But if I go through the 2nd two term in your equality I have $\epsilon^{\dot{b}\dot{a}}\epsilon^{ba}\sigma^{\mu}_{a \dot{a}} = \sigma^{\mu}{}^{b \dot{b}}$ which you claim is equal to $\bar{\sigma}^{\mu,\dot{b}b}$. Doesnt this mean that $\sigma^{T \mu}=\bar{\sigma}^\mu$?

@bolbteppa Isn't this true when the second has spinor indicies fully up and the former has spinor indicies fully done transposed?

It means $\sigma^{\mu}$ (defined with lowered indices) with it's indices raised such that the whole thing is then transposed is then equal to $\overline{\sigma}^{\mu}$, that doesn't mean you can just blindly take transposes of the $\sigma^{\mu}$ to get the $\overline{\sigma}^{\mu}$ via $\sigma^{T \mu} = \overline{\sigma}^{\mu}$

So when you say 'in terms of explicit matricies' are you defining $\sigma^\mu:=\sigma^\mu_{a \dot{a}}$ and $\bar{\sigma}^\mu:=\bar{\sigma}^{\mu, \dot{a} a }$?

Yes

I never knew people did that, I always assumed when someone was defining the matrix, they were writing the components for $A^{a}{}_b$

I explained above why $\sigma^{\mu}$ has to have it's indices down above

Does $\sigma^\mu$ have to have one dotted and one undotted index because it's Hermitian?

I explained why above, you can see exactly why there

@bolbteppa Why shouldnt there be a $\dot{b}$ on the farthest but one index from left

I explained why

I think there should be a dot on the b

No

You haven't explained anything

You've just randomly defined things

Sorry there should be a dot on the lower $b$

and the definition of $\psi'^{\dot{a}}$ should be $\psi'^{\dot{a}} = (M^{-1})^*_{\dot{b}} \ ^{\dot{a}} \psi^{\dot{b}}$ (had two upper $\dot{a}$'s above) and $\psi_{\dot{a}}'$ was missing a prime

@bolbteppa Don't we first have to define $(\sigma^m):= (\sigma^m)_{\alpha \dot{\alpha}}$ in $P=P_m \sigma^m$?

You want to figure out where the indices are in $\sigma^m$, you know $P = p_{\mu} \sigma^{\mu}$ transforms as $P' = M P M^{\dagger}$, you know where the indices are in $M$ and $M^{\dagger}$, you can use them to determine where the indices should be on $\sigma$. That's all there is to it

Yes sorry you are right

It seems a bit weird to have a tensor where half of it transforms in a different rep to the other half

From a representation theory perspective, the fact that $X = X_{a \dot{a}}$ has $2 \times 2 = 4$ components, just like a four-vector, tells you both of these objects are realizations of an irreducible representation of the Lorentz group, i.e. things that representations of the Lorentz group act on, so you can expect a relationship between $x$ and some $X$

$X = X_{a b}$ and $X = X_{\dot{a} \dot{b}}$ have only three components, so a bispinor $(X_{ab},Y_{\dot{a} \dot{b}})$ has 6 components made out of two 3-component objects, i.e. an anti-symmetric 4-tensor which splits into two 3-vectors, an example of which is the electromagnetic field

So you are saying $X_{a \dot{a}}$ is shorthand for something else?

Are we saying $X\in V \otimes W$ where $V$ and $W$ are spinors transforming under the normal and cc representations respectively?

$X_{a \dot{a}}: V \times W \rightarrow \mathbb{C}$

$X_{\dot{a} a }: W \times V \rightarrow \mathbb{C}$

$X^{\dot{a} a }: W^* \times V^* \rightarrow \mathbb{C}$

$X^{ a \dot{a}}: V^* \times W^* \rightarrow \mathbb{C}$

or something

I don't think what I wrote is right

It just means the indices transform under the relevant $M$ that I wrote at the very beginning, so that when you contract against a spinor the whole thing will produce an invariant

hello, i had a question about the experiment that shows the dependence of the gravity-induced phase on the rotation angle. so, in sakurai, he says this effect is purely quantum mechanical. the authors also say this and attribute it to the use of Bragg reflection for velocity selection. i am still just struggling to see why this phenomenon is purely quantum mechanical.

i also thought this experiment is motivated by wanting to show that there is a mass-dependent impact on the trajectory/positions of objects purely in a gravitational field. i do not understand why the effect has to be purely quantum mechanical for this to be the case. btw this is the paper I am referencing. journals.aps.org/prl/pdf/10.1103/PhysRevLett.34.1472

@Relativisticcucumber Like the Aharonov-Bohm effect, it is "purely quantum" because it acts on the phase of the wavefunctions

this notion of relative phase that is measured in interferometers exists classically only for waves, not particles, so it can't be a "classical" effect because classically you can't even say what the effect is supposed to be

In addition to that, the fact that the mass plays a role in the phase shift is a quantum feature

I mean, it's not as if "quantum" has a fixed formal meaning. Sometimes people just mean that the effect vanishes as $\hbar \to 0$ if they say it's quantum :P

The world is quantum

and ruled by quants

Hey guys

I have a problem with methane combustion

In the NO–O3 system for exemple, only 9.6 kJ/mol is required to react and it produces ΔE is −200.8 kJ/mol so here I understand that it can be self-sufficient. But why is the activation energy here so low and different than bonding energies summed ?

Because for methane maybe we only need to break one C-H bond for the free radical to be so reactive that the chemical reaction is taking place in a chain reaction and that the heat released is superior to breaking another C-H bond of another methane molecule and so on

The enthalpy of combustion is 890.8 kJ/mol. That should be enough to break another C–H bond of methane at 439.3 kJ/mol.

Anyway, how did you start the first combustion? The heat that started the combustion is still somewhere. The enthalpy of combustion just feeds more heat to the fire.

I thought that we needed to break all 4 C-H bonds and the 2 O=O bonds in order for the reaction to occur

Well imagine that the heat the starts the combustion is just breaking the initial bonds

But now I read some Stanford course on kinetic chemistry and I understand a bit more than a chemical reaction is before all a succession of elementary reactions

And so instead of breaking the 4 C-H et the 2 O=O bonds, maybe just breaking one C-H is sufficient to start the chain reaction

(i am reading this course : cefrc.princeton.edu/sites/g/files/toruqf1071/files/Files/…)

I try to understand it because I need to explain why fire is self-sustaining

And if we use the basic chemistry bonds model, we find that for methane combustion, the bond breaking energy is 2000 kj/mol and the products are 3000 kj/mol

So we have a heat output of 1000 kJ/mol

And that is just not sufficient to break all 4 C-H bonds and the 2 O=O bonds

With this model, to explain self-sufficiency, it would be required to have and net enthalpy >= activation energy = sum of all breaking energy of reactants bonds

which is not the case

And now with seeing a chemical reaction as a sum of elementary reaction, we can probably see the activation energy to be just the breaking of ONE bond !! Which the exothermic reaction provides and so it is exponentially self-sustaining

$N x N^\dagger$ as a tensor equation should be (ignoring dotted indicies for now) $N_{a}{}^b (\sigma^\mu){}_{b}{}^c N^\dagger{}_c{}^d$

Which gives $-N_a{}^b (\sigma^\mu)_{bc} N^\dagger{}^{cd}$

$N_a{}^b (\sigma^\mu)_{bc} N^\dagger{}^{c}{}_d$

$N_a{}^b (\sigma^\mu)_{bc} N^*{}_d{}^{c}$

Oh everything is fine

But there is no point in having the $\bar{\sigma}$

Its just $(\sigma^\mu)^{\dot{a} a}$ which we precisely have a symbol for

i.e. $(\sigma^\mu)^{\dot{a} a}$

It's like defining $g=g_{\mu \nu}$ and $g^{-1}=g^{ \nu \eta}$. But whats the point?

It already works with the indicies

I mean it may come in handy

You then get into weird unnessacary things like $g_{\mu \nu}=(g^{-1})_{\nu \mu }$ or something

Pauli even avoided this stupid bar thing

@ACuriousMind physics.stackexchange.com/questions/46015/why-quantum-mechanics I'm confused about the top answer here. What is the difference between the Poisson bracket lie Algebra and the Hamiltonian symplectomorphism Lie algebra?

It says the latter is a central extension of the former by the line lie algebra

I only know about one Lie algebra from classica mechanics : The Poisson bracker defined between phase space functions

@RyderRude The Poisson algebra includes constant functions $f(q,p) = c\in\mathbb{R}$, but the Hamiltonian vector field of a constant function is the zero vector field, meaning the map from phase space observables to Hamiltonian vector fields has as its kernel the constant functions (and the constant functions are of course just $\mathbb{R}$)

Thanks

@DIRAC1930 Does $N^{\dagger}$ have dotted or un-dotted indices? Since $N^{\dagger} = (N^*)^T$ it should have dotted indices not undotted indices, so already you've made one big mistake by trying to ignore the $\overline{\sigma}$, another mistake is your $\sigma^{\mu}$ now has two undotted indices which is wrong.

To motivate the bar, in $\sigma^{\mu}$ the dotted index is on the right and all indices are lowered, I don't need to display the indices the convention is fixed. If I start playing with the indices I need to make this clear. My derivation showed that $\overline{\sigma}^{\mu}$ has dotted indices on the left and the indices are raised, the bar makes this clear without displaying the indices, that's all there is to it.

If I want to obtain the $\overline{\sigma}^{\mu}$ starting from a given $\sigma^{\mu}$, I need to multiply my $\sigma^{\mu}$ by certain matrices to end up with the $\sigma^{\mu}$ (i.e. $\overline{\sigma}^{\mu} = (\sigma^y \sigma^{\mu} \sigma^y)^T$), which gives the same matrix I get by writing it in components and using the definition $\overline{\sigma}^{\dot{b} b} = \epsilon^{\dot{b} \dot{a}} \epsilon^{ba} \sigma_{a \dot{a}}$

should be: "to end up with the $\overline{\sigma}^{\mu}$"

@bolbteppa Sorry my expression was just meant to be schematical

Another problem with it is the position of the indices on your $\sigma^{\mu}$, if you write it your way in terms of dots and expand out the $\dagger$ then the indices on the $N$ flip and you break the summation convention when written your way, which is why it's always written the way it usually is

Ignoring our specific case, for normal tensors (i.e. without the cc rep) is it correct to write $S M S^\dagger$ as $S_{a}{}^b M_b{}^c (S^\dagger)_c{}^d$?

Which will be $S_a{}^b M_b{}^c (S^*)^d{}_c$

There's something I'm missing

$S (\sigma^m)_{a}{}^{\dot{c}} S$ should be $S_{a}{}^b (\sigma^m)_{b}{}^{\dot{b}} (S^\dagger)_{\dot{b}}{}^{\dot{c}}$

loweing the index on the RHS should give $S_{a}{}^b (\sigma^m)_{b}{}^{\dot{b}} (S^\dagger)_{\dot{b}}{}_{\dot{c}}$

$=-S_{a}{}^b (\sigma^m)_{b}{}_{\dot{b}} (S^*){}_{\dot{c}}{}^{\dot{b}}$

For a general matrix defined as $M_a \ ^b$, I have no idea what $M^a \ _b$ is (until I start bringing in metrics, but even then it still reduces to the original definition at the end of the day)

The minus sign will disappear if I lower on the LHS

$M^T_a \ ^b = M_b \ ^a$

Yes my expression seems to be what is in texts

I think I was making a mistake before

$S \sigma^m S^\dagger := S_{a}{}^b (\sigma^m)_{b}{}^{ \dot{b}}(S^\dagger)_{\dot{b}}{}^{\dot{c}}$

Here the def of $\sigma^m$ is $(\sigma^m)_b{}^{\dot{b}}$

To find the one with lower indicies

We have $S_{a}{}^b (\sigma^m)_{b}{}^{ \dot{b}}(S^*)^{\dot{c}}{}_{\dot{b}}{}$

That's completely wrong because you are ignoring the definition of $S^{\dagger}$ is such that the indices end up getting flipped so the index convention is not broken, it's only looks broken because you've inserted a transpose there, but by including the transpose we know the index convention will look broken but it isn't, however this misunderstanding has led you to define the indices on $\sigma^m$ incorrectly

I can explain all this stuff to you, but it's taken a day to explain something as simple as the position of the indices on a matrix, I can spend another day on it but this is stuff you'll learn better by doing yourself, I'd suggest saving the questions for bigger things

I have done that when I do $(S^\dagger){}_{\dot{a}}{}^{\dot{b}} = (S^*){}{}^{\dot{b}}{}_{\dot{a}}$

The point is that this whole issue is due to an abuse of notation

I just explained why that is wrong above

The issue is they are using the word matrix when that terminology is only essentially for (1,1) tensors

Which $(\sigma^m)_{a \dot{a}}$ is clearly not

Anyway it reproduces whats in Tong

@bolbteppa This is wrong

Anyway we will have $S_a{}^b (\sigma^m)_{b \dot{b}} (S^*)_{\dot{c}}{}^{\dot{b}}$ when done completely

From which it is defined that $(\sigma^m): = (\sigma^m)_{b \dot{b}}$

i.e. it is a definition

We are going from $S X S^\dagger$ which is a matrix equation

you cant just define $X$ to have lower indicies from the start

Or maybe you can who knows

Yes that seems true

I have no idea what $M^a \ _b$ is

I don't know why people write it though

It's like everywhere

Unless it's another convention

Anyways I'm going to give up on this because it's pointless since everything is just a random convention. Non of the authors state whats a convention, whats a definition etc.

Well if I write $M X$ where $X$ is a vector as $M^a{}_b X^b$ then $X^T M^T$ would be $X_b M^b{}_a$ however this is not generally true, only for certain metric spaces I think

i.e. when $X^T$ is a covector and $X$ is a vector

I think it's true in normal Euclidian case

My $X$ here is a column vector

It's all consistent and it all makes sense

We started with $M \in \mathrm{SL}(2,\mathbb{C})$ and defined it's indices as $M_a \ ^b$ and $M_{\dot{a}} \ ^{\dot{a}}$. Then we found there are matrices, like $\sigma^{\mu}$, with indices in different positions, $\sigma^{\mu}_{a \dot{a}}$, found by studying $P' = M P M^{\dagger}$. Another place where this happens is in studying $\epsilon = M \epsilon M^T$ (which is basically just $\det(M)$ written in $\epsilon$ notation), which tells us where to put the indices on $\epsilon$ (and it's inverse)

The quantum states for electrons, when we consider the Dirac Eq., meaning when we consider relativity in quantum mechanics, represent unbound states?

Is it correct to say that according to Dirac: The vacuum, is the space where all the particles with negative energy eigenstates reside ?

How much energy would it take for us to do this? ^

Why, when we consider postrons, the direction of the momentum is opposite to the direction of the propagation of the fermionic particles (positrons in this case) ?

@imbAF What do you mean by "direction of propagation" if not the direction of momentum/velocity?

well, apparently in what was drawn in our class, for positrons, the direction in which they physically move, is not the same with the one of the momentum

for electrons, that is always the case

that sounds like nonsense to me :P

well it cannot be

because we did several sketches where:

we consider an electron positron collision, and for the electron we have two arrows pointing in the same direction, while for the positron one arrow points in the direction of the region where it collides with the electron, and the other arrow in the opposite direction

I'll draw it

...and what are these "two arrows" you're drawing supposed to mean? If someone just drew two random arrows pointing in opposite directions next to a particle, I would not immediately interpret that to mean "the momentum is opposite to the direction of propagation"

and what kind of diagrams are you drawing anyway? Is this a Feynman diagram with momenta added?

on the topic of smart birds

this bird is basically archimedes

@imbAF that's a Feynman diagram and the arrows sitting right on the lines are not the "direction of propagation"

what are they?

cuz the arrows close the e must point at the direction of the propagation of them right?

@imbAF they're just a marker for whether the particle described by the line is a fermion or its anti-particle

(oh sorry, didn't realise you guys were mid conversations)

but we already have written e^+

isn't that enough to show that?

you have to pick a convention (e.g. arrows pointing to the right are fermions and arrows pointing to the left are antifermions)

@imbAF sure but usually one doesn't write that

Ok, so the arrows in the black lines, show the type of the particle?

@antimony that's actually not unusual for crows!

@ACuriousMind its incredible, they are so smart :)

some of them are probably smarter than some humans haha

In the Dirac sea model, the positron is the hole in the sea, is that correct?

@antimony maybe the only reason we and not them dominate the planet is that we figured out agriculture first ;)

@imbAF yes, but really the Dirac sea is an obsolete concept, it is of no relevance other than historical

But that is what I am doing right now, and I have trouble understanding a simple thing

The way I see it, this model implies that the positron is always present, and if we are able to somehow change the negative energy of the electron into positive, we "make" a positron visible

again, the Dirac sea is not an accurate model

But it's my lecture of this week

it's obsolete - if you find something about it that doesn't make sense, then that's not a problem: No one is claiming it's true anymore!

So while obsolete, is an in between step towards QFT

@ACuriousMind hahah yes because from our agriculture we ended up with the seeds to put into puzzles for the birds to solves

@antimony lol

@ACuriousMind Ok, just for the sake of the argument, help me understand a simple thing

It was said: Hole= absence of an electron with negative charge and E<0

(...but what I meant was that agriculture is more or less the beginning of human civilization as we would recognize it today as the humans settled down near their crop fields)

= relative to the vacuum : presence of a particle with positive electron and E>0=positron

The electron which isn't anymore in the dirac sea "space", is it in the region where all the electrons with positive energy are located? Did it's energy become positive because, let;s say it absorbed some gamma radiation?

@ACuriousMind ahh yes true :) i read that may have happened as they could have noticed plants they ate and discarded on their rubbish pile ended up regrowing the same plants

@imbAF the idea is more that "lifting" an electron from the sea to positive energy corresponds to pair production

Ok, cuz I have one question in between, before I further try and understand how what you said happens in reality

As I wrote, we said, in the lecture: relative to the vacuum : presence of a particle with positive electron and E>0=positron

I understand that, since the electron left, the region where the electron was, now it has 0 charge value, which can be interpreted as if "something" with equal charge to the electron charge but possible took it's place, to bring the total charge value to zero. So I understand the first part of the interpretation, aka particle with positive energy

What i don't understand is, why we assume that is particle has also positive energy ?

@antimony I mean, the hard part probably isn't figuring out how seeds work but to actually benefit from that. The yield of most wild crops is rather low and you need to defend your field year-round from all the other animals that would like to snack on your crops

I understand the positive charge, but why positive energy?

aka particle with positive charge*

@imbAF It's just energy conservation: When you put energy $E$ into an electron with negative energy $-E_0$, you end up with an electron of energy $E-E_0$ and a hole. Since you invested $E$ "into the vacuum" but the resulting electron has only energy $E-E_0$, you must interpret the hole to have energy $E_0$ for energy to be conserved.

with that into vacuum part

I got it

Because once you give the energy to the electron, the electron is "visible" (lack of better words) to the non vacuum region, but it's energy is not equal to the energy spent, it's less then that, which means that energy that is missing goes to the hole, right?

yes, that's the idea

(but again, that's not really how we think about it today)

yes I know, that my pov will change, but for the moment we are going with a historic way as to how things changed

And 2 more things:

@ACuriousMind hmmmm very good point i didn't think of that, it really was the end of nomadic life and the beginning of permanent structures

the energy of the hole is $E_0$ because $E-(E-E_0)=E_0$ which is the difference between the energy given and the one that,we see that the electron received

I assume, that's why its $E_0$