The h Bar

RewCie

5:12 AM

How do you feel Might Guy didn't kill Madara?

Slereah

5:27 AM

Those nlab diagrams are getting out of hand

Slereah

7:30 AM

apparently the world function goes back to at least Ruse : era.ed.ac.uk/handle/1842/30712?show=full

Physics Meta

8:02 AM

0

Q: Why does spacetime bend or curve in the downward direction?

In relativity , when mass bends spacetime in a downward direction. So ultimately we are coming to the conclusion that any mass body is also going down .

discussion

DIRAC1930

1:02 PM

So fields with dotted indices belong to $\overline{V}$ and fields with undotted indices belong to $V$

But why can't I just normally contract between them

As in $V^* \overline{V}$ or something

Where $V^*$ is a vector in the dual space of $V$

ACuriousMind

1:24 PM

"contraction" is the application of some dual vector to a vector

when you write $v_a w^a$, this is $v(w)$ for some $v\in V^\ast$ and $w\in V$. The application $v(w)$ is defined by the definition of what the dual is

there is no such application defined for a conjugate

so it just doesn't mean anything to "try to" contract a vector and its conjugate

DIRAC1930

Okay so I can't have $v^*(w)$ because it's not defined

sorry I mean $\overline{v}(w)$

ACuriousMind

while you might write something like $\bar{v}_a w^a$ for some $\bar{v}\in\bar{V}$ in a particular basis, this isn't "a thing" - there's no basis invariant way to write what is supposed to happen here

DIRAC1930

Ah okay

ACuriousMind

(another reason why coordinate/basis-free notation is great because you can't even write down the things that don't make sense :P)

DIRAC1930

So how come people seem to throw away these rules for $(\sigma_m)_{\alpha \dot{\alpha}}$?

There must be something weird going along with the Pauli matricies

ACuriousMind

1:28 PM

the annoying thing is that people usually work with spinors in a concrete representation

you choose some particular basis in which the $\gamma$ matrices have a particular form, and particular sign conventions

and then you do all sort of manipulations that don't really make sense abstractly but nevertheless lead to useful result in your chosen basis

bolbteppa

There is no distinction between dotted and undotted indices for $\mathrm{SU}(2)$, only for $\mathrm{SL}(2,\mathbb{C})$

DIRAC1930

What's the reason behind there being no distinction?

ACuriousMind

I rant about that in my question about Majoranas - it's extremely hard to find resources that actually explain what's happening here in a linear algebraic way and not in the physicist's "just write down the matrices and compute this stuff" way

@DIRAC1930 the conjugate representation is isomorphic to the original one for SU(2)

while the conjugate representation of the rep $(s_1,s_2)$ of SL(2) is $(s_2,s_1)$, i.e. not generally isomorphic

DIRAC1930

Okay thanks

Would this be easier just to do it from the Clifford algebra?

123

1:56 PM

Hi All...

How Moment of Force $M = r \times F$ equal torque $\tau = I \alpha$ .

In moment of force how $r$ and $F$ take of property rotational inertia.

fqq

3:07 PM

@ACuriousMind you can, e.g. the $\overline{v}(w)$ above makes as much sense and is as clearly wrong as \bar{v}_a w^a$

ACuriousMind

@fqq but $\bar{v}(w)$ doesn't tell you what to compute, it's unclear what its value would be. $\bar{v}_aw^a$ directly carries with it the recipe for how to compute it, so you have no chance to notice it's not actually something that was defined

fqq

$\overline{v}$ should have a dotted index

so if you parse it correctly, you see it's wrong

if you parse $\bar{v}(w)$ incorrectly you can also assign some computation to it

ACuriousMind

fair enough

fqq

but I see your point, many physicist probably would compute the one with indices blindly and would stop at the function notation

bolbteppa

These susy conventions are hard enough, the difficulty really ramps up when you try to do the sugra version

fqq

3:12 PM

to me they are really the same, I take the index notation to also be coordinate free (as in Wald)

DIRAC1930

I still haven't managed to construct a single invariant

Of the form $M_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$

fqq

3:29 PM

@DIRAC1930 I really don't think it is, the representation theory of SU(2) is pretty straightforward but maybe I'm missing something

Slereah

I guess the difference with the relativistic version is that since the spinor space and the conjugate one are isomorphic, the conjugate is an automorphism?

So you just have the same basis

Or you can write the isomorphism explicitly if you're bothered

I don't super know why you have to consider the conjugate space in the first place rly

I assume it's related to that conjugate rep business

but I'm not privvy to the esoteric mysteries of group theory

3

Q: Are the complex conjugate representations $e,\bar e$ and $d,\bar d$ for $\mathfrak{sl}(2,\mathbb{C})$ inequivalent?

Let $\mathfrak{sl}(2,\mathbb{C})$ be the complex Lie algebra of $SL(2,\mathbb{C})$ and $\mathfrak{sl}(2,\mathbb{C})_\mathbb{R}$ be its realification; that is $\mathfrak{sl}(2,\mathbb{C})_\mathbb{R}$ is $\mathfrak{sl}(2,\mathbb{C})$ considered as a real Lie algebra. Let $d$ be an irrep of $\mathf...

abstract-algebra complex-numbers representation-theory lie-groups lie-algebras

something about this I assume

physicsforums.com/threads/su-n-conjugate-representation.477798

also here

DIRAC1930

3:45 PM

I keep getting $M_{\alpha \beta} \psi^*{}^{\alpha}\psi^\beta \rightarrow -M_{\alpha \beta} \psi^*{}^{\alpha}\psi^\beta$ for literally everything and I don't know where I'm going wrong

fqq

@Slereah because you can, it's useful to build real lagrangians and the theories you build with are good

bolbteppa

$\psi^{\dagger} \gamma_0 \psi$ is an invariant under the rotation group, what's wrong with that

DIRAC1930

Yes, I've been trying to get that to work but I'm not doing the transformations correctly or something

bolbteppa

$M = \gamma_0$

Do you know how to do a rotation of a spinor

Slereah

I guess if you want to write it fancily you can use a map $f : \bar{V} \to V$

DIRAC1930

3:47 PM

$(\gamma_0)_{\alpha \beta} \rightarrow F_{\alpha}{}^\gamma F_\beta{}^\delta (\gamma_0)_{\gamma \delta }$ right?

Slereah

and then you have $f(\psi^{\dot{a}} t_{\dot{a}}) = \psi^{a} t_a$

Or somethibng

idk if the isomorphism from the spinor space to its conjugate has a standard symbol

but you can just write that $*$ is an isomorphism

$^* : V \to V$

fqq

@Slereah that looks $\mathbb{C}\to\mathbb{C}$, not $\bar{V} \to V$

DIRAC1930

and then $\psi^*{}^{\alpha} \rightarrow \psi^*{}^\gamma(F^{-1})^*{}_\gamma{}^\alpha$

Slereah

@fqq That's why I included the basis vectors

bolbteppa

Wait, maybe write out the invariance of $\psi^{\alpha} \psi_{\alpha}$ fully

DIRAC1930

3:50 PM

The $F's$ cancel and produce a $-$ sign for everything I try independently of what I choose the $M$ to be

fqq

clearly there's something wrong, but it's right that what you choose for $M$ does not matter, that's the whole point of the indices after all

DIRAC1930

But then I can't get anything invariant unless $M=-M$ which means $M=0$

fqq

no, you are doing something wrong, scalars exist

DIRAC1930

Trying to find what I'm doing wrong with all these different conventions is impossible

bolbteppa

It's certainly a nightmare, you should try to stick to one credible source and learn from that instead of reading around until after you get one down

Doing this non-relativistic stuff is a waste of time, do it the relativistic way first

Slereah

3:54 PM

From what I can see the basic difference between the relativistic and non-relativistic case is that one uses the actual $\mathfrak{su}(2)$ while the other one uses the complexification $\mathfrak{su}_{\mathbb{C}}(2)$

This is why you need the dotted indexes

because the complexified version of the algebra is not isomorphic to its conjugate

bolbteppa

From a non-relativistic quantum mechanical perspective, one finds $\mathrm{SL}(2,\mathbb{C})$ from the representation theory of angular momentum, but because the probability density $|\psi|^2 = |\psi_1|^2 + |\psi_2|^2$ must be invariant under these transformations, we must restrict to it's real form $\mathrm{SU}(2)$.

From a relativistic perspective, the probability density is just the time component of a four-current vector so there's no restriction, we're stuck with $\mathrm{SL}(2,\mathbb{C})$.

The Weyl representations $J_a = \sigma_a$, $K_a = + i \sigma_a$ and $J_a' = \sigma_a$, $K_a' = - i \sigma_a$ form in-equivalent representations of the Lorentz group because any transform that sent $J_a$ into $J_a'$ couldn't send $K_a$ into $K_a'$.

Slereah

4:12 PM

Anyway I guess the lesson to take is that it's kosher to write your spinor product as $(\psi^a)^* \psi_a$

Now you can stop worrying about that and start worrying about the ordering of your Lagrangian in Grassmann variables

bolbteppa

$(\psi^a)^* \psi_a$ is not an invariant, I think that may be the issue

DIRAC1930

Lol

Slereah

@bolbteppa Dag nabbit

bolbteppa

I know what you meant but I held off suggesting it earlier so good opportunity :p

Slereah

What's the lagrangian of the superpoint

bolbteppa

4:17 PM

The usual $\dot{x}^2 - \psi i \partial_t \psi$ right

Slereah

Does that really change much?

I don't think rotation would change under $\partial_t$

bolbteppa

That's actually assuming a massless particle through using the whole einbein stuff

Slereah

Although I guess it might for Galilean transforms

Semiclassical

Here’s something on my mind lately

DIRAC1930

I give up lol

Semiclassical

4:24 PM

How Kepler orbits are taught in intro courses emphasizes conservation of energy and conservation of angular momentum

Which, within context, entirely makes sense

But one thing that forces upon you is a sense of scale: how far are the various planets from the sun

Slereah

Grosche suggests $i\bar{\eta} \dot{\eta}$ indeed

idk if it's the most general one

Semiclassical

And the funny part of this to me: historically, distances are exactly what you didn’t know

Slereah

@Semiclassical would you rather they teach the historical reasoning

Semiclassical

All you could observe was where planets were in the sky, ie, the angular part of spherical coordinates

Slereah

Planets move on those orbits due to esoteric reasons

"the universe itself was an image of God, with the Sun corresponding to the Father, the stellar sphere to the Son, and the intervening space between them to the Holy Spirit."

Teach the controversy

Semiclassical

4:29 PM

For an intro course, no. But it’s just a bit funny how we we simultaneously emphasize Kepler’s role in history conveniently ignoring how limited their tools actually were

Slereah

I'm guessing you could probably try to work out the most general Lagrangian for non-relativistic (free) spinors the same way as the one for point particles

Just find the most general quadratic Lagrangian that's invariant under the Galilean group

Since the Lagrangian also has to be real I guess it somewhat limits the number of possible terms

Can't have any linear term due to this

I guess you're limited to the bilinear terms of $\psi$ and $\dot{\psi}$

idk what's the space algebra for $SO(3)$ but I'm guessing it's similar-ish to the Lorentz group

Although since you have $\sigma_j \sigma_k = \delta_{j k}I + i\varepsilon_{j k \ell}\,\sigma_\ell$, I'm guessing the algebra's basis is just $I$ and the Pauli matrices?

Semiclassical

the amount of stuff which astronomers were able to achieve over the centuries does blow my mind at times

Slereah

@Semiclassical it wasn't easy

For a while they just kind of had to look at the sky and figure it out

Semiclassical

ya

Slereah

Also this was even before the technology for a notebook

if you had to write something down you had to carve a wax tablet or something

Semiclassical

4:39 PM

like, how tf did Kepler come up with this picture: researchgate.net/figure/…

bolbteppa

$I$ and those is the full clifford algebra in 3D,

Slereah

yeah looks like it

@Semiclassical sorcery

bolbteppa

There's a Schrodinger equation for the non-relativistic case, everything passes through it

Semiclassical

@Slereah i think this is related to the following fact: you can write down the Dirac equation in 2+1 spacetime using just 2-by-2 matrices rather than having to go through 4-by-4 gamma matrices

Slereah

I'm not even sure how anyone had the patience to notice that the celestial sphere rotates

But I guess this was before television

Semiclassical

4:41 PM

b/c in that case you only need three gamma matrices rather than the usual 4

bolbteppa

They still haven't figured out that it's all just a big turtles shell rotating about as it shuffles

Semiclassical

and the pauli matrices suffice for those (up to factors of i)

though i don't know what the 'best' choice is there. this came up as a HW problem for a course i'm grading recently

Slereah

I don't even know how we can guarantee that $\bar{\psi} \dot{\psi}$ is real

Semiclassical

e.g. you need $\{\gamma^0,i\gamma^1,i\gamma^2\}$ to be the three Pauli matrices. but what's the nicest choice to assign that?

it can't make a difference at the end, but what's the least tedious choice along the way?

Slereah

I guess it doesn't matter much for the EoM since it turns out linear

But how can we be sure that the kinetic term for spinors is real in the Lagrangian?

Is it always real or only on shell?

Although... I guess the spinor part is all Grassmann stuff so it's all linear

So it's not too hard to work out

it's probably fine

why no $\bar{\psi} \psi$ terms allowed?

is it one of those terms that can just be integrated away

Grosche has some non-relativistic term that's proportional to $[\bar{\psi}, \psi]$

bolbteppa

4:51 PM

I'd say it is but they usually discuss a massless particle

Slereah

It's non-relativistic, isn't the mass a factor of the kinetic term?

from what I can see there are Lagrangians with $\bar{\psi} \psi$ terms, but those are more for potential terms

like the constant magnetic field Lagrangian

Also there seem to be terms in $\bar{\psi} \sigma^\mu \psi$ for EM stuff?

when currents are needed

DIRAC1930

I'm specifically looking for potential terms

Slereah

Well use the spinor and its derivative and their conjugates as fields, and the two possible bilinears

Also since it is grassmann variables I think you can exhaust every possible polynomial terms

rob

5:07 PM

2

A: Anomalous muon magnetic moment and muon catalyzed fusion

When cannonading ghosts, it's very hard to tell what you shot down... Theorists do speculate; but, counterintuitively, they have pretty stringent rules of how to do that, and your question is actually quantitative (!!). For the muon magnetic moment anomaly, we know that could be smaller than one ...

> When cannonading ghosts, it's very hard to tell what you shot down.

I appreciate Cosmas's writing so much.

DIRAC1930

@Slereah I cant get anything to work

Slereah

@DIRAC1930 that's physics for you

DIRAC1930

$\psi^\dagger \psi$ should be one I'm assuming because it's used everywhere in non-rel phys

which would be $\delta_{\alpha \beta} \psi^*{}^\alpha \psi^\beta$ right?

Slereah

I s'ppose

Hm

Why are they mixing spacetime and spin indices

DIRAC1930

What is that dot on top of the $x$?

Slereah

5:15 PM

Time derivative

a less mysterious example

"It is real up to a total time derivative"

Dang

123

In moment of force $M = r \times F$ formula, how we take care of moment of inertia.

Slereah

complex terms swept under the rug of the De Rham cohomology

bolbteppa

ncatlab.org/nlab/files/DermisekDiracField.pdf

Slereah

Question is, does this argument work for QM?

Is $\dot{\bar{\psi}} \dot{\psi}$ illegal

bolbteppa

If you don't do relativity, you don't have the same kind of constrained thinking

The potential can basically be anything to a first approximation

Slereah

5:22 PM

I don't think I've ever seen that term, though

bolbteppa

Relativity and covariance just makes everything easier

Also what are you even doing non-relativistically with fermions except looking at multiparticle systems so you can take your multiparticle Hamiltonian and turn it into second quantized form for fermions

Slereah

What would even be the EoM of that term, $\ddot{\psi} = 0$?

Also wait, are we doing the lazy physicist Lagrangian

bolbteppa

$\psi = \psi_0 + a t + \frac{1}{2} g t^2$ :\

Oh wait, with $g = 0$

Slereah

I thought the proper one was like $$\frac{1}{2} (\bar{\psi} \sigma^\mu \partial_\mu \psi + \psi \sigma^\mu \partial_\mu \bar{\psi})$$

I guess $\ddot{\psi}$ doesn't constrain the equation since it's always zero?

Because Grassmann

although idk if that's true if we add all the other terms

but since $\ddot{\psi}$ is always zero I suspect you can probably get rid of that term via some Grassmann magic

I think technically you need the symmetric one because otherwise the EoM of the spinor and its conjugate will be different?

idk it's one of those things people don't bring up much

$C$ symmetry or whatever

Oh wait, IIRC those two Lagrangians are equivalent under partial integration

nvm

DIRAC1930

5:54 PM

Okay, so far I can get $\epsilon_{\alpha \beta} \psi^\alpha \psi^\beta$ being invariant which I think is a known invariant

according to L&L

Slereah

Seems plausible

DIRAC1930

Now I need to see if I can find one that is real

Slereah

Now you just have to work out uuuh 31 other terms

Well, 16 up to symmetry

Also 8 i guess since psi² = 0

DIRAC1930

Why can't I just have $(\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta + \mathit{c.c.})$ as a real invariant?

DIRAC1930

6:17 PM

To have $M_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$ to be invariant, I effectively need $M$ to satisfy $2M_{\alpha \gamma}(F^T){}^\gamma{}_\beta = M_{\alpha \beta}$

Which doesn't look like is possible

Which I think suggests (in matrix notation) that $2 F^T = \mathbb{I}$ which is impossible

bolbteppa

Where is the 2 coming from

DIRAC1930

$F{}_\alpha{}^\beta F{}_\beta{}^\gamma = 2 F{}_\alpha{}^\gamma$ I think

Since 2 transformations are being done in succession

bolbteppa

6:33 PM

All there is to it is noting, for $M \in \mathrm{SU}(2)$, that $\psi'^a \psi_a' = (\psi M^{-1})^a (M \psi)_a = \psi^a \psi_a$. So $(\psi'^a)^* \psi_a' = [(\psi M^{-1})^a]^* (U \psi)_a \neq \psi^a \psi_a$ because $(U^{-1})^* U \neq I$.

Slereah

You're mixing your notation :p

@bolbteppa wouldn't that argument also hold for $\bar{\psi} \dot\psi$?

Also as mentionned some Lagrangians do have $\bar{\psi} \psi$

Also wait why is $\psi^a$ transformed to $\psi^a M^{-1}$

Wouldn't the transformation here be $$\langle \bar{\psi}, \psi \rangle \to \langle \bar{M\psi}, M\psi \rangle = \langle M^*\bar{\psi}, M\psi \rangle = \langle M^{-1}\bar{\psi}, M\psi \rangle = \langle \bar{\psi}, \psi \rangle $$

replace with $\delta_{ab} \psi^a \psi^b$ for appropriate component

Man, looking up how spinors in $R^3$ work again in Landau and I still don't know

bloody non-commuting things

It does seem that the spinor metric is still $\varepsilon$?

But the inner product is $\psi^1 \psi^{1*} + \psi^2 \psi^{2*}$

bolbteppa

6:56 PM

Oh damn $M = U$ :p

What's $\overline{\psi}$, is it $\psi^{\dagger} \gamma^0$ or $\psi^{\dagger}$

Slereah

Well the conjugate I would assume, but now considering that the musical isomorphism isn't trivial, I'm not sure

bolbteppa

$U^{\dagger} = U^{-1}$

Slereah

Trying to think of a book that may discuss it

Hall maybe

bolbteppa

Need to be careful about whether you're using $\mathrm{SU}(2)$ or $\mathrm{SL}(2,\mathbb{C})$, $\psi^{\dagger} \psi$ is invariant under $\mathrm{SU}(2)$ but not $\mathrm{SL}(2,\mathbb{C})$

DIRAC1930

Isn't the whole point that any term with fully contracted indices is invariant? (for $V$ anyway)

Slereah

7:00 PM

Well we're talking SU(2) here, since we're doing non-relativistic QM I assume?

bolbteppa

There's a discussion of this in L&L in the spinor chapter, they even set up raising and lowering indices for them

Slereah

yeah

DIRAC1930

$M_{\alpha \beta}\psi^\alpha \psi^\beta$ is invariant for any $M$

Slereah

they do seem to imply that the inner product is just the obvious one, but the map between dual spaces makes things a bit trickier to be sure

I thought the metric of a vector space defined its musical isomorphisms

bolbteppa

When you write $M$ that way you assume it transforms as a 'tensor' under $\mathrm{SU}(2)$ transformations

Slereah

7:07 PM

He defines the bilinear form $\psi^1 \phi^2 - \psi^2 \phi^1$ and the norm $\psi^1 \psi^{1*} + \psi^2 \psi^{2*}$

"$\psi^{1*}$ and $\psi^{2*}$ must be transformed as $\psi^2$ and $-\psi^1$ respectively"

ugh

I need to brush up on my spinors I think

Is the dual space isomorphic to the conjugate space???

bolbteppa

7:20 PM

What's the conjugate space

Slereah

Complex conjugate representation

nice cover

it seems like the appropriate book for the topic

bolbteppa

Basically an insane book

It starts by using complex vectors under the Euclidean dot product, fun times

Slereah

Let's find out

bolbteppa

The approach in that book is very confusing because the way he starts doesn't generalize

Pure spinor

In the domain of mathematics known as representation theory, pure spinors (or simple spinors) are spinors that are annihilated under the Clifford action by a maximal isotropic subspace of the space V {\displaystyle V} of vectors with respect to the scalar product determining the Clifford algebra. They were introduced by Élie Cartan in the 1930s to classify complex structures. Pure spinors were a key ingredient in the study of spin geometry and twistor theory, introduced by Roger Penrose in the 1960s. == Definition == Consider a complex vector space...

Buried a hundred pages or so in you find out he's just doing a special case of the normal Clifford thing

Slereah

Can i skip it

bolbteppa

7:28 PM

Unless you want to get into the weeds of one approach I'd say so

But look at the GR chapters in there

DIRAC1930

How do I construct different $M_{\alpha \beta}$'s?

Can it be anything or do they have to be built out of something specific

like is it built out of spinors $\psi_\alpha$?

Slereah

I would guess it's built from the algebra of physical space

DIRAC1930

Might look in the book Roger Penrose says was impenetrable or something

Introduction to Tensors Spinors and Relativistic Wave Equations

bolbteppa

7:45 PM

I certainly closed it multiple times

DIRAC1930

Okay so I made a mistake earlier. To have $M_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$ to be invariant, we require $F^T F =\mathbb{I}$ which is impossible

Slereah

"A vector is said to be isotropic if its scalar square is zero."

DIRAC1930

Something weird is going on when we mix up $V$ and $\overline{V}$

Slereah

ah right

Bloody mathematicians

bolbteppa

A complex vector squared under the Euclidean dot product, madness

DIRAC1930

7:58 PM

Something weird is happening

I think $F^T F = \mathbb{I}$ is the correct constraint which is impossible

Wtf in L&L it says that the way $\psi{}^*{}^\alpha$ transforms is related to $\psi^\alpha$ by arguments involving the change in sign when reversing time.

I think this non-rel stuff is weird

DIRAC1930

8:21 PM

I'm surprised it's difficult to find things to do with all of this online

Slereah

8:51 PM

looking at Chevalley's spinor book

Not the worst but he has terminal mathematician disease

DIRAC1930

If I have $M$ transforming like $M_{\alpha \dot{\alpha}}$ it works I think but that seems like an illegal operation

DIRAC1930

9:08 PM

So I would need a contraction like ${\overline{R}}_{\dot{\alpha}} {S}_\alpha \overline{\psi}{}^\dot{\alpha} \psi^\alpha$ or something which is probably wrong

Slereah

You can tell Chevalley is doing too much because he keeps talking about $\Gamma$-graded algebras

what weirdo does anything but $\mathbb{Z}_n$ graded algebras

also he is terribly wrong because he says "semi-graded" instead of "super"

DIRAC1930

Does the $\dagger$ on the field terms arise when considering the quantum inner product?

As in after cann. quant.

DIRAC1930

10:33 PM

I'm not sure about Landau's argument

So we have $\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta$ being invariant

Then he says $\psi_\beta^*$ transforms like $\psi^\beta$ which is true

And I think that's how he justifies $|\psi^1|^2 + |\psi^2|^2$ being invariant

Actually it works I think

So he has $\psi^\alpha = (\psi^1 ,\psi^2)^\alpha$ essentially

$\psi_\beta^*$ transforms like $\psi^\beta$, so he defines the components of $\psi^\beta$ with the components of $\psi_\beta^*$

imbAF

@ACuriousMind Hey man, may I ask you for your help in thermodynamics regarding something?

DIRAC1930

So $\psi^*_\beta = \psi^\beta= (\psi_1^* ,-\psi_2^*)^\beta$

Sorry that was meant to be $\psi^*_\beta = \psi^\beta= (\psi^*{}^1 ,-\psi^*{}^2)^\beta$ I think

And then when replaced in the $\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta$, you get $\psi^1{}^*\psi^1 + \psi^2{}^* \psi^2$

Does this seem correct?

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