@ACuriousMind Sec VIII in that article is interesting. Nice that they at least explicitly call out the relationship. I do agree that it feels like a sleight of hand fallacy when they say spinors defined as A are SOOOO hard to understand, spinors are actually defined as B!=A, look how easy they are to understand! But it's not the same thing.

Then this sentence raises the question: What if we had a higher dimensional physics? Is it obvious that minimal left ideal (irred rep spinors) would be the physically relevant ones in higher dimensions? Or maybe even grade subalgebra would be imporant.

One counterpoint (that you brought up earlier) is that we have composite physical systems that transform like higher dimensional irreducible representations (like a spin 5/2 electronic state in an atom). How do you describe this in GA?

@naturallyInconsistent that was not my question

I know; I'm specifically bringing your attention to more than just your question

@bolbteppa you seem invested in spelling out a clear answer to this question and I seem invested in getting an answer to this question. If you're willing, I'm interested to discuss in more detail. Maybe we could make another chat room at first to discuss in more detail?

> 'Herein', says Proclus, 'I emulate the Pythagoreans who even had a conventional phrase to express what I mean, "a figure and a platform, not a figure and sixpence", by which they implied that the geometry which is deserving of study is that which, at each new theorem, sets up a platform to ascend by, and lifts the soul on high instead of allowing it to go down among sensible objects and so become subservient to the common needs of this mortal life"

He was the first pure mathematician

@bolbteppa one question for you though: Are you advocating for GA or not? You seem to have said some stuff that harshly criticizes it, but then you have also said stuff that seems to praise it? I think you claim that your answer reconciles the seeming discrepancy between GA and standard spinors, but I'm not sure if you addressed the dimensionality issue that was spelled out more a few hours ago.

@Jagerber48 Notice that the GA rotors, of which spinors are part of, rotate not just spin half stuff, but also vector i.e. spin 1 stuff. There is no reason why it would not be able to construct higher spin stuff and rotate them too. Kinda just like how in standard spinor presentations, the higher spins are constructed on top of spin half.

@Jagerber48 bolbteppa is always having that incredibly dismissive attitude; he is anti-GA, but is kinda forced to be nice because he found that it is technically tolerable, and then he gets into a error state because he continues to want to be dismissive.

@naturallyInconsistent Yes, I see that both the GA spinors and regular spinors can be generalized to higher dimensions, but in higher dimensions there is no longer a 1:1 correspondence between the two. But I'm wondering if that is to be considered a mark against GA spinors or not. @RyderRude also pointed out the question about how does GA describe a composite 5/2 spin object (like a J=2 S=1/2 electronic state in a hydrogen-like atom)

@Jagerber48 Well, it is at least obvious that we should care about irreps. The point of Wigner's classification (the group plague bearer again!) is that we identify "a particle" with an irrep of the Poincaré group, and the fields corresponding to the spin-1/2 irrep on the particles need to transform as Dirac spinors or Weyl spinors

I know about that; have made that clear a few times by now, that as long as the GA even-subalgebra space is always big enough to accommodate and thus include spinors inside them, I'm perfectly fine to just take the spinors as living in some subspace of the GA even subalgebra.

@Jagerber48 I absolutely do not recommend GA it is a big distraction, you don't need to make a separate room half the discussions in here about spinors

@ACuriousMind Right. I think it's an interesting question: How does GA describe a e.g. non-relativistic spin 5/2 particle or composite system? It has dimension 6 so is it part of Cl(6)? Where does that space come from?

@bolbteppa Is your answer to my question "Can spinors be understood/explained without group/representation theory?" yes or no?

@fqq I said whatever I said, resonating with the views of other respected users, but deep down I feel that C&H is right. Initially the problem was that the paper they mentioned used $\langle c^\dagger_p c_q \rangle$ as 2 point correlator while I am familiar with $\langle c_p c^\dagger_q \rangle$ i.e. in the reverse order.

Although this was my question, people made it into a "discrete vs. continuous" subtlety question which I was NOT worried about (not because I understood it: but because I naively ignored such a subtlety)

> The geometrical algebra, therefore, as we find it in Euclid, Books I and II, was Pythagorean.

Turns out Pythagoras invented geometric algebra

@Jagerber48 He said multiple times, no. As did ACM.

@fqq I say "initially" because now the problem is resolved, because another paper by C&H use the same ordering as Peschel(the other reference) used...

My answer is not completely 'no' because you can technically define them as an ideal in a Clifford algebra, and a Clifford algebra is technically just an algebraic object on its own, however it is a terrible idea to take this approach, and my post warns against this in multiple ways

i said it too. there is no concept of spinors without rotations

However if you pick up a book like Chevalley I think this is the approach they take, I can't remember and didn't spend much time on that perspective, they always pull things out of thin air when you take this kind of approach

How did they come to $(9)$ from $(8)$?

@Jagerber48 No, the higher spin representations in general just don't occur inside the Clifford algebra at all. Anyway the detailed discussion of higher dimensions is subtle because spin isn't a single number (but a list of numbers) there anymore.

@Sanjana you can get one from the other easily, no?

@fqq By using anticommutation relations, yes. But the case of $\langle c_k^\dagger c_q \rangle$ should vanish, I thought. Because $c_q | 0\rangle=0$?

No, it's not a vev

@fqq That's what it looks like from Peschel's paper and what ACM and you are saying. And I think I am missing something but Casini-Huerta's paper treat it as one. If you are interested and have some time, look at this paper eqn. (8)

@Sanjana I don't understand what the problem is supposed to be: That you can write an expectation value either as $\langle \psi\vert A\vert \psi \rangle$ or as $\mathrm{Tr}(\rho_\psi A)$ is a completely general fact

the lattice paper you're looking at does this for $\psi$ some Slater determinant and $A = c^\dagger c$, the Casini Huerta paper does this for $\psi$ the vacuum and $A = \psi \psi^\dagger$. What's the issue?

@ACuriousMind How can these two be the same?

who claims they're the same?

My understanding of the Casini-Huerta paper which is probably wrong...

If you are looking at the lattice paper by Peschel it's equation (12) is directly used in Casini Huerta's (12)...

@Sanjana I don't see CH saying anywhere around equation (12) that they would "use" that result

the two derivations/approaches are just very similar because they treat formally analogous cases

@ACuriousMind They say something like that at the beginning of the section.

@Sanjana where?

The very first sentence

"An expression for the reduced density matrix ρV in terms of the two point correlators is known for free Dirac fields [7, 8, 9]." is just a statement of fact

[7] is the lattice paper.

@ACuriousMind To be clear, the question is, the Clifford algebra has $\sum_{k=0}^{2n} {2n \choose k} = (1 + 1)^{2n} = 2^{2n}$ elements in $D=2n$. The even collection contains half of these thus we find $\frac{2^D}{2} = 2^{D-1}$ elements. However the $\gamma^{\mu}$ are $2^{D/2} \times 2^{D/2}$ matrices acting on $2^{D/2}$ column 'vectors'. How can the $2^{D-1}$ elements $\gamma^{m_1 .. m_{2k}}$ describe a $2^{D/2}$ column vector, since $2^{D-1} \neq 2^{D/2}$?

it does not imply that what follows is an exact copy of what is in [7, 8, 9], nor that it uses results from it

it's just what you write when you're about to do a construction you've seen in [7, 8, 9] and you don't want anyone to think of it as original

@ACuriousMind Not everything needs to be said out that clearly :p It is implicit that the paper directly uses results from Peschel's work

@ACuriousMind So how am I supposed to derive (10) in Casini-Huerta if not use Peschel's paper :(

@Sanjana by the same logic as in that paper!

the point is that the two papers consider similar situations, so similar tools will work

(and things seem to get worse since we can sometimes even have $2^{(D/2) - 1}$ spinors...)

@bolbteppa The relation is exactly the "semi-spinor projection" from that paper or the projectors from the eigenchris video

Is what I said a fair summary or am I missing something?

the video even explicitly illustrates those projectors to project the matrices from the Clifford algebra onto their first column, turning them into $2^{D/2}$ vectors

This stuff is a bit of a mess, the author should have included dimensionality counts everywhere

@ACuriousMind wtf! You were just saying this all this time? How silly of me...but tell me two things---if you saw something like $\langle 0| \Psi_i^\dagger(x) \Psi_j(y)|0 \rangle$, that would be zero, right? And what are these $i$ and $j$ thing doing in the Casini/Huerta paper if they have moved onto continuum already?

As I read it, (8.1) is defining a Dirac spinor, then (8.4) is defining it as being made up from a left and right Weyl spinor. I don't think this addresses why you can say a $2^{D/2}$ object is the exact same as a $2^{D-1}$ object, since e.g. $2^{4-1} = 8 \neq 2^{4/2} = 4$ right?

@Sanjana I'm sorry, do I have to read the entire paper for you? The introduction clearly says: "In this paper we will assume implicitly such a regularization (a lattice regularization for example)", i.e. they're not doing continuum theory.

they're not claiming it's "the exact same", they're just talking in a way that suggests it and then at the end throw in a brief remark so that they're technically correct

@ACuriousMind No, no...I thought you already read it, but okay...I get the idea

not exactly a rhetorical trick that inspires confidence, but that's what's happening

@Sanjana Huh? Why would I have read it?

@ACuriousMind because you have a curious mind :p?

Thus it looks like they at least really need to explain this $2^{D/2} \neq 2^{D-1}$ thing

@ACuriousMind Sorry, but will you be sooooooooooooooooooooooooo kind to explain how they came at (9) from (8), in this?? I feel it's a very simple thing, and I wasted nearly 2 hours on it...

'We advocate this approach to spinors as the most straightforward and physically transparent technique available'

To be clear, once you write the $\gamma^{m_1 .. m_{2k}}|0 \rangle$ in terms of oscillators we can see why $2^D$ reduces to $2^{D/2}$ (there are only $D/2$ creation operators in $D=2n$), and then the even subalgebra brings this to $2^{(D/2) - 1}$, but without the oscillators I can't immediately see how you'd resolve this $2^{D/2} \neq 2^{D-1}$ thing

So I'd say what this GA thing is saying is right but it needs more details

@Sanjana you...just perform the trace in the basis where this is diagonal?

you'll have to be a bit more specific about at which point you get stuck

@ACuriousMind In the actual computation of the trace...I am not able to take the trace. I mean it would be helpful if you outline the particular steps--- it seems like a spoon-feeding request,i have no idea what to do

@bolbteppa No, on my reading, 8.1 is a "Hestenes" spinor and either one of the terms in 8.4 is a Dirac spinor.

@Sanjana Perhaps let's start small: Do you recognize the $\frac{1}{\mathrm{e}^{\epsilon_k} + 1}$ expression here?

@bolbteppa Yeah, this is exactly ACM's comment on the first answer that kicked off this entire discussion.

The author is saying a 'Hestenes spinor' is just a Dirac spinor, they are not claiming it's a different thing from the usual spinor, on the first page they say he first made the identification in [3],[4],[5], if it wasn't the usual spinor they couldn't claim this

@Jagerber48 The result of the projection is a Dirac spinor in odd and a Weyl spinor in even dimensions

I've now even added Polchinski's starting perspective on spinors to the answer to make it clear how everything lines up, Weinberg is similar to this as well

how do I know this? Representation theory! (+ accepting the claim that the result of the projection is an irrep/minimal left ideal)

in this answer, we use a separate coordinate system like depicted in the question image. What if we instead used one coordinate system; say, centered at the middle mass?

Even Polchinski talks about the Dirac representation being reducible because the states with an even vs an odd number of oscillators not mixing in even dimensions below (B.1.10)

@Obliv 1. You linked a question, not an answer. 2. You get more complicated formulae, what else do you expect? :P

Oh, how do you link answers again?

every post has a "share" button that gives you a direct link to the post

ah i found it. this answer. So we define $A$ as the force vector and do the inner product with each component?

doesn't $A$ vary on each mass though

@Obliv $A$ is not "the force vector"

I need to make a meme about ACM when someone mislinks a question for an answer :P

the answer organized the three $x_i$ into a vector, but they are not the components of a single vector in ordinary 3d space, this is just an efficient way to write a system of equations

I'm unsure what the first step is to determine normal coordinates of a system such as this one.

If we have only one degree of freedom, then the equation of motion is the "normal" coordinate?

if we have two or more equations of motion due to multiple degrees of freedom, we write the normal coordinates as solutions to the system of equations?

for ex: $m_1 \ddot{\theta_1} = ...$ and $m_2 \ddot{\theta_2} = ...$ you solve this system?

nvm i think i got it

@ACuriousMind no, that's the problem

@Sanjana it has a very famous name

@ACuriousMind It looks like the FD stat distribution function but why would that appear here?

@Sanjana It is Fermi-Dirac!

@Sanjana This is a standard staple of statistical thermodynamics. It comes from us wanting $\text{Tr}\rho=1$. Because you are dealing with fermions, $a^\dagger a$, which is a number operator, can only have eigenvalues of 0 or 1, and thus the results of the exponential is only $e^0=1$ and $e^{\varepsilon_k}$

That then fixes $\mathcal K$ and is precisely the Fermi-Dirac thingy.

@naturallyInconsistent I don't understand the exact math done here, although I have done this. Maybe not using density matrices, that's why; I know density matrices separately...

In stat therm, it, with temperature dependence put back in, is a partition function, typically labelled as $Z$, and $Z=1/\mathcal K$

thing is, if you were as familiar with statistics as research papers expect you to be, you should recognize that, in terms of the $a_k$ operators, this trace is just the expectation value of the number operator, and the expectation value of the number operator for a bunch of Fermions is Fermi-Dirac. So you just do the usual derivation of Fermi-Dirac, with the $\phi(i)$ along for the ride as prefactors

@ACuriousMind How? Right now, I just substituted (7) and (8) in (4). I got a bunch of $a_k$ in the exponential and also outside it...What do I need to get from there to the FD distribution?

@ACuriousMind I know that expectation value of number operator is supposed to give the FD distribution function...but I don't see how to get the Number operator here

$c_i^\dagger c_i$ is the number operator, and there's a $j$ there.

I got it

sorry, it was so simple...

please don't joke about that

@ACuriousMind is this when extending time dimensions? I thought that for general space dimension (but still one time dimension) we obtain the spinor reps using similar clifford algebra methods

$$c_i=\sum_k\phi_k(i)a_k\qquad\implies\qquad c_i^\dagger c_j=\left(\sum_k\phi_k^*(i)a_k^\dagger\right)\sum_\ell\phi_\ell(j)a_\ell=\sum_k\sum_\ell\phi_k^*(i)\phi_\ell(j)a_k^\dagger a_\ell$$

@SillyGoose how is what you said in contradiction to what I said?

and no, no one is extending time dimensions here

@naturallyInconsistent Yeah thanks. I think I was in a dream or something. That was super-simple

Or i am wondering about what you mean by higher spin representations. Was jaeger referencing spin 3/2 and 5/2 and in that sense higher? I think i misread this as higher spacetime dimension

@SillyGoose yes

@Sanjana I don't think it is simple. This is a double sum, and not a number operator yet. Put the Hamiltonian in, and it is a quadruple sum. But there should be a way to see that the quadruple sum turns into a single sum.

The generous reading is correct, for example around (2.17) they use the normal definition of a spinor from the spin group to motivate their definition, I don't think the fact that they are claiming 'Hestenes spinors = Dirac spinors' is controversial, whether they are right that this is the case due to the dimensionality issue seems to be the question

I have now even linked the answer to the highest weight representation theory of angular momentum (i.e. $SU(2)$) from quantum mechanics, that's how deep this rabbit hole goes

Why in single author papers, the authors write "we" instead of "I" like the old times?

because it probably sounds weird to read?

if you're writing something for an audience that isn't just yourself like in a journal, to engage with the reader better it's helpful to use "we" as the subject pronoun

We as in we are both reading this article together, we are both doing this computation together

@Sanjana see academia.stackexchange.com/a/2948/111995

I think first person is more engaging in story-telling than in academic papers in general

how do we get from $\frac{d^2}{dt^2}(x_1 + x_2) = \frac{-k}{m}(x_1 + x_2)$ to $x_1(t) + x_2(t) = 2A_s\cos(\omega_s t + \phi_2)$?

where $\omega_s \equiv \sqrt{\frac{k}{m}}$

oh wait nvm it's an ODE

could also be $Ae^{i\omega t}$ i think

once I have $$x_1(t) = A_s e^{i\omega_s t}+A_f e^{i\omega_f t}\\x_2(t) = A_s e^{i\omega_s t}-A_f e^{i\omega_f t}$$, I just add or subtract them to get the normal coordinates?

wait no u change the initial conditions to find normal modes so like $A_f = 0$ so $x_1(t) = x_2(t)$

so a system with $n$ degrees of freedom has $n$ normal modes?

Hi everyone...How are you, all?

I have a question. Is there a quick trick to tell if a Riemann curvature component is zero and is there a "direct" interpretation of each component of Riemann-Christoffel Curvature tensor?

@ManasDogra What's wrong with Ricci identity? It doesn't give you a component-wise interpretation?

Precisely because the RHS has a sum on one of the curvature tensor indices, and I want a component by component interpretation so that I can tell a story like "$R_{abcd}$ just means do this in the "a" direction, if you then go in "b" direction etc etc then you get this much value for R_{abcd}" or something like that...@Sanjana

what's the difference between a normal mode and normal coordinates?

what are the normal coordinates here for example

@bolbteppa I dunno, 2.17 just says the even grade subalgebra is closed under the Spin(p, q) group (which I understand is just the unit length elements within the even grade subalgebra). That doesn't mean the even grade subalgebra is equal to the Dirac spinor representation. I think from 2.17 you can probably conclude that the even grade sub algebra is a superset oft he Dirac spinor representation?

or in my example $$x_1(t) = ...\\x_2(t) = ...$$ I found these equations by adding/subtracting the equations of motion. Are these the normal coordinates? and by changing the initial conditions we find the normal modes

@Jagerber48 This is now 3 days trying to clean up a mess left by geometric algebra, that should tell you something

why use matrices and linear algebra? If I have a non-symmetric system, even, I can still just rearrange the EoM for normal coordinates and find the normal modes?

We're talking about whether some imprecise claim by GA can be fixed or just needs to be clarified, even though I have already explained how to fix/clarify it, we're just ignoring my answer which is referencing every canonical textbook you can find, in favor of some half-explained definition

(2.17) does imply that, that's what they are saying, they are simply saying because the spinor representation of an orthogonal transformation sends the even set of tensors into itself, so we can ignore the spinor rep and just focus on the set of tensors sent into themselves, again this is already obvious in the standard approach, 2 of my references use this directly

The paper only gives the 2,3 and 4d case because most GA authors just don't understand the normal approach that's why they are doing GA, I'm sorry, they could easily have linked it to all the stuff I mentioned but they ignored it

(This is a general fact, e.g. it generalizes to people doing 'alternative approaches to QM' while not understanding what the usual approach says etc..., the standard approaches are the way they are for a reason, usually you're talking about multiple nobel prize winners fleshing out these theories for their entire lives...)

$2m\ddot{x_1} = -2k x_1 - k(2x_1 + x_2)\\ m\ddot{x_2} = -kx_2 - k(x_2-2x_1)$ does this look like the correct EoM for this system

@naturallyInconsistent yeah i kind of feel like learning is about sifting through useless stuff. i think im building better physics skills even if the content is useless. i think is someone does something in a bad way, it's worthwhile for them to go through that to learn what a better way is. perhaps this is the only true way to learn to do something XD

gaining "maturity" and whatnot

i dont mind it. i think for me life is about finding things to pass the time until i die so learning qft is just as good as the next thing.

@RelativisticCucumber do you know what I'd have to do to solve this problem? I want to try the matrix method but I'm not sure how to set it up

@Obliv the matrix method :D

i'm going off of this "second method"

you want something of the form $M\ddot{\vec{x}} = -K\vec{x}$ right?

I guess, I'm not really sure tbh.

well we have two equations $\{l_1 = r_1, l_2 = r_2\}$

what are $l$ and $r$ here?

we can write this in vector form (recall that basis are orthogonal, so we can identify entries of the same components as equal in a vector equation)

$\begin{bmatrix} l_1 \\ l_2 \end{bmatrix} = \begin{bmatrix} r_1 \\ r_2 \end{bmatrix} \implies l_1 = r_1, l_2 = r_2$

arbitrary left and right hadn sides of an equation

so $l_1$ corresponds to the first particle and $l_2$ corresponds to the motion of the second?

precisely

so we want some equation that looks like $\text{blah1} \cdot \begin{bmatrix} \ddot{x_1} \\ \ddot{x_2} \end{bmatrix} = \text{blah2} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $

where $\text{blah1}$ is the mass and $\text{blah2}$ is the linear force ?

in the cases of springs $k$

exactly (the mass matrix and "$K$ matrix")

OH ok I see now

:D

so if I got my EoM correct I just plug those into this matrix equation?

and solve the determinant?

$\begin{pmatrix}2m & m \end{pmatrix} \begin{pmatrix}\ddot{x_1} \\ \ddot{x_2} \end{pmatrix} = ...$?

er wait that's not how matrix multiplication works

well think about the action of a diagonal matrix on a vector

what does that do and would that help us get the correct left hand side of the EoM?

i think you were alluding to basis vectors earlier but I may need a moment to digest what I'm doing right now

also, writing the inner product as you did is correct, but I think it is more convenient to write $M$ and $K$ as matrices

ok, brb gotta move my car to avoid tickets

but the euclidean inner product (normal dot product) is actually precisely $\langle v, w \rangle = \vec{v}^T \cdot \mathbb{I} \cdot \vec{w}$

which we can further reduce to $\begin{bmatrix} v_1 & 0 \\ 0 & v_2 \end{bmatrix} \cdot \vec{w} $ (in the two dimensional case, but this is general)

@Obliv essentially

you want to solve the equation something of the form $(K - \omega^2M)\vec{x} = 0$ (where we made an ansatz that the $x_1, x_2$ are of the form $e^{i\omega t}$.

in linear algebra terms, we want to solve for the null space of the $K- \omega^2 M$ matrix.

you can see that if $K-\omega^2 M$ is invertible, then we only have the trivial solution $\vec{x} = \vec{0}$

but we want non-trivial solutions. Hence, we suppose that $K- \omega^2 M$ is non-invertible. What constraint does this put on the matrix? Well, a matrix $A$ is non-invertible if and only if its determinant is $0$. We can then use this constraint so solve for the null space, which coincides with the solution space of our original coupled differential equations because of the ansatz we made

This is very similar to solving for eigenvalues of a matrix. However, you can see that it is not precisely the same thing. In particular, when all the masses have the same mass we are solving an eigenvalue problem. Elsewise, it is not precisely an eigenvalue problem.

Because when $m_i = m_j$ for all $i, j$ we have that the mass matrix is precisely $m\cdot \mathbb{I}$. Hence, with our ansatz, the action of the $K$ matrix results in multiplying the constant $m\omega^2$

it is quite cool because you can actually apply perturbation theory (as is usually first presented in a course in quantum mechanics) to the equal mass case.

for example, the canonical first example of coupled oscillators is two equal masses connected by three equal springs. okay this is easy enough to solve. then you move to "weakly coupled" oscillators which is two equal masses and two equal outter spring with constant $k$ and a middle spring with constant $k_2 < < k$.

you can write the $K$ matrix in this case as a $K_0 + k_2V$, a base matrix with easy to solve for eigenvalues (it is diagonal, actually) and a perturbation $k_2V$. then you can use first order degenerate perturbation theory to find the correct normal modes

you can also just solve it normally, but it is cool to see this sort of application

Ignore the blue highlight. Why are they taking the hermitian conjugate of $M_{\mu \nu}$ to be given by the transpose? I understand the situation for rest of the cases, but not this one.

Is it real because it is Euclidean?

@SillyGoose is it really an ansatz if the only functions whose derivatives are constant multiples of itself are $e^{ax}$ , $\cos (ax)$ etc?

@Obliv i guess that's true :)

maybe it is more precisely an "informed guess"

so the matrices are $K,M$ and the vectors would be $\begin{pmatrix}\ddot{x_1} & \ddot{x_2} \end{pmatrix}, \begin{pmatrix}x_1 & x_2 \end{pmatrix}$

right

so we have $\begin{pmatrix}m_1 \\ m_2 \end{pmatrix}\begin{pmatrix}\ddot{x_1} & \ddot{x_2} \end{pmatrix} = -\begin{pmatrix} k_1 \\ k_2 \end{pmatrix} \begin{pmatrix} x_1 & x_2 \end{pmatrix}$?

or in general some "force matrix" (lol this sounds like some made up sci-fi term) in place of the K matrix

which becomes $\begin{pmatrix}m_1 \\ m_2 \end{pmatrix}\begin{pmatrix}\ddot{x_1} & \ddot{x_2} \end{pmatrix} + \begin{pmatrix} k_1 \\ k_2 \end{pmatrix} \begin{pmatrix} x_1 & x_2 \end{pmatrix} = 0$

without the ansatz, is this possible to solve?

you should check to see if your RHS correctly multiplies out to the wanted two right hand sides of the EoMs

well it multiplies out to $-k_1x_1 - k_2 x_2$ I think

@Obliv you could certainly solve it as two differential equations. but i think that this matrix method sort of hinges on the fact that you can reduce teh problem to solving for a null space of a linear tranfsormation

@Obliv don't you have two EoM, though?

well matrix multiplication gives you one answer tho right

how else can i get two solutions on the rhs

the equation you wrote above is a single differential equation. inner product sends two vectors to a scalar, so you have a single equation at the end of it. but we in general want to condense a system of equations by using matrices

@Obliv well recall that a vector equation is equivalent to writing a system of $n$ equations

I do wish I could recall my bunk linear algebra course I took in the spring

1 moment as I look that up

my bad, it'd be $\begin{pmatrix}-k_1x_1 & -k_2 x_2 \\ -k_2 x_1 & -k_2x_2 \end{pmatrix}$

nvm it'd be $- \begin{pmatrix} k_1 x_1 & k_1 x_2 \\ k_2 x_1 & k_2 x_2 \end{pmatrix}$

so then we have $\begin{pmatrix} m_1 \ddot{x_1} & m_1 \ddot{x_2} \\ m_2 \ddot{x_1} & m_2 \ddot{x_2} \end{pmatrix} + \begin{pmatrix} k_1 x_1 & k_1 x_2 \\ k_2 x_1 & k_2 x_2 \end{pmatrix} = 0$

wait that's 4 eq. of motion

wait why does your second mass have two spring forces acting on it?

in the EoM you wrote above

because I'm a silly goose

but let's not focus on that wrong EoM, i'd like to learn this general solution method

here is using your particular case to illustrate what i mean by writing it using matrices

oh ok so we want a 2x2 * a 2x1 to get a 2x1

yes

I thought for some reason we needed a 2x1 * 1x2

we want a vector equation at the end of this all because a vector equation is a nice way to write a system of equations. and, these spring problems with our nice guess are just solving systems of equations

is the force matrix 2x2 to allow for the most general case of a system with 2 degrees of freedom?

i.e a spring on either side of each mass

you need as many dimensions as masses whose motion you are tracking, if that is what you mean by d.o.f.

dof is quite a loaded term is why i said what i said in the way i did above. e.g. dof in statistical mechanics means the number of energy terms quadratic in position or velocity

consider a system of $N$ masses on springs. then you want to solve a system of $N$ equations, right? After all, you are interested in the motion of each and every one of the $N$ masses.

I almost don't want to know, but what if we added another spatial dimension

how does this scale I guess

idk if u can even do matrices with 1 dim. Ok well for 3 mass case we have $\begin{pmatrix} m_1 & 0 & 0 \\ 0 & m_2 & 0 \\0 & 0 & m_3 \end{pmatrix} \begin{pmatrix} \ddot{x_1} \\ \ddot{x_2} \\ \ddot{x_3} \end{pmatrix} = -K\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$ I think, where $-K$ is a 3x3 matrix

idk what to do for the 2 dim. case.. add another set of equations for the $\ddot{y}$ equations I guess

with the component of the gravitational force in whatever the system is, maybe an incline plane or something.

so like $\begin{pmatrix} m_1 & 0 & 0 \\ 0 & m_2 & 0 \\0 & 0 & m_3 \end{pmatrix} \begin{pmatrix} \ddot{y}_1 \\ \ddot{y}_2 \\ \ddot{y}_3 \end{pmatrix} = -G\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}$

since these forces are already independent of each other, we can solve these equations separately maybe..

Ugh

If I want to learn the weird pythagorean ideas looks like I may have to read Aristotle's Metaphysics

@Supersymmetry You have to be careful with the terminology in the infinite-dimensional case, the ordinary idea of "lifting" and "inducing" representations between group and algebra does not work as straightforwardly in infinitely many dimensions

which is to say: I don't understand the question - with the proper definitions of "unitarizable admissible representation" of a Lie algebra and "unitary representation" of a Lie group, there is a bijection between them just like in the finite-dimensional case, I don't know what you mean by "sl(2,C) has no infinite-dimensional irrep that lifts to SL(2,C)"

@Supersymmetry The answer is: no, the Lie algebra action alone is not enough, you need an admissible representation resp. a $(\mathfrak{g},K)$-module

I don't quite see what this has to do with representations

the half-plane is not a vector space, so the action of sl(2,C) is not a representation

There is no Lie theorem for arbitrary actions

you need a representation to have the "Lie integration" between the algebra action and the group action

how are you "exponentiating the map" if you're not acting on a vector space?

the exponential is defined for linear operators on a vector space, not arbitrary transformations of arbitrary spaces

...you should have mentioned you're talking about $L^2$ and not the half-plane itself :P

still, for integration to a unitary representation of the group, you need a unitarizable $(\mathfrak{g},K)$-module, not an action of $\mathfrak{g}$ alone

so unless you can find an action of the maximal compact SU(2) on your $L^2$-space and show this is a $(\mathrm{g},K)$-module in the sense of admissible representations, there is no "integrating" this action

infinite-dimensional representation theory is subtle and you will usually not find answers as clean as in the finite-dimensional case

I have no intution at all for this case

@Supersymmetry Yes. Note that there might be more than one choice of a $K$-representation for a given action of $\mathfrak{g}$, or there might be none. Note also that if you're doing stuff at this level of rigor, you should carefully think about the Lie algebra action, since in general for infinte-dimensional representations, the Lie algebra will only act on a dense subspace of K-finite vectors

(simple example: the representation of U(1) as the shift operator on $L^2(\mathbb{R})$ has as its Lie algebra generator $\partial_x$, which only acts on those elements of $L^2$ whose derivative is square-integrable)

so if you actually have a well-defined Lie algebra action on the whole space, we might suspect you cannot actually turn this into a $(\mathfrak{g},K)$-module