I Physical meaning of the highest root / weight

[fresh_42]

TL;DR 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?

 

[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?

 

[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
https://www.physicsforums.com/threads/su-3-multiplet.860348/post-5399569

 

[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

 

[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.

 

[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.

 

[fresh_42]

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 don't 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]

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.

 

[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?

 

[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.

 

[fresh_42]

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]

Mathematically it is forced by the finite dimension of the representation space.

What physical property does this finite dimension correspond to?

 

[PeterDonis]

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.

 

[A. Neumaier]

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.

 

[fresh_42]

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]

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.

 

[PeterDonis]

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 anyone 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.

 

[martinbn]

I still don't understand why there could be a physical meaning to the highest weight? The ordering of weights depends on a choice that, to me, seems to be unrelated to physics. So how can it have physical meaning? It is almost like asking what is the physical meaning of the x axis.

 

[strangerep]

I still don't understand why there could be a physical meaning to the highest weight?

Returning to the $SO(3)$ example, in any given representation the (absolute value of the) highest eigenvalue of the $J_z$ operator relates to the value of total angular momentum (squared), i.e., the eigenvalue $j(j+1)$ of $J^2$. This eigenvalue distinguishes that particular representation from all others. Different eigenvalues of $J^2$ correspond to distinct representations of $SO(3)$, hence to distinct classes of physical systems, each distinguished by its fixed value of total angular momentum.

 

[A. Neumaier]

I still don't understand why there could be a physical meaning to the highest weight? The ordering of weights depends on a choice that, to me, seems to be unrelated to physics. So how can it have physical meaning? It is almost like asking what is the physical meaning of the x axis.

Yes, it is exactly like that. If you have a constant magnetic field in x direction, the x-axis has a physical meaning.

 

[martinbn]

Returning to the $SO(3)$ example, in any given representation the (absolute value of the) highest eigenvalue of the $J_z$ operator relates to the value of total angular momentum (squared), i.e., the eigenvalue $j(j+1)$ of $J^2$. This eigenvalue distinguishes that particular representation from all others. Different eigenvalues of $J^2$ correspond to distinct representations of $SO(3)$, hence to distinct classes of physical systems, each distinguished by its fixed value of total angular momentum.

I agree with all this. But my comment is about the choice that doesn't seem physical to me. If you have a group or an algebra, the set of roots and simple roots is fixed, but the set of positive ones requires a choice. The ordering of the weights in a given representation depends on that choice, and which is highest and which not depends on the choice as well. For example if you have two simple roots $\alpha$ and $\beta$, which are opposite of each other (i.e. $\alpha=-\beta$) you can choose either one to be the positive and the other the negative. The choice will determine what is your highest and lowest vectors in you representation. The choice is arbitrary and gives equivalent descriptions. In higher rank there is even more freedom of choice. So what is the physical meaning of that.

Yes, it is exactly like that. If you have a constant magnetic field in x direction, the x-axis has a physical meaning.

But is there a physical meaning to the positive direction of the axis? It is just a name. And in higher rank when you don't have a linear order, then what is the physical meaning of any of the orderings?

 

[Tendex]

But is there a physical meaning to the positive direction of the axis? It is just a name. And in higher rank when you don't have a linear order, then what is the physical meaning of any of the orderings?

Well, a particular convention, say of orientation in the x-axis example, is not physical but the need to choose one convention does appear to be a physical consistence requirement, this determines certain orderings, other conventions determine others.

 

[strangerep]

For example if you have two simple roots $\alpha$ and $\beta$, which are opposite of each other (i.e. $\alpha=-\beta$) you can choose either one to be the positive and the other the negative.

I interpreted "highest" as tacitly meaning "highest magnitude", i.e., "highest absolute value". At least, that makes more physical sense to me.

 

[A. Neumaier]

is there a physical meaning to the positive direction of the axis? It is just a name.

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

in higher rank when you don't have a linear order, then what is the physical meaning of any of the orderings?

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.

 

[Tendex]

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

The physical meaning lies on the consistent choice of what is labeled positive and negative spin, (or highest weight more generally for the OP question) not the difference between the labels themselves which are conventional. Otherwise measurements couldn't be meaningfully compared.

 

[A. Neumaier]

The physical meaning lies on the consistent choice of what is labeled positive and negative spin, (or highest weight more generally for the OP question) not the difference between the labels themselves which are conventional. Otherwise measurements couldn't be meaningfully compared.

If one has a constant magnetic field and the coordinate system is chosen such that the field vector points in the positive z-direction then there is a standard convention for how to label the spin components. This is not different from all other conventions which provide physical meaning to abstract objects.

 

[Tendex]

If one has a constant magnetic field and the coordinate system is chosen such that the field vector points in the positive z-direction then there is a standard convention for how to label the spin components. This is not different from all other conventions which provide physical meaning to abstract objects.

Yes, the consistent choice of a coordinate system aligned with the way a measurement is performed provides physical meaning by allowing to compare different observations and make predictions based on local approximations.

 

[martinbn]

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?

 

[A. Neumaier]

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.

 

[samalkhaiat]

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.

 

[samalkhaiat]

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.

 

[samalkhaiat]

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.

 

[samalkhaiat]

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".

 

[samalkhaiat]

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.

 

[A. Neumaier]

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.

 

[samalkhaiat]

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]

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?

 

[samalkhaiat]

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.

Then what about representations which are not the adjoint representation,

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?

 

[fresh_42]

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?

 

[martinbn]

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?

 

[samalkhaiat]

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.

 

[fresh_42]

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.