The Mathematics of Growth Mindset

@ACuriousMind the ultimate path i have in mind is: given a representation $\pi: G \to GL(\mathcal{H})$, $\pi(\gamma(t))\lvert \psi \rangle$ where $\lvert \psi \rangle \in \mathcal{H}$.

so i am thinking that it is the paths in the lie group + choice of representation that induce the paths in hilbert space

@ACuriousMind if representation theory is failing to live up to what it was advertised to do (for physical theories)

A lot of things fail in the infinite dimensional case

ncatlab.org/nlab/show/U%28%E2%84%8B%29

the horror

@SillyGoose What was it "advertised to do"? By whom?

@SillyGoose I don't understand what this is supposed to have to do with the Schrödinger equation

The solution to the Schrödinger equation is the time evolution operator $U : \mathbb{R} \to \mathrm{GL}(H)$. What is the generic $G$ in your notation supposed to be?

@ACuriousMind i was thinking: 1) what is a time-dependent state in QM. 2) it is a path in Hilbert space parameterized by $t$ satisfying a certain equation and initial condition; so, the abstraction of this situation is to have the parameterized image of some representation of a group act on some initial state.

@ACuriousMind in some physics context, $G$ would be the group generated by the algebra of operators for that context i believe.

@SillyGoose I don't understand why you think such a representation of some group would be the abstraction of path in Hilbert space. A path in Hilbert space is just a map $\mathbb{R}\to H$, where do representations enter here?

well the map is supposed (as in assumed) to be implemented via some family of maps $\phi_t: \mathcal{H} \to \mathcal{H}, \ t \in I$

and then well maybe just to add more structure, suppose that these maps live in the image of some representation

@SillyGoose "generated" in what sense? Again, this is infinite-dimensional setting, ordinary Lie theory does not apply

@SillyGoose we just call that $U(t)$ usually, but sure - that's just the time evolution operator

I still don't understand where this abstract $G$ or representations enter

hm well i guess i wanted to start with some mathematical ingredients of quantum mechanics and then see what can be done with them, broadly speaking.

and i wanted to consider the simplest sort of case. e.g. finite dimensional Hilbert space. so perhaps also a finite dimensional Lie group

and the ingredients that came to mind were a Hilbert space, a Lie group, and a representation of that Lie group over the Hilbert space

why a Lie group?

what is that group physically?

well it should contain "physical transformations"

so it's a (potential) symmetry group, okay

what does it have to do with the Schrödinger equation?

i assume a Lie group has some properties that make it amenable to modeling such physical transformations with

@SillyGoose you don't need to "assume" this - there is a physical meaning to the axioms of Lie groups (i.e. being a group and being smooth)

if you want to do mathematical physics you should always strive to understand why the mathematical structures we're looking at are the right ones, i.e. why are Lie groups so special compared to generic groups in modelling physical transformations?

well a particular inquiry was: given this data, can one come up with a finite number of criteria that selects "dynamic relations" (differential equations in terms of parameters and vectors of the Hilbert space)

@SillyGoose I don't know what that means

i have a very superficial notion of why Lie groups are perhaps more appropriate for physical transformations, but I do not have an understanding of the diffe g to perhaps attain an actual understanding at this moment

The Schrödinger equation in a finite-dimensional setting isn't even really a differential equation

sure, you may write it as $\partial_t \psi = \mathrm{i}H\psi$, but the solution $\psi(t) = \mathrm{e}^{\mathrm{i}Ht}\psi(0)$ to this is straightforward and the exponential easily performed in finite dimensions

the physics isn't in coming up with the Schrödinger equation, it's in choosing the physically correct $H$

@SillyGoose The argument is straightforward: Many physical transformations - like translations, and rotations - are parametrized by smooth (in the sense that you can take differentials with respect to them) parameters - e.g. the distance or the angle. Lie groups are precisely the mathematical structure you get when you make it axiomatic that all transformations are either like that or discrete (like reflection).

hm well i was reading a q recently posted here and it made me think of what motivates the Schrödinger equation. so that was the motivation going into this inquiry

the Schrödinger equation is just the Hilbert space version of the (wholly classical) statement that the Hamiltonian is the generator of time translation

hm but is that what they knew when they were writing all this down at the beginning of quantum theory?

depends on who you mean by "they" and by "knew" :P

but e.g. Dirac with his canonical quantization of replacing Poisson brackets by commutators certainly understood this

The people making quantum theory didn't even know what a group was

at least early on

i am not sure who wrote down or thought of the Schrödinger equation first

since the name of something doesn't always match the history of the thing

Depends what you call the Schrödinger equation I guess

Schrödinger and Heisenberg themselves? Probably not so much, but if you try to understand the genesis of old quantum theory through a mathematical lens, you're not going to be happy

i am curious of how one would come to the Schrödinger equation without thinking of in terms of generators of time translation

and it precisely turns out to be that object

@SillyGoose doesn't just about any intro to QM book contain some argument for that?

my "it's the generator of time translation" is usually unavailable in that context because (sadly) we don't expect people going into QM to know Hamiltonian mechanics

Early quantum theory was mostly like 1) What if there's a quantization constraint on mechanics (the Bohr-Sommerfeld rule) 2) What if things are waves

I don't recall either of the two QM books I used motivating the Schrödinger equation

A lot of early QM as wave stuff is what people learn in pop science

i dislike the wave stuff greatly xD

griffiths behavior

oh wait none of the three QM books I have read parts of motivated the Schrödinger equation; to my recollection

in any case, to underscore what Slereah's saying again: You will not understand the history of QM through the lens of modern mathematical frameworks. The modern concepts of groups, representations etc. only entered with Wigner and Weyl, and at first were much maligned as a plague

asking "How did they come up with that equation in the first place?" is a valid question, but the answer will not lie in talking about Lie groups and Hilbert spaces

Not really a specific thing too really, group theory was a pretty highly specialized thing back then

hm i guess i do not know when group theory became more "standard" material

This was before the trend of trying to assign abstract structures to all things in math

@SillyGoose Well it was because of QM in physics :p

And I guess relativity to some degree too

well i guess it is surprising that classical mechanics and quantum mechanics share this group theoretic structure

Lorentz group and all that

@SillyGoose which structure, exactly?

but it took a while to become a standard thing

a group alone isn't such a weird thing - it is more or less just the formalization of what a physical transformation needs to obey to be consistent

these notions of observables as elements of some algebra, which then generate some parameterized family of group element, which then transforms states of the system

I think calling it a group was a bad idea imo

The old term "Group of transformations" was a bit more clear

@SillyGoose The question is: How else could it work, really?

the power of Lie theory is that as soon as you talk about smoothly parametrized transformations, you are more or less already committed to ending up with the theory of Lie groups and algebras

@ACuriousMind Don't you also need time independence

So that $U_{s}U_{t} = U_{s + t}$

@Slereah sure, but in essence if your parametrization doesn't behave like that, what's the point of the parameter?

It's the clock

⏰

realistically, usefully, you're going to end up with some version of Lie theory like 99% of the time

@ACuriousMind if this was the case, shouldn't people have considered Lie theory for physics since the days of classical mechanics?

Lie wasn't even born

but i mean mathematical physicists probably existed back then too

Physicists didn't even exist

Just natural philosophers

@SillyGoose Again, you need to be careful - the modern idea of math and mathematical physics is much younger than you seem to assume

and the idea of a group didn't exist

hm

Like you don't really get the notion of a group properly until the 19th century, and even then it was a very niche topic

and Lie himself worked only at the end of the 19th century - not a lot of time for anyone to consider "classical mechanics and Lie theory" before quantum mechanics started

Some weird math field for Galois theory or whatever

i suppose perhaps it is during hilbert's age that the sort of "formal" mathematics we do now appeared?

It has been a process

hi

Like starting in the 19th century with the formalization of mathematics with Frege and al., Hilbert and company, the Bourbaki for the notion of structuralism, etc

we have today a tendency to believe that classical mechanics must have been worked out in its mathematical formalism before people started on QM, this is not true - the "mathematization" of physics towards the mathematical frameworks we use today really started only in the 20th century, when quantum theory was already known

Nobody wrote the way you'd read in a mathematical physics textbook in the early 20th century

@SillyGoose yes, but "Hilbert's age" was already the turn of the century, i.e. right before the discovery of QM!

there isn't really an age of "mathematical classical mechanics"

I mean there kind of was, but it was very different from today

Like this is what people thought was ALL OF MATH in the 19th century :

bleb

If you want to see how people tried to formalize classical mechanics in the 19th century it was shit like this

archive.org/details/principlesofmech00hertuoft

well in form it looks like a modern math textbook xD

Also there was the whole Positivism and Instrumentalism thing going on in physics during the early 20th century

Pretty influential to how physics was formalized

what will the hbar 100 years in the future say about the sort of physics done today

Well there's no lack of modern trends in physics :p

remains to see if it will stick though

what do physicists do when the representations of the symmetry group of the physical theory are not completely reducible?

in what situation would that happen?

all the groups you ever encounter are semi-simple, just like all functions are smooth

is there a reason that it would never happen

Isn't it the case for all Lie groups

Oh Heisenberg group isn't apparently

@SillyGoose I simply don't know of a single example where a non-semi-simple group would be relevant in this sense, so why would physicists have a procedure for that case? :P

what would be the physical meaning of a not completely reducible representation? it seems like this would mean a state belonging to the representation space just cannot be labeled with the quantity that would be associated to irreps of the group being represented

is there an experiment to test that a system really evolves via the Schrödinger equation?

i.e. given that the system has been certified to initially begin in some state $\psi$, that at every $t$ along some time interval, $\psi(t)$ evolves as predicted by the Schrödinger equation

I guess you need to check for unitary evolution

@SillyGoose essentially all quantum experiments are tests of the Schrödinger equation

So if a particle appears twice or disappears

Bad sign

if our understanding of the Schrödinger equation was wrong, we'd constantly get the wrong results, starting with the double-slit

hm

separately, is it necessary to have a zero vector to do quantum mechanics?

i.e., could we reproduce the theory with a modified Hilbert space $\mathcal{H} \backslash \{0\}$

@SillyGoose What then would an operator with eigenvalue 0 produce acting on its eigenstate?

I mean if you want you can use the projective Hilbert space, which doesn't have it

But the projective Hilbert space isn't a vector space

It would be quite nasty to deal with

i am okay with the resulting theory no longer using vector spaces

well wait then what do operators with eigenvalue $0$ send rays to in the projective version of things?

There aren't any

You just have operators on the projective states

there are no operators over projective space with eigenvalue $0$?

or what is there aren't any responding to

I mean what would it be

It doesn't have a zero

i agree with that. but then what does the corresponding linear algebra situation get mapped to in the projective situation

e.g. suppose we have a spin-1 system. in the lienar algebra situation, the $S_z$ operator has one zero eigenvalue. what happens to this operator in the projective situation

The Hilbert space and it's projectivization aren't isomorphic

well maybe i should just ask what a spin-1 system looks like in terms of projective terms

There's an injection of the Hilbert space in it's projectivization

you should already know the archetypal projective space, i.e. the Bloch sphere

And the same applies for its algebra of operators

Or for the simplest case

The projective Hilbert space of C is just a point

And it's only operator is the identity

i see it would be an injection, but i am wondering how the data that a certain state has spin-z value as $0$ is preserved through passing from hilbert space to projective hilbert space

Doesn't matter too much for QM since there are no physical states that are 0

i mean i could just ad hoc say that $0$ eigenvalue in the linear algebra translates into a non-zero element $a$ in the spectrum of the operator over projective space, or something to this effect, and just use this operational dictionary, but that doesn't seem okay

I don't understand what you're trying to do/what the point of this line of questioning is

hm but i mean the fact that a state is a "definite state with value $\lambda$" should be encoded somewhere

indeed we need not ever map a state into the zero vector

Quantum mechanics - with its operators and Hilbert spaces and so on - works. Why change it?

but shouldn't it work in the projective version as well

I don't really understand what the "projective version" is - yes, you can also talk about the projective Hilbert space, but that doesn't mean you forget all about the original space

no one claims that you can do QM solely by talking about the projective space

i certainly do not want to do quantum mechanics using this projective business, but I am just wondering what this $0$ eigenvalue turns into in the projective version of things

that's the thing - it's not another "version" of QM

it's just an aspect of it

in the context of the projective space we usually don't talk much about the observables, but about the unitaries generated by them

again, think of the Bloch sphere: We talk about how the rotations move the point on the sphere around, but usually not so much about the action of some random spin operator $S_i$ itself

but that seems strange because we look for projective representations and then pass into the algebra and then do etc. but at the outset we are dealing with projective representations of the algebra (of observables) then, aren't we? and the "etc." part is building a bridge between projective representations and normal representations of some other group over the non-projective Hilbert space.

i mean to say that one seems to talk about a sort of equivalence between the projective and linear picture of things even at the level of the algebras (at least in terms of their representations) for a case like $SO(3)$

there is no notion of a projective representation of an algebra

it's the groups that have projective representations

but if i make a bijection between projective representations of a group and representations of an algebra, these data then seem linked together

or like if i establish a way to get from one to the other and vice versa*

yes, because Lie groups and Lie algebras are linked

you have projective representations of some Lie group = linear representation of universal cover = linear representation of algebra

but then it seems like this link should allow me to start with data from one end (operator $A$ has a zero eigenvalue) and see what it turns into on the other end

I don't know what that means

you can construct the projective representation of the group from the linear representation of the algebra, via exponential maps etc.

it does not imply that any element of the algebra ends up directly acting on elements of projective space

how is the exponential map defined in that setting

@SillyGoose Hm? I mean just the usual exponential map in Lie theory that turns representations of the algebra into representations of the group

sorry linear operators can exist on non vector spaces too

but idk if addition exists on the projective space

i have only read Hall (which only looks at Lie matrix groups) :P so I don't know the formal definition of the exponential map in Lie theorey

well, then you should read a non-matrix text on Lie theory if that bothers you :P

but this issue is entirely disconnected from anything about projective spaces or physics

but then i don;t understand why formalize any notions of QM in projective spaces at all

it seems much more appropriate to define things within the existing construction (Hilbert space) rather than introduce (seemingly) useless names like "projective Hilbert space"

I'd wager that plenty of physicists go their entire live without ever encountering a projective space

it's just one of the ways to explain why we need to look at projective representations, i.e. why half-spin representations are "allowed" in QM even though we're still using the rotation group

well now it seems much better to simply speak of states as unit vectors in Hilbert space.

you can talk about the irrelevance of the global phase factor and hence define projective representations without ever explicitly mentioning the words "projective space"

but even that is explained by defining states as equivalence classes of vectors in Hilbert space

It is pretty rare to encounter a projective space in physics texts

And when you do they typically don't even say it explicitrly

We tend to prefer vector spaces because they are easier to handle

They're just big arrays of numbers that you can mostly manipulate component-wise

Much easier to do operations on them

but even in vector space quantum mechanics i don't see when one would ever need to compute $S_z \lvert 0 \rangle$.

@ACuriousMind well you mentioned this as a potential issue with removinng the zeror vector

@SillyGoose how else would you answer the question "what's the spin-z value of $\lvert 0\rangle$?"

It's a vector space, you're gonna get zero vectors once in a while

Like if you do a projection in a basis

Some elements are gonna be zero

hm but if all one ever eventually does is take expectation values or modulus squares, then one can carry around such quantities until then

and then it all reduces to scalar multiplication, a zero of which is well-defined

@ACuriousMind well i could still say it satisfies $S_z \lvert 0 \rangle = 0 \lvert 0 \rangle$, perhaps. the right hand side need not be defined to be something

The big secret is that most of physics is done with mathematical structures that are way too big for this purpose, but are easier to use than the bare ones structure

@SillyGoose how can you write down something that's not defined? What does it mean?

The shortest path between two truths is through a lie, as the saying goes

if you want to be mathematically rigorous, you can't just throw that overboard as it suits you

well the quantity is well-defined once i do a "genuine" calculation (expectation value or etc.)

so why does the intermediate step (which would not be done as an isolated computation) need to be well-defined

how would you know that $S_z\lvert 0\rangle = 0\lvert 0\rangle$ if the r.h.s. is ill-defined?

like, usually, I have something like $S_z$ as a matrix and then we just compute its application to the vector

but you're saying you think it's smarter to just pretend we know the answer without having any way to calculate it? :P

I really don't get what you're after here

note that an operator's range is the vector space... so we r destroying all the definitions by making $S_z |0\rangle$ undefined

well i don't think this is a smarter way to do things, the zero vector is convenienet as slereah said

it is just a curiosity

i feel like there should be a way to deduce the third eigenvector of a $3 \times 3$ nondegenerate hermitian matrix given its other two eigenvectors via checked orthogonality

like if i have the two eigenvectors, then I should be able to find a third eigenvector orthogonal to both of them

and this eigenvector has eigenvalue of the third eigenvalue

but i mean maybe some of the results that go into this logic go bad if you remove the vector space structure (via removing the zero vector)

Well for a start if you don't have a vector space you're not gonna have a matrix

Matrices are there because they're linear maps on a vector space

well what ever object $S_z: \mathbb{C}^3 \backslash \{0\} \to \mathbb{C}^3 \backslash \{0\}$ would be; "almost-linear"

how wud u define additive inverses without an additive identity

suddenly the negative of a vector has no meaning

consider the expectation value of diag[-1, 1] on the state [1,1]

well to clarify i don't think this is a nice or clean or good way to do quantum mechanics. i was just wondering if it was literally possible to do quantum mechanics having removed the zero vector from Hilbert space

@RyderRude the expectation value is well defined since the "inner products" are well defined

The coordinates on a projective space are the homogeneous coordinates, which are basically just the vector space version of it 😔

there are also the so-called "local octant coordinates" (via Geometry of Quantum States: An Introduction to Quantum Entanglement by Ingemar Bengtsson and Karol Życzkowski)

@SillyGoose yeah...bad example

also i mean physically the zero vector is not a well-defined object, so it is physically unmotivated to pass through these physically not well-defined objects in order to perform a computation convenienetly

without the 0 vector, we need to write things like -1v and 0v, because -v is meaningless and 0v cant be simplified further

let's see if this causes problems

check : what's v + 0v? assuming 1v=v is an axiom, we have v+0v=(1+0)v=v

this means 0v is now proved to be the zero vector

so we have to exclude 1v=v as an axiom

modifying vector spaces gets too ugly

i think u might need to just look up a projective space formulation of qm, which completely abandons the vector space instead of modifying it

How famous is this pseudoscientific theory of everything discussed here? Did you guys ever hear of this?

nope, not me

@Sanjana i got to know about it from that video

Apparently if you look at the projectivization functor, it only maps linear operators with a trivial kernel

from the inner product axioms, we hav $\langle v|v-1v\rangle=0$

this means v and v-1v are now orthogonal non zero vectors and they can b used as part of a basis. or rather, the whole notion of a basis probably gets screwed up

the ugliness spreads everywhere