Srednicki Bar Notation: Understanding (39.8)

[LAHLH]

Hi,

Srednicki says in (38.14) that for any combination of gammas \bar{A}\equiv \beta A^{\dag}\beta. This is fine, and I can work at such relations as (38.15), like \bar{\gamma^{\mu}}=\gamma^{\mu} and so on.

We also have for spinors \bar{u_s}\equiv u^{\dag}_s\beta, and for the Dirac field \bar{\Psi}=\Psi^{\dag}\beta

Oh and believe that the following law holds \overline{AB}=\bar{B}\bar{A}.

Now my questions is, how does Srednicki obtain the first equality in (39.8)? namely

\left[ \bar{u_s}\gamma^{0}\Psi\right]^{\dag}=\overline{\bar{u_s}\gamma^{0}\Psi\right}

I can't see how this follows from any of his prior definitions. We know from the previous chapter that if A is some combo of gammas then \bar{A}=A, but no relations concerning \bar{A}=A^{\dag} for certain combinations.

 

[LAHLH]

I think I can prove the final line of (39.8) without taking the route he does through the bar and out again:

\left[ \bar{u_s}\gamma^{0}\Psi\right]^{\dag}=\Psi^{\dag}\left(\gamma^{0}\right)^{\dag}\left(\bar{u_s}\right)^{\dag}

Then use hermiticity of gamma nought, and the definition of \left(\bar{u_s}\right)^{\dag}= (u^{\dag}_s\beta)^{\dag}=\beta^{\dag}(u^{\dag}_s)^{\dag}=\beta u_s

to get:

\Psi^{\dag}\left(\gamma^{0}\right)^{\dag}\left(\bar{u_s}\right)^{\dag}=\Psi^{\dag}\gamma^{0}\beta u_s

Finally noting that numerically gamma nought and beta are the same and the definition of \bar{\Psi}=\Psi^{\dag}\beta, leads to:

=\bar{\Psi}\beta u_s
=\bar{\Psi}\gamma^{0} u_s

which seems to be Srednicki's end result anyway. Although I know that beta and gamma nought have different index structure, so perhaps I have been careless with liberally throwing around beta instead of gamma nought etc, and maybe this is why Srednicki takes the route he does.

 

[matonski]

I think it's because \bar{u_s}\gamma^{0}\Psi is now just a 1x1 matrix since all indices are contracted. In that case, barring and adjointing are the same thing.

 

[LAHLH]

Hi, yes I think you're right actually, since if you apply bar to a c-number called say c, you get \bar{c}=\beta c^{\dag} \beta=c^{\dag} \beta^2=c^{\dag}

But what about his second line following this then. It's very tempting to use \overline{AB}=\bar{B}\bar{A}. To obtain:

<br /> \overline{\bar{u_s}\gamma^{0}\Psi\right} =\overline{\Psi}\bar{\gamma^{0}}\overline{\bar{u_s}}=\overline{\Psi}\gamma^{0}u_s<br />

But I don't think this is consistent. The initial bar over the whole product is presumably of the type \overline{A}=\beta A \beta. As this is the type that means we can equate h.c. to barring for a c-number as you mentioned, and also the type of barring such that the identity \overline{AB}=\bar{B}\bar{A} actually works.

But \bar{u_s}\equiv u^{\dag}_s\beta and \bar{\Psi}=\Psi^{\dag}\beta. So I don't think it's accurate to just naively cancel the bar over the spinor.

 

[LAHLH]

doing this long handed:

\overline{\bar{u_s}\gamma^{0}\Psi\right} =\beta\left(\bar{u_s}\gamma^{0}\Psi\right)^{\dag}\beta=\beta\Psi^{\dag}(\gamma^{0})^{\dag}(\bar{u_s})^{\dag}\beta=\beta\Psi^{\dag}\beta\beta(\gamma^{0})^{\dag}\beta\beta(\bar{u_s})^{\dag}\beta<br />

So now if \bar{\Psi}\equiv \beta\Psi^{\dag}\beta and \bar{u_s}\equiv\beta u^{\dag}_s\beta then things would work out. Since \beta(\bar{u_s})^{\dag}\beta=\beta(\beta(u_s})^{\dag}\beta)^{\dag}\beta=u_s. And we'd end up with Srednicki's \overline{\Psi}\bar{\gamma^{0}}\overline{u_s}.
But this is not the way the bar is defined for the field and spinors.

quite confused...

 

[matonski]

\bar{c}=\beta c^{\dag} \beta

Actually, I don't think this is quite correct. I think barring only means something special for spinors and matrices. For c-numbers, I think it means just the same thing as adjoint.

Thus,

\overline{\bar{u_s}\gamma^{0}\Psi\right} =\beta\left(\bar{u_s}\gamma^{0}\Psi\right)^{\dag}\beta =\beta\Psi^{\dag}(\gamma^{0})^{\dag}(\bar{u_s})^{\dag}\beta

I think this first step is wrong. Also, you end up getting into problems with multiplying a 4x4 matrix by a row vector in the wrong order as you do have things like \beta\Psi^{\dag} there.

 

[LAHLH]

By adjoint to you mean h.c. ?

so how does he get to the next line then? why did he bother changing the dagger into a bar in the first place, if the meaning is equal.

 

[matonski]

Yes, I mean hermitian conjugate. I think the reason he changed from the dagger to the bar is precisely the reason you pointed out above. So he can simply do this derivation:

<br /> \overline{\bar{u_s}\gamma^{0}\Psi\right} =\overline{\Psi}\bar{\gamma^{0}}\overline{\bar{u_s}}=\overline{\Psi}\gamma^{0}u_s<br />

 

[LAHLH]

Yes, I mean hermitian conjugate. I think the reason he changed from the dagger to the bar is precisely the reason you pointed out above. So he can simply do this derivation

Yes but if the bar doesn't mean \bar{A}\equiv \beta (A)^\dag\beta for this quantity then:

1) how does he get to the second line, since the identity \overline {AB}=\bar{B}\bar{A} only holds when barring is defined as \bar{A}\equiv \beta (A)^\dag\beta
2) Even if barring is defined as \bar{A}\equiv \beta (A)^\dag\beta, then this would be mean the \overline{\Psi} of line two was \beta\Psi^{\dag}\beta not the usual \Psi^{\dag}\beta. Similarly the usual definition of barring for spinors is \bar{u_s}=u^{\dag}_s\beta, but for the "u bar barred" you would obtain i.e. \overline{\bar{u_s}} the over bar is defined in the sense \bar{A}\equiv \beta (A)^\dag\beta so the two bars do not cancel, they are different types of barring.

I mean you could say bar here is just hermitian conj, and use the reversing rule for the h.c. of a product, then fiddle about chaning gammas to betas and so forth like I did in post #2 etc etc, but this makes the whole introduction of the bar nothing more than a relabelling of dagger symbol to a bar symbol, and the whole introduction of bar isntead of dagger is pointless, so I don't believe this is what he intends. But I just can't see to find a consistent way of getting to line 2 using the various barring definitions.

 

[matonski]

There is nothing inconsistent. For a 1x1 matrix, bar = dagger. In fact, it has to be this way for everything to be consistent. If u and v are spinors, then
<br /> \overline{ \bar u v} = \left[ \bar u v \right]^\dag = \left[ u^\dag \gamma^0 v \right]^\dag = v^\dag (\gamma^0)^\dag u = v^\dag \gamma^0 u = \bar v u<br />

Thus, you can just say
<br /> \left[ \bar u v \right]^\dag = \overline{ \bar u v} = \bar v u<br />

Similarly,
\overline{\bar u_s \gamma^0 \Psi} = \left[ \bar u_s \gamma^0 \Psi \right]^\dag = \left[ u_s^\dag \gamma^0 \gamma^0 \Psi \right]^\dag = \left[ u_s^\dag \Psi \right]^\dag = \Psi^\dag u_s = \Psi^\dag \gamma^0 \gamma^0 (\gamma^0)^\dag \gamma^0 u_s = \bar \Psi \overline{\gamma^0} u_s<br />

Thus, you can just say
<br /> \left[ \bar u_s \gamma^0 \Psi \right]^\dag = \overline{\bar u_s \gamma^0 \Psi} = \bar \Psi \overline{\gamma^0} u_s = \bar \Psi \gamma^0 u_s<br />

The reason the bar notation is so useful is because it makes the Lorentz transformation properties of spinors very transparent.

 

[Avodyne]

Note that matonski is taking \beta=\gamma^0, as most books do (Srednicki and Weinberg are the exceptions).

 

[LAHLH]

There is nothing inconsistent. For a 1x1 matrix, bar = dagger. In fact, it has to be this way for everything to be consistent. If u and v are spinors, then
<br /> \overline{ \bar u v} = \left[ \bar u v \right]^\dag = \left[ u^\dag \gamma^0 v \right]^\dag = v^\dag (\gamma^0)^\dag u = v^\dag \gamma^0 u = \bar v u<br />

Thus, you can just say
<br /> \left[ \bar u v \right]^\dag = \overline{ \bar u v} = \bar v u<br />

Similarly,
\overline{\bar u_s \gamma^0 \Psi} = \left[ \bar u_s \gamma^0 \Psi \right]^\dag = \left[ u_s^\dag \gamma^0 \gamma^0 \Psi \right]^\dag = \left[ u_s^\dag \Psi \right]^\dag = \Psi^\dag u_s = \Psi^\dag \gamma^0 \gamma^0 (\gamma^0)^\dag \gamma^0 u_s = \bar \Psi \overline{\gamma^0} u_s<br />

Thus, you can just say
<br /> \left[ \bar u_s \gamma^0 \Psi \right]^\dag = \overline{\bar u_s \gamma^0 \Psi} = \bar \Psi \overline{\gamma^0} u_s = \bar \Psi \gamma^0 u_s<br />

The reason the bar notation is so useful is because it makes the Lorentz transformation properties of spinors very transparent.

Thanks alot, this has made it all click into place. Most appreciated!

 

[LAHLH]

On another related note if I may trouble you again. In deriving the first line of equation (39.16), you get

\{b_s(\vec{p}),b^{\dag}_{s&#039;}(\vec{p}&#039;)\}=\int\,\mathrm{d}^3 x\mathrm{d}^3 y\, e^{-ipx+ip&#039;y}\left(\bar{u}_s(\vec{p})\gamma^{0}\Psi(x)\overline{\Psi}(y)\gamma^{0}u_{s&#039;}(\vec{p}&#039;)+\overline{\Psi}(y)\gamma^{0}u_{s&#039;}(\vec{p}&#039;) \bar{u}_s(\vec{p})\gamma^{0}\Psi(x)\right)

Srednicki seems to equate this with:

\{b_s(\vec{p}),b^{\dag}_{s&#039;}(\vec{p}&#039;)\}=\int\,\mathrm{d}^3 x\mathrm{d}^3 y\, e^{-ipx+ip&#039;y}\left(\bar{u}_s(\vec{p})\gamma^{0}\{\Psi(x)\overline{\Psi}(y)\}\gamma^{0}u_{s&#039;}(\vec{p}&#039;) \right)

It doesn't seem obvious to me how the second term in my first equation, can be commuted around in such a way as to enable writing is in the Srednicki form of my second equation. I mean how can the Psi's just commute with the gamma's and spinors like this, since the product \bar{u}_s(\vec{p})\gamma^{0}\Psi(x) is a complex number as we noted previously, and of the form row*(4x4)*column etc. I just can't see how these can be commuted around like this.

 

[matonski]

There are implicit matrix indices everywhere. If you write them out explicity, the only things that don't commute are \Psi and \Psi^\dag. The indices take care of the order of the matrices. Working backwards (I wrote the s and s' upstairs to make room for the matrix indices on the spinors):

<br /> \begin{align*}<br /> \bar{u}^s(\vec{p})\gamma^{0}\{\Psi(x)\overline{\Psi}(y)\}\gamma^{0}u^{s&#039;}(\vec{p}&#039;) <br /> &amp;= \bar{u}^s_a(\vec{p})\gamma^{0}_{ab}\{\Psi_b(x)\overline{\Psi}_c(y)\}\gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;) \\<br /> &amp;= \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \Psi_b(x)\overline{\Psi}_c(y) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;)<br /> + \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \overline{\Psi}_c(y) \Psi_b(x) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;) \\<br /> &amp;= \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \Psi_b(x)\overline{\Psi}_c(y) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;)<br /> + \overline{\Psi}_c(y) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;) \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \Psi_b(x) \\<br /> &amp;= \bar{u}^s(\vec{p})\gamma^{0} \Psi(x)\overline{\Psi} (y) \gamma^{0} u^{s&#039;} (\vec{p}&#039;)<br /> + \overline{\Psi}(y) \gamma^{0} u^{s&#039;}(\vec{p}&#039;) \bar{u}^s(\vec{p})\gamma^{0} \Psi(x) <br /> \end{align*}<br />

 

[LAHLH]

There are implicit matrix indices everywhere. If you write them out explicity, the only things that don't commute are \Psi and \Psi^\dag. The indices take care of the order of the matrices. Working backwards (I wrote the s and s' upstairs to make room for the matrix indices on the spinors):

<br /> \begin{align*}<br /> \bar{u}^s(\vec{p})\gamma^{0}\{\Psi(x)\overline{\Psi}(y)\}\gamma^{0}u^{s&#039;}(\vec{p}&#039;) <br /> &amp;= \bar{u}^s_a(\vec{p})\gamma^{0}_{ab}\{\Psi_b(x)\overline{\Psi}_c(y)\}\gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;) \\<br /> &amp;= \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \Psi_b(x)\overline{\Psi}_c(y) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;)<br /> + \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \overline{\Psi}_c(y) \Psi_b(x) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;) \\<br /> &amp;= \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \Psi_b(x)\overline{\Psi}_c(y) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;)<br /> + \overline{\Psi}_c(y) \gamma^{0}_{cd} u^{s&#039;}_d(\vec{p}&#039;) \bar{u}^s_a(\vec{p})\gamma^{0}_{ab} \Psi_b(x) \\<br /> &amp;= \bar{u}^s(\vec{p})\gamma^{0} \Psi(x)\overline{\Psi} (y) \gamma^{0} u^{s&#039;} (\vec{p}&#039;)<br /> + \overline{\Psi}(y) \gamma^{0} u^{s&#039;}(\vec{p}&#039;) \bar{u}^s(\vec{p})\gamma^{0} \Psi(x) <br /> \end{align*}<br />

Thanks again!

 

[naima]

My understandig of the bar notation remains problematic.
let us go back to the initial question.
If we overbar the whole integral we get
b barred = b dagged
Barring is not always equal to dagging.
so what is barring?

it also exists for 2 component spinors : https://www.physicsforums.com/showthread.php?t=84642"

 

[LAHLH]

Well matonski elucidated barring is equalivent to h.c. for a complex number, but for say a 4 component spinor, then \bar{u}_s(\vec{p})\equiv u^{\dag}_s(\vec{p})\beta (similarly for the Dirac field say). For a 4x4 matrix M barring is defined as \overline{M}\equiv\beta(M)^{\dag}\beta. These are just definitions, why they are useful is another matter. See http://en.wikipedia.org/wiki/Dirac_adjoint for some of their motivation or maybe Srednicki.

 

[naima]

yes but here we have not a c-number but an annihilaton operator acting on Fock space.

 

[LAHLH]

yes but here we have not a c-number but an annihilaton operator acting on Fock space.

Yes, so from my original product it would actually be some sum of b and d operators etc.

Well I think the point is that it's in some sense a 1x1 matrix, and barring has no special definition, it's just h.c.. It's a little like the b and d operators act on a different space (namely the Fock space) to the spinors and gammas,so for some intents and purposes they are just like c numbers. I believe this is why you can commute them with the gammas and spinors and things at other points in Srednicki too.

This is just my understanding of it anyway, I think someone else may be able to say something more precise...