The h Bar

1:35 AM

"Sorry to keep asking random questions here. I am trying to figure out why a question on physics.stackexchange is not mainstream..." Have you figured it out yet?

Physics Meta

2:45 AM

0

Q: MathJax Usage for Questions by New Users

While I agree that MathJax makes things convenient and far more readable, it seems unfair to demand fresh new users to learn it before posting a question, which is something of a common occurence here. A lot of these questions usually recieve quick downvotes and are sometimes closed. I wonder, if...

John Rennie

5:31 AM

@Him It seems mainstream to me, and I've just reopened the question. We have a few users who seem to be unnecessarily zealous about closing questions and I find myself voting to reopen quite a lot of questions these days.

Slereah

8:35 AM

Can't physicists work on spinor pedagogy so that physics communities don't spend most of their time explaining spinors

bolbteppa

Once they finish with spinors it'll start up again with twistors

Mr. Feynman

@Slereah It seems someone got addicted to spinors :P

Slereah

9:04 AM

A lot of physics resources talk about spinors in term of flags, but for some reason no math people do

I'm not sure why

flags are a pretty classic math thing

ACuriousMind

where are you finding these resources? :P

I've never seen anyone but you talk about spinors as flags

but I have to admit I don't really seek out spinor literature, I'm very happy with the representation-theoretic approach and think the rest of you are just overcomplicating things :P

Slereah

@ACuriousMind It's not the most common description

But it is out there

I think it originated from Penrose

ACuriousMind

of course it does

Slereah

The original little penrose flags

ACuriousMind

honestly everything I know from Penrose that's not his work with Hawking seems overcomplicated and/or pointless to me (exhibit A: Penrose graphical notation)

Slereah

9:14 AM

How do you feel about twistors

ACuriousMind

never understood their point

bolbteppa

This talks about the flags a bit, as does this

Slereah

Yeah I have found a few

But they just basically say that you can describe a spinor as an oriented flag

Never really a discussion of how that works out in a more general context

Does that mean that spin groups are related to the parabolic subgroups that preserve flags???

Apparently there's some papers on that topic but they seem pretty austere

bolbteppa

Slereah

Also can you just have particles that are represented by more general flags than $(0,1,2)$

bolbteppa

9:29 AM

Associated to the vector $\vec{r}$, you can imagine two vectors $\vec{x}_1$ and $\vec{x}_2$ orthogonal to $\vec{r}$ and to each other (i.e. $\vec{x}_1 \cdot \vec{x}_2 = 0$) which generate the plane at the end of the flagpole. Setting $\vec{x}_3 = \vec{x}_1 + i \vec{x}_2 = (x,y,z)$ we have $\vec{z}^2 = x^2 + y^2 + z^2 = 0$.

Now we enter the realm of Cartan's spinors, where $z^2 = 4 \frac{(x - iy)}{2} \frac{(- x - iy)}{2}$ tells us $z = - 2 \xi_0 \xi_1$ (choosing the minus arbitrarily). Thus the spinor components characterize the tangent plane orthogonal to $\vec{r}$, where the choice $\vec{x}_1 + i \vec{x}_2$ fixes an ordering/orientation.

Slereah

Yeah that's basically the usual spiel on flags

also spinors have a sign associated to them in addition to a flag

bolbteppa

Instead of displaying the plane tangent to the vector $\vec{r}$, he summarizes it using the flag, where the angle $\alpha$ characterizes the orientation in the tangent plane (presumably you use the right-hand rule in here somewhere) where $\alpha$ tells you whether you chose $\vec{x}_1 + i \vec{x}_2$ or $\vec{x}_2 + i \vec{x}_1$

You set $\vec{r} = \vec{x}_1 \times \vec{x}_2$ to fix the orientation

Slereah

I guess the thing that's been on my mind a lot lately is how important is it that we use vector spaces in general

There's tons of geometric objects that are not vector spaces

ACuriousMind

I don't understand why I would do any of this instead of just working with the normal 2d complex vector in $\mathbb{C}^2$ with $\mathrm{SU}(2)$ acting on it

bolbteppa

Or maybe $\alpha$ tells you something else I'm not sure

Slereah

9:36 AM

might be more pedagogical idk

ACuriousMind

like, yes, I can see why these representations are equivalent but I don't see the point, drawing flags or passing to Cartan's matrices just seems needlessly complicated to me

bolbteppa

@ACuriousMind Where is $\mathrm{SU}(2)$ in any of this, you have to show $\mathrm{SU}(2)$ is even relevant.

ACuriousMind

@bolbteppa Wigner's theorem tells me QM should have projective representations of symmetry groups, and projective representations of the rotation group are representations of SU(2)

I find it much harder to explain why these little flags should have any relevance for quantum physics, too!

Slereah

@ACuriousMind No more than spinors, I'd say!

Flags also have reps of the rotation group acting on them!

ACuriousMind

yes, of course, otherwise these things wouldn't be an equivalent way to talk about spinors

bolbteppa

9:40 AM

Why are the irreps of $\mathrm{SO}(3)$ the same as those of $\mathrm{SU}(2)$? It's because of this picture ultimately, you are just dodging this because it's too hard and relying on things which ultimately reduce back to all this without making it explicit

ACuriousMind

I can prove all the Lie theory involved here without ever drawing a flag :P

Slereah

How unamerican

ACuriousMind

I'm not saying this isn't an equivalent way to do it, I'm just saying I don't see the point of doing it this "geometric" way

Slereah

Question is, why are those flags in particular the ones that are like that

ACuriousMind

when we have Lie theory and algebra that work very generally for all groups even ones not connected to spatial rotation

Slereah

9:42 AM

If you have any other flag object, how are their parabolic group

ACuriousMind

@Slereah I mean, as you said since they're "made of vectors" they have to have some representation of the rotation group on them

bolbteppa

You can prove the Lie theory but there is always going to be a step where you pull a rabbit out of a hat like postulating the matrix $X = x^i \sigma^i$ out of thin air, or just noticing the fact that the Lie algebra is the same, without any indication as to why the Lie algebra's are the same, which ultimately reduces to this picture

Slereah

You can have "flags" of arbitrary construction, though

You can have vector-plane-volume flags if you want

ACuriousMind

@Slereah ...in 3d?

Slereah

why is the rotation group not important for those

@ACuriousMind Well in 4D :p

Technically all flags in 3D are also of this form, since all flags have a 0D object and an nD object, but they are trivially all the same

The 3D object of a 3D flag is just the whole space

ACuriousMind

9:45 AM

that's why I asked

@Slereah consider that in 4d you aren't considering 4d rotations

you're considering the Lorentz group

Slereah

yeah idk how things work in arbitrary dimensions

Also physicists use "flag" in a difference sense than mathematicians I thikn

Math people "flags" are sequences of subspaces

they don't have magnitudes

idk what the object is when the objects do have magnitudes

bolbteppa

As far as I can see, the vector $\vec{r}$ is actually supposed to be the 'axis' in the 'axis-angle' decomposition of a rotation, which is one of the eigenvectors of that rotation. Then the tangent plane $\{\vec{x}_1,\vec{x}_2\}$ at the end of $\vec{r}$ is supposed to be the plane of complex conjugate eigenvectors of the rotation.

So these three vectors characterize your rotation, however we only need 2 of the vectors so we can ignore $\vec{r}$, but we have to fix an orientation of the plane in order to reproduce $\vec{r}$, so we get these oriented complex vectors $\vec{z} = \vec{x}_1 + i \vec{x}_2$.

Noticing that $\vec{z}^2 = x^2 + y^2 + z^2 = 0$ let's us find an equivalent description of $\vec{z}$ in terms of the spinor $(\xi_0,\xi_1)$ defined via $z = - 2 \xi_0 \xi_1$, where $x$ and $y$ can be expressed in terms of $\xi_0,\xi_1$.

Instead of using $\vec{x}_1$ and $\vec{x}_2$, I think the flag is a way to say we can use $\vec{r}$ and this label $\alpha$ which tells us the orientation of the plane

Then from this you can derive the link to $\mathrm{SU}(2)$ by asking how changes in $\vec{r}$ (i.e. rotations of $\vec{r}$) sends the old spinor to a new spinor, which you show is via an $\mathrm{SU}(2)$ transform

Slereah

What is the group of transformation of those flags anyway

bolbteppa

But if this picture is right, I really have no idea how to extend this picture to higher dimensions visually, but mathematically that's what the big Cartan discussion all last week was doing

Slereah

Yeah it's one of those things I can't really find

the general picture

bolbteppa

9:51 AM

@DIRAC1930 have a read of all that ^

ACuriousMind

@Slereah Actually, I don't think this is related to the Minkowski/Euclidean split. It's rather that you have Weyl spinors in all even dimensions (regardless of signature), so for $d$ odd you can always construct the spinors in $d+1$ as pairs of Weyl spinors, which will tend to look a lot like the spinors in $d$

Slereah

I tried finding papers that may discuss such things but anything I find looks like this

math.ucla.edu/~merkurev/papers/spind16.pdf

it hurts my eyes

bolbteppa

I think these are different kinds of flags than the flagpole thing

Slereah

As I said yeah kinda

"flag" means "sequences of subspaces" here

So it's kind of the difference between a vector and a distribution

I hope it's not up to me to work this out

ACuriousMind

well, drawing a flag in 3d does define a vector-plane subspace sequence (the vector is the pole, the plane is the flag on the pole)

Slereah

10:01 AM

I mean sure, but that is slightly different informations

The stabilizer of flags will not leave the spinors invariant since they could change norm

bolbteppa

This stuff is such a mess

A rotation sends the $(x,y,z)$ to an $(x',y',z')$ satisfying $F = \vec{z} \cdot \vec{z} = x^2 + y^2 + z^2 = x'^2 + y'^2 + z'^2 = \vec{z}' \cdot \vec{z}' = 0$ for $\vec{z} = \vec{x}_1 + i \vec{x}_2$. In terms of this $\vec{r} = \vec{x}_1 \times \vec{x}_2$ picture, a rotation sending $\vec{z}$ into $\vec{z}'$ is a rotation in the plane around $\vec{r}$, i.e. is it not just a $\mathrm{U}(1) = \mathrm{SO}(2)$ rotation twisting around $\vec{r}$...

Then your full $\mathrm{SU}(2)$ is also moving $\vec{r}$ to a new position. I have no idea what's going on now...

Slereah

yeah it is a bit strange

I guess it may be worth looking into how the flags relate to the Clifford algebra

bolbteppa

10:16 AM

That's some more madness: From how I defined $z = - 2 \xi_0 \xi_1$ we get the relations $\eta_0 = \xi_0 z + (x - i y) \xi^1 = 0$ and $\xi_1 = \xi_0 (x + i y) - \xi_1 z = 0$ which is $X \xi = 0$ for $X = \begin{bmatrix} z & x - i y \\ x + i y & - z \end{bmatrix} = x^i \sigma^i$ which magically satisfies $X^2 = \vec{z}^2 I = 0$ which we can use to show the $\sigma^i$ satisfy the Clifford algebra

So the $\vec{z} = \vec{x}_1 + i \vec{x}_2$ fixes an orientation which the $\alpha$ of the flag is supposed to tell us so that we can find $\vec{r} = \vec{x}_1 \times \vec{x}_2$, then the $(\xi_0,\xi_1)$ can be derived from $\vec{z}^2 = 0$, so the spinor tells us what the flag value is by tracing all this back, then we derive this $X \xi = 0$ and $X^2 = \vec{z}^2 I = 0$ thing from all this to get the link to Clifford algebras...

Physics Meta

0

Q: Why accepting edits with edit privilege should not be immediate?

If a user suggest an edit, two users with access to the review queue are necessary to validate the edit, even if the first user has the "edit applied immediately" privilege. Why is that so? Users with edit privileges can override the edit, do the same edit themselves, and get it applied immediate...

feature-request review suggested-edits

bolbteppa

"The crucial property is that under the action of a rotation, the direction of the spinor changes just as a vector would, and the flag is carried along in the same way is if it were rigidly attached to the ‘flag pole’. A rotation about the axis picked out by the flagpole would have no effect on a vector pointing in that direction, but it does affect the spinor because it rotates the flag"

Slereah

I should dig out that Penrose article to see if he had more to say on the topic

He also talks about it in his twistor book

I wonder if it's possible to classify all possible geometric objects

there are a lot

bolbteppa

10:32 AM

The $\alpha$ represents an angle rotating around $\vec{r}$, i.e. it lets us locate the vector $\vec{x}_1$ so that we can then form $\vec{r} = \vec{x}_1 \times \vec{x}_2$.

So different $\alpha$'s represent different starting orientations, that makes more sense now

Slereah

Calling it an angle is a little bit weird since that involves a choice of a starting vector, but I guess the space of all plane intersecting that vector is parametrized by an angle

Actually half angle since that's a projective thing, but that would be the orientation of the flag going in too maybe?

idk

bolbteppa

Yeah, you are basically talking about an orthogonal triad/basis where $\vec{r}$ is one of the vectors, and we must have $||\vec{r}|| = ||\vec{x}_1|| ||\vec{x}_2||$ (from $\vec{r} = \vec{x}_1 \times \vec{x}_2$) so we set $||\vec{x}_1|| = ||\vec{x}_2||$ (which is needed/crucial when working out $\vec{z}^2 = 0$). Thus all we need is the angle around $\vec{r}$ where $\vec{x}_1$ is located, which is a $\mathrm{U}(1) = \mathrm{SO}(2)$ rotation around the axis given by $\vec{r}$

Slereah

Aren't the multivectors in a Clifford algebra essentially a direct sum of various vectors, planes, etc

which would have the set of flags as a subset

mb it is related

Does something interesting happen to a flag if you use the Clifford product on it

bolbteppa

Instead of the word 'flag', the word/letter 'L' would make more sense, i.e. you have the letter L with the top starting at the origin, the | part of the L is your vector, and the - part of the L is (unit normal parallel to) the second vector $\vec{x}_1$, it can twist 360 degrees around itself

Slereah

The original paper called them axes :p

with axe blades

bolbteppa

10:51 AM

It makes a bit of sense that a 3D vector $\vec{r}$ can be represented on a complex plane via stereographic projection, and that when you add the fourth parameter $\alpha$ you can represent these using a ratio of complex numbers on the complex plane

in-theory.net/what-is-a-spinor

That's interesting

They move the flag to the $z$ axis for some reason, and express the spinor components related to it

imbAF

11:04 AM

In statistical mechanics, multiplicity (also called statistical weight) refers to the number of microstates corresponding to a particular macrostate of a thermodynamic system.

What is that quality of a microstate, which can decide whether it corresponds to a particular macrostate or no? It's internal energy ?

Or a set of state vairables such as, T,V,P?

Slereah

There are many kinds of macrostates

It is typically temperature + a bunch of intensive/extensive pairs

Like pressure/volume

imbAF

so in the most cases, state variables

One more thing: If a process is quasi-static, (can it be translated to reversible?) then for the internal energy we say $dU=-PvD+TdS$, whereas if it's non-quasistatic (irreversible ?) $du=\delta W + \delta Q$ ?

bolbteppa

So basically there are two ways to associate a complex plane to this. One way is starting from $\vec{r}$, then choosing a vector $\vec{x}_1$ orthogonal to $\vec{r}$ that satisfies $||\vec{x}_1|| = \sqrt{||\vec{r}||}$, thus the only question is the orientation of $\vec{x}_1$ around $\vec{r}$ (given by $\alpha$ w.r.t. some reference axis...). Then introducing vector $\vec{x}_2$ orthogonal to those two, with $||\vec{x}_2|| = \sqrt{||\vec{r}||}$, we have $\vec{r} = \vec{x}_1 \times \vec{x}_2$.

We then have the two 3D vectors $\vec{x}_1$ and $\vec{x}_2$ living in a 2D plane whose origin is at $\vec{r}$, and we can locate the points that these vectors end at in this plane using two complex numbers in the plane, which we find by forming $\vec{z} = \vec{x}_1 + i \vec{x}_2$ and using $\vec{z}^2 = 0$ to find a pair of complex numbers.

The second way is via this stereographic projection stuff, which gives a different complex plane, and again the whole 3-component vector plus angle specifies four real parameters which can be described by two complex numbers in that stereographically projected plane

Slereah

11:29 AM

the penrose paper on flags was "Structure of Space-Time" in Batelle Rencontre

I should scan it

I'm not sure it's available online

bolbteppa

In higher dimensions, instead of fixing a $3-2 = 1$ dimensional line in the axis-angle decomposition, a rotation fixes an $n-2$ hyperplane, thus rotating two axes at a time. So I can still see a $\vec{r} = \vec{x}_1 \times \vec{x}_2$ type thing going on (obviously no real cross product here, but $\vec{x}_1 \cdot \vec{x}_2 = 0$ generalizes, as does $||\vec{x}_1|| = \vec{x}_2||$) where $\vec{z}^2 = 0$ holds, i.e. $F = (x^0)^2 + x^1 x^{1'} + ... + x^{\nu} x^{\nu'} = 0$ in $n = 2 \nu + 1$ dimensions

bolbteppa

12:02 PM

7

A: Rotation in Higher Dimensions

One may show that a general rotation $R\in SO(N)$ in $N\geq 2$ spatial dimensions can be composed $$R ~=~ R_1\circ \ldots\circ R_{k} $$ of at most $k=[\frac{N}{2}] $ pairwise commuting rotations $$R_1,\ldots, R_{k}~\in~ SO(N)$$ that each leaves a co-dimension-2 subspace invariant (although not n...

So you have $k = \lfloor \frac{2 \nu + 1}{2} \rfloor = \nu$ pairwise commuting rotations, and an isotropic $\nu$-plane associated to spinors...

Slereah

12:58 PM

Good way to spot the limey bastards here

user 726941

Whelp, the language is named after them.

Slereah

It's not called limeysh

user 726941

We've got higher mountains to climb.

As Voltaire once said, writing is the painting of the voice.

Slereah

1:49 PM

what's the good book on clifford algebras

I had one and I forget the title

Physics Meta

3:00 PM

0

Q: How do i get the moderator to provide more informative comments on question closure?

My question has been closed with the comment that it is a "homework-like" or "check my work" question. I have read the posts recommended by the moderators to address the issues and edited the question several times. However, the question remains closed with the only comment being "Original close ...

support closed-questions

Slereah

3:28 PM

31

Q: Are Methodological Assumptions of StackExchange Fundamentally Flawed?

I looked here for an answer while writing a paper on evidence and scientific inference. I then saw the bold claims made by the website that the process goes as follows: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top What reason do we have to thi...

epistemology philosophy-of-science social-epistemology

ACuriousMind

3:51 PM

@Slereah "But when I'm doing philosophy I don't want to get at what we agree about, I want to get at what's true." Ah, yes, philosophy, where famously people never disagree about what "true" even means :P

Slereah

Sorry @ACuriousMind, stackexchange has been proven philosophically wrong

bolbteppa

There's different approaches to Clifford algebras, pedantic formal boring ones, or ones for physicists

ACuriousMind

@Slereah I mean...I'm not jumping to defend the existing voting mechanisms as always selecting the "best" answers :P

It's just that this question pretends to make some sort of philosophical argument when it's just pointing out the one extremely obvious flaw with any systems of popular vote :P

Slereah

@bolbteppa Response to

imbAF

4:17 PM

If we consider a matter wave propagating into a double potential wall, one after the other, then for the first wave, the initial one, we will have a reflected and transmitted wave, then for the transmitted wave that propagates into the region where the first potential v1 is active, will we also have a reflected and transmitted wave, once it reaches the region of the potential V2?

ACuriousMind

why wouldn't you have that?

imbAF

for the same reason you have that once the initial wave propagates towards the first barrier, which is a region with different potential energy

shouldn't this quantum process, somehow resemble to when we consider a wave propagating into regions with different refractive indices?

Well If I am having that, then the reflected wave in the region where V1 acts on, will also be split in a transmitted one and a reflective one, meaning, we will have a bunch of reflected and transmitted waves within each region. So how does one solve the problem of find the coef. of transmission and reflection, by using the probability density current ?

ACuriousMind

I'm not sure I understand the question

yes, the wave will reflect/transmit at each potential barrier

in principle you compute that in exactly the same way as you compute it for a single barrier

imbAF

I see

One more follow up question in that computing part

We considered the case of one barrier

rob

Also related: the blueberry Earth paper by our own Anders Sandberg, who wrote a serious answer to a silly question here, and posted his analysis to arXiv because the question was closed. I learned about this paper driving down the road listening to "RadioLab" on NPR, who interviewed Sandberg and thought we were party poopers for closing the question. — rob ♦ 54 secs ago

imbAF

4:31 PM

and we tried to solve the problem, by using mathematica. The question I have is

rob

Just in case "physics stack exchange trivia" comes up at the pub

imbAF

for the incident wave, we used the formula for the probability density current, in order to find the incoming current. Which is understandable

ACuriousMind

@rob Heh, "Blueberry Earth" sounds like a band name in some experimental music genre

imbAF

@ACuriousMind When we tried to find the probability current density of the incoming wave, which was of the form $\psi_{inc}=e^{Ikx} + R e[-Ikx]$, we said that $R$ and $R^* $ are both zero, or go to zero. Any reason for this assumption ?

ACuriousMind

the wavefunction is not a "probability current density"

it's a probability amplitude, its square is a probability density, and the probability current (density) is $j = -\mathrm{i}(\psi^\ast \partial_x \psi - \psi \partial_x\psi^\ast)$

I'm not sure why you're talking about a current density in this context

imbAF

4:40 PM

In the code I am looking at, the incoming current was equalized to the probability current density which is $\vec j=\frac{1}{2mi}[\psi^*\nabla \psi - \psi \nabla \psi^*]$ (assuming that $\hbar =1$)

ACuriousMind

your wavefunction for a setup with different potential barriers should be a piecewise wavefunction that looks like $T_i \mathrm{e}^{\mathrm{i}k_i x} + R_i\mathrm{e}^{-\mathrm{i} k_i x}$ in each region with constant potential, and if your wave is coming from the left then in the right-most region you have $R_i = 0$ because there's no barrier to the right to reflect from so there's no reflected part there

imbAF

I was just going to ask why theres no reflection once a wave makes it;s way to the other end, but you answered that xD

ACuriousMind

The rest of the $T_i$ and $R_i$ are usually determined by matching the different pieces via boundary conditions much like Wiki does it for the rectangular potential here

imbAF

I was able to do that part, and find expressions for the coef. multiplying the exponentials

The thing I am struggling with is, finding refl. and trans. coeficient

ACuriousMind

uh, they are essentially the same things?

my calling them $T_i$ and $R_i$ isn't an accident :P

imbAF

4:47 PM

yeah you are right

so in a way, i look at each region individually

"disregarding" the rest of the setup

imbAF

5:33 PM

@ACuriousMind I believe that $T_i$ and $R_i$ in each region, if summed, should amount to 1, right?

Mr. Feynman

$|T|^2+|R|^2=1$

imbAF

Yes, but I mean for every region, since I am having 3 potential barriers one after the other

DIRAC1930

5:56 PM

What is wrong with deriving the covariant derivative through $\partial_m V^n \mathbf{e}_n = (\partial_m V^n ) \mathbf{e}_n + V^n(\partial_m\mathbf{e}_n)$ and then defining $\partial_m\mathbf{e}_n = \Gamma^k_{mn}\mathbf{e}_k$ i.e. that it is in the same vector space?

DIRAC1930

6:27 PM

@bolbteppa I think the reason behind what Cartan is saying must be simple otherwise he would have elaborated more on it

bolbteppa

There's nothing wrong with getting the covariant derivative that way it's in plenty of books

DIRAC1930

Ah okay thanks

What benefit does thinking about parralel transport give (derived for example in LL) instead of the method above (I'm learning GR properly for the first time)

bolbteppa

6:56 PM

This way, you see a covariant derivative is nothing but an elementary derivative, but where the basis vectors now change, and then parallel transport amounts to studying 'homogeneous' vector fields, which satisfy $\vec{A}(\vec{r}+d\vec{r})=\vec{A}(\vec{r})$, and this immediately leads to $V^n_{;m} = 0$.

DIRAC1930

Ah okay thanks

bolbteppa

7:14 PM

The problem is, if you work exclusively in components, the question is how you end up with the extra term, completely ignoring the basis vectors. You do know the differential of the components, $dV^m$, do not transform like a tensor (except for linear transformations), so you know something is missing, thus we need to come up with some waffle to justify ending up with the same result as above, so you have to put that second term in the above derivative into words without invoking basis vectors.

Mr. Feynman

Shouldn't using the EoM in the Hamiltonian lead to problems?

DIRAC1930

@bolbteppa Thanks, I was thinking this might be the case

bolbteppa

Not sure what you mean to use the EoM in the Hamiltonian

Mr. Feynman

Reading about the quantization of the EM field (but this should go for any classical field), I've seen books writing the energy $\mathcal{E}\sim\int \frac{1}{c^2}|\dot{\vec{A}}|^2+|\nabla\times\vec{A}|^2 d^3\vec{r}$ and then by means of Fourier mode expansion and the EoM writing it as $\sim\sum_{k,\sigma} k^2(a_{k,\sigma}^*a_{k,\sigma}+a_{k,\sigma}a_{k,\sigma}^*)$ so we can easily apply the quantization rule

Without using the EoM (by using that $\omega=ck$) some terms would not disappear. So, I know that is the correct Hamiltonian by other ways but I'm confused about this one

bolbteppa

7:31 PM

When you write the energy, you are assuming the fields involved are the fields which solve the equations of motion

Mr. Feynman

This way I'm only considering on-shell energy, though

bolbteppa

Why would you want to consider anything other than on-shell energy. Even when you derive the conservation laws via Noether's theorem you use the equations of motion

Mr. Feynman

7:48 PM

Because in that case you are proving that the function $(q,\dot{q})\mapsto E(q,\dot{q})=\dot{q}\partial_{\dot{q}}L-L$ is constant along the solutions of the EoM. Now we're doing this to have a Hamiltonian/energy in terms of "creation" and "annihilation" (not quantized yet), so I'm puzzled we need to use the EoM just to recast the Hamiltonian in a different form

bolbteppa

7:58 PM

All we're doing is noting that the solution of the free Maxwell equations breaks up into a sum of plane waves, and showing that the energy also breaks up into a sum involving individual energies associated to each of the plane waves, and you can also show each of the waves behave like harmonic oscillators etc...

Another way to see that you absolutely have to use the equations of motion is to just check whether the energy-momentum tensor is conserved, noting you get the expression for $\mathcal{E}$ from the energy-momentum tensor of the Maxwell Lagrangian. If you are not allowed to use the eom you will not get a conserved energy-momentum tensor

Mr. Feynman

I've seen other sources only performing a Fourier expansion of $A(\vec{r},t)=\frac{1}{\sqrt{V}}\sum_{\vec{k},\sigma}q_{\vec{k}}^{(\sigma)}(t)\epsilon_{\vec{k}}^{(\sigma)}e^{i\vec{k}\cdot\vec{r}}$ and after that using this in the lagrangian of the EM field to derive the hamiltonian $H=\sum_{\vec{k,\sigma}}c^2|p_\vec{k}^{(\sigma)}|^2+k^2|q_\vec{k}^{(\sigma)}|^2$ without using explicitly the EoM (i.e. solving $\ddot{q}+c^2k^2q=0$)

And then rewrite it in the form above

Maybe I'm confused about where (and if) this other method uses the EoM

bolbteppa

Even in the non-Lagrangian formalism, the expression for $\mathcal{E}$ is derived by starting from the Maxwell equations (i.e. the eom)

Mr. Feynman

Mhh, that's also true

Which is basically what you've said above about the stress-energy tensor or Noether theorem

On the other hand the part that puzzles me is the following: consider the simplest possible example, the harmonic oscillator. If we write $H(p,q)=\frac{p^2}{2}+\omega^2\frac{q^2}{2}$. Here you only need to define $a\sim q+ip$, you don't need the EoM to write it as $H(p,q)\sim a^*a$

Maybe things are working because I'm already in the Hamiltonian formalism, let me check if I can figure out something starting from energy in the Lagrangian formalism

imbAF

9:32 PM

How can one interpret the transition rate between states? Does it show how the probability of transitioning between states, changes in value per unit of time? In other words, with the help of Γmn
we can find moments in time when the transition has higher probability of taking place and moments when it has not?

But it's actually constant. Meaning the system can jump in all the states, which have energy that belongs to an interval, with the same probability?

bolbteppa

10:41 PM

@DIRAC1930 Have a look at this (P.4-7)

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