Why do we need Hermitian generators for observables in quantum mechanics?

[fresh_42]

fresh_42 submitted a new PF Insights post

How to Tell Operations, Operators, Functionals and Representations Apart

Continue reading the Original PF Insights Post.

 

[S.G. Janssens]

Fresh, thank you for this work, which I believe fits in your more general - and much welcomed - effort to familiarize interested physics students with the rigorous mathematical underpinnings of their field.

I feel qualified to comment only on the section titled "Linear and Non-Linear Operators, Functionals and Hilbert spaces". Here are some comments, mostly minor. Quotes from your text are in italic.

1. If we drop the requirement of a dot product, and only want to consider a complete real or complex vector space with a norm, then we call it Banach space. Another important example are Lebesgue spaces, which deal with measurable subsets of $R^n$.

I find writing another here a bit confusing. People unfamiliar with $L^p$-spaces might think that Lebesgue spaces are something else than just particular cases of Banach spaces ($1 \le p \le \infty$) or Hilbert spaces ($p = 2$).

2. Now a linear operator $L$ is simply a linear map between vector spaces: $L:V \to W$. The elements of $V$ on which $L$ is defined is the domain of $L$ and its image, i.e. the elements of $W$ which are hit by $L$ is the codomain or range of $L$.

Since at the end of this section you give examples of bounded as well as unbounded operators (and since these classes of operators require substantially different techniques for their analysis, as you of course know), I would have preferred to write $L : D(L) \subseteq V \to W$ where $D(L)$ is the domain of $L$, with $D(L) = V$ in case $L$ is bounded (e.g. the Volterra operator).

In this notation, for me the codomain of $L$ would be $W$ itself, while the range of $L$ would be $R(L) := LD(L)$.

3. In the non-linear case, operators are (non-linear) functions between normed (in general infinite dimensional) vector spaces (Banach spaces), and (non-linear) functionals are those which map to $\mathbb{R}$ or $\mathbb{C}$. Sometimes entire classes of operators are considered, e.g. compact operators, which map bounded sets to those whose closure is compact.

The occurrence of the last line in a paragraph on nonlinear operators unjustly suggests that compact operators are only considered in a nonlinear context.

4. I think if I were to be unfamiliar with functional analysis, this presentation is very brief. The text could benefit from some specific references that you like, preferably per topic of discussion. This way, it really becomes an initiation for the curious student.

Hopefully you find these comments more useful than irritating.

 

[fresh_42]

Thanks @Krylov for the careful reading.

ad 1) Changed phrasing a bit.
ad 2) Well, "codomain" was easy to correct, but the rest was a bit tricky in order to keep it simple and not to go into details. Not sure if it's better now though. It is always a big temptation to me to say more and more about certain subjects, historically as well as technically. All the doors I have knocked on and din't open hide so much mathematical beauty, that it's hard not to enter.
ad 3) Swapped to the next paragraph.
ad 4) I added some sources, but I admit that my library doesn't contain a lot of functional analysis, resp. in the wrong language. E.g. I have a lovely book by Courant and Hilbert (1924), Mathematical Methods of Physics I, but this is probably not a book to recommend nowadays. If you know of a "must read", I would appreciate to add it to the list.

So thank you, once more.

 

[strangerep]

functionals certain operators

Typo? Should it be "functionals are certain operators" ?

 

[fresh_42]

Typo? Should it be "functionals are certain operators" ?

A case of lost in translation. Being a list I thought a repetition of the verb wouldn't be necessary as long as it is still valid from the first part and until a new one is needed. However, I'm not sure how the English grammar deals with these cases and I applied mine. I'll add the repetition.

 

[strangerep]

[...] I'm not sure how the English grammar deals with these cases [...]

Heh, I'm not sure either -- although English is my first language.

To my ear, there's a difference between everyday English, and highfalutin English.

 

[UsableThought]

Typo? Should it be "functionals are certain operators" ?

Being a list I thought a repetition of the verb wouldn't be necessary as long as it is still valid from the first part and until a new one is needed.

Actually, @fresh_42, you're correct in your understanding of that particular convention for listing. However 1) it is a convention more common to UK and continental English than to U.S. English; and 2) it can become somewhat awkward if noun phrases are involved that require extra parsing by the reader. Also 3) listing of any sort is helped by what copyeditors call the "serial comma," which is a comma used in any list of three or more members; it comes after "and" and before the final member. This helps the reader understand that the last two members of the list aren't joined by that "and" into a separate category, but rather, are both members of the overall list. Thus rather than "X, Y and Z", which subtly implies that Y and Z are somewhat closer to each other than to X, better to write "X, Y, and Z".

So with all that in mind, you could indeed write your list like this -

Operators are functions, functionals certain operators, and functional analysis deals with operators on spaces where the elements are functions.​

- and it would just barely skate by as grammatically correct. However, "functionals certain operators" will likely cause many readers to hesitate and re-parse in order to recognize "certain operators" as a complete noun phrase; and even then some may not understand what you mean by "certain" since the word has potentially more than one meaning. Moreover, the last element in the list is a much longer phrase than the first two, which may cause additional hesitation and re-reading. So better in this case to go with @strangerep's suggestion.

Also, to introduce your list, better in this case to use a colon rather than a period in the preceding sentence - a very minor point, but helpful to alert the reader to what's coming; e.g.

Operators are subject to functional analysis. The terms are a bit confusing:​

I imagine that's more than enough pedantic copyediting for today . . .

 

[S.G. Janssens]

Hello Fresh, thank you for your reply, and you are welcome.

ad 4) I added some sources, but I admit that my library doesn't contain a lot of functional analysis, resp. in the wrong language. E.g. I have a lovely book by Courant and Hilbert (1924), Mathematical Methods of Physics I, but this is probably not a book to recommend nowadays. If you know of a "must read", I would appreciate to add it to the list.

I don't know if this would violate one rule or the other, but I personally would not object to foreign language references, as long as there are also references in English. It just so happens that some very good mathematics books are written in German, Russian, etc. and not all of them have been translated.

Speaking about this, a well-known German author whose works have been translated into English is Eberhard Zeidler. His Nonlinear Functional Analysis and its Applications four-part series is no less than encyclopedic, but individual chapters are remarkably readable. His two-part Applied Functional Analysis is a good introduction that contains linear as well as non-linear material.

There is the classical book by Reed and Simon, Functional Analysis. @micromass has its front cover on his shirt.

There are many more personal favorites, but these may be interesting specifically for physics students.

 

[fresh_42]

@UsableThought Thanks, for the trip to English grammar. Yes, the Oxford comma is a discussion on its own. In German the rule is a different one: Commas only in listings without conjunctions like and, or if the conjunction is the start of a complete sentence with subject and predicate. The and substitutes the comma if used as separator only.

@Krylov Thanks for the sources. I added them to the list, although Zeidler was a bit of work to do with his many volumes. On the other hand I found that especially Zeidler is a very valuable source. This way the list got a bit long, but in comparison to the importance of the subject, a recommendable list of textbooks should be a good thing. And if some decided to use the links I integrated to buy a book, then it's even good for us.

 

[Stephen Tashi]

Can we give a good mathematical answer to the frequently asked question "What is the physical significance of operators?" ? (e.g. https://www.physicsforums.com/threads/what-are-operators.919653/ )

The answers to that question in the physics section say to study more physics. Since we are in the general math section, perhaps we can take a more abstract view. We are faced with the mathematical problem of modelling a phenomena that is probabilistic, but whose probabilistic nature is, in some sense, deterministic. Does this lead to some natural choice of the appropriate mathematical tools?

 

[S.G. Janssens]

We are faced with the mathematical problem of modelling a phenomena that is probabilistic, but whose probabilistic nature is, in some sense, deterministic. Does this lead to some natural choice of the appropriate mathematical tools?

I would say: "yes". As soon as the state of a system (e.g. the wave function) is not determined by finitely many numbers, infinite-dimensional function spaces naturally occur. Since, unlike $\mathbb{R}^n$ or $\mathbb{C}^n$, these function spaces generally have non-equivalent norm topologies, the study of their (functional) analytic properties - as well as the properties of the functions (i.e. operators or functionals) defined on them - becomes non-trivial and linear algebra alone no longer suffices.

Indeed, in the infinite-dimensional state space in your example, states evolve deterministically.

 

[fresh_42]

Can we give a good mathematical answer to the frequently asked question "What is the physical significance of operators?" ? (e.g. https://www.physicsforums.com/threads/what-are-operators.919653/ )

I think the most obvious operator is the differential operator. It occurs naturally in physics, as an approximation of first order, as well as the description of change. More interesting is the historical development. As I see it, modern functional analysis originates in the calculus of variation, i.e. the need to solve eigenvalue and stability problems - both very physical. Dieudonné (History of Mathematics 1700-1900) writes:

Around 1880 mathematicians of various areas felt the need for a new analysis with functions of "infinitely many variables". The common idea of functions from one set to another, generalizing Dirichlet's definition of a function began to be melted with the set theoretic interpretation of mathematics, even if the clarity of Dedekind's definition was missed by some. Now if the elements of these sets were functions (in the common sense), curves or surfaces etc. one calls the functions [between those sets] operators, and functionals, if there values are in $\mathbb{R}$ or $\mathbb{C}$. A classic example for functionals is the integral $f \mapsto \int_a^b p(x)f(x)dx$ .

Courant and Hilbert (Methods of Mathematical Physics, Vol. I, 1924) do not use the term operator. They speak of functionfunctions (French: fonction de ligne ou fontionnelle) in their chapter "Fundamental Facts of Calculus of Variations".

By functionfunction a quantity or function is understood, which does not depend on a certain number of independent, in certain boundaries deliberate variables, but depend on the course of one or more in certain boundaries deliberate functions, which represent the independent variables. The simplest example is the length of a curve ...

Then they investigate a couple of examples:

• Length of a curve on a surface.
• Lighttime.
• Area of a rotational surface.
• Potential energy of an elastic stick.
• Solution of an under determined differential equation.
• Potential energy of a uniformly spanned, elastic membrane.
• Potential energy of an elastic board.

 

[TeethWhitener]

This might be the wrong place to post this, but I was wondering about your inner automorphism example. In defining a representation $\varphi$ (equation 5), you specifically look at the action of a group $G$ on a vector space $V$. But then in the inner automorphism example, you have the group action of $G$ on itself: $G\times G \rightarrow G$, with the representation $Ad: G \rightarrow GL(\mathfrak{g})$. But given the initial definition of a representation, shouldn't this be $Ad: G \rightarrow GL(G)$? Why do we get to substitute the Lie algebra $\mathfrak{g}$ for the Lie group $G$? Or alternatively, why doesn't the group $G$ act on the vector space $\mathfrak{g}$?

 

[fresh_42]

This might be the wrong place to post this, but I was wondering about your inner automorphism example. In defining a representation $\varphi$ (equation 5), you specifically look at the action of a group $G$ on a vector space $V$. But then in the inner automorphism example, you have the group action of $G$ on itself: $G\times G \rightarrow G$, with the representation $Ad: G \rightarrow GL(\mathfrak{g})$. But given the initial definition of a representation, shouldn't this be $Ad: G \rightarrow GL(G)$? Why do we get to substitute the Lie algebra $\mathfrak{g}$ for the Lie group $G$? Or alternatively, why doesn't the group $G$ act on the vector space $\mathfrak{g}$?

To be able to speak of group action, all we need is - of course - a group $G$ and a set $X$ where the group acts on, resp. operates on. This is a function $G \times X \longrightarrow X$. From here there can be various examples, depending on the nature of $X$. As we have a group operation, $X=G$ is a natural example. So $(g,x) \mapsto L_g(x) = g\cdot x$ or $(g,x) \mapsto \iota_g(x) = g\cdot x \cdot g^{-1}$ are two possible actions of a group on itself. The latter defines an (inner) automorphism $x \mapsto gxg^{-1}$ for every $g$, i.e. we have a function $G \longrightarrow \operatorname{Aut}(G)$. This can always be done.

Now if $X=V$ is a vector space, we have another situation, because a group is no vector space. (And usually the groups under consideration are not the additive groups which build the vector space.) First the operation $(g,v) \mapsto g.v \in V$ has to be defined. For an operation it is required that $1.v=v$ and $(g\cdot h).v = g.(h.v))$ hold. These properties are the same as to say $g \mapsto (v \mapsto g.v)$ is a representation (of the group $G$ on the vector space $V$) or $G$ operates on $V$ or $G \longrightarrow GL(V)$ is a group homomorphism: three languages for the same thing. If we have a matrix group (like orthogonal or Hermitian transformations) of a vector space $V$ we get a representation in a natural way: matrix times vector = vector.

As to your question about the adjoint represenation. First of all: $GL(G)$ doesn't make sense. The "L" stands for linear and "G" doesn't carry a linear structure. We can say $\operatorname{Inn}(G)$ for inner automorphisms, or $\operatorname{Aut}(G)$ for simply automorphisms. In the case of conjugations like above we have $G \longrightarrow \operatorname{Inn}(G) \subseteq \operatorname{Aut}(G)$ by the map $g \mapsto \iota_g$.

Let's now go back to a matrix group $G$ which simply means $G \subseteq GL(V)$ for some vector space $V$, i.e. $G$ consists of some kind of regular (here finite dimensional) matrices which apply to vectors from $V$. As above we have an operation $g \mapsto \iota_g$ where $\iota_g(x)=gxg^{-1}$ is the normal matrix multiplication of matrices $g,x,g^{-1}$. In the language of group homomorphisms this means $G \longrightarrow \operatorname{Aut}(G) \subseteq \operatorname{Aut}(GL(V))$.

If $G$ is a matrix group with Lie algebra $\mathfrak{g}$, we have the two things given needed: a group $G$ and a vector space $\mathfrak{g}$ on which the group acts. The action, resp. operation is given by $(g,X) \mapsto gXg^{-1}$ for $g\in G\; , \;X\in \mathfrak{g}\,.$ In cases where $G$ and $\mathfrak{g}$ are given, it is usually easy to see, that this defines a representation. The general case ($G$ an analytic group and $\mathfrak{g}$ its left-invariant vector fields) needs a little bit more care (cp. Varadarajan [11]). This operation is induced by the conjugation in $G$, that is the group conjugations $\iota_g$ can be extended to a conjugation of the vector fields and is called adjoint representationt of the Lie group $G$.

... you have the group action of $G$ on itself: $G\times G \rightarrow G$, with the representation $Ad: G \rightarrow GL(\mathfrak{g})$

isn't precisely true. I have a group action (called conjugation, $\iota\, : \, G \times G\longrightarrow G$ ) on itself (no vector space), and I have a group action (called adjoint, $\operatorname{Ad}\, : \,G \times \mathfrak{g}\longrightarrow \mathfrak{g}$) on its tangent space (vector space, resp. representation space). These are two different operations, and only the second is called a representation, although it is induced by the first, but it's another operation.

Although we have matrices here, i.e. linear transformations of a vector space $V$, these vectors don't play any role so far. And of course this is confusing, as $\mathfrak{g}$ are also matrices of the same dimension, which also apply to vectors of $V$. However, this isn't even the end here, as $\mathfrak{g}$ also operates on the vector space $\mathfrak{g}$ by Lie left multiplication of the Lie algebra $\mathfrak{g}$, i.e.
$$
\mathfrak{g} \times \mathfrak{g} \longrightarrow \mathfrak{g}\\
(X,Y) \longmapsto \operatorname{ad}(X)(Y) = [X,Y]
$$
is a Lie algebra homomorphism, which is also called adjoint representation, as $\operatorname{ad}X$ are inner derivations of $\mathfrak{g}.$

All three are connected by

• Matrix group operation: conjugation induces ...
• Lie group operation: adjoint $\operatorname{Ad}$ connected via the exponential map to ...
• Lie algebra operation: adjoint $\operatorname{ad}$

I have tried to explain this mess earlier here:
https://www.physicsforums.com/insights/representations-precision-important/

 

[TeethWhitener]

Thanks for your response. There's a lot to unpack here. I'm sure I'll have other questions.

 

[fresh_42]

Thanks for your response. There's a lot to unpack here. I'm sure I'll have other questions.

Yes, we haven't talked about the representations of the Lie algebras. E.g. $\mathfrak{g} = \mathfrak{su}(2) \cong \mathfrak{sl}(2)$ for which all possible representations on vector spaces are known. Here we get a Lie algebra homomorphism $\mathfrak{sl}(2) \longrightarrow \mathfrak{gl}(V)$ with a finite dimensional vector space $V$ of some dimension. The adjoint case $\operatorname{ad}$ with $V = \mathfrak{g}$ is only one example.

But perhaps you can tell me what is meant, if physicist speak of the adjoint representation: the one of the group, $\operatorname{Ad}\,,$ or the one of the Lie algebra, $\operatorname{ad}\,?$ And what are those generators? Are they basis vectors of weight spaces, basis vectors of a vector space $V$ where the group or the algebra acts on (which?), are they generating orbits, or are they the the group elements by which the conjugations are defined? Or are they the linear mappings $\operatorname{Ad}(g)$ or $\operatorname{ad}(X)$? Which one? I read it so often but didn't find a mathematical explanation, yet. What is it and where lives it, that is generated here? I really want to know. I know it is some animal in the zoo, but which one? And then there are those Pauli matrices $\sigma$, where actually $i \cdot \sigma$ is the tangent vector. Not that it makes a difference for the concept, but it is confusing.

 

[TeethWhitener]

I'm probably not the best person to answer this, because physicists do tend to get quite sloppy in their notation, and it's confusing to me as well. But I think typically you look for matrix representations of the infinitesimal generators of a Lie group (which I've always assumed must be elements of the corresponding Lie algebra). Since these are matrices, you can get a matrix representation of the corresponding Lie group simply by taking the matrix exponential (I'm pretty sure an additional requirement here is connectedness of the Lie group). As you can see, my knowledge of how all this hangs together is spotty at best.

 

[fresh_42]

I'm probably not the best person to answer this, because physicists do tend to get quite sloppy in their notation, and it's confusing to me as well. But I think typically you look for matrix representations of the infinitesimal generators of a Lie group (which I've always assumed must be elements of the corresponding Lie algebra). Since these are matrices, you can get a matrix representation of the corresponding Lie group simply by taking the matrix exponential (I'm pretty sure an additional requirement here is connectedness of the Lie group). As you can see, my knowledge of how all this hangs together is spotty at best.

This would mean, the physicists strike back. If mathematicians call operations: actions, representations or even homomorphisms, then physicists call their tangent vectors, (left invariant) vector fields, tangent bundles, covariant derivative or infinitesimal generators, depending on the corresponding point of view? Fair enough.

 

[TeethWhitener]

Ok, thanks so much for your insight. I have another stupid question: If I understand correctly, the adjoint representation of $g\in G$ is given by $gXg^{-1}$, where $X\in \mathfrak{g}$. Does this mean that each element $X\in \mathfrak{g}$ generates its own unique representation of the group $G$? In other words, if we change $X$, does that give us an entirely new representation of the group? Or is it the other way around: the action of a single element $g$ on all of the $X\in \mathfrak{g}$ gives a representation of $G$, so that each group element generates a unique representation of $G$?

 

[fresh_42]

Ok, thanks so much for your insight. I have another stupid question: If I understand correctly, the adjoint representation of $g\in G$ is given by $gXg^{-1}$, where $X\in \mathfrak{g}$. Does this mean that each element $X\in \mathfrak{g}$ generates its own unique representation of the group $G$? In other words, if we change $X$, does that give us an entirely new representation of the group? Or is it the other way around: the action of a single element $g$ on all of the $X\in \mathfrak{g}$ gives a representation of $G$, so that each group element generates a unique representation of $G$?

No need to consider single elements here. It is all in all only one representation, the adjoint representation of the group $G$. This means

$\begin{align*} G &\longrightarrow GL(\mathfrak{g}) \\ g &\longmapsto (X \mapsto \operatorname{Ad}_g(X)=gXg^{-1})\\ &\\ &\text{ or written as action }\\ &\\ (G \times \mathfrak{g}) &\longrightarrow \mathfrak{g}\\ (g,X ) &\mapsto g.X := gXg^{-1} \end{align*}$

One says $\operatorname{Ad}$ is the adjoint representation of the group $G$ on $\mathfrak{g}$.
This means, a single group element $g\in G$ is represented by the single map $\operatorname{Ad}_g$ which is an action on $\mathfrak{g}$. But this is only one element of the entire representation, which means all of $\{\,\operatorname{Ad}_g \,\vert \, g\in G\,\}$ build the representation $\operatorname{Ad}$ because it is the same rule for all elements.

And this is the point where this "infinitesimal generator" terminology loses me: is it $X \, , \,\operatorname{Ad}_g$ or $\operatorname{Ad}_g(X)\,?$ "Infinitesimal" indicates a tangent space, but $X \in \mathfrak{g}\, , \,\operatorname{Ad}_g \in GL(\mathfrak{g}) \subseteq \mathfrak{gl}(\mathfrak{g})$ and $\operatorname{Ad}_g(X) \in \mathfrak{g}$ so all of them are "infinitesimal". And if it's a tangent anyway, why do we have to use another name than vector or basis vector? And what is here generated? "Generator" indicates a kind of function, but $X$ and $\operatorname{Ad}_g(X)$ are tangent vectors, which can be viewed as directional derivative, and $\operatorname{Ad}_g$ is a function $\mathfrak{g} \rightarrow \mathfrak{g}$, so this "hint" is worthless.
And to complete the confusion, it is normally not told whether the adjoint representation of $G$ or the one of $\mathfrak{g}$ is meant.

If we consider a single element $g \in G$, then $\{\,\operatorname{Ad}_g(X) = gXg^{-1} \,\vert \, X \in \mathfrak{g}\,\}$ is the orbit of $g$ with respect to the adjoint representation or equivalently according to the adjoint operation ( of $G\,$) on $\mathfrak{g}$.
If the adjective "adjoint" is not mentioned, then we don't have $\mathfrak{g}$ as representation space (where the elements of $G$ act on) but an arbitrary, usually finite dimensional vector space $V$, i.e. $G \longrightarrow GL(V)$. Such a representation is called irreducible, if $V$ cannot be written as $V=U_1 \oplus U_2$ with $g.U_i \subseteq U_i$ for all $g \in G\,.$
In this case, however, care is requested. If $G$ is a matrix group, say $G \subseteq GL(\mathbb{R}^n)$ then we may have different operations, resp. representations here:

• the natural one on $\mathbb{R}^n$ given by matrix times vector $g_{ij}.(r^j)=g_{ij}r^j$
• again the conjugation on itself (matrix multiplication)
• or another arbitrary one, e.g. if in general representations are investigated, without further conditions. Then we have $G \longrightarrow GL(V)$ and thus matrices (from our matrix group $G\,$) which are represented by other matrices of a complete different dimension, namely by those of $GL(V)$. The case $V=\mathfrak{g}$ and $\operatorname{Ad}$ is an example for one of those.

 

[TeethWhitener]

Ok, I think it's slowly starting to come into focus for me. I just need to play around with it for a little while. Thanks for your help.

 

[Greg Bernhardt]

Great article Fresh!

 

[samalkhaiat]

...
And this is the point where this "infinitesimal generator" terminology loses me:

In physics, an infinitesimal generator (mostly of a symmetry group) is related to what mathematicians call a derivation or vector field. We always speak of infinitesimal generators of a group and its representations. So, in this sense $\mathfrak{g}$ is the infinitesimal generator of $G$, $\mbox{ad}$ is the infinitesimal generator of $\mbox{Ad}$, and more generally, for simply connected Lie group $G$, $\mbox{d}\pi$ is the infinitesimal generator of $G$ in the representation $\pi$.

The group of automorphisms ( $t \to \alpha_{t}$) of a $C^{*}$-algebra, $\mathcal{A}$, has an infinitesimal generator given by $$\delta (A) = \lim_{t \to 0} \frac{\alpha_{t}(A) - A}{t} ,$$ with natural domain $$D(\delta) =\{ \forall A \in \mathcal{A} \ | \ \lim_{t \to 0}t^{-1}(\alpha_{t}(A) - A) \ \mbox{exist} \} .$$

Clearly $\delta (A)$ is a *-derivation, i.e., linear operation which commutes with complex conjugation/adjoint, $\left(\delta (A) \right)^{*} = \delta ( A^{*})$, and satisfies the Leibnitz rule. Indeed, writing $$\frac{1}{t}\left( \alpha_{t}(AB) - AB\right) \equiv \frac{1}{t}\left( \alpha_{t}(A) - A\right) \cdot B + \alpha_{t}(A) \cdot \frac{1}{t} \left( \alpha_{t}(B) - B\right) ,$$ and taking the limit, we get $$\delta(AB) = \delta (A) \cdot B + A \cdot \delta (B) .$$

In QM, the algebra $\mathcal{A}$ is given by $B(\mathcal{H})$, the algebra of bounded operators on separable Hilbert space. In this case one can show that for every 1-parameter group of automorphisms of $B(\mathcal{H})$ (continuous in the weak operator topology), there exists a 1-parameter group of unitary operators $U(t) = e^{iHt}$ (continuous in the weak topology) such that $$\alpha_{t}(A) \equiv A(t) = e^{iHt} \ A \ e^{-iHt} .$$ Here, we have the so-called inner derivation $$\delta_{H}(A) = i[H , A],$$ with domain $$D(\delta_{H}) = \{ \forall A \in B(\mathcal{H}) \ | \ [H,A] \ \mbox{is dense} \} .$$ When the derivation $\delta$ is inner derivation $\delta_{X}$, we call $X$ the infinitesimal generator. The same extends to every representation $\rho$ of a $C^{*}$-algebra on the Hilbert $\mathcal{H}_{\rho}$, $$\rho(\delta_{H}(A)) = i [H_{\rho} , \rho (A)], \ \ \rho (A) \in D(\delta) .$$ In this case, we call $H_{\rho}$ the infinitesimal generator of the group of automorphisms in the representation $\rho$.

 

[fresh_42]

In physics, an infinitesimal generator (mostly of a symmetry group) is related to what mathematicians call a derivation or vector field. We always speak of infinitesimal generators of a group and its representations...

Thank you. It drove me nuts what exactly a generator is meant to be. On all occasions I have seen it here, nobody ever explained it and in 90% of the cases I got the impression, that the poster himself wasn't sure. I have a 400 pages thick book about Lie groups, but Varadarajan doesn't mention "generator" at all. Maybe in the context of one-parameter subgroups, not quite sure. I bookmarked your post so I can look it up for details next time I'll try to understand someone speaking of generators or quote it as reference here. But I think I've understood the point. I'm coming from the algebraic side of it, and in this contexts, they are simply vectors and what matters alone is, in which vector space, algebra, vector field or bundle. An information which is basically never given explicitly.

Maybe I'll also ask you about the $i$ which mathematically turns a homomorphism into an anti-homomorphism but nobody ever cares. I understand that it doesn't change representations qualitatively, nevertheless I find it a bit sloppy, e.g. when Pauli matrices are used.

 

[samalkhaiat]

Maybe I'll also ask you about the $i$ which mathematically turns a homomorphism into an anti-homomorphism but nobody ever cares. I understand that it doesn't change representations qualitatively, nevertheless I find it a bit sloppy, e.g. when Pauli matrices are used.

When $G$ is a (Lie) symmetry group of the physical system, the dynamical variables of the system transform in some unitary representation $\rho$ of $G$, and the “infinitesimal generators” in the representation $\rho$ are (either directly or indirectly via Noether theorem) represent the observables of the system, i.e., the real quantities (numbers) that we measure in the Lab. In quantum mechanics, observables are represented by Hermitian operators/matrices. So, we need those generators to be Hermitian not the anti-hermitian that are used in the mathematical literatures: In some coordinates $\alpha^{a}$ on the group manifold we may use the exponential map and write $$\rho(g) = e^{\alpha^{a} \left( d \rho (X_{a})\right)} \ .$$ But $d \rho (X_{a})$ is not good for us, so we instead define the hermitian generators by $$i \ d \rho (X_{a}) = J_{a} ,$$ and rewrite $$\rho (g) = e^{- i \alpha^{a} J_{a}} \ .$$ In field theory, Noether theorem dresses these $J_{a}$’s by the dynamical pair $(\varphi^{r} , \pi_{r})$, giving us the so-called Noether charges. These are Lorentz scalar and time-independent (i.e., conserved) objects given by $$Q_{a} = \int_{\mathbb{R}^{3}} d^{3}x \ \pi_{r} (x) J_{a} \varphi^{r}(x) \ .$$ In this case, and to annoy you more, we actually call the Noether charges the generators of the symmetry group of the transformations in the unitary representation $\rho (G)$. The reason for term infinitesimal generators is natural in this case because these $Q_{a}$’s do actually generate the correct infinitesimal transformations on the dynamical variables $$\delta_{a}\varphi^{r} (x) = [i Q_{a} , \varphi^{r}(x)] ,$$ and satisfy the Lie algebra of the group $G$ $$[Q_{a} , Q_{b}] = i C_{ab}{}^{c} Q_{c} \ .$$