# Physical meaning of the highest root / weight

• I
It gives the difference of positive and negative spin, hence has a physical meaning.

For higher rank models, one has usually contributions to the Hamiltonain that break the symmetry along a physically determined chain of subgroups, which induce for many groups a unique labelinǵ of the roots. In some cases there are missing label parameter poblems, in which case it is not so clear how to assign meaning to a chosen labeling.
That might be a good answer to the original question in this thread. Where can we see some of the higher rank models?

Where can we see some of the higher rank models?
The book
• F. Iachello,Lie algebras and applications, Springer, Berlin 2006
is a good introduction to the physics of higher rank Lie algebras. There is lots of other work by Iachello on the interacting boson model, dynamical symmetries, chains of Lie algebras, and other related stuff. They cite other relevant work in the area.

fresh_42
I observed a certain behavior of maximal eigenvectors in mathematics, not in physics, and wanted to understand, what makes the end of the ladder so special and how it is physically described.
I would say that the proton (and the neutron) is special. Define the operators $$E_{+} = |p \rangle \langle n|, \ \ \ \ E_{-} = E_{+}^{\dagger} = |n \rangle \langle p|,$$$$H = \frac{1}{2} \left(|p \rangle \langle p|- |n \rangle \langle n|\right) ,$$ where $|p\rangle$ and $|n \rangle$ are orthonormal basis in 2-dimensional Hilbert space. Clearly, these operators satisfy the Cartan form of $\mathfrak{su}(2)$: $$\big[ H , E_{\pm} \big] = \pm E_{\pm} , \ \ \ \big[ E_{+} , E_{-} \big] = 2H ,$$ and the root-space is one-dimensional. Also, $E_{+}|p\rangle = 0$, implying that the proton is the highest weight state with $H|p\rangle = \frac{1}{2}|p\rangle$.
The fact that one cannot go beyond the state $|p\rangle$, mathematically, means that the representation space is 2-dimensional. Physically, it means that there is no other particle in nature with the same iso-spin, hyper-charge (i.e., quarks content), and the Lorentz’s Casimirs $( m , J^{\pi})$ as those of the proton and the neutron (considering that the iso-spin group is a good symmetry).

If I consider combined roots as superposition of basic roots, then I see no reason why the ladder ends at all.
I don’t understand what you mean. For any semi-simple algebra, one can show that the roots form finite “strings”: $$\vec{\alpha} , \vec{\alpha} - \vec{\beta} , \vec{\alpha} - 2 \vec{\beta} , \cdots , \vec{\alpha} - 2\frac{\vec{\alpha} \cdot \vec{\beta}}{\vec{\beta} \cdot \vec{\beta}} \vec{\beta} .$$ The same argument shows that the weights also form finite strings: $$\vec{w} - n \vec{\alpha}, \vec{w} - (n-1)\vec{\alpha}, \cdots , \vec{w} , \vec{w} + \vec{\alpha}, \cdots , \vec{w} + 2 \frac{\vec{w} \cdot \vec{\alpha}}{\vec{\alpha} \cdot \vec{\alpha}} \vec{\alpha} ,$$ where $\alpha$ is any root and $$2 \frac{\vec{w} \cdot \vec{\alpha}}{\vec{\alpha} \cdot \vec{\alpha}} = \mbox{integer}.$$ In an irreducible representation there will be a number of Weyl orbits, each with a dominant weight. The most positive of the dominant weights is called the highest weight $j$ of the irreducible representation and satisfies $E_{\alpha}|j\rangle = 0, \ \alpha > 0$.

Truecrimson and dextercioby
What physical property does this finite dimension correspond to?
There is no “physical property”. However, one can give a physical meaning to certain n-dimensional irreducible representation, if one (for example) think of it as a multiplet of n particles having the same mass, spin and parity.

The thing I'm not crystal clear about lies in the emphasis on "highest weight" by more mathematically oriented authors. Ballentine doesn't even mention that phrase.
This is only because he works with $SU(2)$, the simplest spin group. Even with more sophisticated groups, one can use the tensor methods to deal with lower-dimensional irreducible representations. But imagine using the tensor methods to construct the states of the irreducible representations $[27] \mbox{and} [35]$ of $SU(3)$! The roots diagrams become very convenient when dealing with higher-dimensional irreducible representations. However, the most important aspect of the canonical Cartan form is the fact that it (the Cartan form) reduces, considerably, the number of the structure constants of the algebra. Namely from $\frac{1}{2}n^{2}(n-1)$ 3-index constants $f^{a}{}_{bc}$ to $\frac{1}{2}(n - l)(n - 1)$ 2-index constants $N_{\alpha \beta}$ and $\alpha^{(i)}$. In fact, the reduction is even greater because the $N_{\alpha \beta}$ is determined by the $\alpha^{(i)}$, and the $\alpha_{i}$ are determined in terms of $l$ fundamental eigenvalues.

Truecrimson and strangerep
The number of different values of the spin, minus one.
What number is that? Did you mean to write $2J + 1$? But this number is not a "physical property".

The ordering of weights depends on a choice that, to me, seems to be unrelated to physics.
In fact, when Heisenberg discovered iso-spin symmetry, he put the proton and the neutron in the “wrong” order.

Tendex
Mathematically it is forced by the finite dimension of the representation space.
What physical property does this finite dimension correspond to?
The number of different values of the spin, minus one.
What number is that? Did you mean to write $2J + 1$? But this number is not a "physical property".
No. The discussion was about the physical meaning of the dimension ##2j+1## of the irreducible representation with spin ##j##. Actually, the minus 1 was spurious (corrected).; the dimension is the number of different spins.

No. The discussion was about the physical meaning of the dimension ##2j+1## of the irreducible representation with spin ##j##. Actually, the minus 1 was spurious (corrected).; the dimension is the number of different spins.
I thought the discussions was about a general semi-simple algebra. The number $2j+1$ is particular to $\mathfrak{su}(2)$ where the irreducible representations are characterized by one (half) integer $j$. For the flavour symmetry group $SU(3)$, an irreducible representation of $\mathfrak{su}(3)$ is characterized by two numbers $(p,q)$, with $$\mbox{dim}(p,q) = \frac{1}{2}(p+1)(q+1)(p+q+2).$$ The physical meaning (even though the question used the phrase “physical property”) of the finite-dimensional $(p,q)$ is the following: Assuming that $SU(3)$ is a good symmetry, and for most of the irreducible rep's $(p,q)$, there are $\mbox{dim}(p,q)$ number of particles (Baryons and mesons) with the same (Lorentz Casimirs) $(m,J^{\pi})$ in the irreducible representation $(p,q)$.

fresh_42
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I thought the discussions was about a general semi-simple algebra.
Yes. My question was rather simple. (Level "I" to allow an answer at all, and not level "A" since I didn't want to talk about Hamiltonians.) Which physical properties corresponds to highest roots and maximal weights of irreducible, finite dimensional representations of simple Lie algebras?

So far I saw: maximal spin for ##(SU(2),\operatorname{ad})##, maximal hypercharge (?), electric charge for ##(SU(3),\operatorname{ad})##, and a lot of trivia about root systems in general, which is pure mathematics and off topics here. But already mass - in case it joins the list - makes me wonder, as I thought that mass could always be increased arbitrarily.

Then what about representations which are not the adjoint representation, and which although finite dimensional, can be of arbitrary dimension? What does their highest weights mean?

What about the ##(SU(5),\operatorname{ad})## as GUT candidate. All of a sudden maximal spin isn't a highest root anymore, and neither are the others. Instead we have new end points in the root system. Which physical property corresponds to those roots?

Yes. My question was rather simple. (Level "I" to allow an answer at all, and not level "A" since I didn't want to talk about Hamiltonians.) Which physical properties corresponds to highest roots and maximal weights of irreducible, finite dimensional representations of simple Lie algebras?

So far I saw: maximal spin for ##(SU(2),\operatorname{ad})##, maximal hypercharge (?), electric charge for ##(SU(3),\operatorname{ad})##, and a lot of trivia about root systems in general, which is pure mathematics and off topics here. But already mass - in case it joins the list - makes me wonder, as I thought that mass could always be increased arbitrarily.
Well there is no shift operator for the mass. The Hermitian irreducible representations of the Poincare algebra are infinite dimensional.

That is not a problem: Since the adjoint representation $A(x) = e^{T(x)}$, $T(x)^{a}_{b} = x^{c}f^{a}_{cb},$ of the group is faithful modulo the centre $Z$, the adjoint representation $T(x)$ of the algebra is faithful modulo the centre $c$ of the algebra. So, if the group centre is discrete, the centre of the algebra vanishes, and $T(x)$ is a faithful representation of the algebra. This is the case for semi-simple algebras. So, in this case, there is no loss of generality in using the adjoint representation for the construction of the Cartan form.

What about the ##(SU(5),\operatorname{ad})## as GUT candidate. All of a sudden maximal spin isn't a highest root anymore, and neither are the others. Instead we have new end points in the root system. Which physical property corresponds to those roots?
Well, yes. “symmetry” groups are used in different contexts. Who knows the actual symmetry of nature?

dextercioby and Tendex
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The Hermitian irreducible representations of the Poincare algebra are infinite dimensional.
What does Hermitian mean in this context. Say we have ##\varphi: \mathfrak{P}\longrightarrow \mathfrak{gl}(V)##, which is the condition on ##\varphi##?. There are certainly finite dimensional ##V## and irreducible ones among them. Which condition forces ##V## to be inifnite dimensional?

I don't want to be the curmudgeon of the form, but I think the discussion about the roots was needed. In fact I think it is still not resolved, except for the reference A. Neumaier gave, no one tried to clarify it.

@fresh_42 why don't you give us the definition of highest weight vector that you use, because I do think that there is a choice it depends on. And that choice is a matter of convention. Why do you think it doesn't matter?

Which condition forces ##V## to be inifnite dimensional?
The spectrum of the Casimir operator $P^{2} = m^{2}$ is continuous. The unitary representations of the Poincare group $U(a, \omega) = e^{i(a^{\mu}P_{\mu} - \frac{1}{2}\omega^{\mu\nu}M_{\mu\nu})}$, with Hermitian operators $(P_{\mu} , M_{\mu\nu})$ are all infinite-dimensional.

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I don't want to be the curmudgeon of the form, but I think the discussion about the roots was needed. In fact I think it is still not resolved, except for the reference A. Neumaier gave, no one tried to clarify it.

@fresh_42 why don't you give us the definition of highest weight vector that you use, because I do think that there is a choice it depends on. And that choice is a matter of convention. Why do you think it doesn't matter?
https://www.amazon.com/dp/0387900535/?tag=pfamazon01-20
page 72, section 13.4. Saturated sets of weights.

https://www.amazon.com/dp/0387900535/?tag=pfamazon01-20
page 72, section 13.4. Saturated sets of weights.
It says
We say that a saturated set ##\Pi## has a highest weight ##\lambda (\lambda\in\Lambda^+)## if ##\lambda\in\Pi## and ##\mu\prec\lambda## for all ##\mu\in\Pi##.

Both ##\prec## and ##\Lambda^+## (defined on pages 47 and 67) depend on the choice of a base of the root system.

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Both ##\prec## and ##\Lambda^+## (defined on pages 47 and 67) depend on the choice of a base of the root system.
See theorem 20.2 and its corollary. We wouldn't talk about weights if they were as arbitrary as you pretend they are. They all come from the Killing form or another trace form.

Of course we do not need isospin, charge and hypercharge as basis operators of ##SU(3)##. But why should we not? Maximality is given in any finite dimensional case. To ask what it physically means is in my opinion a natural question. Change of basis is a pure distraction and senseless. The question is the same in any basis system: Which physical quantities make the ladders end?