Understanding Hermitian Conjugates of $\psi^{\dag}\bar{\sigma}^{\mu}\chi$

[LAHLH]

Hi,

If we start with $$\psi^{\dag}\bar{\sigma}^{\mu}\chi$$ and take its Hermitian conjugate:

$$\left[\psi^{\dag}\bar{\sigma}^{\mu}\chi\right]^{\dag}=\left[\psi^{\dag}_{\dot{a}}\bar{\sigma}^{\mu\dot{a}c}\chi_{c}\right]^{\dag}$$

I'm basing this on Srednicki ch35 (p219 in my edition). His next line is,

$$=\chi^{\dag}_{\dot{c}}(\bar{\sigma}^{\mu a\dot{c}})^{*}\psi_{a}$$

I can't understand how he's managed to use the transpose part of the dagger, to change which index of $$\sigma$$ is dotted. At first I thought well the index on $$\chi$$ has gone from $$c$$ to $$\dot{c}$$ so maybe he's just relabelling dummies, but then this of course wouldn't explain why the tranpose part has been used and we're left only with *.

I might have expected,
$$=\chi^{\dag}_{\dot{c}}(\bar{\sigma}^{\mu a\dot{c}})^{\dag}\psi_{a}$$
$$=\chi^{\dag}_{\dot{c}}(\bar{\sigma}^{\mu \dot{c}a})^{*}\psi_{a}$$
Could anyone spell it out for me? thanks

 

[arkajad]

Dotted index means "it transforms under the complex conjugated representation matrix". Therefore complex conjugation always changes a dottted index into an undotted one and vice versa. Instead of a * there should be "T" - transpose, and all is OK.

 

[tiny-tim]

Hi LAHLH!

From my Penrose and Rindler (Volume I, §2.5, p. 107) …

T^A' is anti-isomorphic with T^A, the anti-isomorphism being expressed by (2.5.8):

λκ^A + µω^A = τ^A

↔ λ*κ*^A' + µ*ω*^A' = τ*^A'​

 

[LAHLH]

Dotted index means "it transforms under the complex conjugated representation matrix". Therefore complex conjugation always changes a dottted index into an undotted one and vice versa. Instead of a * there should be "T" - transpose, and all is OK.

So are you saying this is an errata in Srednicki and he should have T instead of *?

 

[arkajad]

It looks like that. Because with this small correction all the rest is clear.

 

[LAHLH]

It looks like that. Because with this small correction all the rest is clear.

hmm, I would agree with you, but you said complex conjugation always turns a dotted in undotted and vice versa, but according to the rest of Srednicki it is Hermitian conjugation that turns $$\left[\psi_a\right]^{\dag}=\psi^{\dag}_{\dot{a}}$$

So, $$\left[\psi^{\dag}_{\dot{a}}\bar{\sigma}^{\mu\dot{a}c}\chi_{c}\right]^{\dag} =\chi^{\dag}_{\dot{c}}(\left(\bar{\sigma}^{\mu\dag}\right)^{a\dot{c}})\psi_{a}$$

 

[arkajad]

Hermitian conjugation involves, in particular, complex conjugation, so it should be so.

 

[LAHLH]

Hermitian conjugation involves, in particular, complex conjugation, so it should be so.

huh? obviously I know h.c. is T and *. The point is that you said c.c. takes dotted to undotted, whereas Srednicki is saying h.c. takes dotted to undotted and vice versa. This makes a lot of difference in trying to derive the above.

Anyway things are not working out,

$$\left[\psi^{\dag}_{\dot{a}}\bar{\sigma}^{\mu\dot{a}c}\chi_{c}\right]^{\dag} =\chi^{\dag}_{\dot{c}}(\left(\bar{\sigma}^{\mu\dag }\right)^{a\dot{c}})\psi_{a} =\chi^{\dag}_{\dot{c}}(\left(\bar{\sigma}^{\mu* }\right)^{\dot{c}a})\psi_{a}<br /> \neq\chi^{\dag}_{\dot{c}}(\left(\bar{\sigma}^{\mu a\dot{c}}\right)^{*}\psi_{a}$$

If it was solely c.c. that took dotted to undotted perhaps things would work out here (still assuming that Srednicki * should really be a T, and is an errata), but then this seems at odds with the rest of Srednicki where he says it is h.c. that does this function, at least on single index objects.

 

[arkajad]

But you did not yet take into account the hermicity of Pauli's matrices. They are Hermitian. Which means (skipping the mu index):

$$(\bar{\sigma}^{a\dot{c}})^\dag=\bar{\sigma}^{a\dot{c}}$$

or, if you wish,

$$\bar{\sigma}^{a\dot{c}}=\bar{\sigma}^{c\dot{a}}$$

P.S. I noticed this subject has been https://www.physicsforums.com/showthread.php?t=281207" .

 

[LAHLH]

But you did not yet take into account the hermicity of Pauli's matrices. They are Hermitian. Which means (skipping the mu index):

$$(\bar{\sigma}^{a\dot{c}})^\dag=\bar{\sigma}^{a\dot{c}}$$

Surely it would mean

$$(\bar{\sigma}^{a\dot{c}})^\dag=\left(\bar{\sigma}^{\dag}\right)^{\dot{a}c}=\bar{\sigma}^{\dot{a}c}$$

Where the last equality follows from hermiticity.

 

[arkajad]

All this dot notation may be confusing. It is not the symbol $$sigma$$ that is hermitian, but the whole matrix $$\bar{\sigma}^{a\dot {c}}$$. Therefore

$$(\bar{\sigma}^{a\dot{c}})^\dag=\bar{\sigma}^{a\dot {c}}\quad \mbox{added later: that is wrong}$$

That is the starting point. Srednicki could have used just that. But he did not, he started manipulating indices of sigma and got confused and confused the reader without explaining. I hope this will be fixed in the next draft.

Once you realize Paul' matrices are Hermitian, then you may start, if you wish so, playing with indices - how can that be written using the potentially confusing index notation. Srednicki is skipping the discussion of these issues. But you can find them discussed somewhere else. For instance in "Theory of spinors: an introduction", by Carmeli, p. 88 you will find Eq. (5.34):

$$\sigma^\mu_{AB'}=\bar{\sigma}^\mu_{B'A}$$

In Carmeli bar denotes complex conjugation. This is much better. Srednicki is using * for complex conjugation. To be less confusing I should have followed Carmeli and write

$$\bar{\sigma}^{a\dot{c}}=\bar{\sigma}^{*c\dot{a}}$$

 

[LAHLH]

All this dot notation may be confusing. It is not the symbol $$sigma$$ that is hermitian, but the whole matrix $$\bar{\sigma}^{a\dot {c}}$$. Therefore

$$(\bar{\sigma}^{a\dot{c}})^\dag=\bar{\sigma}^{a\dot {c}}$$

I really am confused now, haha, sorry. As I understand it $$\bar{\sigma}^{\mu a\dot{c}}$$ is just the $$\mu$$'th element of the vector of matrices $$\bar{\sigma}$$ for example $$\bar{\sigma}^{0 (1\dot{1})}}$$ is just 1 (since it is the top left element of the ident matrix)

Now when we take dagger it is w.r.t the spinor indices, and $$\mu$$ is a specific number, meaning we are just using hermiticity of the pauli matrices and the ident when we say sigma is Hermitian.

Hermiticity means that if we have a matrix $$A^{\dag}_{ij}=A^{*}_{ji}$$ thus using this definition on say the $$\mu=1$$ element

$$(\bar{\sigma}^{1 a\dot{c}})^\dag=\bar{\sigma}^{* 1 \dot{c}a}$$

Then of course something has to happen, to flip dots $$\bar{\sigma}^{* 1 c\dot{a}}$$ So I arrive at,
$$(\bar{\sigma}^{1 a\dot{c}})^\dag=\bar{\sigma}^{* 1 c\dot{a}}$$I know I'm missing something here, but Srednicki does exactly the same thing in 36.1 so I don't think his * instead of T is a genuine typo.

 

[arkajad]

Alright, it is confusing. Without indices it would be clear. We have $$(\bar{\sigma}^\mu)^\dag =\bar{\sigma}^\mu,$$ therefore

$$(\psi^\dag\bar{\sigma}^\mu\chi)^\dag=\chi^\dag\bar{\sigma}^\mu\psi.$$

Now we add indices, and we add with the last before last line in Srednicki. As for other manipulations - what rules are being used - it is not clear to me now. I will try to clarify it to myself first.

 

[arkajad]

I noticed that everywhere before sigmas has the second index dotted. In the formula we are discussing, the first index becomes dotted. That means we are using either conjugate sigma or transposed sigma - which is the same, because sigmas are hermitian. So I would first write it as:

$$\psi^\dag \sigma^T \chi$$

and then, during calculations:

$$(\sigma^{T\dot{a}c})^\dag=(\sigma^{c\dot{a}})^\star=\sigma^{\star\dot{c}a}=\sigma^{T\dot{c}a}$$

Where in the last equality I am using the fact that sigmas are hermitian, therefore $$\sigma^*=\sigma^T.$$

I believes that Srednicki assumes that we are supposed to read from the position of the indices whether we are using the sigma or its transpose (=complex conjugate).

 

[Sojourner01]

I thought I'd stick my oar in here since I also find these problems in general extremely challenging.

For one, I've never even seen these 'dotted indices' before. The author of my QFT textbook doesn't use them.

In reference to arkajad's second-to-last post: What is the reasoning behind moving $$\chi$$ to the far left? This doesn't seem implicit to me in the expression. Does this imply that $$\chi$$, whatever it is, commutes with $${\sigma}^\mu\psi$$?

edit: Apparently I need to review my basic complex algebra. It appears that this is a fundamental property of Hermitian conjugation, which I'd forgotten about.

 

[arkajad]

For one, I've never even seen these 'dotted indices' before.

Dotted indices are used to indicate the transformation law when you change the basis in the two-dimensional complex space, usually by a SU(2) matrix. When you use matrix A you use undotted index, when you use its complex conjugate - you use dotted one. Sigma transform usually as

$$\sigma\mapsto A\sigma A^\dag$$

so the first index is undotted, the second is dotted.

Another abstract algebraic way is to consider sigmas as elements of the tensor space

$$V\otimes{\bar V}$$

where $$V$$ is $$\mathbf{C}^2$$, and $$\bar{V}$$ is its complex conjugated space. This last concept is abstractly defined in some linear algebra books, but not that often. Usually only complex dual space is defined, but not the complex conjugate.

 

[Sojourner01]

Thanks arkajad, you are clearly a patient teacher. Spelled out explicitly, the dotted index rule makes perfect sense and it seems that as usual, there's a subset of authors who simply can't be bothered to express their derivations rigorously. I'm displaying my ignorance once again, though in my defence, it's not entirely my fault. In my entire history of education (for the record, up to Masters' level physics), nobody has ever bothered to explain what a tensor or a tensor space is. I've had to discover this entirely by myself - after graduating.

 

[LAHLH]

I noticed that everywhere before sigmas has the second index dotted. In the formula we are discussing, the first index becomes dotted. That means we are using either conjugate sigma or transposed sigma - which is the same, because sigmas are hermitian.

But sigma bar is defined with it's first index dotted, not vice versa in 35.19, so not sure what you mean by previously it always had its second index dotted? Also in 35.20 it is the first index dotted. So I don't see how he can be implicitley assuming that this arrangement of dotted/undotted indices is the transposed version.

 

[arkajad]

But sigma bar is defined with it's first index dotted...

Aargh! You are right. Back to my decoding again.

 

[arkajad]

Using matrix notation I would write the definition of $$\bar{\sigma}$$ as
$$\bar{\sigma}={\bar\epsilon}\sigma^T\epsilon^T.$$

Sigma bar is also hermitian. Therefore we have the formula that it is equal to its transposed complex conjugate:

$$\bar{\sigma}^{\dot{a}c}=\bar{\sigma}^{*\dot{c}a}.$$

We calculate from the definition of $${}^\dag={}^{*T}$$

$$(\bar{\sigma}^{\dot{a}c})^\dag=(\bar{\sigma}^{*a\dot{c}})^T=\bar{\sigma}^{\dag\dot{c}a}=\bar{\sigma}^{\dot{c}a},$$
Where I have used the fact that $$\bar{\sigma}^\dag=\bar{\sigma}.$$

 

[LAHLH]

Using matrix notation I would write the definition of $$\bar{\sigma}$$ as
$$\bar{\sigma}={\bar\epsilon}\sigma^T\epsilon^T.$$

Sigma bar is also hermitian. Therefore we have the formula that it is equal to its transposed complex conjugate:

$$\bar{\sigma}^{\dot{a}c}=\bar{\sigma}^{*\dot{c}a}.$$

We calculate from the definition of $${}^\dag={}^{*T}$$

$$(\bar{\sigma}^{\dot{a}c})^\dag=(\bar{\sigma}^{*a\dot{c}})^T=\bar{\sigma}^{\dag\dot{c}a}=\bar{\sigma}^{\dot{c}a},$$
Where I have used the fact that $$\bar{\sigma}^\dag=\bar{\sigma}.$$

On your second to last equality why have you swapped the indices too. You had a * and a T and endedup with a dagger, which given that the dagger=*T, means nothing should have changed?

 

[Avodyne]

Hi,

If we start with $$\psi^{\dag}\bar{\sigma}^{\mu}\chi$$ and take its Hermitian conjugate:

$$\left[\psi^{\dag}\bar{\sigma}^{\mu}\chi\right]^{\dag}=\left[\psi^{\dag}_{\dot{a}}\bar{\sigma}^{\mu\dot{a}c}\chi_{c}\right]^{\dag}$$

I'm basing this on Srednicki ch35 (p219 in my edition). His next line is,

$$=\chi^{\dag}_{\dot{c}}(\bar{\sigma}^{\mu a\dot{c}})^{*}\psi_{a}$$

I believe he's treating sigma as just a set of numbers. Thus, the operators (the fields) get hermitian conjugated (and switch order), and the numbers get complex conjugated. On a field, hermitian conjugation changes a dotted index to undotted (and vice versa), and so the explicit indices on the sigma have also been changed to match.

 

[vela]

I believe he's treating sigma as just a set of numbers. Thus, the operators (the fields) get hermitian conjugated (and switch order), and the numbers get complex conjugated. On a field, hermitian conjugation changes a dotted index to undotted (and vice versa), and so the explicit indices on the sigma have also been changed to match.

That's how I read it too. If you want to reverse the indices, you'd have to write

$$\chi^\dag_{\dot{c}} [((\bar{\sigma}^{\mu})^\mathrm{T}})^{\dot{c}a}]^* \psi_a$$

 

[arkajad]

On your second to last equality why have you swapped the indices too. You had a * and a T and endedup with a dagger, which given that the dagger=*T, means nothing should have changed?

Yes. Because $$\bar{\sigma}$$ are hermitian.

 

[LAHLH]

I believe he's treating sigma as just a set of numbers. Thus, the operators (the fields) get hermitian conjugated (and switch order), and the numbers get complex conjugated. On a field, hermitian conjugation changes a dotted index to undotted (and vice versa), and so the explicit indices on the sigma have also been changed to match.

So if I were to write this out slowly step by step, justifiying each step it would look like:

$$\left[\psi^{\dag}_{\dot{a}}\bar{\sigma}^{\mu\dot{a}c}\chi_{c}\right]^{\dag}$$
$$=\left[\chi_{c}\right]^{\dag}\left[\bar{\sigma}^{\mu\dot{a}c}\right]^{\dag}\left[\psi^{\dag}_{\dot{a}}\right]^{\dag}$$

Now I change the $$\dag$$ to a * on the sigma bracketts because I think of each as a number rather than a matrix at this step? leading to:

$$=\left[\chi_{c}\right]^{\dag}\left[\bar{\sigma}^{\mu\dot{a}c}\right]^{*}\left[\psi^{\dag}_{\dot{a}}\right]^{\dag}$$

Now since I'm about to change the c to $$\dot{c}$$ and the $$\dot{a}$$'s to a, I also do the same relabelling of "dummy indices" type thing on the sigma too? leading to,

$$=\chi^{\dag}_{\dot{c}}\left[\bar{\sigma}^{\mu a\dot{c}}\right]^{*}\psi_{a}$$

Which is Srednicki's line 2 of 35.29. Now I need to use hermiticity of the sigmas (i.e. for some reason now actually think of them as matrices again?), which leads to $$\left(\bar{\sigma}^{\mu a\dot{c}}\right)^{*}=\left(\left[\bar{\sigma}^{\dag}\right]^{\mu a\dot{c}}\right)^{*}=\left(\left[\bar{\sigma}^{*}\right]^{\mu \dot{c}a}\right)^{*}=\left(\bar{\sigma}^{\mu \dot{c}a}\right)$$

So now plugging this in I arrive at:$$=\chi^{\dag}_{\dot{c}}\left(\bar{\sigma}^{\mu \dot{c}a}\right)\psi_{a}$$
$$=\chi^{\dag}\bar{\sigma}^{\mu}\psi$$

which is of course Srednicki's final line.

So this all seems to work out nicely, but I'm not really sure I understand why we can think of them as just c numbers as one stage then we treat them as matrices the next, and how exactly the relabelling of dummies if indeed that is what's going on, can occur like this.

My understanding at the moment is when the dagger is outside the symbol e.g. $$\left[\bar{\sigma}^{\mu a\dot{c}}\right]^{\dag}$$, we are thinking of what's in the square bracketts as just a complex number so change the dagger to a star. But if we write $$\left[\bar{\sigma}^{\dag}\right]^{\mu a\dot{c}}$$ then we're thinking of the different matrix $$\bar{\sigma}^{\dag}$$

Thanks a lot for the help

 

[LAHLH]

just wondering if people agree with my derivation above, or if I'm still missing something?

 

[arkajad]

As for me, it seems you are thinking alright, but I am not really sure. I have proposed my way of writing, which I adjusted from Carmeli, and which I like, and it works.

 

[Avodyne]

just wondering if people agree with my derivation above, or if I'm still missing something?

Yes, I agree. And I don't think there's anything unusual about treating matrices this way. For example, consider

$$\chi^\dagger A\psi = \chi_i^* A_{ij}\psi_j$$

where $\psi_i$ and $\chi_i$ are column vectors of complex numbers, and $A_{ij}$ is a matrix (and repeated indices are summed). This expression is just a number, so its hermitian conjugate is the same as its complex conjugate:

$$(\chi^\dagger A\psi)^\dagger = (\chi^\dagger A\psi)^*$$

Evaluating the left-hand side by symbolic manipulation, we get

$$(\chi^\dagger A\psi)^\dagger=\psi^\dagger A^\dagger\chi$$

Evaluating the right-hand side by explicitly writing it out, we get

$$(\chi^\dagger A\psi)^*=(\chi_i^* A_{ij}\psi_j)^*=\chi_i A_{ij}^*\psi_j^*$$

But now we can rearrange this to

$${}= \psi_j^*A_{ij}^*\chi_i = \psi_j^*(A^{T*})_{ji}\chi_i = \psi_j^*(A^\dagger)_{ji}\chi_i=\psi^\dagger A^\dagger\chi$$

The differences with spinor indices are that (1) there are two kinds, dotted and undotted, and we have to keep track of which is which, and (2) conjugation (hermitian or complex) transforms one kind into the other.

EDIT: final expression in last equation now corrected

 

[LAHLH]

$${}= \psi_j^*A_{ij}^*\chi_i = \psi_j^*(A^{T*})_{ji}\chi_i = \psi_j^*(A^\dagger)_{ji}\chi_i=\chi^\dagger A\psi$$

This is very helpful, thanks. Just want to clarify your very last equality, did you mean $$\psi_j^*(A^\dagger)_{ji}\chi_i=\psi^\dagger A^\dagger \chi$$

 

[Avodyne]

Oops! Yes, I've now corrected the original post.