $$

Thus one has :

  • $(s_+,s_-)=(0,0)\Rightarrow d=1$ scalar
  • $(s_+,s_-)=(1/2,0)\Rightarrow d=2$ spinorial chiral (not sure of the english terminology)
  • $(s_+,s_-)=(0,1/2)\Rightarrow d=2$ spinorial antichiral
  • $(s_+,s_-)=(1/2,1/2)\Rightarrow d=2$ vectorial
  • ...

And again there is an isomorphism between $o(1,3)$ and $sl(2,\mathbb{C})$.


So my question is : what is the correct algebra and representation to describe a spin one-half fermion (an electron for instance) ? From what I understand I could choose between $su(2)$, $o(3)$, $o(1,3)$ or $sl(2,\mathbb{C})$ but with a given representation.

Same question for a spin one boson ? higher spins... ?


My guess would be that :

  • spin 1/2 fermion is described by the $su(2)$ in the fundamental representation (labeled by a single box within the Young tableau formalism)
  • spin 1 boson is described by the $su(2)$ with the symmetrical combination of two fundamental representation (two box in a single row)
  • spin 3/2 is described by the $su(2)$ with the symmetrical combination of two fundamental representation and one antisymmetrical one(two boxes in the first row and one box in the second)


I know that this question is confused... and so am I... My goal is to undersand the link between the physical particles, the $su(N)$ algebra (for any $N$), the representation of the algebra and the Young tableaus. So instead of answering the question above, it might be easier to explain my those links...


Thanks !

  • $\begingroup$ You know that particle is described by the unitary irreducible representations of the Poincare group. Also, the field is described by non-unitary irreducible representation of the Lorentz field. $\endgroup$
    – user8817
    Aug 29, 2013 at 23:12
  • 1
    $\begingroup$ By the spinor $\psi_{a_{1}...a_{n}\dot {a}_{1}...\dot {a}_{m}}$, which is symmetric in dotted and undotted indices, and by it's transformation law you can build the fundamental irreducible representation $\left(\frac{n}{2}, \frac{m}{2}\right)$ of the Lorentz group, which is satisfied to field with spin $s = \frac{n + m}{2}$. $\endgroup$
    – user8817
    Aug 29, 2013 at 23:15
  • 1
    $\begingroup$ Then, if you use discrete transformation operator, like operator of space inversion, you will see, that it mixes $\left( \frac{n}{2}, \frac{m}{2}\right)$ and $\left( \frac{m}{2}, \frac{n}{2}\right)$ representations. So you must get their direct sum. $\endgroup$
    – user8817
    Aug 29, 2013 at 23:21
  • 1
    $\begingroup$ Then, if the field $\psi_{a_{1}...a_{n}\dot {a}_{1}...\dot {a}_{m}}$ satisfies the equations $$ (\partial^{2} + m^{2})\psi_{a_{1}...a_{n}\dot {a}_{1}...\dot {a}_{m}} = 0, $$ $$ \partial^{\dot {b}a}\psi_{aa_{1}...a_{n - 1}\dot {a}\dot {a}_{1}...\dot {a}_{m - 1}}, $$ it realizes the irreducible representation of the Poincare group with spin $s$ and mass $m$. $\endgroup$
    – user8817
    Aug 29, 2013 at 23:25
  • 1
    $\begingroup$ So it is obviously that you can describe an arbitrary representation by the irreducible combinations of the spinors representation (symmetrical spinor tensors). For example, an arbitrary 4-vector corresponds to $$ x^{\mu} \to X_{a \dot {a}} = (\sigma^{\mu})_{a \dot {a}}x_{\mu}, $$ antisymmetric tensor of rank 2 corresponds to $$ M_{\mu \nu} \to h_{\alpha \beta \dot {\alpha} \dot {\beta }} = -\varepsilon_{\alpha \beta}(\tilde {\sigma}^{\mu \nu})_{\dot {\alpha }\dot {\beta }}M_{\mu \nu} + \varepsilon_{\dot {\alpha}\dot {\beta }}(\sigma^{\mu \nu})_{\alpha \beta }M_{\mu \nu }, $$ etc. $\endgroup$
    – user8817
    Aug 29, 2013 at 23:31

1 Answer 1


Disclaimer: There is nothing about Young tableau in this answer; I realize after rereading your question that your primary question might regard Young tableau; my apologies if this is useless to you.

As a matter of notation, if Lie algebras $\mathfrak g_1$ and $\mathfrak g_2$ are isomorphic, then let's write $\mathfrak g_1\cong\mathfrak g_2$.

In physics, we are interested in the four Lie algebras you listed because the quantum fields and states of particles transform under representations of groups having those algebras.

The spin state of a particle of spin $s$ transforms under the irreducible spin $s$ representation of $\mathrm{SU}(2)$, and we use it's Lie algebra $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ to analyze its representations.

In a Lorentz-invariant Qauntum field theory with Hilbert space $\mathcal H$, we assume in physics that there is a unitary representation $U$ of the Lorentz group $SO(1,3)$ acting on $\mathcal H$ which represents the Lorentz symmetry. This representation acts on the fields $\phi$ in the theory by conjugation \begin{align} \phi(x) \to U(\Lambda) \phi(x) U(\Lambda)^{-1} \end{align} If $\phi$ is the quantum field describing a certain kind of particle, then this conjugation action will do something special to the indices of the field via a certain finite-dimensional representation of the Lorentz group. The finite-dimensional, irreducible representations $D^{(s_1, s_2)}$ of the Lorentz group are labeled with two "spins" as you have indicated (see below for how this works). But quantum fields don't necessarily transform under on these these.

Example. Dirac Fermion

The electron is a Dirac Fermion. If we denote the corresponding quantum field by $\Psi^a$, then this means that there is a finite-dimensional representation $\rho_\mathrm{D}$ of the Lorentz group such that \begin{align} U(\Lambda) \Psi^a(x) U(\Lambda)^{-1} = \rho_\mathrm{D}(\Lambda)^a_{\phantom ab}\Psi^b(\Lambda^{-1} x) \end{align} It turns out that the Lie algebraic Dirac representation can actually be written as a direct sum of the left and right handed Weyl spinor representations representations of the Lorentz algebra \begin{align} D^{(\frac{1}{2},0)}\oplus D^{(0,\frac{1}{2})} \end{align}

In general, the indices (target space) of the field corresponding to each kind of particle in a given Lorentz-invariant field theory with transform under a particular finite-dimensional representation of the Lorentz group.

Lie algebraic gymnastics

As you note \begin{align} \mathfrak{su}(2) \cong\mathfrak{so}(3) \end{align} so both of these algebras equivalently describe the spin state of every particle. In particular, the state of a spin $s$ particle is described by the spin-$s$ irreducible representation $D^{(s)}$ of these algebras.

Next, note that for any $n$, we have an isomorphism $\mathfrak{sl}(n, \mathbb C)\cong\mathfrak{su}(n)_\mathbb C$ In other words, $\mathfrak{sl}(n,\mathbb C)$ is isomorphic to the complexification of $\mathfrak{su}(n)$. In particular, for $n=2$ we have \begin{align} \mathfrak{sl}(2, \mathbb C)\cong\mathfrak{su}(2)_\mathbb C \end{align} What the heck do complexifications have anything to do with anything you ask? Well, recall that when we construct the representations of $\mathfrak{su}(2)$, we usually do so by defining the ladder operators \begin{align} J_{\pm} = J_1 \pm iJ_2 \end{align} These ladder operators are not actually elements of $\mathfrak{su}(2)$, they are, in fact, elements of its complexification, which, by the isomorphism above is really just the same as $\mathfrak{sl}(2,\mathbb C)$! So what we're really doing with ladder operators, is we're determining irreducible representations of $\mathfrak{sl}(2,\mathbb C)$, and then we're using the following theorem (see Hall's Lie Groups, Lie Algebras, and Representations proposition 4.6):

Let $\mathfrak g$ be a real Lie algebra, then every finite-dimensional complex representation $\pi$ of $\mathfrak g$ has a unique extension to a complex-linear representation $\pi_\mathbb C$ of $\mathfrak g_\mathbb C$ given by \begin{align} \pi_\mathbb C(X+iY) = \pi(X) + i\pi(Y) \end{align} Furthermore, $\pi_\mathbb C$ is irreducible if and only if $\pi$ is irreducible.

This theorem shows that in order to make an exhaustive study of finite-dimensional representations of $\mathfrak{su}(2)$, it suffices to study instead the complex-linear representations of $\mathfrak{sl}(2,\mathbb C)$.

Next, note that, $\mathfrak{o}(1,3)$ is not isomorphic to $\mathfrak{sl}(2,\mathbb C)$. In fact, you can kind of see from the representation theory that there are two "copies" of the spin $s$ representation in the representations $(s_+, s_-)$ of $\mathfrak{o}(1,3)$ to which you refer. In fact, there is an isomorphism \begin{align} \mathfrak{o}(1,3)_\mathbb C \cong \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C) \end{align} To see this in what we do as physicists, recall that in studying the representations of $\mathfrak{o}(1,3)$, we start with the rotation and boost generators $J_i, K_i$, where $i=1,2,3$, and we form the complex linear combinations \begin{align} A_i = \frac{1}{2}(J_i + iK_i), \qquad B_i = \frac{1}{2}(J_i - iK_i) \end{align} which now an "A-copy" and a "B-copy" of the $\mathfrak{sl}(2,\mathbb C)$ algebra; \begin{align} [A_i, A_j] = i\epsilon_{ijk} A_k, \qquad [B_1, B_j] = i\epsilon_{ijk}B_k, \qquad [A_i, B_j] = 0 \end{align} Why do we do all of this? Well, the original motivation is that we want to study the representations of $\mathfrak{o}(1,3)$ which are important because the particle states in relativistic quantum field theories, for example, basically transform under these representations. But then we notice that if we pass to the complexification of $\mathfrak{o}(1,3)$ (like with did in going from $\mathfrak{su}(2)$ to $\mathfrak{sl}(2,\mathbb C)$ when studying spin $s$ irreps), then the algebra splits into a direct sum of an algebra with itself, namely $\mathfrak{sl}(2,\mathbb C)$ whose representations we are already quite familiar with from our analysis of spin! This allows us to almost immmediately write down the finite-dimensional, complex-linear irreps of $\mathfrak{o}(1,3)_\mathbb C$, and therefore those of $\mathfrak{o}(1,3)$.

  • $\begingroup$ thanks for your answer ! i don't have time to read it today but i'll do that carefully next week ! $\endgroup$
    – PinkFloyd
    Aug 30, 2013 at 11:17
  • $\begingroup$ I have a few questions regarding what you wrote. I will split them in different comment that i will label, so you can answer them more easily. $\endgroup$
    – PinkFloyd
    Sep 2, 2013 at 12:24
  • $\begingroup$ 1. If i don't use quantum field theory, namely i work with non-relativistic quantum mechanics, is it correct to assume that i don't need the $O(1,3)$ group to describe my system ? $O(3)$ group will suffice ? $\endgroup$
    – PinkFloyd
    Sep 2, 2013 at 12:26
  • $\begingroup$ 2. as the algebra of $o(3)$ and $su(2)$ are isomorphic, one can use the irreductible representation of $su(2)$ to describe the spin state of a particle ? $\endgroup$
    – PinkFloyd
    Sep 2, 2013 at 12:29
  • $\begingroup$ 3. I agree with what you wrote about Weyl spinors, but as it comes from the Lorentz group irreductible representations, I don't need (as I'm not a quantum-field physicist) to consider Dirac fermions ? $\endgroup$
    – PinkFloyd
    Sep 2, 2013 at 12:32

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