View Full Version : Questions on section 3.1 GSW
Rene Meyer
May19-04, 11:08 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hi,\n\nI was working through chapter 3 of GSW, and got some questions. Maybe\nsomeone who did this before can answer or discuss them with me.\nHere we go...\n\n1) on 3.1.10\n------------\n\nHow can GSW restrict themselfes to infinitesimal diffeomorphisms when\ncalculating (3.1.10). They have to do so, because (3.1.6) is only\nvalid in that limit. But I think the group G of diffeomorphisms is\nmuch more bigger, and the Faddeev-Popov-determinants should be\ncalculated for finite diffeomorphisms also.\n\n2) again 3.1.10\n---------------\n\nComparing with 3.1.6, there is a factor 2 missing in 3.1.10, right?\n\n3) On the ghost action\n----------------------\n\nIn 3.1.11, the ghost action is normalized with -1/pi. First, the sign:\nIs the path integral here already written in euclidian space? Second:\nHow do they arrive at this normalization? And when calculating\n(3.1.35) out of (3.1.31) I actually get 1/2pi and not 1/pi, because\n\nsqrt(g) g^ab = ((0,1), (1,0))\n\nThis is because the determinant g = det g_ab = 1/4 exp(2phi), and g^ab\n= 2 exp(-phi)((0,1),(1,0))\n\nDid anyone calculate this on his own?\n\nAnother problem I have is the sequence of the integration in the\nfermionic path integral (3.1.13): Shouldn\'t this be reversed? At least\nit is the way how Polchinski defines his path integral in Appendix A\nof his book.\n\n4) On 3.1.15\n------------\n\nAnother very basic one: I get\n\nd_+- = 1/2(d_tau -+ i d_sigma)\n\nThats because of sigma^+- = tau +- i sigma\ninvert to sigma = 1/2i(sigma^+ - sigma^-).\n\n5) On (3.1.29)\n--------------\n\nWhat has been done in the last equality here?\n\nI hope this is not too much. Thanks in advance.\n\nRené.\n\n--\nRené Meyer\nStudent of Physics & Mathematics\nZhejiang University, Hangzhou, China\n\n_________________________________________ ______________________________________\nWeb page of SPS: http://schwinger.harvard.edu/~sps/\nPosted via: http://groups.google.com/groups?group=sci.physics.strings\n^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi,
I was working through chapter 3 of GSW, and got some questions. Maybe
someone who did this before can answer or discuss them with me.
Here we go...
1) on 3.1.10
------------
How can GSW restrict themselfes to infinitesimal diffeomorphisms when
calculating (3.1.10). They have to do so, because (3.1.6) is only
valid in that limit. But I think the group G of diffeomorphisms is
much more bigger, and the Faddeev-Popov-determinants should be
calculated for finite diffeomorphisms also.
2) again 3.1.10
---------------
Comparing with 3.1.6, there is a factor 2 missing in 3.1.10, right?
3) On the ghost action
----------------------
In 3.1.11, the ghost action is normalized with -1/\pi. First, the sign:
Is the path integral here already written in euclidian space? Second:
How do they arrive at this normalization? And when calculating
(3.1.35) out of (3.1.31) I actually get 1/2pi and not 1/\pi, because
\sqrt(g) g^{ab} = ((0,1), (1,0))
This is because the determinant g = det g_{ab} = 1/4 \exp(2phi), and g^{ab}= 2 \exp(-\phi)((0,1),(1,0))
Did anyone calculate this on his own?
Another problem I have is the sequence of the integration in the
fermionic path integral (3.1.13): Shouldn't this be reversed? At least
it is the way how Polchinski defines his path integral in Appendix A
of his book.
4) On 3.1.15
------------
Another very basic one: I get
d_+- = 1/2(d_{tau} -+ i d_{sigma})
Thats because of \sigma^+- = \tau +- i \sigma
invert to \sigma = 1/2i(\sigma^+ - \sigma^-).
5) On (3.1.29)
--------------
What has been done in the last equality here?
I hope this is not too much. Thanks in advance.
René.
--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
__{_______________________________________________ ______________________________}
Web page of SPS: http://schwinger.harvard.edu/~sps/
Posted via: http://groups.google.com/groups?group=sci.physics.strings
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Lubos Motl
May19-04, 03:32 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Wed, 19 May 2004, Rene Meyer wrote:\n\n> How can GSW restrict themselfes to infinitesimal diffeomorphisms when\n> calculating (3.1.10). They have to do so, because (3.1.6) is only\n> valid in that limit. But I think the group G of diffeomorphisms is\n> much more bigger, and the Faddeev-Popov-determinants should be\n> calculated for finite diffeomorphisms also.\n\nSorry if I misunderstand what you are saying, but this text of yours looks\nto me like a conceptual misunderstanding of the whole Faddeev-Popov\ntechnique. The Faddeev-Popov determinant is *always* evaluated with\nrespect to the *generators* of your continuous symmetries - i.e. from the\ntransformation rules under the infinitesimal transformations\n(diffeomorphisms, in this case). I even don\'t know what you could ever\nmean by the "Faddeev-Popov determinants for finite transformations".\n\nThe determinant counts the volume element of the changes of your\nconstraints - the quantities that you want to set to zero - under the\ninfinitesimal transformations, because only the infinitesimal\ntransformations matter. For example, delta(f(x)) equals sum_i\ndelta(x-x_i) / |f\'(x_i)| where the sum goes over the zeros x_i of the\nfunction f. Note that the factor 1/|f\'(x_i)| only depends on the first\nderivative of the constraint, i.e. on its behavior under the\ninfinitesimal, linearized transformations. The Faddeev-Popov determinant\nis a higher-dimensional generalization of 1/|f\'(x_i)|, i.e. the\ngeneralization to the case of many constraints and many generators, but it\nis still true that only the variations under the infinitesimal linearized\ntransformations matter.\n\nThe infinitesimal transformations form a Lie algebra, and a Lie algebra\ncontains all local information about the associated Lie group - the group\nof finite transformations. Locally, the infinitesimal transformations (Lie\nalgebra) and the finite ones (Lie group) are exactly the same thing -\nthese words "exactly the same thing" are meant to replace your words "much\nmore bigger", and the fact that the Lie algebra encodes essentially\neverything about the Lie group is an important fact, a building fact\nexplaining why we ever consider the objects such as the Lie algebras (and\ninfinitesimal variations) at all.\n\nThese theorems are why it is always enough to work with the infinitesimal\ntransformations; you can get the finite ones by composing the\ninfinitesimal ones. It is enough to prove that an action is invariant\nunder the infinitesimal transformations of some kind, in order to know\nthat it is invariant under all finite transformations connected to the\nidentity. This is the relation between the Lie groups and their Lie\nalgebras.\n\nWell, some global subtleties - discrete subgroups and identifications, as\nwell as the components of the gauge group that are topologically\ndisconnected from the identity - may add some extra complications of a\ncombinatorial (discrete) type, but these additional subtleties are not\nwhat the Faddeev-Popov procedures/determinants were developed for. The FP\ndeterminant should count all analytical, continuous and field-dependent\nfactors, and the canonical clean example where the FP rules apply without\nmodifications are the topologically trivial ones.\n\n> 2) again 3.1.10\n> ---------------\n> Comparing with 3.1.6, there is a factor 2 missing in 3.1.10, right?\n\nAllan Adams might have a newer version of GSW, but 3.1.10 there says that\nthe functional derivative equals 2 times the covariant derivative of the\ndelta-function, which I believe means that the correct factor of 2 is\nthere.\n\n> 3) On the ghost action\n> ----------------------\n> In 3.1.11, the ghost action is normalized with -1/pi.\n\nThe normalization of the action that contains 1 term only is a pure\nconvention, and you can absorb any factor of 1/pi of this type into the\ndefinition of "c" and/or "b". This redefines the path integral by a\nuniversal (divergent) numerical constant only, but the overall\nnormalization of the path integral is a convention-dependent and\ncomplicated issue anyway.\n\nThe question whether there is an "i" or not depends on your convention\nwhether b,c are Hermitean, or antihermitean fields (well, you can easily\nredefine b,c by "i" if you need) - the Minkowski Lagrangian density must\nalways be Hermitean, and the Euclidean one is just a continuation if the\nMinkowski density to "i.t". This makes the Euclidean Lagrangian density\n*antihermitean* for the terms with an odd number of time derivatives plus\ntime-valued indices (e.g. the instanton action density), but most terms\nremain Hermitean.\n\nYou can get a natural normalization of the action once you choose a\nnormalization of the parameters of the infinitesimal transformations.\nOnce you choose a coefficient of the b.dc term in the action, the\nnormalization of all other terms is determined. It is one of the possible\nconventions to normalize all these actions with 1/pi - this is what we\noften do for many bosonic actions, and this normalization is chosen for\nthe sake of unity. These pi\'s can appear from the bosonic gaussian\nintegrals: int exp(-x^2) is sqrt(pi), which really means that you get it\nin the other way. But if the b,c fields were normalized in any other way,\nyou would have asked the same question, so there is no universal answer.\n\n> And when calculating\n> (3.1.35) out of (3.1.31) I actually get 1/2pi and not 1/pi,\n\nI seem to agree, but we might be missing another factor of 2 somewhere...\n\n> Another problem I have is the sequence of the integration in the\n> fermionic path integral (3.1.13): Shouldn\'t this be reversed?\n\nWhat do you exactly want to be reversed? The order of dc and db? In the\nworst case, it could change the overall sign of the path integral (if the\nEuler character - the regularized number of points - of the spacetime is\nodd, whatever it means), but the overall numerical normalization of the\npath integral is a matter of convention anyway.\n\n> 4) On 3.1.15\n> ------------\n> Another very basic one: I get\n> d_+- = 1/2(d_tau -+ i d_sigma)\n\nSo do they, at least in Allan\'s copy of GSW.\n\n> 5) On (3.1.29)\n> --------------\n> What has been done in the last equality here?\n\nThe last line of (3.1.29) was not obtained from the previous one, but from\nthe first line. The commutator of two covariant derivatives, acting on a\ntensor, is always and identically equal to the Riemann tensor contracted\nwith with all the indices of the tensor. The components of the Riemann\ntensor are rho.R(2) (up to signs and factors of 2 or 1/2 that I don\'t want\nto fine-tune right now), and the factor "n" comes because you are summing\n"n" identical terms (because "n" is the number of indices of the tensor,\nand all of them contribute the same thing).\n_________________________________________ _____________________________________\nE-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/\neFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)\n^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Wed, 19 May 2004, Rene Meyer wrote:
> How can GSW restrict themselfes to infinitesimal diffeomorphisms when
> calculating (3.1.10). They have to do so, because (3.1.6) is only
> valid in that limit. But I think the group G of diffeomorphisms is
> much more bigger, and the Faddeev-Popov-determinants should be
> calculated for finite diffeomorphisms also.
Sorry if I misunderstand what you are saying, but this text of yours looks
to me like a conceptual misunderstanding of the whole Faddeev-Popov
technique. The Faddeev-Popov determinant is *always* evaluated with
respect to the *generators* of your continuous symmetries - i.e. from the
transformation rules under the infinitesimal transformations
(diffeomorphisms, in this case). I even don't know what you could ever
mean by the "Faddeev-Popov determinants for finite transformations".
The determinant counts the volume element of the changes of your
constraints - the quantities that you want to set to zero - under the
infinitesimal transformations, because only the infinitesimal
transformations matter. For example, \delta(f(x)) equals sum_i\delta(x-x_i) / |f'(x_i)| where the sum goes over the zeros x_i of the
function f. Note that the factor 1/|f'(x_i)| only depends on the first
derivative of the constraint, i.e. on its behavior under the
infinitesimal, linearized transformations. The Faddeev-Popov determinant
is a higher-dimensional generalization of 1/|f'(x_i)|, i.e. the
generalization to the case of many constraints and many generators, but it
is still true that only the variations under the infinitesimal linearized
transformations matter.
The infinitesimal transformations form a Lie algebra, and a Lie algebra
contains all local information about the associated Lie group - the group
of finite transformations. Locally, the infinitesimal transformations (Lie
algebra) and the finite ones (Lie group) are exactly the same thing -
these words "exactly the same thing" are meant to replace your words "much
more bigger", and the fact that the Lie algebra encodes essentially
everything about the Lie group is an important fact, a building fact
explaining why we ever consider the objects such as the Lie algebras (and
infinitesimal variations) at all.
These theorems are why it is always enough to work with the infinitesimal
transformations; you can get the finite ones by composing the
infinitesimal ones. It is enough to prove that an action is invariant
under the infinitesimal transformations of some kind, in order to know
that it is invariant under all finite transformations connected to the
identity. This is the relation between the Lie groups and their Lie
algebras.
Well, some global subtleties - discrete subgroups and identifications, as
well as the components of the gauge group that are topologically
disconnected from the identity - may add some extra complications of a
combinatorial (discrete) type, but these additional subtleties are not
what the Faddeev-Popov procedures/determinants were developed for. The FP
determinant should count all analytical, continuous and field-dependent
factors, and the canonical clean example where the FP rules apply without
modifications are the topologically trivial ones.
> 2) again 3.1.10
> ---------------
> Comparing with 3.1.6, there is a factor 2 missing in 3.1.10, right?
Allan Adams might have a newer version of GSW, but 3.1.10 there says that
the functional derivative equals 2 times the covariant derivative of the
\delta-function, which I believe means that the correct factor of 2 is
there.
> 3) On the ghost action
> ----------------------
> In 3.1.11, the ghost action is normalized with -1/\pi.
The normalization of the action that contains 1 term only is a pure
convention, and you can absorb any factor of 1/\pi of this type into the
definition of "c" and/or "b". This redefines the path integral by a
universal (divergent) numerical constant only, but the overall
normalization of the path integral is a convention-dependent and
complicated issue anyway.
The question whether there is an "i" or not depends on your convention
whether b,c are Hermitean, or antihermitean fields (well, you can easily
redefine b,c by "i" if you need) - the Minkowski Lagrangian density must
always be Hermitean, and the Euclidean one is just a continuation if the
Minkowski density to "i.t". This makes the Euclidean Lagrangian density
*antihermitean* for the terms with an odd number of time derivatives plus
time-valued indices (e.g. the instanton action density), but most terms
remain Hermitean.
You can get a natural normalization of the action once you choose a
normalization of the parameters of the infinitesimal transformations.
Once you choose a coefficient of the b.dc term in the action, the
normalization of all other terms is determined. It is one of the possible
conventions to normalize all these actions with 1/\pi - this is what we
often do for many bosonic actions, and this normalization is chosen for
the sake of unity. These \pi's can appear from the bosonic gaussian
integrals: \int \exp(-x^2) is \sqrt(\pi), which really means that you get it
in the other way. But if the b,c fields were normalized in any other way,
you would have asked the same question, so there is no universal answer.
> And when calculating
> (3.1.35) out of (3.1.31) I actually get 1/2pi and not 1/\pi,
I seem to agree, but we might be missing another factor of 2 somewhere...
> Another problem I have is the sequence of the integration in the
> fermionic path integral (3.1.13): Shouldn't this be reversed?
What do you exactly want to be reversed? The order of dc and db? In the
worst case, it could change the overall sign of the path integral (if the
Euler character - the regularized number of points - of the spacetime is
odd, whatever it means), but the overall numerical normalization of the
path integral is a matter of convention anyway.
> 4) On 3.1.15
> ------------
> Another very basic one: I get
> d_+- = 1/2(d_{tau} -+ i d_{sigma})
So do they, at least in Allan's copy of GSW.
> 5) On (3.1.29)
> --------------
> What has been done in the last equality here?
The last line of (3.1.29) was not obtained from the previous one, but from
the first line. The commutator of two covariant derivatives, acting on a
tensor, is always and identically equal to the Riemann tensor contracted
with with all the indices of the tensor. The components of the Riemann
tensor are \rho.R(2) (up to signs and factors of 2 or 1/2 that I don't want
to fine-tune right now), and the factor "n" comes because you are summing
"n" identical terms (because "n" is the number of indices of the tensor,
and all of them contribute the same thing).
__{_______________________________________________ _____________________________}
E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Rene Meyer
May21-04, 02:04 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Wed, 19 May 2004 15:32:48 -0400, Lubos Motl wrote:\n> The determinant counts the volume element of the changes of your\n> constraints - the quantities that you want to set to zero - under the\n> infinitesimal transformations, because only the infinitesimal\n> transformations matter. For example, delta(f(x)) equals sum_i\n> delta(x-x_i) / |f\'(x_i)| where the sum goes over the zeros x_i of the\n> function f. Note that the factor 1/|f\'(x_i)| only depends on the first\n> derivative of the constraint, i.e. on its behavior under the\n> infinitesimal, linearized transformations. The Faddeev-Popov determinant\n> is a higher-dimensional generalization of 1/|f\'(x_i)|, i.e. the\n> generalization to the case of many constraints and many generators, but it\n> is still true that only the variations under the infinitesimal linearized\n> transformations matter.\n\nOK, I understand now. Actually thats a point that is not really\nemphasized in the QFT I read. Do you know any source where the Faddeev\nPopov technique is treated in a more strict way?\n\n>> 2) again 3.1.10\n>> ---------------\n>> Comparing with 3.1.6, there is a factor 2 missing in 3.1.10, right?\n> Allan Adams might have a newer version of GSW, but 3.1.10 there says that\n> the functional derivative equals 2 times the covariant derivative of the\n> delta-function, which I believe means that the correct factor of 2 is\n> there.\n\nOK. I have the first version of GSW here, so the typos aren\'t fixed\nyet.\n\n>> 3) On the ghost action\n>> ----------------------\n> The question whether there is an "i" or not depends on your convention\n> whether b,c are Hermitean, or antihermitean fields (well, you can easily\n> redefine b,c by "i" if you need) - the Minkowski Lagrangian density must\n> always be Hermitean, and the Euclidean one is just a continuation if the\n> Minkowski density to "i.t". This makes the Euclidean Lagrangian density\n> *antihermitean* for the terms with an odd number of time derivatives plus\n> time-valued indices (e.g. the instanton action density), but most terms\n> remain Hermitean.\n\nSo if b and c both are hermitian or antihermitian, c delta b is\nhermitian. But if only one of them is hermitian or antihermitian,\nthen one needs to include a "i", right? But what about the c and b\nswitching places in the adjoint:\n\n(c delta b)^dagger = (+-) b^dagger delta^T c^dagger\n\nc and b are Grassmann numbers, thus the adjoint should bring up\nanother minus sign?\n\n> You can get a natural normalization of the action once you choose a\n> normalization of the parameters of the infinitesimal transformations.\n\nWhat do you mean here?\n\n>> And when calculating\n>> (3.1.35) out of (3.1.31) I actually get 1/2pi and not 1/pi,\n> I seem to agree, but we might be missing another factor of 2 somewhere...\n\nAgreeing with me or with the book?\n\nRené.\n\n--\nRené Meyer\nStudent of Physics & Mathematics\nZhejiang University, Hangzhou, China\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Wed, 19 May 2004 15:32:48 -0400, Lubos Motl wrote:
> The determinant counts the volume element of the changes of your
> constraints - the quantities that you want to set to zero - under the
> infinitesimal transformations, because only the infinitesimal
> transformations matter. For example, \delta(f(x)) equals sum_i
> \delta(x-x_i) / |f'(x_i)| where the sum goes over the zeros x_i of the
> function f. Note that the factor 1/|f'(x_i)| only depends on the first
> derivative of the constraint, i.e. on its behavior under the
> infinitesimal, linearized transformations. The Faddeev-Popov determinant
> is a higher-dimensional generalization of 1/|f'(x_i)|, i.e. the
> generalization to the case of many constraints and many generators, but it
> is still true that only the variations under the infinitesimal linearized
> transformations matter.
OK, I understand now. Actually thats a point that is not really
emphasized in the QFT I read. Do you know any source where the Faddeev
Popov technique is treated in a more strict way?
>> 2) again 3.1.10
>> ---------------
>> Comparing with 3.1.6, there is a factor 2 missing in 3.1.10, right?
> Allan Adams might have a newer version of GSW, but 3.1.10 there says that
> the functional derivative equals 2 times the covariant derivative of the
> \delta-function, which I believe means that the correct factor of 2 is
> there.
OK. I have the first version of GSW here, so the typos aren't fixed
yet.
>> 3) On the ghost action
>> ----------------------
> The question whether there is an "i" or not depends on your convention
> whether b,c are Hermitean, or antihermitean fields (well, you can easily
> redefine b,c by "i" if you need) - the Minkowski Lagrangian density must
> always be Hermitean, and the Euclidean one is just a continuation if the
> Minkowski density to "i.t". This makes the Euclidean Lagrangian density
> *antihermitean* for the terms with an odd number of time derivatives plus
> time-valued indices (e.g. the instanton action density), but most terms
> remain Hermitean.
So if b and c both are hermitian or antihermitian, c \delta b is
hermitian. But if only one of them is hermitian or antihermitian,
then one needs to include a "i", right? But what about the c and b
switching places in the adjoint:
(c \delta b)^dagger = (+-) b^{dagger} \delta^T c^{dagger}
c and b are Grassmann numbers, thus the adjoint should bring up
another minus sign?
> You can get a natural normalization of the action once you choose a
> normalization of the parameters of the infinitesimal transformations.
What do you mean here?
>> And when calculating
>> (3.1.35) out of (3.1.31) I actually get 1/2pi and not 1/\pi,
> I seem to agree, but we might be missing another factor of 2 somewhere...
Agreeing with me or with the book?
René.
--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
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