# Physical meaning of the highest root / weight

• I
Mentor

## Summary:

Simple Lie groups and their algebras in QM, physics of the root system

## Main Question or Discussion Point

As some simple Lie groups and their algebras are essential for our current understanding of QM, I wondered if especially the highest positive (or likewise lowest negative) root can be explained physically. The roots are the weights of the adjoint representation. Are their physical meanings behind other representations, too?

## Answers and Replies

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martinbn
Summary:: Simple Lie groups and their algebras in QM, physics of the root system

As some simple Lie groups and their algebras are essential for our current understanding of QM, I wondered if especially the highest positive (or likewise lowest negative) root can be explained physically. The roots are the weights of the adjoint representation. Are their physical meanings behind other representations, too?
What is the physical meaning of the representation itself?

Truecrimson
samalkhaiat
Summary:: Simple Lie groups and their algebras in QM, physics of the root system

As some simple Lie groups and their algebras are essential for our current understanding of QM, I wondered if especially the highest positive (or likewise lowest negative) root can be explained physically. The roots are the weights of the adjoint representation. Are their physical meanings behind other representations, too?
Look up the roots diagrams for $SU(3)$ in particle physics literatures. Or see

dextercioby
samalkhaiat
What is the physical meaning of the representation itself?
Representation is a pair $(V , \pi )$ consisting of a vector space $V$ and a homomorphism $\pi : G \to \mbox{GL}(V)$. The mathematical objects (states and/or operators) which describe the physical systems (particles and/or fields) are elements of $V$ and the homomorphism $\pi$ determines the properties of physical systems such as spins, iso-spins, colours etc.

In physics, we distinguish between the above defined representations and a (often non-linear) realization of the group $G$ on some manifold $M$. The latter is described by the group action $\varphi : G \times M \to M$, $(g,m) \mapsto \varphi (g , m)$ which satisfies $$\varphi (e , m) = m ,$$$$\varphi (g_{1}g_{2} , m) = \varphi \left( g_{1} , \varphi (g_{2} , m) \right) .$$ This is the case when the Lie group $G$ breaks down spontaneously to one of its Lie subgroup $H$ resulting in the appearance of Goldstone bosons (as local coordinates on the space $M = G/H$) which transform non-linearly under $G$

dextercioby, vanhees71 and fresh_42
strangerep
What is the physical meaning of the representation itself?
Here's my (far more low-brow) explanation...

Distinct representations of a (physically relevant) symmetry group correspond to distinct types of physical entities e.g., according to their spin, mass, and other quantum numbers.

Currently we know of several physically relevant symmetry groups, each of whose representations characterize a particular (sub)class of types of physical entities.

Many people have tried, and keep trying (so far unsuccessfully), to find suitably larger symmetry groups (or indeed more general mathematical symmetry concepts) which could subsume the currently known symmetry groups.

vanhees71
martinbn
You are explaining how representations are used in physics, is that what is meant by their physical meaning? My question had the intention to prompt the OP to explain what he means by physical meaning.

strangerep
My question had the intention to prompt the OP to explain what he means by physical meaning.
Oh, sorry. That wasn't obvious to me. I'll try to resist such explanatory urges in future.

martinbn
Oh, sorry. That wasn't obvious to me. I'll try to resist such explanatory urges in future.
The explanations will always be useful to someone.

Spinnor
Mentor
Oh, sorry. That wasn't obvious to me. I'll try to resist such explanatory urges in future.
Neither was it to me. I had a real question looking for an actual answer. If I had wanted to read a textbook I would have done so.

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. If I consider combined roots as superposition of basic roots, then I see no reason why the ladder ends at all. So something in my naive understanding is fundamentally wrong, hence I was looking for a correction. To ask me anything instead doesn't make sense. I am already insecure.

This isn't the homework section and a game of questions and counter-questions is ridiculous. I didn't answer because I wanted to test how useful or useless PF really is for people who look for answers. It was revelatory.

The more I appreciated your answers. Thank you.

martinbn
@fresh_42 I just wanted more detail in your question. If it is so inappropriate ask for my posts to be deleted. In any case the question is not clear to me. Do you have an example from physics? May be that will make it easier to understand.

strangerep
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. If I consider combined roots as superposition of basic roots, then I see no reason why the ladder ends at all. So something in my naive understanding is fundamentally wrong, hence I was looking for a correction. [...]
Hmm. In that case, I'll risk asking a (genuine) related question of my own in the hope that further clarifications might emerge.

I think I understand the quantum angular spectrum reasonably well, best learned from Ballentine sect 7.1, imho. In his treatment, positivity of the (abstract) Hilbert space inner product shows quickly that the $J^2$ eigenvalue $\beta$ is greater than or equal to $m^2$, where $m$ is the eigenvalue of the $J_z$ operator. A little playing with the ladder operators $J_x \pm i J_y$ shows that $m$ must take discrete values, whose range is constrained by $\beta$. That, in turn, determines the allowed values of $\beta$ (aka $j(j+1)$), thus classifying the (unitary) irreducible representations.

So far, so good. In merely a couple of pages, he's derived some extremely important features of theoretical QM (unlike, e.g., in Peter Woit's QM book where he seems to take vastly longer to reach the same conclusions about the QM angular momentum spectrum -- which makes me wonder if I'm missing something important).

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. I get that, in the so(3) case, the highest value of $m$ in a given representation is the $j$ value of that representation, and values of $j$ classify the unireps.

I also understand (I think) that the term "weight" means a "generalized eigenvalue", in the sense that, e.g., $$Hv ~=~ \lambda(H) v ~,$$where $H$ is an element of the Cartan subalgebra, with $v\in V$ a vector in the representation carrier space, and $\lambda$ is an eigenvalue specific to $H$. So I understand "highest weight" to mean "highest eigenvalue" for a particular operator in a particular representation. But is there really anything more to (the essence of) the "weight" concept than that? (For physics, that is.)

martinbn
Hmm. In that case, I'll risk asking a (genuine) related question of my own in the hope that further clarifications might emerge.

I think I understand the quantum angular spectrum reasonably well, best learned from Ballentine sect 7.1, imho. In his treatment, positivity of the (abstract) Hilbert space inner product shows quickly that the $J^2$ eigenvalue $\beta$ is greater than or equal to $m^2$, where $m$ is the eigenvalue of the $J_z$ operator. A little playing with the ladder operators $J_x \pm i J_y$ shows that $m$ must take discrete values, whose range is constrained by $\beta$. That, in turn, determines the allowed values of $\beta$ (aka $j(j+1)$), thus classifying the (unitary) irreducible representations.

So far, so good. In merely a couple of pages, he's derived some extremely important features of theoretical QM (unlike, e.g., in Peter Woit's QM book where he seems to take vastly longer to reach the same conclusions about the QM angular momentum spectrum -- which makes me wonder if I'm missing something important).

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. I get that, in the so(3) case, the highest value of $m$ in a given representation is the $j$ value of that representation, and values of $j$ classify the unireps.

I also understand (I think) that the term "weight" means a "generalized eigenvalue", in the sense that, e.g., $$Hv ~=~ \lambda(H) v ~,$$where $H$ is an element of the Cartan subalgebra, with $v\in V$ a vector in the representation carrier space, and $\lambda$ is an eigenvalue specific to $H$. So I understand "highest weight" to mean "highest eigenvalue" for a particular operator in a particular representation. But is there really anything more to (the essence of) the "weight" concept than that? (For physics, that is.)
I dont think there is anything more to it. Also note that heighest is dependent on a choice of positive roots. How can that have physical meaning!

A. Neumaier
2019 Award
highest is dependent on a choice of positive roots. How can that have physical meaning!
It is meaningful when the rotation symmetry is broken, as in a constant magnetic field in z-direction.

vanhees71
strangerep
[...] Also note that heighest is dependent on a choice of positive roots. How can that have physical meaning!
I don't understand this remark in the context of quantum angular momentum. There (iiuc) the highest weights correspond to eigenvalues of the operator $J^2$ which are unambiguously non-negative, and therefore physically meaningful (since the squared magnitude of angular momentum cannot sensibly be negative).

Or am I still missing something?

vanhees71
strangerep
It is meaningful when the rotation symmetry is broken, as in a constant magnetic field in z-direction.
Ah, good -- I was hoping you'd visit this thread.

But, alas, although I understand that a constant magnetic field breaks rotation symmetry, I don't understand at all how/why this relates to whether or not a "choice of positive roots" is physically meaningful in the unbroken case. Could you pls explain further?

martinbn
I don't understand this remark in the context of quantum angular momentum. There (iiuc) the highest weights correspond to eigenvalues of the operator $J^2$ which are unambiguously non-negative, and therefore physically meaningful (since the squared magnitude of angular momentum cannot sensibly be negative).

Or am I still missing something?
No, it was about represetations, not angular momentum.

vanhees71
Mentor
I think I understand the quantum angular spectrum reasonably well, best learned from Ballentine sect 7.1, imho. In his treatment, positivity of the (abstract) Hilbert space inner product shows quickly that the $J^2$ eigenvalue $\beta$ is greater than or equal to $m^2$, where $m$ is the eigenvalue of the $J_z$ operator. A little playing with the ladder operators $J_x \pm i J_y$ shows that $m$ must take discrete values, whose range is constrained by $\beta$. That, in turn, determines the allowed values of $\beta$ (aka $j(j+1)$), thus classifying the (unitary) irreducible representations.
Here is the point I try to understand. Where do those restrictions come from? What do they mean? Why can't we just add more and more energy to an experiment and observe ever higher excitations?
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. I get that, in the so(3) case, the highest value of $m$ in a given representation is the $j$ value of that representation, and values of $j$ classify the unireps.
Highest weight is just the end of the ladder in an arbitrary irrep, highest or maximal root in case of the adjoint representation. Mathematically it is forced by the finite dimension of the representation space.
I also understand (I think) that the term "weight" means a "generalized eigenvalue", in the sense that, e.g., $$Hv ~=~ \lambda(H) v ~,$$where $H$ is an element of the Cartan subalgebra, with $v\in V$ a vector in the representation carrier space, and $\lambda$ is an eigenvalue specific to $H$.
Yes. Simultaneous to all $H$.
So I understand "highest weight" to mean "highest eigenvalue" for a particular operator in a particular representation. But is there really anything more to (the essence of) the "weight" concept than that? (For physics, that is.)
Yes, it is the point at which certain operators push the vectors from the ladder. This has to have a physical description.

PeterDonis
Mentor
2019 Award
Mathematically it is forced by the finite dimension of the representation space.
What physical property does this finite dimension correspond to?

PeterDonis
Mentor
2019 Award
Why can't we just add more and more energy to an experiment and observe ever higher excitations?
In the case of spin, the higher rungs on the ladder don't correspond to higher energies; all of the eigenstates are degenerate (at least, they are in the absence of an external field). So moving up and down the ladder does not correspond, physically, to moving to higher or lower energies.

dextercioby
A. Neumaier
2019 Award
although I understand that a constant magnetic field breaks rotation symmetry, I don't understand at all how/why this relates to whether or not a "choice of positive roots" is physically meaningful in the unbroken case. Could you pls explain further?
The eigenvalues of $J^2$ determine the different irreducible representations (up to isomorphisms). The positive roots are the end points of the ladders in each particular of these representations. The ladder depends on picking a direction in 3-space., though the result (the finite number of rungs, and hence the dimension of the representation space) is independent of it.

What physical property does this finite dimension correspond to?
The number of different values of the spin.

it is the point at which certain operators push the vectors from the ladder. This has to have a physical description.
It determines the extremal possible values of the spin in a given irrep.

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strangerep
Mentor
The number of different values of the spin, minus one.
But isn't this a circular reasoning? It is finite because spins are, and spins are finite because the irrep is.

martinbn
But isn't this a circular reasoning? It is finite because spins are, and spins are finite because the irrep is.
For compact groups irr.reps. are finite dinesional.

A. Neumaier
2019 Award
But isn't this a circular reasoning? It is finite because spins are, and spins are finite because the irrep is.
My answer was a consequence of what was already reasoned, and gave the physical significance you asked for.
The reasoning is not circular if done correctly: The fact that there is a bound on the eigenvalues forces the ladder to be finite, hence the existence of a highest root. Using this, one finds the possible values of the Casimir $J^2$. From this follow the possible dimensions of the irreps.

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PeterDonis
Mentor
2019 Award
spins are finite because the irrep is
No, spins are finite because of the physical fact that there is a limit to how much spin angular momentum you can add to a particular quantum system. Ladder operators in the spin case correspond to adding or subtracting a unit of spin angular momentum. Using finite dimensional irreps is a reflection of that physical fact, not a cause of it.

strangerep
I think I understand the quantum angular spectrum reasonably well, best learned from Ballentine sect 7.1, imho. In his treatment, positivity of the (abstract) Hilbert space inner product shows quickly that the $J^2$ eigenvalue $\beta$ is greater than or equal to $m^2$, where $m$ is the eigenvalue of the $J_z$ operator. A little playing with the ladder operators $J_x \pm i J_y$ shows that $m$ must take discrete values, whose range is constrained by $\beta$. That, in turn, determines the allowed values of $\beta$ (aka $j(j+1)$), thus classifying the (unitary) irreducible representations.
Here is the point I try to understand. Where do those restrictions come from? What do they mean? Why can't we just add more and more energy to an experiment and observe ever higher excitations?
(I'm not sure whether this question has already been answered sufficiently, so I'll just say a few more things.)

I was talking here in the context of quantum angular momentum and nothing else. I.e., we consider only the group $SO(3)$ acting on Hilbert space. There is no Hamiltonian so we aren't talking about energy.

We start with the usual $so(3)$ (or $su(2)$) generators $J_x, J_y, J_z$ and assume they, and their UEA (universal enveloping algebra), act as self-adjoint operators on a Hilbert space. The task is then to classify the set of all such (non-isomorphic) Hilbert spaces compatible with this assumption, and how the $J$'s act on them.

We start with a maximal set of mutually commuting operators, i.e., $J^2 := J_x^2 + J_y^2 + J_z^2$, and (say) $J_z$. Since $J^2$ and $J_z$ commute, they share a common set of eigenvectors, denoted $|\beta,m\rangle$, where $$J^2 |\beta,m\rangle ~=~ \beta |\beta,m\rangle ~;~~~~~~ \mbox{and}~~~ J_z |\beta,m\rangle ~=~ m|\beta,m\rangle ~.$$(I'm using units such that $\hbar=1$, for brevity.)

From the definition of $J^2$ we have $$\langle \beta m|J^2 |\beta,m\rangle ~=~ \langle \beta,m|J_x^2|\beta,m\rangle ~+~ \langle \beta,m|J_y^2|\beta,m\rangle~+~ \langle \beta,m|J_z^2|\beta,m\rangle ~.$$ Since all the $J$'s are self-adjoint, we have, e.g., $$\langle \beta,m|J_x^2|\beta,m\rangle ~=~ \Big( \langle \beta,m| J_x^\dagger \Big) \; \Big(J_x |\beta,m\rangle\Big) ~\equiv~ \Big\|J_x |\beta,m\rangle\Big\|^2 ~\ge~ 0$$because the inner product of a Hilbert space vector with itself cannot be negative. Therefore $\beta \ge m^2$.

With a little more work, we can partition the set of all Hilbert spaces that carry $so(3)$ unirreps into distinct subspaces based on the values of $\beta$. This is so because there is no operator in the current scenario which fails to commute with $J^2$. I.e., for this class of physical situations, there is no operator that can map between eigenvectors with different values of $\beta$. Thus we have achieved the 1st step in our desired classification of the Hilbert spaces applicable to this scenario: each such space is associated with a different eigenvalue of $J^2$, and one cannot form physically sensible superpositions of eigenvectors having different values of $\beta$. (The physics terms for this are "superselection rule", or "superselection sectors".)

For the next step, we pick any one of these Hilbert spaces, i.e., pick a value of $\beta$, denote the space $H_\beta$, and then recognize that the eigenvectors of $J_z$ span $H_\beta$. Then analyze the action of $J_x,J_y$ on those eigenvectors to determine whether there is a finite or infinite set of them. A little investigation involving how the ladder operators $J_x \pm iJ_y$ act on $J_z$ eigenvectors, and remembering the earlier constraint $\beta\ge m^2$ shows that $H_\beta$ is finite dimensional. At some point, one realizes that it's more convenient to define a $j$ variable via $\beta=j(j+1)$, and we find $-j \le m \le +j$, which determines the dimension of $H_\beta$ (now renamed as $H_j$).

For more sophisticated physical scenarios, e.g., the hydrogen atom, one also has, besides angular momentum, a Hamiltonian operator (corresponding to conserved energy), and an operator corresponding to the Laplace-Runge-Lenz vector (also conserved). The overall Lie algebra is more complicated, but eventually one can deduce the energy levels of the H-atom (and the allowable values of total angular momentum allowed at each energy) by essentially similar techniques as sketched above.

HTH.