# Regarding spinor technology

1. Sep 7, 2014

### kau

I got this while I was reading spinor indices manipulating in Sredinicki's qft
in case of spinor representation we get a relation like the following one:

$[(ψ_a^{'}(0))^{\dagger},M^{\mu \nu},]= [(S^{\mu \nu}{}_R)_{a^{'}}{}^{~b^{'}}](ψ_{b {'}}(0))^{\dagger}$
where $a^{'}$ and $b^{'}$ represent right handed spinor and unprimed version of it represent left handed spinor. $M^{\mu \nu}$ is lorentz group generator.
and ψ with dragger and without dragger represent right and left handed spinor respectively.
if I take hermitian conjugate of above
$[M^{\mu \nu},(ψ_{a}(0))]= [(S^{\mu \nu}{}_R)_{a^{'}}{}^{~b^{'}}]^* ψ_{b}(0)$
my question is why we don't need to take transpose of $S^{\mu \nu}$,why taking only complex conjugate is enough???even we are not changing spinor index from primed(dotted in Sredinicki's book) to unprimed(undotted).it's a matrix afterall.. also the ordering in the right hand side of eqn 2 should be interchanged since they are matrix and we are taking hermitian conjugate of them.. (ref is sredinicki's qft page no 211 eqn no 34.15 and 34.16).
also to link between a vector and two spinor indexed wave function (where one is left handed and other is right handed) Sredinicki introduced an object like $σ ^{\mu}{}_{a a^{'}}$ my question is - it's a number or matrix.. because I suspected that Sredinicki assumed commutating property of it in deriving 35.15 in his book.
Thanks,
Kau.

Last edited: Sep 7, 2014
2. Sep 9, 2014

### muppet

I suspect that here the $M^{\mu\nu}$ are operators that act on Hilbert space, which is different to matrices that act in a vector or spinor space with which the indices on e.g. a Dirac field are associated. (Think the angular momentum operator in some abstract representation, and not a matrix!). The tensor indices on M here really indicate that you have a set of operators that get shuffled into each other under Lorentz transformations.

When you take the Hermitian conjugate of the first line, you're basically just taking the hermitian conjugate of some operator which happens to be expressed as a linear combination of other operators, so you hermitian conjugate the operators and complex conjugate the coefficients.

I hope that helps.

Last edited: Sep 9, 2014
3. Sep 9, 2014

### Avodyne

Yes, this is correct. Suppose we have $N$ operators $A_i$, $i=1,\ldots,N$, and also $N$ operators $B_i$. Suppose that they are linearly related,
$$B_i = \sum_{j=1}^N c_{ij}A_j$$
where $c_{ij}$ is a complex number. Taking the hermitian conjugate of this relation yields
$$B_i^\dagger = \sum_{j=1}^N c^*_{ij}A^\dagger_j$$
The $c$ matrix gets complex conjugated, but not transposed. (If this is not clear, try setting $i$ to a particular value, like $i=1$.)

In the OP's formula, $M$ and $\psi$ are operators, but the $S$ coefficients are complex numbers.

4. Sep 11, 2014

### kau

But you see the lorentz indices S matrix got... what about them. Also the thing is if you take dragger then spinor indices S matrix has got those $a^{'}$ and $b^{'}$ they should be a and b under taking hermitian conjugation. Isn't it? I am convinced that ok If $M$ and $ψ$ are operator then they should go change under hermitian conjugation. But since s is like a co efficient so it doesn't change. Bt you convince me that $S$ is indeed like a coefficient here. I can suggest you to look at Sredinicki before answering it since I am assuming many things that is done there and not writing them in details here. anyway if you are already done with it feel free to clear my doubts, i am eagerly waiting.

5. Sep 11, 2014

### ChrisVer

It's a vector: $\sigma^{\mu} = (1, \sigma^{1}, \sigma^{2}, \sigma^{3})$ whose components when they act on spinors are represented by matrices with elements:
$(\sigma^{\mu})_{a \dot{a}}$
there are commutation relations of this, since it has the Pauli-$\sigma$ matrices...

6. Sep 11, 2014

### ChrisVer

yes, after taking the conjugate the spinor indices change from the (0,1/2) to the (1/2, 0) representation.
However in what you've written, the indices haven't changed yet -the conjugate isn't written inside the indices.]

7. Sep 11, 2014

### kau

ok... that i understood but my problem is see in deriving 35.15 he multiplied the previous eqn by $(\sigma^{/rho})_{a \dot{a}}$ and it seems to me there he didn't care about the ordering of these stuff... that confused me.. please checkit and tell me the reasn..

8. Sep 11, 2014

### ChrisVer

$(g^{\mu \rho} \delta^{\nu}_{\tau} - g^{\nu \rho} \delta^{\mu}_{\tau}) \sigma^{\tau}_{a \dot{a}} + i (S^{\mu \nu}_{L})_{a}^{~~b} \sigma^{\rho}_{b\dot{a}} + i (S^{\mu \nu}_{R})_{\dot{a}}^{~~\dot{b}} \sigma^{\rho}_{a \dot{b}} =0$

Now multiplying with the $\sigma_{\rho c\dot{c}}$:
$(g^{\mu \rho} \delta^{\nu}_{\tau} - g^{\nu \rho} \delta^{\mu}_{\tau}) \sigma_{\rho c\dot{c}} \sigma^{\tau}_{a \dot{a}}+ i (S^{\mu \nu}_{L})_{a}^{~~b} \sigma_{\rho c\dot{c}} \sigma^{\rho}_{b\dot{a}} + i (S^{\mu \nu}_{R})_{\dot{a}}^{~~\dot{b}} \sigma_{\rho c\dot{c}} \sigma^{\rho}_{a \dot{b}}=0$

The first terms are the same as in Schre, so I think you have problem with the others?
It's a matter of algebra to show that you can move them...
$g_{\mu \nu} \sigma^{\mu}_{a\dot{a}} \sigma^{\nu}_{b \dot{b}} = 2 \epsilon_{ab} \epsilon_{\dot{a} \dot{b}}$
(this can be proven by taken the completeness relation for the sigma matrices.)

$g_{\mu \nu} \sigma^{\mu}_{a\dot{a}} \sigma^{\nu}_{b \dot{b}} = 2 \epsilon_{ab} \epsilon_{\dot{a} \dot{b}} = (a \leftrightarrow b) \& (\dot{a} \leftrightarrow \dot{b})= g_{\mu \nu} \sigma^{\nu}_{b \dot{b}}\sigma^{\mu}_{a\dot{a}}$

that's how you can do that
so
$\sigma_{\rho c\dot{c}} \sigma^{\rho}_{a \dot{b}}=g_{\tau \rho} \sigma^{\tau}_{ c\dot{c}} \sigma^{\rho}_{a \dot{b}}=g_{\tau \rho} \sigma^{\rho}_{a \dot{b}}\sigma^{\tau}_{c\dot{c}}=\sigma^{\rho}_{a \dot{b}}\sigma_{\rho c\dot{c}}$

Last edited: Sep 11, 2014
9. Sep 11, 2014

### kau

Thanks.. thanks a lot buddy.. for explaining it clearly.. I got it. well as you have gone through these so it would be good for me to keep you questioning on doubts I will get during reading this book. hope you won't mind to explain. When you will be free answer to this ques.. see I still have doubt on the first question why even under hermitian conjugation the spinor dotted indices didn't flip and though it's carrying $\mu$ $\nu$ we are only taking complex conjugate..why???? why $S$ is just a coefficient[\B]... next, if $\sigma^{\mu a\dot{c}}$ is a matrix (now I am convinced that it's a matrix) then explain me this
$[/psi^{\dagger}\bar{\sigma}a\dot{c}) \chi]^ {\dagger}=[\psi^{\dagger}{}_{\dot{a}}\bar{\sigma}^{a\dot{c}}) \chi]^ {\dagger}{}{c}= \chi^{\dagger}{}_{\dot{c}} (\bar{\sigma}^{\mu a \dot{c}})^{\ast} \psi_{a}=\chi^{\dagger}{}_{\dot{c}} ( \bar{\sigma}^{\mu \dot{c} a}) \psi_{a}$
show me here the $\dagger$(hermitan conju) of $\sigma$ matrices has been properly taken and how do they use hermiticity of pauli matrices here.
Now tell me why
$U(\lambda)^{-1} \psi_{a} U(\lambda)= L^{~~c}{}_{a} (\lambda) \psi ^{b} (\lambda^{-1} x)$ but in case of scalar field the right hand side was simple $\lambda$ matrices only which was acting on $\phi (x)$ and that was equivalent to writing $\phi(\lambda ^{-1}x)$ but in spinor case we are writing the argument on right hand side as $(\lambda^{-1} x)$ but we are multiplying by $L^{~~c}{}_{a} (\lambda)$ quantity. why???? I can understand now spin is there so this extra thing.. but is there other reason to see this in this way. please don't explain it in terms of detail group theory. because we still don't have background on representation stuff, this is in our next sem. tell me if there is any hand waving way to view this.
third one.... we have this $U(1+\delta\omega)= I + \frac{i}{2} \delta\omega_{\mu\nu} M^{\mu \nu}$ .... we also have $L^{~~B}{}_{A} (1+{\delta \omega})= \delta ^{~~B}{A}+\frac {i }{2}\delta\omega_{\mu\nu} (S^{\mu \nu})^{~~B){}_{A}}$ my ques is $(S^{\mu \nu})^{~~B}{}_{A}$ quantity isn't same as $M^{\mu \nu}$ I know they are not... $M^{\mu \nu}$ are lorentz generators and carrying no spinor index.. form of $U(1+\delta\omega)$ remain unchanged whether ir's a vector or scalar or spinor field... but trnasformation matrix changes like for lorentz indices it's $\lambda ^{\rho}{}_{\tau}$ and for spinors it's $L^{~~B}{}_{A}$ ... but their function is same to give lorentz transformation.. isn't it???
also look at it
$\lambda ^{\rho}{}_{\tau}= \delta ^{\rho}{}_{\tau}+\frac {i}{2} \delta\omega_{\mu\nu} (S^{\mu \nu})^{~~\rho{}_{\tau}}$ To me the function $(S^{\mu \nu})^{\rho}{}_{\tau}$ and $(S^{\mu \nu})^{~~B}{}_{A}$ are same but one is the vectorial representation of lorentz transformation and other one is giving the spinorial representation of lorentz transformation .. that's it but it's just different representation otherwise they are same. isn't it???? ..
lastly in eqn 35.21 the kind of summation rule Srednicki talked about is little fuzzy to me.. I got the idea but I didn't get how this comes from 35.21 and 35.22 in his book ...

10. Sep 11, 2014

### kau

11. Sep 11, 2014

### ChrisVer

For the $S$ I think Avodyne's post is pretty clear... the only think conjugation will do on $S$ is to change the representation on which it acts on (from the bar you are going to the unbar- or from left you are going to right). The conjugation doesn't happen on Lorentz indices but on spinor ones, that's why:
$[(S^{\mu \nu})^{b}_{a}]^{*}= [(S^{\mu \nu})^{*}]^{\dot{a}}_{\dot{b}} \in (Right Repr)$
That's what I wrote... the conjugation has not yet acted on the indices themselves- it's written outside.
For example a spinor in (1/2,0) repr has indices:
$\chi^{a}$ whereas the $(\chi^{a})^{\dagger}= \bar{\chi}^{\dot{a}}$ belongs to the (0,1/2). That means that they transform differently under lorentz transformations...the one goes with $(\sigma^{\mu \nu})^{b}_{a}$ while the other is going with $(\bar{\sigma}^{\mu \nu})^{\dot{a}}_{\dot{b}}$... In written form it means that under lorentz transformations:
$\chi_{a}' = M_{a}^{~~b} \chi_{b}$
whereas the complex conjugate (transforming in (0,1/2)) has:
$\bar{\chi}_{a}' = (M^*)_{\dot{a}}^{~~\dot{b}} \bar{\chi}_{\dot{b}}$
More can be found in Ch.3 of M. Drees, R. Godbole, P. Roy book -the theory and phenomenology of Sparticles. In general I haven't ever read Schredn's book, but I'm familiar with spinor indices by susy.

As for your other question, I don't understand where's the problem.
first of all your indices for the sigma bar are wrong in the first place... for the sigma bar the dotted are left and the undotted is right. Then take the conjugate of and see by yourself what you are finding...
P#20

The next one is very difficult I guess to explain without group theory... Without representations and stuff, the only way you can learn how a spinor transforms is by learning it parrot-like.
A scalar in general is a field, which doesn't transform under lorentz transformations. That's why there is no Lorentz transformation acting on PHI.
A vector field, is a field which transforms in the vector representation of the Lorentz group, that's why the transformation is similar to the coordinate transformations.
A spinor on the other hand transforms in the spinor reprs... That is equivalent in using the L in front of Psi...
One actual example about this, I guess you came across in QM, where you did rotations and stuff... Are the generators L and $\sigma$ the same in normal QM? they both generate rotations, but they do that on different spaces.
However with lorentz transformation the coordinates also change, and that's why you are bringing them back to your initial x in the argument, in all cases.

I don't understand your rest questions. In 35.10 -35.13 however Schred is giving those expressions...

Last edited: Sep 11, 2014
12. Sep 11, 2014

### ChrisVer

ah I didnt read the last question...
However it doesn't make sense the way you wrote it... [35.20-21] are the ones you are asking about...
Well, try to follow the summation rule for the spinor indices....undotted go like this \ and dotted go like that /.

One fast way is then to see that (leaving out factors):
$S_L = \sigma^{\mu \nu} = (\sigma^{\mu} \bar{\sigma}^{\nu} - \sigma^{\nu} \bar{\sigma}^{\mu})$
because the Lorentz transformation for the left handed spinors is like that:
$D_L = e^{-\frac{i}{2} \omega_{\mu \nu} \sigma^{\mu \nu}}$ with the $sigma^{\mu \nu}$ difined as the one I gave you... D_L corresponds to the matrix M I wrote above, which is a 2x2 complex matrix, element of SL(2,C) which is the universal covering of SO(3,1).
Now following exactly the rule of summations you have:
$S_L = (\sigma^{\mu} \bar{\sigma}^{\nu} - \sigma^{\nu} \bar{\sigma}^{\mu})= (\sigma^{\mu}_{a \dot{b}} \bar{\sigma}^{\nu~~ \dot {b} c} - \sigma^{\nu}_{a \dot{b}} \bar{\sigma}^{\mu~~\dot{b} c})= (\sigma^{\mu \nu})_{a}^{~~c}$
In order to see the positions of the $\sigma^{\mu}$ and $\bar{\sigma}^{\mu}$ spinor indices, you can apply some transformation on left and right spinors...combining the result you will see that they are of the form:
$\sigma^{\mu}_{x \dot{x}}$ and $\bar{\sigma}^{\mu~~\dot{x}x}$
otherwise the indices won't be contracted well...

In general in order to do it like Schrednicki does, you have to do what is happening... he multiplies with the "spinor-space" metric epsilon... so what he is doing is raising an index of S_L and S_R...

13. Sep 11, 2014

### ChrisVer

In post #11, I gave a wrong reference to posts of some other thread... Now it's corrected [2 became 20]- I hadn't checked I was sent in the 2nd page.

14. Sep 11, 2014

### kau

15. Sep 11, 2014

### kau

thanks.. yeah it's clear to me now.. I got the thing xactly what you got.. but I had doubts.. but it's clear now. As I am learning these stuff first time so getting things a bit more difficult than what it actually is.. any way thanks a lot.. but though just reading these spinor chapters from chapter 32 to 36 I didn't find any reasn why should i keep track of that kind of contraction ... bt i guess it would be clearer as i go further.. i will keep posting my doubts..

16. Sep 11, 2014

### ChrisVer

No I have the book alright...It's just I never studied from that...

For sigma yes, I think it's right.. I don't really remember how to prove everything rigorously about the sigma matrices, coz I'm following "rules"... So for example for the complex conjugate of the the $\bar{\sigma}$ I remember that I can take it's transpose (interchange the indices) and then take it's complex value- thus it will change the representation of the indices, and the dotted will become undotted and vice versa. Of course it's a matter of rule that I made for myself to remember... the proof can be done by checking it straightforwardly...as in the post I referred to.

Well to me it's always more obvious on how to use the indices:
$\theta_{a} \rightarrow \theta_{a}'= (D_L)^{~~b}_{a} \theta_{b}$
Whereas for the dotted:
$\bar{\theta}^{\dot{a}} \rightarrow \bar{\theta'}^{\dot{a}} =( D_L^*)_{~~\dot{b}}^{\dot{a}} \bar{\theta}^{\dot{b}}$

See I keep the D's indices so that the "product" rule is fine.

The summation rule is a matter of convention. In case you want to follow the most bibliography convention, you have to learn that rule... some others change the summation rule in the other way for the left and right handed, but if I recall well they must also change their spinor metrics... The overall change will be a minus sign in the end, that's why...
because: $\theta^{a} \theta_{a} = - \theta_{a} \theta^{a}$
so I guess the most natural thing for them to do, is put a minus sign in the spinor metric $\epsilon$...

17. Sep 11, 2014

### ChrisVer

Now for where that is useful...I guess everywhere you deal with grassmann variables; eg Weyl spinors... which anticommute...
That is as you can see the rigorous way to make lorentz transformations of spinors etc, and also those things are widely used in the supersymmetry theories... [that's the background from where I answer to you].