Weights in SU(3)

Question

I'm a physicist trying to understand the concept of weight vectors in SU(3). In my professor's notes, weights are defined as the eigenvalues of the elements of the basis, which can be simultaneously diagonalized as $H_i$, i.e., they commute. It is stated that there exist some "weight vectors" formed by these weights, and their dimension is the rank of the Cartan subalgebra $l$. However, I don't understand this.

First of all, why is their dimension the rank of the Cartan subalgebra? I can understand that there are $l$ eigenvalues, but in this case, I think there will be only one "weight vector," and I know that there are 3 for SU(3).

Secondly, I don't see why $H_i\left|E_\alpha\right\rangle=\alpha_i\left|E_\alpha\right\rangle$ where $E_\alpha$ are the states not corresponding to the Cartan.

I would greatly appreciate any clarification; I'm a bit lost in the theory of group representations.

Answer 1

There seems to be some confusion about what a weight vector is as compared to the weight itself. I'll try to lay this all out in standard mathematical form but be aware that some of the physics notation/conventions will vary so you may have to translate. I am also going to start off working over $\mathbb{C}$ as for a compact Lie group/algebra the roots and root spaces are not real.

Let $\mathfrak{g}$ be a complex semisimple Lie algebra (in your example we would take $\mathfrak{g}=\mathfrak{sl}(3,\mathbb{C})$) and pick a Cartan subalgebra $\mathfrak{h}$. The crucial feature of a Cartan subalgebra is that it can be simultaneously diagonalised (for any representation of $\mathfrak{g}$). What that means in practical terms is that any representation $V$ has a basis where each element is an eigenvector for every single element of $\mathfrak{h}$. Let $v$ be such an element of $V$ so given $H\in \mathfrak{h}$ there is some $\lambda\in \mathbb{C}$ such that $H(v) = \lambda v$. Then let $\omega$ be a map from elements of $\mathfrak{h}$ to $\mathbb{C}$ (so $\omega \in \mathfrak{h}^*$)sending each element to its eigenvalue for $v$, i.e. $H(v) = \omega(H)v$. We call $\omega$ a weight and $v$ a weight vector corresponding to $\omega$. As I mentioned we can find a basis of weight vectors for $V$ but some of these can have the same weight. Note also that a scalar multiple of a weight vector will still be a weight vector as well.

Important distinction here: the weights are vectors in the mathematical sense (they live in a vector space $\mathfrak{h}^*$) but they are not the same as the weight vectors (which live in a different vector space $V$)

So the weights of a Lie algebra depend on a choice of representation but it turns out that all possible weights live on a nice lattice in $\mathfrak{h}^*$. That is they are integer linear combinations of certain "fundamental" weights. That takes us a bit far afield but to bring it back answering the first question: the weights span the whole of $\mathfrak{h}^*$ which has the same dimension as $\mathfrak{h}$. Meanwhile, the weight vectors span the representation $V$ instead which has whatever dimension it has.

The most important representation of a Lie algebra is the one it has on itself, usually known as the adjoint representation. Specifically this means that $X(Y)$ is now more usually written $[X,Y]$. First we should note that since the Cartan subalgebra is abelian all its elements are weight vectors for the $0$ weight: $[H_1, H_2] = 0$. The rest of the Lie algebra is spanned by non-zero weight vectors but when talking about the adjoint representation we usually call the non-zero weights roots. So the root vectors are the $E_\alpha$ that you refer to with corresponding weight (aka root) $\alpha$. Feeding that in to our definition above we are saying $$[H,E_\alpha] = \alpha(H)E_\alpha.$$ I think this is what your equation is supposed to be saying (although it is also throwing in bra-ket notation so I am not 100% sure).

Edit: The other possibility is that they are using the Killing form to identify $\mathfrak{h}$ with $\mathfrak{h}^*$. This is quite common and you often see roots referred to as elements of $\mathfrak{h}$ as a result. In that case I would assume that they have defined $H_i$ as the element such that $\alpha_i(H) = (H_i,H)$ for all $H\in \mathfrak{h}$. But this still doesn't really make that equation you have written make sense.

Answer 2

• The dimension of $\mathfrak{su}(3)$ is 8.
• Everything must be done in the complexification of $\mathfrak{su}(3)$ which is $\mathfrak{sl}(3)$. The matrices of $\mathfrak{su}(3)$ have imaginary eigenvalues so you won't be able to find eigenvectors in the adjoint representation of $\mathfrak{su}(3)$.
• A Cartan subalgebra has dimension $2$, so the rank $l$ is $2$.
• There are $6$ roots and $2 + 6 = 8$.
• Of the $6$ roots, $3$ are positive and $3$ are negative.
• Of the $3$ positive roots, $2$ are simple and the third is their sum. $2$ simple roots as the rank $2$.
• In an irreducible representation there is a single (up to scalar multiplication) highest weight vector $v_\lambda\ne 0$ for the highest weight $\lambda$.
• The number of linearly independent weight vectors is the dimension of the representation.
• There are two linearly independent lowering operators corresponding to the two simple roots.
• Starting with the highest weight vector you repeatedly apply the lowering operators to get a basis to the representation of eigenvectors simultaneously diagonlizing the Cartan subalgebra.

Concretely, let $v_1=(1,0,0), v_2 = (0,1,0), v_3=(0,0,1)$. Take

• $h_1 = {v_1}^\top v_1 - {v_2}^\top v_2$
• $h_2 = {v_2}^\top v_2 - {v_3}^\top v_3$
• $L_\alpha = {v_1}^\top v_2$ and $L_{-\alpha}={v_2}^\top v_1$
• $L_\beta = {v_2}^\top v_3$ and $L_{-\beta}={v_3}^\top v_2$
• $L_{\alpha+\beta} = [L_\alpha, L_\beta]$ and $L_{-\alpha-\beta} = [L_{-\beta}, L_{-\alpha}]$

The $h_1, h_2$ span a Cartan subalgebra, $\alpha(h_1)=2, \alpha(h_2)=-1, \beta(h_1)=-1, \beta(h_2)=2$. $L_\alpha, L_\beta$ are the two raising operators, and $L_{-\alpha}, L_{-\beta}$ the two lowering operators.

$[h_*, L_\alpha]=\alpha(h_*)L_\alpha$, $[h_*, L_\beta]=\beta(h_*)L_\beta$ and $[h_*, L_{\alpha+\beta}]=(\alpha(h_*)+\beta(h_*))L_{\alpha+\beta}$

$[L_{\alpha},L_{-\alpha}]=h_1, [L_{\beta},L_{-\beta}]=h_2$ and $[L_{\alpha+\beta},L_{-\alpha-\beta}]=(h_1+h_2)$

Some examples:

For a root or weight $\lambda$ denote by $(n,m)$ the values $n=\lambda(h_1),m=\lambda(h_2)$.

For the defining representation of $\mathfrak{sl}(3)$, the vectors ${v_1}^\top, {v_2}^\top, {v_3}^\top$ are weight vectors. We have $h_1 {v_1}^\top = {v_1}^\top$ and $h_2 {v_1}^\top = 0$, so ${v_1}^\top$ is a weight vector of weight $(1,0)$. Similarly, ${v_2}^\top$ is a weight vector of weight $(-1,1)=(1,0)-(2,-1)=(1,0)-\alpha$. ${v_3}^\top$ is a weight vector of weight $(0,-1)=(1,0)-(1,1)=(1,0)-(\alpha+\beta)$. So ${v_1}^\top$ is a highest weight vector, ${v_2}^\top = L_{-\alpha}{v_1}^\top$ and ${v_3}^\top=L_{-\beta} L_{-\alpha}{v_1}^\top=L_{-\alpha-\beta}{v_1}^\top$. The representation type is $(1,0)$ which is the highest weight.

Corresponding to the $(1,0)$ representation there is a representation $(0,1)$ which is obtained from $(1,0)$ via an outer automorphism of $\mathfrak{sl}(3)$ that swaps $h_1, h_2$ and $\alpha, \beta$.

For the adjoint representation, the highest weight vector is $L_{\alpha+\beta}$ with weight $(1,1)$ which is also the type of the representation. The other weights are $(2,-1), (-1, 2), (0, 0), (0, 0), (-2, 1), (1, -2), (-1, -1)$ with weight vectors $L_\alpha, L_\beta, h_1, h_2, L_{-\alpha}, L_{-\beta}, L_{-\alpha-\beta}$ respectively.

It can be shown that there exists an irreducible representation of type $(n,m)$ for all non-negative integers $n, m$ which is unique up to equivalence. The representations $(n,m)$ and $(m,n)$ are related by the outer automorphism described above.

Answer 3

$\newcommand{\Lie}[1]{\mathfrak{#1}}$ $\newcommand{\bra}{\langle}$ $\newcommand{\ket}{\rangle}$ $\newcommand{\ad}[1]{\text{ad}_{#1}}$ $\newcommand{\m}[1]{\pmatrix{#1}}$

There is a lot to unpack here. Let's get to the answers step-by-step (I swear this is not written by ChatGPT):

What is a weight?

Suppose $D: \Lie{g} \to \text{End}(V)$ is an irreducible representation of $\Lie{g}$ on the space $V$. Let us denote by $H_{i}$ both the elements of $\Lie{h}$ and their action in the representation space $V$. Since $H_i$ as operators on $V$ commute with each other (because they form a representation of $\Lie{g}$), one can choose such basis in $V$ that $H_{i}$ are all diagonal. For every vector of that basis, $|e_{\mu}\ket$, there exists a tuple $(\mu_{1}, ..., \mu_{l})$ of eigenvalues of $|e_{\mu}\ket$ under the action of $H_{i}, i=1,...,l$. That tuple is called the weight vector $\vec{\mu}$. The weight vector lives in $l$-dimensional vector space. Each basis vector $|e_{\mu}\ket$ has its own weight vector. A vector in representation $(D, V)$ that has weight vector $\vec{\mu}$ is denoted $|\vec{\mu}, D\ket$. Note that:

1. The dimension of the weight vector space equals the number of generators in the Cartan subalgebra, because each weight vector is just a tuple of eigenvalues of different basis elements of Cartan subalgebra.

That answers

why is their dimension the rank of the Cartan subalgebra?

2. The total number of weight vectors equals the dimension of representation $(D, V)$. Some weight vectors may coincide and/or be equal to zero.

Indeed, each matrix $H_{i}$ is diagonal. The first weight vector corresponds to the first row of all matrices, the second - to the second row of all matrices, e.t.c. So the statement

but in this case, I think there will be only one "weight vector," and I know that there are 3 for SU(3).

is untrue at two levels. First, the tuples of eigenvalues may be different for different $|e_{\mu}\ket$, which gives you different weight vectors. Second, the number of weight vectors is not a property of the algebra, it is a property of specific representation. I think you confuse the number of weights in general with the number of weights in the defining representation (see below).

3. The representation is usually chosen in such a way that weight vectors are real (i.e. all eigenvalues of actions of Cartan subalgebra elements are real, for instance, when the representation is by Hermitian operators).

Now let's remove the confusion about roots and weights and answer the second part of the question:

Secondly, I don't see why $H_{i}|E_{α}⟩=α_{i}|E_{α}⟩$ where $E_{α}$ are the states not corresponding to the Cartan.

To answer this question, we must first understand what $|E_{\alpha}\ket$ really are. Previously we've introduced the notion of a weight for an arbitrary representation. Let us consider the specific case of the adjoint representation. I remind that the adjoint representation is the action $\ad{}:\Lie{g} \to \text{End}(\Lie{g})$ of a Lie algebra on itself, which is defined by $\ad{A}B:=[A,B]_{\Lie{g}}$.

Denote as $E_{\alpha}$ generators of $\Lie{g}$ that do not lie in the Cartan subalgebra $\Lie{h}$. To obtain $H_{i}|E_{\alpha}\ket = \alpha_{i}|E_{\alpha}\ket$, we need to choose a special basis in $\Lie{g}$ (remember that $\Lie{g}$ is a vector space of itself). Namely, let us choose such a basis in which $H_{i}$ are all diagonal in the adjoint representation. Since $[H_{i}, H_{j}]=0$, and the adjoint representation is a representation, we have $[\ad{H_{i}}, \ad{H_{j}}]=0$ (which can be checked explicitly). But that means $\ad{H_{i}}$ are simultaneously diagonalizable as matrices on $\Lie{g}$. Hence, we can choose such a basis in $\Lie{g}$ that

$$\ad{H_{i}}E_{\alpha}=[H_{i}, E_{\alpha}]=\alpha_{i}E_{\alpha}$$

where $\alpha_{i}$ is an eigenvalue dependent on $i$. The crucial consequence of this fact is that, with such a choice of basis in $\Lie{g}$, $[H_{i}, E_{\alpha}]=\alpha_{i}E_{\alpha}$ in every representation, not just in the adjoint, since $[,]$ here is the bracket in the Lie algebra itself, and every representation by definition conserves the bracket.

For each $E_{\alpha}$, $\alpha_{i}$ form the weight vector of $E_{\alpha}$ in the adjoint repesentation. Weight vectors in the adjoint representation are called roots. Unlike generic weight vectors, roots are uniquely defined for a given algebra (up to orthonormalization of the basis vectors in $\Lie{g}$ with respect to the Killing form $(A, B)=\text{Tr}(\ad{A} \ad{B})$). Remembering what we've learned about the general properties of weights, we conclude:

1. The dimension of roots equals the rank of Cartan subalgebra
2. The total number of roots equals the dimension of the Lie algebra
3. The root vectors corresponding to the elements of Cartan subalgebra are all zero (because $\ad{H_{i}}|H_{j}\ket = [H_{i}, H_{j}]=0$). Moreover, the converse is true: if the root of an element of $\Lie{g}$ equals $0$, then this element lies in the Cartan subalgebra.

Now, consider arbitrary representation $(D, V)$ and a vector $|\vec{\mu}, D\ket$ in this representation. Note that diagonalizations of $\Lie{g}$ and of the representation space $V$ are two different operations and are independent, so we have both commutation relation $[H_{i}, E_{\alpha}]=\alpha_{i}E_{\alpha}$ and existence of weight vectors $|\vec{\mu}, D\ket$ such that $H_{i}|\vec{\mu}, D\ket = \mu_{i}|\vec{\mu}, D\ket $ What is then (if any) the weight vector of $E_{\alpha}|\vec{\mu}, D\ket$?

$$H_{i}E_{\alpha}|\mu, D\ket = [H_{i}, E_{\alpha}]|\vec{\mu}, D\ket + E_{\alpha}H_{i}|\vec{\mu}, D\ket$$ $$=\alpha_{i}E_{\alpha}|\vec{\mu}, D\ket + \mu_{i}E_{\alpha}|\vec{\mu}, D\ket$$ $$=(\mu+\alpha)_{i}E_{\alpha}|\mu, D\ket$$

Hence, there are two options:

1. $E_{\alpha}|\vec{\mu}, D\ket$ is a vector with weight $(\vec{\alpha}+\vec{\mu})$, i.e. $|\vec{\mu}+\vec{\alpha}, D\ket$ (here we use "weight" instead of "weight vector" to evade repeating the word "vector" too often).
2. $E_{\alpha}|\vec{\mu}, D\ket=0$ (in such situation we say that $\vec{\alpha}+\vec{\mu}$ "is not a weight").

That answers your second question. $E_{\alpha}$ acts as "raising operator" which increases the weight $\vec{\mu}$ by the root $\vec{\alpha}$, or annihilates the state.

Let's briefly clarify the tangentials and the meaning of all this. First, the defining representation of $SU(3)$ is the one in which Gell-Mann matrices are the generators. The Gell-Mann matrices $T_{3}$ and $T_{8}$ form the representation of the Caratan subalgebra, hence, weight (and root) vectors for $SU(3)$ are two-dimensional. The dimension of the defining representation is $3$. Hence, you have three weight vectors in the defining representation. Let us calculate them explicitly:

$$H_{1}=T_{3}=\frac{1}{2}\pmatrix{1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0}, \ \ \ H_{2}=T_{8}=\frac{1}{2\sqrt{3}}\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2}$$

The basis vectors in the defining representation are $e_{1} = \m{1 \\ 0 \\ 0}$, $e_{2} = \m{0 \\ 1 \\ 0}$, and $e_{3} = \m{0 \\ 0 \\ 1}$.

The first weight vector is thus $\left(H_{1}e_{1}, H_{2}e_{1}\right) = \left( \frac{1}{2}, \frac{1}{2\sqrt{3}} \right)=\vec{\mu}_{1}$, the second is $\vec{\mu}_{2}=\left(-\frac{1}{2}, \frac{1}{2\sqrt{3}} \right)$, and the third is $\vec{\mu}_{3}=\left(0, -\frac{1}{\sqrt{3}}\right)$

They turn out to form a regular triangle. Since weight vectors are obtained one from another by adding (and subtracting, via accordingly defined "lowering operators" which in the case of Hermitian representations are just conjugates of the rising operators), root vectors for $SU(3)$ can be obtained as differences between weight vectors, there are six of them, + 2 zero vectors that correspond to the Cartan subalgebra, giving you a total of 8, as expected (the adjoint representation is 8-dimensional). The 6 nontrivial root vectors span the vertices of a regular hexagon.

As mentioned before, roots are unique for a given Lie algebra, while weights are not. It is never the case that all roots are linearly independent: as Euclidean vectors, they form a highly symmetric structure (reflection of a root system with respect to any plane to which a root is a normal vector does not change it). One can choose "simple roots" that generate the whole root lattice. Using these simple roots and theorems about them, one can classify irreducible finite-dimensional representations of arbitrary simple Lie algebras, using "the method of highest weight": you sort your weight vectors in lexicographical order, call the "most positive" weight vector "the highest weight", show that the corresponding state is annihilated by all raising operators, and then descend from that state to all others using "lowering operators", constructed from the raising operators.

A familiar example of this procedure is given by the algebra of operators of angular momentum: the Cartan subalgebra is $L_{z}$, it is 1-dimemsional. The rising and the lowering operators are $L_{+}$ and $L_{-}$. The weight vector is one-dimensional: it is the eigenvalue of $L_{z}$, the projection of angular momentum to the $z$-axis. The highest weight vector $|j, m\ket$ corresponds to the state with with maximal projection of angular momentum $m$, and the action of $L_{+}$ annihilates it. By applying successively $L_{-}$ to the highest weight state, you obtain vectors of lower and lower weight, until you reach $|j, -m\ket$, which is annihilated by $L_{-}$, and you've spanned the whole irreducible representation of the algebra. $j$ is actually the eigenvalue of the Casimir operator $L^2$ which enumerates different irreducible representations.