Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Right now I am having this silly difficulty from the following:

  1. BTW, Conformal dimension/scaling dimension is -ve of mass dimension ..right?
  2. In p-63 of Magoo, after 3.15 eq, they said a.) $\phi$ is dimensionless..why? b.) They, after getting length dimension of boundary field as $\Delta d$, referred to 3.13 to comment that O operator has conformal dimension $\Delta$. I know its right..But as in the lhs of 3.13, the exponent must be dimension less, it seems O should have length or conformal dimension to be $-\Delta$..

I know I am making a silly mistake here..Thanks for answering.

share|improve this question
2  
You'll get much better help if you ask a question that's independent of a particular reference. You'll also understand your own question better by translating the difficulties you're having with the book into a form that someone else can read and comment on. –  Mark Eichenlaub Jan 20 '11 at 21:11
    
Yeah I know..I am not voting for my question either..But it was giving me some unhappiness for some time..that's why..sorry again..but a little answer is welcome. –  user1349 Jan 20 '11 at 21:44
1  
@user13 If the question is important to you, wouldn't it be worth it to put in the effort to make it presentable? –  Mark Eichenlaub Jan 20 '11 at 21:48
    
Now that I know what MAGOO stands for (thanks @lubos), this question sounds a lot less silly than the title would indicate. Please include the full reference for everyone's benefit. –  user346 Jan 21 '11 at 2:18
    
Never mind. I took care of it. –  user346 Jan 21 '11 at 2:20
show 3 more comments

1 Answer 1

up vote 2 down vote accepted

Just to be sure, MAGOO is the AdS/CFT Bible

http://arxiv.org/abs/hep-th/9905111

whose authors are usually listed alphabetically. Concerning your questions,

  1. Conformal dimension or scaling dimension or mass dimension are meant to be the same thing in the context of conformal field theories in more than 2 dimensions. In 2 dimensions, one may distinguish the left-moving (holomorphic) and right-moving (antiholomorphic) dimensions or their sum - but the separate chiral "dimensions" are usually called "weights", anyway. The dimension pretty much counts the exponent of "kilogram" in the unit of the corresponding operator - except that in quantum mechanics, the exponent is often fractional. The dimensions are usually positive - they get mapped to the energy which is positive as well. Operators have positive dimensions of mass. If your cryptic "-ve" meant that there is a minus sign, then there is no minus sign. If you prefer to use units of length, then their powers are negative, but you should switch to masses as the base whose powers count the dimension.

  2. a) The field $\phi$, as opposed to $\phi_0$, is dimensionless because it is an actual field in the AdS space (the bulk), unlike $\phi_0$ that is defined on the boundary. They mean that it is dimensionless under the conformal transformations of the boundary - and that's true because the conformal transformations are realized as simple isometries in the bulk, so they can't rescale the AdS field $\phi$ - at most, they move it to another point. Also, on that page, one deals with particular finite modified boundary conditions for $\phi$ at the boundary of the AdS space so it must be finite even at $z=0$.

  3. b) In equation (3.15), the left hand side is dimensionless. The right hand side has a power of $\epsilon$ whose dimension is $length^{d-\Delta}$ because $\epsilon$ has units of length (on boundary), and $\phi_0$ whose dimension must be $length^{\Delta-d}$ as a consequence, to get a dimensionless product. In equation (3.13), the exponent on the left hand side has $d^4x$ whose dimension is $length^d$ - note that $d=4$ for $AdS_5$; $\phi_0$ whose dimension is $length^{\Delta-d}$ as I said in the previous sentence - note that the $d$ term in the exponent cancels; and $O$ whose dimension must therefore be $length^{-\Delta}$ which means $mass^{\Delta}$. As I said, the standard base for counting dimensions is mass, so we also say that $O$ has dimension $\Delta$. So I suspect that your sign error is simply caused by the point 1) - namely by your incorrect assumption that dimensions refer to the powers of length.

When we say "dimension" without extra specifications, we mean the power of the mass, not length. It's a healthy convention because the operators end up having non-negative dimensions. You may want to remember the dimensions of basic operators in 4 dimensions. The identity operator is always dimensionless (and $x$-independent): the dimension is 0. Bosonic scalar fields $\phi_0$ and potentials $A_\mu$ have dimension 1, their field strength has dimension $2$, fermions like the Dirac $\psi$ have dimension 3/2, and all terms in the Lagrangian density have dimension $4$. These are classical dimensions; each derivative adds $1$ to the dimension. Products of these operators have dimensions that are sums of the dimensions of the factors - except that quantum mechanics also adds "anomalous dimensions" to this classical form of the dimension, so that the total dimensions may have fractional terms that are proportional to powers of $g$ etc.

share|improve this answer
    
ok..so let me just verify something. suppose a field \phi(x) under conf.trans. x->\lambda x, goes to \lambda^h \phi(\lambda x). Then h is the conf. dim..right? But isn't it only possible if \phi(x) has some (length scale)^h power in it? I know that length^h means mass dim. '-h'. But if mass dim. and conf. dim are same then is for the above transformation, the conf. dim. '-h', and not h? –  user1349 Jan 20 '11 at 22:14
    
You're mixing active and passive transformations (and/or you are mixing transformations of coordinates and fields) in a strange way. If $x_{old}$ goes to $x_{new}=\lambda x_{old}$, then the new $\phi_{new}(x_{new})$ is equal to $\lambda^{-h} \phi_{old}(x_{old}/\lambda)$ and $h$ is the conformal dimension. There is no contradiction, at any rate, the dimension is the power of the mass in the units of the operator. Imagine $\phi(y)=y$ itself, then you get dimension minus one, $h=-1$. –  Luboš Motl Jan 20 '11 at 22:36
    
@user1349 please include latex in single dollar signs for inline math and double dollars for displaymath. –  user346 Jan 21 '11 at 2:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.