The h Bar

ACuriousMind

20:00

I'm very expljcitly not doing statistical mechanics here, I'm just explaining why you'll find isolated atoms in ground states

Silly Goose

right let's stick to an isolated atom

Relativisticcucumber

i thought a justification for the fact that the ground state is occupied in the hydrogen atom case is because that's the most probable as calculated by statistical mechanics? and this method only necessitates that you find the energy eigenstates, but doesnt involve any qft? @ACuriousMind

Silly Goose

i get the ground state explaination

Relativisticcucumber

@ACuriousMind lol o no

Mr. Feynman

I remember the cheap explanation I was given before QFT, which was like "nature tends to minimize energy so excited atoms decay to the ground state" and I was like "wth, how can eigenstates decay?"

life was worse without QFT

ACuriousMind

20:01

@Mr.Feynman I was hoping no one would notice that right now :P

Mr. Feynman

I really wanted to be that guy

ACuriousMind

@Relativisticcucumber no, I really don't think we should be arguing statistically here

Silly Goose

but let's say i observe that the hydrogen atom never occupies superpositions of energy eigenstates at any time $t$

then i don't get how the mechanism you described explains this

Relativisticcucumber

but i thought this is literally the point of quantum stat mech -- to show you which states should be occupied? is that not so?

Silly Goose

because your mechanism describes how a hydrogen atom will eventually end up in a ground state from whatever initial state

ACuriousMind

20:03

@SillyGoose why would you say that

that's just not true :P

Mr. Feynman

@SillyGoose energy eigenstates are a complete set, what does that mean?

ACuriousMind

I'm just explaining the mechanism for why you shouldn't expect an isolated atom that's been sitting around for a while to be in anything but the ground state

if you just bombarded that atom with radiation, for instance, all bets are off as to what its state is, it's probably not an energy eigenstate :P

Silly Goose

well idk i thought decoherence tells you that in particular environments the pointer states of atoms are energy eigenstates

so sure at small time steps you might see something thats a superposition of energy eigenstates

but in general you can only ever observe energy eigenstates

ACuriousMind

@Relativisticcucumber I'm not sure what you mean exactly but I'm fairly confident that a single atom is not a statistical system. There's no thermodynamic limit here, it's just a single system.

@SillyGoose it may well be that there's some argument like that but that's not what I'm saying

I'm just saying the ground state is the one state the atom will always evolve towards

Silly Goose

hm well i understand (at least to a superficial level) the argument you are making for the claim you are making

about H ending up in ground state

ACuriousMind

20:06

so there's no mystery in finding atoms in these ground states

Mr. Feynman

@SillyGoose Why are you uncomfortable with hydrogen lying in the ground state after a long enough time (I'm catching up)?

ACuriousMind

and I see no need to make any farther-reaching claim that that about the atom always being in an energy eigenstate or something like that

imbAF

What does it mean to define $s^\mu$ (four spin???) in the rest frame of the polarisation vector $\vec S$ $\hat S^\mu=(0,\vec S)^T$ ?

ACuriousMind

many textbook exercises make it look like systems are often in energy eigenstates but that's because you can usually compute nice things for that case, not because anyone would claim this is how the real world always is

the real world is messy, our theories are idealizations and exercises often simplifications

@imbAF you really need to get out of this habit of asking questions where no one except you has the necessary context to understand them :P

Silly Goose

@Mr.Feynman i didn't know of any mechanism before but i think if it is true that QFT tells you from first principles that excited energy eigenstates decay with high probability towards the ground state, then i can be comfortable with it

imbAF

20:09

What am I not describing ?

Mr. Feynman

@ACuriousMind That reminds me of @MoreAnonymous posting papers with unknown notations :P

Silly Goose

@ACuriousMind but isn't condensed matter theory research crazy about ground states

Relativisticcucumber

@SillyGoose is it time for fermi's golden rule

ACuriousMind

@SillyGoose I mean, many systems have mechanisms like the atoms where they can get rid of energy if they're not in their ground state!

Mr. Feynman

@SillyGoose So is your problem that an atom lying in an energy eigenstate different from the ground state should lie there forever?

ACuriousMind

20:11

@imbAF Well: Polarization vector of what? What do you mean by "defining $s^\mu$"? That's just random notation, not an explanation of what $s^\mu$ is or how you're defining it

Silly Goose

@Mr.Feynman well in textbook quantum i expect this to happen, in QFT all bets are off for what i expect. i was told that fluctuations in the EM field can lead to a sort of up and down (weighted down) path towards the ground state, which i can sort of buy

ACuriousMind

you really shouldn't assume that the notation your particular text uses is universal

Silly Goose

because i expect the EM field to not really be a definite value (well it's an operator field?) field in QFT

imbAF

Basically $(\gamma^\mu p^\mu - m)U(p,s)=0$. $U(p,s)$ is a spinor for the solution of the DE with positive energies, momentum $p^\mu$ and Spin $s^\mu$.
Then, the half a.. text says: We define $s^\mu$ (four spin???) in the rest frame of the polarisation vector $\vec S$ $\hat S^\mu=(0,\vec S)^T$ ?

ACuriousMind

I have no idea what that means

imbAF

20:13

Ok forget it

ACuriousMind

so if this is all your text says you're right to be confused but I suspect you skipped something :P

imbAF

Do solutions to the DE, the spinors, span a basis?

ACuriousMind

basis of what

imbAF

basis in the sense that

Mr. Feynman

@SillyGoose It would be true if the $H=p^2/2m+V(r)$ were the Hamiltonian, as those you have there are its eigenstates

imbAF

20:14

orthonormality and completness

are satisfied

ACuriousMind

orthonormality with respect to what inner product and completeness inside what space

Silly Goose

@Mr.Feynman right, but ACM mentioned there is an EM interaction term

which probably changes thangs

Mr. Feynman

the complete problem also involves an EM magnetic field and you need to account for its Hamiltonian too and most importantly the interaction part between matter and EM field. One treats the interaction perturbatively, and finds that atoms do decay

imbAF

@ACuriousMind Yes, that's what I am trying to know

ACuriousMind

@imbAF you can't just ask if something is a "basis", you have to specify the vector space you're looking at

imbAF

20:15

That;s the whole point

One second I'll try to write down what happens

hopefully that will make things clearer

@ACuriousMind how do I use slash notation in latex?

is it possible?

feynman slash

Mr. Feynman

Under the assumption that the perturbation (interaction) is not strong enough to change the space of states, your states are the tensor product between the space of states of hydrogen and that of the EM field

Relativisticcucumber

wait is it true that we can use fermis golden rule to see how probable the decay you mentioned is? coincidentally im also learning this rule right now @ACuriousMind

Mr. Feynman

@Relativisticcucumber Sure

That's how you calculate it perturbatively

Relativisticcucumber

ok great

Mr. Feynman

Using Fermi Golden Rule you find the decay rate per unit time

Its inverse is the so-called lifetime

Relativisticcucumber

20:19

okay i see thanks !

Mr. Feynman

wait

we need to cook

Silly Goose

hm so what mathematical object is the space of all states of the EM field?

Relativisticcucumber

yay one thing resolved today ┏(・o･)┛ ┏(・o･)┛

ACuriousMind

@imbAF it's annoying on SE because MathJax has no simple command for it

Silly Goose

another hilbert space?

Mr. Feynman

20:20

a Fock space

imbAF

@ACuriousMind It's ok

ACuriousMind

@SillyGoose I don't really like that phrasing because we don't call the state spaces of ordinary quantum mechanics "the space of all states of the position operator", do we?

the fields are operators on the state space, not objects that have state

Mr. Feynman

I mean, that's a common jargon though :P

ACuriousMind

yes, the phrase is common enough that I can't call it wrong

which is why I just say I don't like it :P

Mr. Feynman

@SillyGoose The easiest way to think about it is in terms of the occupation number representation (secondquantizationACMforgivemeIknowyouloathethiswords), i.e. the space spanned by states specified by the number of "photons" in each frequency

Silly Goose

20:23

hm i guess i was mainly confused by how one can "tensor product" a state of a system and a "state of a field"

imbAF

So, starting with the DE for a particle at rest we get 4 solutions. 2 for positive energy and 2 for negative one. Now, we jump to the case of DE for a free particle in motion. The DE (one version of it is $(i\hbar\gamma^mu \partial_\mu -mc)\psi=0$.
We say that the solution is of the form $\psi=U(p)(\phi \chi)^T$. Now if we consider the positive solutions only, and try to find an expression we end up with two linearly independent solutions $U^{(1)}(p)$ and $U^{(2)}(p)$. Same can be said when considering the case for negative energy, two linearly idependent solutions are $V^{(1)}(p)$ and $V^…

with $\alpha$ and $\beta$ being called spinor indices, whatever that means

ACuriousMind

I really think whatever you're reading should explain that

or you should get a different book :P

because I can't really explain all the theory of spinors in a few chat messages :P

imbAF

It does not

and it's not even a book

it's a script

Mr. Feynman

using the searchbar of this chat, one can find an entire book about spinors

imbAF

of some time ago, and I am revising it

Ok, a different question

ACuriousMind

20:27

but maybe have a look around section 3.5 in these notes

bolbteppa

The script is likely based off a book which explains things better

imbAF

In this notes, I noticed terms such: Lorentz transformation, Spinor transformation and Parity transformations

ACuriousMind

I think Weigand explains this rather well with a minimum of overhead

imbAF

the last 2 act directly on a spinor right?

ACuriousMind

parity transformations can act on anything :P

imbAF

20:28

While the LT acts on what we call lorentz covariant quantity

ACuriousMind

and I don't really know in what context one would specifically call something a "spinor transformation"

instead of, say, a Lorentz transformation acting on a spinor or in the spinor representation

Mr. Feynman

@ACuriousMind By any chance, was he your Professor?

imbAF

But it's not a lorentz transformation directly though , is it?

Because

ACuriousMind

@Mr.Feynman I took his QFT courses, yes

Mr. Feynman

me feeling like Sherlock Holmes rn

imbAF

20:29

$\psi'(x'^\mu)=S(\Lambda)\psi(x^\mu)$

I don't think

$S(\Lambda)$ is the same as $\Lambda^\mu_\nu$

ACuriousMind

and to this day every time I think about something in basic QFT I first look into these notes because it helps jog my memory to see this stuff exactly how I first learned it

@imbAF yes, that's exactly what I mean by "a Lorentz transformation acting on a spinor or in the spinor representation"

imbAF

why people say spinor representation

what other representation is there

ACuriousMind

because that's what it is?

Mr. Feynman

Oh yeah, I know that feeling. I also have old sets of notes I sometimes check

imbAF

in the context of Dirac

equation

Mr. Feynman

20:31

It also happens that seeing them again I realize that I've gained a deeper understanding over the years

imbAF

what exactly representation encompasses?

ACuriousMind

it's just the standard meaning of a group/algebra representation

imbAF

@bolbteppa It is. Next year fall I am continuing masters in Theoretical physics, and hopefully we will revisit QFT

bolbteppa

The Lorentz transformations form a group, you can consider different representations of this group. The $\Lambda^{\mu}_{\nu}$ matrices act on the $x^{\mu}$ vector components, but the $\psi(x)$ are a different kind of 'vector' to the $x^{\mu}$ vectors, in a different 'space', so you need to consider representations of the $\Lambda$ as new matrices that correctly act on the $\psi(x)$ 'vectors', these turn out as $S(\Lambda) = e^{i\omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu}]/8}$ (up to a sign iirc)

imbAF

@ACuriousMind Thanks for the link and also do you have any notes on a proper clean solution to the DE for a particle at rest or in movement when using the +--- convention ?

bolbteppa

20:37

In other words, a spinor is a 'column vector' $\psi(x)$ in a space where the Clifford matrices $\gamma^{\mu}$ act on these vectors, which is different to the space of vectors $x^{\mu}$ that the $\Lambda^{\mu}_{\nu}$ act on.

ACuriousMind

@imbAF I think the notes I linked use that convention?

bolbteppa

@imbAF this is where people used to first ever see the Lorentz transformation of a spinor

imbAF

@bolbteppa And that is why we say Spinor representation. The LT have changed in such a way, that now, the group elements can act on spinors

@bolbteppa Yes, that is the exact case were, in my notes, we consider these $S(\Lambda)$. But it wasn't mention that these represent a new representation of the LT.

naturallyInconsistent

@SillyGoose I think I can help with this convo. As you have gotten from ACM, every excited state in QM actually would decay in QFT to the ground state, and so, really, any long-time expected detected state is a cartesian product of ground states of some sort. However, if you stick purely within the disastrous realm of QM, then one thing you can prove is that a superposition of two different energy eigenstates has the tendency to decay, whereas just one energy eigenstate would not.

imbAF

Is the decay, the result of the oscillation, when considering the superposition?

naturallyInconsistent

20:46

In fact, an olde style argument would be that the quantum field fluctuations in the energy of a system induces a slight superposition from the initially purely excited state, and that then causes the QM style decay.

Yes, the argument from within QM is that the superposition gives rise to the oscillation of the correct frequency needed for the decay.

But again, QM is a disastrous mess and really ought not to be taken seriously. It is really QFT that is real.

ACuriousMind

but don't worry, QFT is a disastrous mess, too!

imbAF

is the solution to the DE what one can call a quantum field?

naturallyInconsistent

Anyway, because of the limitations of QM, if we were to work purely within the framework of QM, then you have to treat seriously the case that excited states also get long lifetimes, somewhat akin to the metastable long resonances that are so common in statistical thermodynamics. Feynman nicely put it: in infinite time limit, all gases leak from their containers, so physics has to contend with "long enough that the transients have died out, but short enough to not destroy basic constraints"

bolbteppa

There's a good picture of what's going on, something like this:

0

Q: Lorentz transformation of a (relativistic) wave and a field operator

My question follows chapter 10 of Tung's Group Theory book, in particular definitions 10.11 and 10.12. Let $\mathcal{R}[\Lambda]$ be an $n\times n$ matrix representation of $L_+^{\uparrow}$ and a wave-function $\{\Psi^a\}$, which transforms (actively) under Lorentz as $$\Psi\xrightarrow{\Lambda}{...

imbAF

I understand the notations

not the axis though

Relativisticcucumber

20:54

meow @naturallyInconsistent

naturallyInconsistent

Now, if we do an experiment whereby you set the initial state to some superposition of different energy eigenstates, and then let the system evolve through a region that is not pinned down enough, then the system can easily pick up random relative phase. Even if each detected state is a superposition, the statistical aggregate of such a situation will be a classical mixture of energy eigenstates. That is what is likely to appear in simple experiments and so we must study them.

jigglypuff @Relativisticcucumber

Relativisticcucumber

i have a question about lasers -- say we have some medium which is an EM field. so my understanding is that this medium is used to excite the atoms, and that we want to achieve population inversion so that the atoms can decay, emitting photons at a faster rate than they are being excited, such that they amplify the light. in this case, i have read there can be spontaneous or stimulated emission, but what is the "stimulator" in the stimulated emission here? [...]

[...] and since light is coherent only when stimulated emission occurs, does that mean that the spontaneous emission rate is low in this case? i was thinking as per the earlier discussion with feynman and acm, the rate for spontaneous is high, so would this not influence the coherency of the light?

naturallyInconsistent

That is, if we are not in the QFT long long time limit in which all excited states have decayed away, then in the intermediate time limit, the QM style energy eigenstates nonsense is an important beginner's understanding of what is happening.

bolbteppa

Basically, if $\Psi(x)$ is a 'vector field', one arrow at each point $x$ in a frame $K$, then in a new rotated $K'$ frame, the 'vector' field $\Psi'$ in this frame at some new point $x' = \Lambda x$, is a 'rotated' version of the old field $\Psi$ at the old point $x$ via $\Psi'(x') = R[\Lambda]\Psi(x)$, where some $R[\Lambda]$ is a 'representation' of the rotation $\Lambda$

naturallyInconsistent

@Relativisticcucumber The rate of spontaneous is NOT high. It is, by definition, constant.

@Relativisticcucumber The incoming "mean-field" photons is the stimulator.

Relativisticcucumber

21:00

@naturallyInconsistent well i mean i thought it was not negligible

naturallyInconsistent

@Relativisticcucumber By the time you actually end up having the laser pulse, the N of photons is so big that the +1 is negligible. There are still some parts where it is not negligible, since we really definitely need (N+1)/N > 1 which is not possible if the +1 is not there, but it is quite negligible compared to N

ACuriousMind

@Relativisticcucumber the stimulator is just the spontaneous emission by another excited state

imbAF

@bolbteppa I see. But one thing, why a representation of the rotation and not the rotation itself?

Relativisticcucumber

oh no i think im back to an earlier question -- so if we have an electron in an excited state and we bombard it with some energy, i dont see why it should decay? as feynman said we can appeal to "lower energy" but thats not quite a mechanism?

@naturallyInconsistent i see -- okay i guess i can buy that it is, in fact, not so substantial

naturallyInconsistent

@imbAF A "rotation itself" is an abstract notion that is not captured by maths. Every mathematical pinning down of how the rotation behaves is a representation.

Relativisticcucumber

21:06

@ACuriousMind but then given what @naturallyInconsistent said, doesnt that mean we are limited by the rate of spontaneous emission which is not so substantial?

imbAF

Ahaa

Relativisticcucumber

or am i misunderstanding?

naturallyInconsistent

@Relativisticcucumber This is a very deep question that is geniusly deduced by Kirchoff, of the idiotic Kirchoff laws of electric circuits fame.

ACuriousMind

@Relativisticcucumber there's a lot of atoms in a laser :P

imbAF

so the matrix which represents rotation around i.e x axis, is a rrepresentation of rotoation

naturallyInconsistent

21:07

@imbAF Yes, that is a vector-matrix, spin-1 representation of a rotation.

Relativisticcucumber

@ACuriousMind okay i think i see

naturallyInconsistent

Whereas if you have a spinor, you need the spin-half representation of a rotation to represent what the rotation is doing to the spinor

bolbteppa

A rotation/Lorentz transformation acts on the $x^{\mu}$ vectors, the $\Psi(x)$ (i.e. functions of $x$) live in a different space and are simply different animals, e.g. a spinor in 11 dimensions is a 32 component column vector (off the top of my head), you need to figure out how to represent $\Lambda$ in this space (if it's even possible, which it turns out to be, i.e. these weird objects we call spinors turn out to be representations of the Lorentz group)

imbAF

I am not familiar with the definition @naturallyInconsistent , isn't the rotation in euclidian space represented via a matrix?

ACuriousMind

@naturallyInconsistent Hm? afaik, it was Einstein who first deduced stimulated emission, Kirchhoff died before the advent of QM

naturallyInconsistent

21:09

@imbAF You can represent it that way, but that is not the smartest way to do it. For example, we arrived at the rotation matrix long after Hamilton first realised that the correct quaternion way to rotate a vector is a two-sided rotation scheme.

@ACuriousMind Correct, but Einstein was using Kirchoff's argument, basically verbatim.

ACuriousMind

huh

bolbteppa

Another way to say it: $x^{\mu}$ in $D$-dimensional Minkowski space is a $D$-dimensional column vector $x^{\mu}$, a Dirac spinor $\Psi(x)$ is a $2^{D/2}$-dimensional column vector $\Psi_{\alpha}(x)$, and the matrices $M_{\alpha}^{\beta}$ that act on $\Psi_{\alpha}(x)$ are different to $\Lambda$, amazingly it turns out we can find matrices $S(\Lambda)_{\alpha}^{\beta}$ that act on the $\Psi_{\alpha}$ which can be interpreted like Lorentz transformations, i.e. that are 'representations' of them

naturallyInconsistent

@ACuriousMind @Relativisticcucumber the central idea behind Kirchoff's argument is that we need to achieve thermodynamical stability. A hot body in isolation can radiate, hence spontanous. It can also absorb. But if it is absorbing heat/light from anywhere, then at an equilibrium temperature, it must emit as much heat/light as it is absorbing. Hence stimulated emission must be necessary.

Einstein took this argument and derived the relations between A B and C coefficients, which is why Einstein is celebrated as the guy ushering in the age of the lasers.

imbAF

@bolbteppa I think I might understand what you mean. The LT $\Lambda^\mu_\nu$'s represent different geometrical activities (I don't know what is a term for boosts,rotations, mirroring). Now, when $\Lambda^\mu_\nu$ acts of a four vector, it's acting on a 4 dimensional space. The spinors, are in a different dimensional space, so we must find a way how to represent different geometrical activites i.e boost in x, but in this space were the spinors reside. Am I correct?

bolbteppa

Yes, the $\Lambda$ act on $D$ vectors $x$, the (Dirac) spinors are $2^{D/2}$ vectors in a different space (it just so happens that in $D=4$ the number of components of a Dirac spinor is also 4), it appears as a miracle that we can find new $2^{D/2} \times 2^{D/2}$ matrices which act on the $2^{D/2}$ spinors $\Psi_{\alpha}(x)$ that we can interpret as though they were the same Lorentz transformations we'd use to transform the $x$ via $\Lambda$, despite living in a completely different space

Mr. Feynman

21:17

@Relativisticcucumber beware that I used quotation marks because the "nature minimizes energy" is not really an argument here :P

naturallyInconsistent

It is also important to note that Einstein derived the relations semi-classically, which means that when the Dirac computed the correct quantum version of this, he did not need to compute all 3 coefficients (Einstein already derived that 2 of them are the same, and the ratio between the leftover), he only needed to compute one of them. The last one to compute is actually pretty difficult and require QFT proper, but within QM the absorption one is easy to compute.

@Mr.Feynman The speedy vegetable's question stands on its own, even if you think it is going too far off your tangent.

Relativisticcucumber

@Mr.Feynman hehe

naturallyInconsistent

@Relativisticcucumber did you understand how my answer from Kirchoff's argument is the correct answer to your question?

oh, I got starred again and I didn't realise! Yay!

imbAF

@bolbteppa Ok I fully understand what is happening now. But out of curiousity, how did, whoever they were, detected or came to the conclusion that Spinors belong to a different space than a four vector? I am familiar with the proof of the DE being lorentz covariant. I don't know if it is from there were people got the hint that solutions to the DE and a four vector do not belong in the same space

naturallyInconsistent

@imbAF This part is easy. Stern and Gerlach's experiment can be used to work out the mathematics of spinors, and then matching with how Dirac's equation would deal with the 4-component spinors and the representation of rotation for them, we can then compare and contrast with the 4-vector representation of rotation, and immediately discover that they must be different

i.e. the matrices they get are different rotation matrices

Relativisticcucumber

21:25

@naturallyInconsistent i think so. so you mean that it was observed that a hot body at equilibrium temp can absorb heat, thus it has to emit heat as well due to being at equilibrium temp already and we go from there to work out the rates and whatnot?

imbAF

You mean the matrices acting on the ket's and those on the solutions to the DE look different?

bolbteppa

$\Psi(x)$ is a function of $x$, i.e. $x$ is a point in space-time, $\Psi(x)$ is just some function of space-time, in general it could be anything, i.e. you are attaching to each point of space-time some new quantity $\Psi(x)$, there is no reason why this thing you attach to each point of space-time has to relate to $\Lambda$ in any convenient way, or in some new frame relate to how it looked in the old frame in any convenient way

imbAF

@naturallyInconsistent and this is the case also for boosts? or mirroring?

naturallyInconsistent

@Relativisticcucumber Yes. In particular, if an electron in a system is in an excited state and you bombard it with energy just equal to one of the decay modes it could get into, then the probability of it decaying by that particular mode must increase.

Relativisticcucumber

@naturallyInconsistent i see. that is interesting

naturallyInconsistent

21:28

@imbAF The rotations and the boosts and the mirroring are all working on the same representation bits and so yes. The following is a complication you should ignore: However, rotations don't care about a thing called chirality/helicity while boosts do, so there is a pair of possible boost schemes per rotation scheme.

imbAF

@bolbteppa I didn't get this part "..or in some new frame relate to how it looked in the old frame in any convenient way"

Relativisticcucumber

okay back to masers for me

toodaloo!

naturallyInconsistent

@Relativisticcucumber It is one of the more mind-blowing stuff and so I got rather excited when I first learnt it. Very sad that my peers were too confused to catch the wow moment.

ok it is 5:30am and I need to try and get some shut-eye again

imbAF

@naturallyInconsistent so it's sorta like counter propagating circular polarized light, which give you linear polarized light. Since chirality/helicity (I believe spin direction relative to momentum) for rotations plays no roles, it needs to vanish or something, and perhaps that can be done via a combination of boosts?

naturallyInconsistent

@imbAF Close but no cigar. When you finally get to learning about Weyl spinors, you will see that they entertained that there are two different possibilities for the 2-component spinor extension of the spinor rotation scheme to get to boosts, and only by combining both can you write down the Dirac 4-component spinor. This is far too ahead for you right now.

imbAF

21:34

Yes it is xD

naturallyInconsistent

IIRC, ACM was saying that Weyl and Wigner and Lie and a few other people tried to hard to bring group theory to quantum theory, and everybody was annoyed by them. However, what they did is the correct thing, and made things so much clearer, so we have no choice but to later learn what they did.

imbAF

Well, since Weyl was a mathematician, obliviously his work would bring clarity to the problem

It's what happens when you use crisp math in physics

and don't freestyle at every possible moment

naturallyInconsistent

well, as a theorist by training, I have absolutely no idea what my maths colleagues are even trying to say, even in the rare case they are using a lot of effort to try to say something to meow.

I'm already studying from their textbooks as much as I can.

The chasm between theo phys and maths phys is yuuge

Mr. Feynman

@naturallyInconsistent is that a reference to

imbAF

I am not saying that math alone is more important than physics. A bunch of numbers without meaning in our reality, are useless. So, both are important, math so that everything is clear and can be easily understood, while physics, to interpretate math

Mr. Feynman

21:42

Feb 11 at 22:34, by ACuriousMind

@Mr.Feynman there's a subtlety here where you can motivate density matrices that "classical" way (and indeed I did it that way when talking to the Lorentzian vegetable above) but entanglement means that quantum mechanics actually has a way for this "lack of knowledge" to arise in a natural way

ACuriousMind

@naturallyInconsistent It was called the group plague ("Gruppenpest"), see hsm.stackexchange.com/a/173/3797

Mr. Feynman

@naturallyInconsistent reading "meow" melts my heart everytime

naturallyInconsistent

@Mr.Feynman mew mew~

@Mr.Feynman yeppuu~

imbAF

@naturallyInconsistent what do you call, what's the term for rotations, boosts, mirroring? For example , we say the lorentz group, group of ?

naturallyInconsistent

@imbAF Group of Lorentz transformations, yes, or Poincaré group

Mr. Feynman

21:46

I don't know if you watch anime, but it's a common trope to have characters that have feline manners or just say "nyan"(=japanese for "meow") at the end of words and sentences

imbAF

@naturallyInconsistent since Spinors reside in another space, compared to four vectors, this space were spinors reside, must have a basis. The elements of this basis are eigenstates of something?

naturallyInconsistent

@Mr.Feynman I am very much the weeaboo too, but even as an East Asian, the Japanese are still very weird compared to the rest of East Asia. It is also the case that they are quite unique in how their language accommodates such changes at the end of sentences. At some point I couldn't bear even 2x anime and had to go to manga source to avoid the slow drag, though.

bolbteppa

If I have some function of space-time $\Psi(x)$ in some frame $K$, and I perform a Lorentz transformation to some new frame $K'$, it's not even clear if I can always talk about the same function in this new frame, if I can and I denote it as $\Psi'(x)$, then the relationship between $\Psi'$ in the new frame and the $\Psi$ in the old frame could be anything, I have no idea how to relate them in general. This is why we talk about representations of the Lorentz group, then we have easy answers

naturallyInconsistent

@imbAF Good question, I don't remember. But between Dirac gamma matrices and the standard basis of electron/positron spin-up/down spinors, that would be the answer. Note that there is a choice involved---say, I could have chosen the chiral version, or the mass version.

bolbteppa

We can postulate a linear relationship between $\Psi'$ and $\Psi$ rather than some crazy non-linear relationship between them (if it exists)

ACuriousMind

21:54

@imbAF Vector spaces don't have "a basis", they have infinitely many possible bases you can choose from.

you can't talk about "the basis" of spinor space

there are several popular choices, but it's not as if the basis would be unique in any sense

naturallyInconsistent

@ACuriousMind oops, my bad then. mew mew

anyway, it is 6am and im completely out of ice cream, so good night yall

ACuriousMind

spinor space is just mathematically a vector space

it's not some kind of new mathematical object, it's just a vector space that carries a specific representation of the Lorentz group

bolbteppa

Yes the spinors obviously have a basis in their $2^{D/2}$ vector space, i.e. you can expand a spinor in some basis, where there are infinitely many bases as usual...

imbAF

But how can you not have a basis. In order to have one, you need orthonormality and completeness. How can you not have an unique basis? If you don;t , than how can you express a vector in this space, if the basis is not unique ?

ACuriousMind

what

I think you are confused about how bases work

bolbteppa

21:59

In other words, from trying to 'linearize' the Klein-Gordon equation, you discover that in order to do this you need to introduce the $\gamma^{\mu}$ matrices, with indices $(\gamma^{\mu})_{\alpha}^{\beta}$ to write a consistent equation, these matrices act on 'vectors' $\Psi_{\beta}(x)$ in some new vector space, and it turns out for consistency they have to have $2^{D/2}$ components, i.e. $2^{D/2}$ basis vectors in this space

ACuriousMind

I didn't say the space doesn't have a basis, I said it doesn't have a unique one where it would be clear which basis you're talking about just by saying "the basis"

imbAF

Isn't the basis, the set of vectors,which you can use to express any other vector in the space?

ACuriousMind

just like any other vector space

@imbAF no, it is a set of vectors

there's nothing unique about it

bolbteppa

We are now writing a second book on spinors in here :p

imbAF

like the basis kets of a hilbert space?

ACuriousMind

22:00

I really don't know why you keep using the definite article there

bolbteppa

This is enough, have a think about what was said above, the rest is obvious, e.g. this basis thing is obvious

ACuriousMind

but if you don't understand what I'm saying then you really should brush up on elementary linear algebra

imbAF

I mean

100 010 and 001 are a basis in 3D space

what are other basis vectors, other than (1,0,0) (0,1,0) and (0,0,1) in euclidian space

The 3D space, if I am not mistaken is a vector space. And basis vectors are the e_x,e_y and e_z. Isn't this an unique basis?

ACuriousMind

what do you think the definition of a basis is

you can literally take any matrix with determinant not equal to zero and its columns will be a basis

imbAF

linearly independent and completeness. Any other vector can be expressed as a linear combination of what we call the basis

ACuriousMind

22:08

there's infinitely many bases for any given vector space, that should be the first thing you learn in linear algebra

e.g. you can just rotate your basis vectors

any rotation angle yields a new possible basis

imbAF

@bolbteppa while $\Lambda^\mu_\nu$ acts on $x^\nu$ and $S(\Lambda)$ on $\Psi(x^\mu)$, the parity operator can act on both?

Mr. Feynman

@naturallyInconsistent I will gladly have a chat about manga in the future but now I'm tired. Good nyanight

bolbteppa

This (section 2.3) discusses parity

You basically have enough of a basis to read something like this now

Where I'll just note that instead of $\Psi'(x') = \Psi'(\Lambda x) = S(\Lambda) \Psi(x)$ they are just considering $\Psi'(x) = S(\Lambda) \Psi( \Lambda^{-1} x)$ which is hopefully obvious

imbAF

22:32

@bolbteppa ok thanks

tbh this $\Psi'(x) = S(\Lambda) \Psi( \Lambda^{-1} x)$ is a bit confusing. Why is he using $\Psi'(x) $ instead of $\Psi'(x') $ ?

bolbteppa

23:38

@imbAF Just a convention that lets you talk about the LT of $\Psi(x)$ at $x$ without going to some new point $x$ (though it still means you have to relate it to some point $\Lambda^{-1} x$)

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