Question on calculating central charge for bosonic QFT
Nicholas-Joshua
I'd really appreciate help with this if you have the time:
For the free bosonic field, how does one determine the central charge (c=1
in this case) from the expression for the Virasoro algebra in terms of mode
operators (instead of determining it from the OPE)?
I.e., using L_{m} = 1/2 \sum_{n=-\infty}^{n=\infty} : \alpha^{\mu}_{m-n}
\alpha_{\mu n}:
Thanks a lot!
Nic
Aaron Bergman
Nicholas-Joshua wrote:
You know the commutation relations for the alphas, so take the
expressions for two of the Ls in terms of the alphas and commute
them. If you choose your Ls wisely, you should be able to see
the central charge in your answer.
--
Aaron Bergman
<http://zippy.ph.utexas.edu/~abergman/>
Charles Torre
> For the free bosonic field, how does one determine the central charge (c=1
> in this case) from the expression for the Virasoro algebra in terms of mode
> operators (instead of determining it from the OPE)?
> I.e., using L_{m} = 1/2 \sum_{n=-\infty}^{n=\infty} : \alpha^{\mu}_{m-n}
> \alpha_{\mu n}:
>
Off the top of my head, I believe you can find this "old fashioned"
derivation in the book "Superstring Theory" by Green, Schwarz and
Witten; in the book "Principles of String Theory", by Brink and
Henneaux; and in the classic review article by Scherk, Rev. Mod. Phys.
47, 123, (1975). Experts might have better references, but these are
pretty easy to follow in my recollection.
charlie
Lubos Motl
Dear Nic,
the Virasoro algebra in terms of the Fourier modes L_n is
[L_m,L_n] = (m-n) L_{m+n} + c (m^3-m) / 12 delta_{m,-n}
See, for example, (2.6.19) in Polchinski's book, and the derivation of
it using contour integrals is before (2.6.19). Let me use another
approach.
The linear term proportional to "-c m" can be removed by an additive shift
of L_0. The only term that would be left afterwards would be proportional
to m^3 which is essentially guaranteed by dimensional analysis - this
power m^3 is equivalent to 1/z^4 in the OPEs.
OK, now we can approach your question. You see that the central charge can
be determined from the commutators [L_m,L_n] assuming that m=-n. Moreover,
you see that m=-1 or 0 or 1 annihilates the central term because it is
proportional to m^3-m. You must take at least L_2.
So you're invited to calculate [L_2,L_{-2}].
> I.e., using L_{m} = 1/2 \sum_{n=-\infty}^{n=\infty} : \alpha^{\mu}_{m-n}
> \alpha_{\mu n}:
It won't be too difficult. Let me write \alpha^{\mu}_{k} as "ak" to save
space. Note that the commutators like [L_2,L_{-2}] are always a sum over
"mu", and therefore they're proportional to the number of bosons
(dimensions) - therefore "c" is proportional as well. You want to study
the contribution of one boson to "c" which is c=1 as we will see.
The commutator [L_2,L_{-2}] is
[... + a4 a-2 + a3 a-1 + a2 a0 + a1 a1 / 2,
... + a-4 a2 + a-3 a1 + a-2 a0 + a-1 a-1 / 2]
The normal ordering was very easy for L_2 and L_{-2} - there are no pure
numbers coming from the normal ordering. The pure number terms from this
commutator come from the double-Wick-contracted terms where you evaluate
both [a,a] commutators appearing in the product of four "a"'s.
Let me use a simpler strategy, assuming that you believe me the form of
the commutator. Let's act with [L_2,L_-2] on the vacuum state |0>.
Obviously the term L_{-2} L_{2} annihilates the vacuum because L_{2} does,
and you end up with
[L_2,L_-2] |0> = L_2 L_{-2} |0>
That's easy because the only term surviving in L_{-2}|0> is
a-1 a-1 / 2 |0> (###)
and the only term from L_{2} that converts this back to |0> times a
constant is "a1 a1 / 2". "a-1" and "a1" are creation and annihilation
operators of a harmonic oscillator, and because we act on the state (###)
that is the second excitation,
a1 a1 / 2 a-1 a-1 /2 |0> =
1/4 (a1 (1 + a-1 a1) a-1 |0> =
1/4 (a1 a-1 + a1 a-1 a1 a-1) |0> =
1/4 (1 + a-1 a1 + (1- a-1 a1)^2 ) |0>
Once we get a1 next to |0>, it vanishes (annihilation operator), and
therefore we obtained
1/2 |0>
at the end. It should be equal to (m^3-m).c/12, but because m=2, m^3-m
equals 6, and 6/12 equals one half, and you see that exactly matches when
c=1 - assuming that we used the definition when L_0 annihilates the vacuum
(which means that the linear central term "cm/12 . delta" appears in the
commutator).
All the best
Lubos
______________________________________________________________________________
E-mail: lu...@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
phone: work: +1-617/496-8199 home: +1-617/868-4487
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Superstring/M-theory is the language in which God wrote the world.
Rene Meyer
> Dear Nic,
> the Virasoro algebra in terms of the Fourier modes L_n is
> [L_m,L_n] = (m-n) L_{m+n} + c (m^3-m) / 12 delta_{m,-n}
> The linear term proportional to "-c m" can be removed by an additive shift
> of L_0. The only term that would be left afterwards would be proportional
> to m^3 which is essentially guaranteed by dimensional analysis - this
> power m^3 is equivalent to 1/z^4 in the OPEs.
> So you're invited to calculate [L_2,L_{-2}].
> at the end. It should be equal to (m^3-m).c/12, but because m=2, m^3-m
> equals 6, and 6/12 equals one half, and you see that exactly matches when
> c=1 - assuming that we used the definition when L_0 annihilates the vacuum
> (which means that the linear central term "cm/12 . delta" appears in the
> commutator).
While calculating the Virasoro Algebra for myself, I just encountered
a question related to this post of Lubos:
How can L_0 annihilate the vacuum and the central charge be c=d for
all d? It seems to me that this is only right in d=26, as for any d,
we only have the demand
L_0|phys> = a|phys>
for some normal ordering constant a, that can be found to be (d-2)/12
after regularization. But the formula d/12(m^3-m) for the conformal
anomaly should be valid for all d.
Rene.
Lubos Motl
> How can L_0 annihilate the vacuum and the central charge be c=d for
> all d? It seems to me that this is only right in d=26, as for any d,
> we only have the demand
>
> L_0|phys> = a|phys>
>
> for some normal ordering constant a, that can be found to be (d-2)/12
> after regularization. But the formula d/12(m^3-m) for the conformal
> anomaly should be valid for all d.
Dear Rene, you seem to be talking about too many different things
simultaneously, and therefore your question is not clear. A particular CFT
has a central charge. If your CFT is made of "d" bosons - describing
spacetime dimensions - then its central charge is "c=d", more or less by
definition. The central charge of the sum of two CFTs is the sum of the
central charges.
In string theory, we eventually want to consider c=0 conformal field
theories only, and because the necessary bc-ghosts carry c=-26, we must
cancel them by 26 bosonic dimensions. Nevertheless, you can talk about the
CFT with a nonzero central charge, i.e. with a different dimension. The
operators L_n constructed from a CFT with a central charge "c" simply will
give you the c-number c(m^3-m)/12 for any "c" - this is how the central
charge is defined (as the coefficient of the c-number in the Virasoro
algebra, or equivalently the 1/z^4 term in the T(z)T(0) OPEs), and as I
said above, the normalization is chosen in such a way that a single boson
has c=1.
These are absolutely universal facts. You might be also confused about the
difference between "d" and "d-2" in some of your formulae? The light-cone
gauge description describes the physical spectrum only, and it has "d-2"
transverse sets of oscillators. In the light cone gauge, you don't have
any bc-ghosts. On the other hand, the covariant treatment leads you to
include all "d" directions, and kill the unphysical polarizations using
the BRST tools and the bc-ghosts that carry c=-26.
______________________________________________________________________________
E-mail: lu...@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)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Thomas Larsson
me...@web.de (Rene Meyer) wrote in message news:<d664a6da.04020...@posting.google.com>...
>
> How can L_0 annihilate the vacuum and the central charge be c=d for
> all d? It seems to me that this is only right in d=26, as for any d,
> we only have the demand
>
> L_0|phys> = a|phys>
>
> for some normal ordering constant a, that can be found to be (d-2)/12
> after regularization. But the formula d/12(m^3-m) for the conformal
> anomaly should be valid for all d.
>
The central extension is necessary for unitarity. It is well known that
the Virasoro algebra only has unitary irreps for a discrete set of values
0 <= c <= 1 (more precisely, c = 1 - 6/m(m+1) ) and the continuum c > 1.
A nice review of CFT, with emphasis on a physically successful application,
can be found in http://www.arxiv.org/abs/hep-th/9108028 .
Rene Meyer
On Fri, 13 Feb 2004 19:23:25 +0000 (UTC), Lubos Motl wrote:
>> How can L_0 annihilate the vacuum and the central charge be c=d for
>> all d? It seems to me that this is only right in d=26, as for any d,
>> we only have the demand
>>
>> L_0|phys> = a|phys>
>>
>> for some normal ordering constant a, that can be found to be (d-2)/12
>> after regularization. But the formula d/12(m^3-m) for the conformal
>> anomaly should be valid for all d.
> Dear Rene, you seem to be talking about too many different things
> simultaneously, and therefore your question is not clear. A particular CFT
> has a central charge. If your CFT is made of "d" bosons - describing
> spacetime dimensions - then its central charge is "c=d", more or less by
> definition. The central charge of the sum of two CFTs is the sum of the
> central charges.
Till now I got this right, too, this means I understood the argument
with shifting the constant before m while calculating the Virasoro
Algebra. But what do you mean by the "sum" of two CFTs? Summing their
actions?
> CFT with a nonzero central charge, i.e. with a different dimension. The
> operators L_n constructed from a CFT with a central charge "c" simply will
> give you the c-number c(m^3-m)/12 for any "c" - this is how the central
> charge is defined (as the coefficient of the c-number in the Virasoro
> algebra, or equivalently the 1/z^4 term in the T(z)T(0) OPEs), and as I
> said above, the normalization is chosen in such a way that a single boson
> has c=1.
What are OPEs?
René.
--
René Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China