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Transformation of generator under translations

  1. Jun 18, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Let us study the subgroup of the Poincare group that leaves the point ##x=0## invariant, that is the Lorentz group. The action of an infinitesimal Lorentz transformation on a field ##\Phi(0)## is ##L_{\mu \nu}\Phi(0) = S_{\mu \nu}\Phi(0)##. By use of the commutation relations of the Poincare group, we translate the generator ##L_{\mu \nu}## to a nonzero value of ##x##: $$e^{ix^{\rho}P_{\rho}} L_{\mu \nu} e^{-ix^{\sigma}P_{\sigma}} = S_{\mu \nu} - x_{\mu}P_{\nu} + x_{\nu}P_{\mu}\,\,\,\,\,\,\,\,\,(1),$$ where the RHS is computed using the Baker-Campbell-Hausdorff formula to the second term. Then we can write the action of the generators $$P_{\mu} \Phi(x) = -i\partial_{\mu}\Phi(x) \,\,\,\,\text{and}\,\,\,\,L_{\mu \nu}\Phi(x) = i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})\Phi(x) + S_{\mu \nu}\Phi(x)\,\,\,\,\,(2)$$

    I understand that the LHS of eqn (1) describes the transformation of ##L_{\mu \nu}## under a space time translation. What I want to understand is how they obtained eqn (2).

    2. Relevant equations
    ##\Phi'(x') = (\text{Id} - i\omega_{g}T_{g})\Phi(x)##

    3. The attempt at a solution
    I am trying to understand how we are able to write the action of the generators like that using eqn (1). If we consider the transformation of the fields under a Lorentz transformation then $$\Phi'(x') = (\text{Id} - iL_{\mu \nu}\omega^{\mu \nu})\Phi(0+x)$$ using the equation in relevant equations. Now expand ##\Phi(0+x) \approx \Phi(0) + x^{\mu}\partial_{\mu}\Phi(0)##.

    For a scalar field, ##\Phi'(x') = \Phi(x)## and so we have ##\Phi(x) = \Phi(0) + x^{\mu}\partial_{\mu}\Phi(0) + iL_{\mu \nu}\omega^{\mu \nu}\Phi(0) + iL_{\mu \nu}\omega^{\mu \nu}x^{\sigma}\partial_{\sigma}\Phi(0)##. I can't see a way to connect this to eqn (2) exactly.

    If I now use the fact that the coordinates change ##x^{\rho} = \omega^{\rho}_{\,\,\,\nu}x^{\nu}## then the above becomes, ignoring the last term quadratic in the parameter, $$\Phi(x) = \Phi(0) - \omega^{\rho}_{\,\,\,\nu}x^{\nu}\partial_{\rho}\Phi(0) + iL_{\mu \nu}\omega^{\mu \nu}\Phi(0)\,\,\,\,(3)$$ Now antisymmetrise over the components of ##\omega##, add the result to (3) gives $$2\Phi(x) = 2\Phi(0) + \omega^{\rho \nu}(x_{\rho}\partial_{\nu} - x_{\nu}\partial_{\rho})\Phi(0) + 2i\omega^{\mu \nu}S_{\mu \nu}\Phi(0)$$

    So I am making contact with the orbital and intrinsic generators, but I can't quite get an exact result to match with eqn (1) in order to read off the explicit forms for the generators ##P_{\mu}## and ##L_{\mu \nu}##

    Many thanks.
     
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  3. Jun 19, 2014 #2

    strangerep

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    Let's focus first just on a simple piece of this problem, i.e., $$P_{\mu} \Phi(x) = -i\partial_{\mu}\Phi(x) ~. $$Do you understand how to obtain this? (If so, give a sketch.)
     
  4. Jun 22, 2014 #3

    CAF123

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    Hi strangrep, sorry for late reply.
    Yes I think I get the overall idea but, as I point out below, there are a few mathematical intricacies that I do not understand.

    Consider the expansion of the new field at the new position x, ##\Phi'(x')##. Then $$\Phi'(x') \approx \Phi(x) + \omega_{a}\frac{\delta F}{\delta \omega_a}(x)\,\,\,\,(1)$$ But also $$\Phi'(x') \approx \Phi(x') - \omega_a \frac{\delta x^{\mu}}{\delta \omega_a} \frac{\delta \Phi(x')}{\delta x^{\mu}} + \omega_{a} \frac{\delta F}{\delta \omega_a}(x')\,\,\,\,(2)$$ F was defined to be a function of the field ##F=F(\Phi(x)) := \Phi'(x')##
    Eqn (2) is not so clear right now, but here are my thoughts:

    Translations form an abelian group and so admit only 1D irreps with which the fields transform. Therefore, ##\Phi'(x') = \Phi(x)##. So under a translation ##x'^{\mu} = x^{\mu} + a^{\mu}##, $$\Phi'(x') = \Phi(x) = \Phi(x' - a) \approx \Phi(x') - \omega_{a}\frac{\delta \Phi(x')}{\delta\omega_a}$$ Sub into (1). Then I was wondering why there is a prime on the x at the end of (1). Again, my thoughts being that we have simply defined a function numerically equivalent to one in the unprimed system: ##F' = F'(\Phi(x')) \equiv F(\Phi(x))##. If all that is correct, then I can move on.

    It is now just a case of subbing in the above into the generic transformation of fields: $$\Phi'(x) - \Phi(x) = -\omega_a G_a \Phi(x) \Rightarrow \Phi'(x') - \Phi(x') = -i\omega_a G_a \Phi(x')$$ by sending x → x'. Sub in results, F is trivial as noted above and we get the result.

    My question is how to obtain the same result using what was written in the OP.
    Many thanks.
     
  5. Jun 22, 2014 #4

    strangerep

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    So let 's review... ##\Phi(x)## is (I presume) a quantum field. You are given these equations:
    $$L_{\mu \nu}\Phi(0) ~=~ S_{\mu \nu}\Phi(0) ~~~~~~~~~~ (0)$$ $$e^{ix^{\rho}P_{\rho}} L_{\mu \nu} e^{-ix^{\sigma}P_{\sigma}} ~=~ S_{\mu \nu} - x_{\mu}P_{\nu} + x_{\nu}P_{\mu} ~~~~~(1) $$and you want to understand how to obtain:
    $$L_{\mu \nu}\Phi(x) ~=~ i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})\Phi(x) + S_{\mu \nu}\Phi(x) ~~~~~ (2)$$

    So... do you understand that
    $$\Phi(x+a) ~=~ e^{ia^\rho P_\rho} \Phi(x) e^{-ia^\sigma P_\sigma} ~~~~~~ (4)$$ ?

    If so, you should be able to apply [an adaptation of] eq(4) to [both sides of] eq(0), then use eq(1) to arrive at eq(2).
     
  6. Jun 23, 2014 #5

    CAF123

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    I can write ##\Phi(0+x) = e^{ix^{\rho} P_{\rho}}\Phi(0)e^{-ix^{\sigma}P_{\sigma}}## And then also ##\Phi(0+x) \approx \Phi(0) +x^{\mu}\partial_{\mu}\Phi(0) = \Phi(0) + \omega^{\mu}_{\,\,\,\nu}x^{\nu}\partial_{\mu}\Phi(0) = \Phi(0) + \omega^{\mu \nu}x_{\nu}\partial_{\mu}\Phi(0)##.
    I could antisymmetrise over omega now to obtain a term ##x_{\nu}\partial_{\mu} - x_{\mu}\partial_{\nu}##. I think doing some sort of taylor expansion is the only way to bring in the partial derivative terms and I think the above way is the correct way?

    Then I could multiply ##\Phi(0+x) = \Phi(x)## above by ##L_{\mu \nu}## but that would yield ##\Phi(0)## terms on the RHS and not ##\Phi(x)## as I require.

    What is the adaptation you alluded to? I can write the transformation of the generator itself as ##L'_{\mu \nu} = e^{ix^{\rho}P_{\rho}}L_{\mu \nu}e^{-ix^{\sigma}P_{\sigma}}## but I am not sure what to do with this.

    Thanks.
     
  7. Jun 23, 2014 #6

    strangerep

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    It's a lot simpler than that. Just work out the following expression:
    $$e^{ix^{\rho}P_{\rho}} L_{\mu \nu} \Phi(0) e^{-ix^{\sigma}P_{\sigma}} ~=~ ?$$It involves no more than 1 or 2 lines. Don't overthink it. And ignore what I called eq(0) in my post #4.

    Hint: insert a ##1## between ##L_{\mu \nu}## and ##\Phi(0)##.
     
  8. Jun 24, 2014 #7

    CAF123

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    I see, many thanks. I thought they actually derived ##P_{\mu} = -i\partial_{\mu}## in the actual derivation which is why I was so keen to always do a taylor expansion.

    I have a question about the RHS of your eqn(1). Here is my working:
    From Hausdorff formula, ##e^{ix^{\rho}P_{\rho}}L_{\mu \nu}e^{-ix^{\sigma}P_{\sigma}} = L_{\mu \nu} - [L_{\mu \nu}, x^{\sigma}P_{\sigma}] + ...## I can rewrite the second term like ##[L_{\mu \nu}, x^{\sigma}P_{\sigma}] = x^{\sigma}[L_{\mu \nu}, P_{\sigma}] + [L_{\mu \nu}, x^{\sigma}]P_{\sigma} = ix^{\sigma}(\eta_{\sigma \nu}P_{\mu} - \eta_{\sigma \mu}P_{\nu}) + [L_{\mu \nu}, x^{\sigma}]P_{\sigma}## using the commutation relations of the Poincare group.

    Then I can write it like ##ix_{\nu}P_{\mu} - ix_{\mu}P_{\nu} + (L_{\mu \nu}x^{\sigma} - x^{\sigma}L_{\mu \nu})P_{\sigma}##

    Now what I did in my previous attempt at this was to now set ##L_{\mu \nu} \rightarrow S_{\mu \nu}## and so the second term vanishes since ##S_{\mu \nu}## only acts on the fields.
    But I no longer like what I did here because, besides the fact that the resulting term is incorrect by a factor of i, I have no justication of why the single ##L_{\mu \nu}## term on the RHS should be solely ##S_{\mu \nu}##.

    Thanks.
     
  9. Jun 25, 2014 #8

    strangerep

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    Shouldn't that be ##\cdots - [L_{\mu \nu}, i x^{\sigma}P_{\sigma}]## ?

    ##S_{\mu \nu}## acts on spin indices (which you haven't shown explicitly on ##\Phi##). Is that what you meant?

    See 1st comment above.

    I must say that I don't much like the way the original question is formulated. It would be nicer if ##L## were replaced by ##J##, with ##L## reserved for the orbital part. But at the same time, I don't want to confuse you (too much).

    It might be better to take a step back and review section 7.2 of Ballentine on pp164-166. Do you have (or can you access) a copy of Ballentine?
     
    Last edited: Jun 25, 2014
  10. Jun 25, 2014 #9

    CAF123

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    Yes of course, thanks. That corrects the error later on with the stray i.

    Yes, and if I let ##L_{\mu \nu} \rightarrow S_{\mu \nu}## in that commutator, I get that ##S_{\mu \nu}x^{\sigma} - x^{\sigma}S_{\mu \nu} = 0## since S here does not act on an entity with any spin indices. My trouble is why can I simply send ##L_{\mu \nu} \rightarrow S_{\mu \nu}## on the RHS without also doing it on the LHS.

    Just so that you know, later on in the book I am following, the same derivation is done for the dilation operator. The result is ##e^{ix^{\rho}P_{\rho}}De^{-ix^{\sigma}P_{\sigma}} = D + x^{\sigma}P_{\sigma}## There the 'intrinsic' term (D) that appears on the RHS is also on the LHS. Then the book writes ##D\Phi(x) = (\Delta' - ix^{\sigma}\partial_{\sigma})\Phi(x)##, where ##\Delta'## has the analogous meaning as ##S_{\mu \nu}## in the other case. So it seems there is a lack of care for notation here, unless I am misunderstanding something.

    Also, I read in one of the books I have that each 'full' generator of a symmetry transformation can be regarded as being composed of a spacetime part and an internal part. I believe that the space time part transforms the coordinates in the space and the internal part transforms the field. Is that correct? It just seems strange because, in the case at hand, we have the space time part acting on ##\Phi## too (not just ##S_{\mu \nu}##) in eqn (2), so it made me wonder if I understand the book right. Or perhaps it is a case that since the field is a function of the coordinates,##\Phi = \Phi(x)##, it transforms under the space time part too.

    You did not confuse me and in fact I was thinking the same at some point last week, but just tried to deal with the notation. If we send L to J then in your eqn(2), it is easy to see that J=L+S, L being the orbital angular momentum.

    I took a trip this morning to the uni library and found an older version of Ballentine, I think. 7.2 of that edition does not include anything on the field theory discussion but instead more or less describing the components L and S of the total angular momentum. Is that what you have me look at?
     
    Last edited: Jun 25, 2014
  11. Jun 25, 2014 #10

    strangerep

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    This is all from a book?? Geez, you should have given a specific reference in your opening post. Might have saved a lot of time and guesswork.

    I'll respond to the dilation stuff after you give me the reference (assuming I can access the book, of course).

    There's some different cases here. E.g., one could have (say) a 3-vector valued field ##f^i(x)## (aka a "spin-1" field) on spacetime. A spatial rotation must then act (somehow) on the coordinates, but also on the ##i## 3-vector index. Another example is that one could have a field which behaves as a scalar under Lorentz transformations, but is a doublet under an internal (gauge) symmetry group (e.g., the ##SU(2)_L## isospin group). People don't always write out the indices explicitly -- they expect you to keep track of it all in your head (which is difficult when learning, and can still be difficult later -- I often need to drop back to fully explicit index notation to make sure I understand something properly).

    More on this below...

    Yes. I just wanted you to see a clearer treatment of how a spatial rotation is decomposed into 2 parts when one is not dealing with a scalar valued field. E.g., in Ballentine's case (ii) on p165, he shows in eqn(7.19) how the generic rotation operator ##\mathbf R## is decomposed into a matrix ##D## which acts on the components of the wave function (my ##f^i## above is essentially the same idea), and a differential operator ##R^{-1}## which acts on the coordinates ##x##.

    His ##D## is the exponentiated spin operator -- see eqn(7.21).

    In the general case there might not be 3 components -- that's specific to the spin-1 case. For other spin values, there's a different number of components (corresponding to the range of the ##m## quantum number -- see the list at the bottom of p162).
     
    Last edited: Jun 25, 2014
  12. Jun 26, 2014 #11

    CAF123

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    Sorry about that. I am concentrating on two chapters of the big yellow book on Conformal Field Theory by Di Francesco, Mathieu and Senechal that develop the symmetry transformations/action invariance under transformations and generators of the conformal group. To be precise these are:
    Starting 2.4 P36 - 42.
    And for what we are talking about at the moment,
    Chapter 4, 4.2 P.100.

    And I think I understand the notation, but would appreciate any further comments you have.
    $$e^{ix^{\rho}P_{\rho}}L_{\mu \nu}\Phi(0)e^{-ix^{\sigma}P_{\sigma}} = e^{ix^{\rho}P_{\rho}}S_{\mu \nu}\Phi(0)e^{-ix^{\sigma}P_{\sigma}} = e^{ix^{\rho}P_{\rho}}S_{\mu \nu}e^{-ix^{\alpha}P_{\alpha}}e^{ix^{\gamma}P_{\gamma}}\Phi(0)e^{-ix^{\sigma}P_{\sigma}}$$ The first equality is also equal to ##L_{\mu \nu}\Phi(x)## so the result follows provided that ##S_{\mu \nu} x^{\sigma} - x^{\sigma}S_{\mu \nu}=0## which it should be since S does not act on x, so the term overall vanishes.
     
    Last edited: Jun 26, 2014
  13. Jun 26, 2014 #12

    strangerep

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    Hmm. Have you already mastered ordinary QFT?? This book looks like a difficult read if you haven't. More like a graduate text.

    Not the clearest exposition I've ever seen.

    I would have said the result follows because ##S_{\mu\nu}## commutes with the translation generators ##P_\mu## (but also because it does indeed commute with ##x##, which is a parameter here).

    So I guess we're done here?
     
  14. Jun 27, 2014 #13

    CAF123

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    Actually, not at all. I am doing a project right now and I am only focusing on a very small subset of the book (as I said in the last post, only from 2.4 to end of chap 2 and beginning of chap 4. )These sections deal mostly with the conformal group in the classical field theory. So that I may get the bigger picture for future courses on QFT when I take them, can you recommend any good books that you yourself perhaps worked through/ found useful that I can use alongside this book?
    Some others here have told me that it is not the book I should be using for field theory because it is specialized. I have sought other online resources, but a book would be ideal. Thanks.

    Could you explain why your first statement is true? Using the Haussdorff formula, I arrive at ##e^{ix^{\sigma}P_{\sigma}}S_{\mu \nu}e^{-ix^{\rho}P_{\rho}} = S_{\mu \nu} - [S_{\mu \nu}, ix^{\sigma}P_{\sigma}]##
    The latter term is equal to ##i(x^{\sigma}[S_{\mu \nu},P_{\sigma}] + [S_{\mu \nu}, x^{\sigma}]P_{\sigma})##. Now, if what you say is true both terms here are zero. But that would eliminate one of the terms in the final expression.

    One of the commutation relations of the Poincare group is $$[P_{\rho}, L_{\mu \nu}] = i(\eta_{\rho \mu}P_{\nu} - \eta_{\rho \nu}P_{\mu})$$ I thought we would obtain the correct commutation relation for ##S_{\mu \nu}## by simply sending ##L_{\mu \nu} \rightarrow S_{\mu \nu}## in that formula.

    Yes, just a quick query above left over. Do you have a copy of the book? Could you help with the paragraph on P.101 at the top? I was wondering what it means to say '##\Phi(x)## belongs to an irreducible representation of the Lorentz group..'. and how they obtained that '##\tilde{\Delta}## is simply a number, manifestly equal to ##-i\Delta##'.

    Thanks.
     
    Last edited: Jun 27, 2014
  15. Jun 27, 2014 #14

    strangerep

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    What is the project? I presume you were given some kind of summary or abstract to work from?

    I can't really recommend anything for conformal field theory, since (afaict) it doesn't have much practical use in fundamental physics. I tend to group it with string theory and SUSY. I've heard that it has uses in condensed matter, but that's not my area.

    "Field theory" covers a lot of ground. (And have you noticed my signature line?)

    It depends what level of knowledge you've reached right now. For classical EM, Jackson is my goto book. For more advanced optics and coherent states, Mandel & Wolf is the bible. For ordinary QFT,... well,... the series of textbooks by Greiner (and various co-authors) got me started. Then Peskin & Schroeder for more advanced treatment and 1-loop calculations. Weinberg is considered the QFT "bible", but it's very difficult. Magiorre is also worth a look. Hendrik van Hees (aka "vanhees71" on PF) has an extensive script on QFT, so if you study from that you can always get help here from the author. Zee gives a useful overview of path integral methods in QFT, but is not likely to teach you how to perform the difficult integrals needed for cross-sections.

    TBH, I wouldn't advise trying anything more advanced than Greiner or Peskin & Schroeder until you've mastered the way that Ballentine develops ordinary QM (which involves representation theory at a more introductory level).

    [Edit:]You might also find some early sections of Greiner's "Field Quantization" helpful for stuff about classical fields in Lagrangian/Hamiltonian formulation.

    Ah, well,... this is a bit subtle and you're not the first person to be perplexed about that. OK... (deep breath....)

    Look again at that section of Ballentine I mentioned earlier: i.e., eqn(7.19) in case(ii) on p165. The bold R on the lhs is an operator on Hilbert space. But the ##D## and ##R^{-1}## on the rhs specific to the current representation. Urk -- now I have to explain what "representation" means...
    For now, I'll just point out that, under a rotation, a vector transforms differently from a scalar. (The scalar remains unchanged, whereas the components of the vector get mixed around.) One says that scalars and vectors correspond to different representations of the rotation group: the same (active) rotation in physical 3D does different things to scalars than it does to vectors. The intrinsic spin part of angular momentum captures this distinction: scalars are called spin-0, vectors are spin-1, etc.

    Now suppose we perform a translation in 3-space. The coordinate origin changes of course, but the distinction between "scalar" and "vector" remains. (Imagine translating a vector anchored "here" to become a vector anchored "there". It doesn't stop being a vector -- its "vectorness" is an intrinsic property.)

    Let's take a step back. And I'll use ##J_{\mu\nu}## for total angular momentum. The commutation relations for the Poincare algebra, e.g., ##[P_\lambda, J_{\mu\nu}] = \cdots## are true regardless of which representation we're acting on. One says that these are the abstract elements of the Poincare algebra. But when you write something like ##P_\lambda = -i\partial_\mu##, you're implicitly specializing the abstract ##P_\lambda## to the form it takes when acting on a representation made of wave functions. One says that the Hilbert space pf square-integrable wave functions "carries a representation of ##P_\lambda## in the form of a differential operator". This distinction between an abstract element of a Lie algebra, and it's specific form in a certain representation is one the most important things to grasp in all of modern physics.

    Now comes a more difficult part: a vector-valued wave function is actually constructed as a tensor product of a representation of the rotation group (this is the spin part) and a representation of the translation group (the x part). In tensor products like this, operators acting on one part of the product are blind to the other -- that's the bottom-line reason why the spin operator commutes with the translation operator. But... have you encountered tensor products yet in your QM studies? If not, I suppose the above will seem quite obscure. If so, you'll have to be satisfied (for now) with thinking about how the intrinsic "vectorness" or "scalarness" of a physical entity doesn't change if you move from "here" to "there".

    When we write ##J_{\mu\nu} = L_{\mu\nu} + S_{\mu\nu}##, what's really happening is this: we want a representation of the rotation group that acts on vector-valued functions of ##x##. But this involves a tensor product space, so we must find representations that act on each part independently, and then form the product. Look at Ballentine's eqns (7.17), (7.20) and (7.21). Putting them together, we have
    $$e^{i\theta \hat n \cdot J/\hbar}
    ~=~ e^{i\theta \hat n \cdot L/\hbar} \, e^{i\theta \hat n \cdot S/\hbar} ~.$$But L and S act on different parts of the tensor product (by construction), hence they commute. So we can put them back inside a single exponent, leading to the formula ##J_{\mu\nu} = L_{\mu\nu} + S_{\mu\nu}##.

    Strictly speaking, we should be writing something like
    $$e^{i\theta \hat n \cdot J/\hbar}
    ~=~ e^{i\theta \hat n \cdot L/\hbar} \, \otimes \, e^{i\theta \hat n \cdot S/\hbar} ~,$$to emphasize that we're working with a tensor-product representation here. But physicists rarely do that. They just keep in mind that the different parts operate on different parts of the overall wave function.

    So, when you said that you simply put ##J_{\mu\nu} \to S_{\mu\nu}##, you're ignoring the spatial dependence. I.e., you're ignoring one term of the tensor product. Certainly, for the specific case ##x=0##, the orbital term is zero, but you can't ignore it in general.

    That's why I didn't like the notation in that book: it thoroughly glosses over most of the really important foundational stuff I sketched above.

    [Edit #2] The situation is different depending on whether we're dealing with the non-relativistic (Galilean) or relativistic (Poincare) case. In the former, total angular momentum decomposes covariantly into orbital and spin parts (meaning that each part retains its identity under Galilean transformations. But for Poincare, Lorentz boosts can mix the orbital and spin parts, hence the decomposition is not really meaningful. (Are you familiar with the Pauli-Lubanski spin vector?)

    In the conformal case, we deal only with massless fields, and things are different again. There are various subtleties even in ordinary (Poincare) QFT when trying to construct massless quantum fields.

    So some of what I said above might not be quite right for the CFT case. I must look at more of that CFT book to check -- but right now it's my bedtime.
    [End Edit #2]

    I hope you start to see from my explanation above that the question "what is a representation" is a very big one. For current purposes, you can probably translate it to something like ##\Phi## is a scalar, or spinor, or vector, or (etc), under Lorentz transformations.

    They're explanation is poor. They should have said that the 1st commutator in (4.29), together with Schur's lemma means that, ##\tilde\Delta## must be a multiple of the identity. (The "multiple" is denoted ##-i\Delta##.) Hence the lhs of the 2nd commutator in (4.29) vanishes. Hence the rhs of that commutator (i.e ., ##-i\kappa_\mu##) vanishes also.

    BTW, you should probably move some of these larger non-HW questions to the quantum forum where others can (probably) give a wider perspective than I can. CFT is not really my area, but there's other people on PF who know a lot about it.
     
    Last edited: Jun 28, 2014
  16. Jun 28, 2014 #15

    CAF123

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    Yes, I have a list of aims that more or less get me looking into the basics of the theory before looking into an application. I am focusing on the conformal group in the classical field theory realm since I am yet to take any course on QFT.

    Many thanks. I will keep them all in mind and perhaps look into Greiner because your edit is also relevant to what I have studied in the project.

    A scalar transforms trivially under the Lorentz group, so we can map the scalar to the trivial representation (map to 1) which would encapsulate its trivial transformation nicely. Any scalar representation can likely be mapped to 1, e.g a particle possessing spin-0 transforms like a scalar.

    So, the translation part encodes the orbital dependence of the overall wave function and the spin part encodes the spin value of the wave function. I.e if we consider the wave function of an electron possessing a total spin s=1/2 and some arbritary orbital dependence ##\Phi(x)## then the full wavefunction is a tensor product ##\Phi'(x) = |s=1/2 \rangle \otimes \Phi(x)##. The other use for tensor products I know of is that it naturally leads to the Clebsch-Gordon series decomposition for the tensor product of two spin representations.
    It makes sense conceptually.

    To make contact with the example I posted above, are you saying that the ##e^{i\theta \hat n \cdot L/\hbar}## term is the same as my ##\Phi(x)## and the ##e^{i\theta \hat n \cdot S/\hbar}## is the same as my ##|s=1/2\rangle##?

    Intuitively it at least makes sense but if I use this knowledge in my commutator it simply vanishes: $$e^{ix^{\rho}P_{\rho}}L_{\mu \nu}\Phi(0) e^{-ix^{\sigma}P_{\sigma}} = e^{ix^{\rho}P_{\rho}}S_{\mu \nu}\Phi(0)e^{-ix^{\sigma}P_{\sigma}} = e^{ix^{\rho}P_{\rho}}S_{\mu \nu}e^{-ix^{\gamma}P_{\gamma}}e^{ix^{\alpha}P_{\alpha}}\Phi(0)e^{-ix^{\sigma}P_{\sigma}}$$
    Then ##e^{ix^{\rho}P_{\rho}}S_{\mu \nu}e^{-ix^{\gamma}P_{\gamma}} = S_{\mu \nu} - [S_{\mu \nu}, ix^{\sigma}P_{\sigma}] = S_{\mu \nu} - i(x^{\sigma}[S_{\mu \nu}, P_{\sigma}] + [S_{\mu \nu}, x^{\sigma}]P_{\sigma})##

    So the first term in the reexpressed commutator, given the above discussion, will vanish and so will the second since x is just a parameter of the translation.


    Ok, would 'irreducible' be used just so that we may use Schur's lemma?

    Yes, that makes more sense to me.

    Thanks, and thanks for taking the time to write a long and insightful post.
     
    Last edited: Jun 28, 2014
  17. Jun 28, 2014 #16

    strangerep

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    As I was almost ready to go beddy-bye, I had a horrible feeling that some of the things I said might not be quite right in the context of the conformal group. See my "Edit #2" in previous post.

    I'll respond to the other stuff tomorrow (my time).

    In the meantime, list your project aims in more detail. I can't help feeling that jumping into the conformal group (for anything) is unwise if you haven't yet achieved proficiency in the Poincare case.
     
  18. Jun 28, 2014 #17

    CAF123

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    Studying the conformal group in d>2 dimension, writing explicit representations of the generators/ studying the algebra, note isometries in D+2 dimensional space and isomorphism with SO(d+1,1), application in electrodynamics.
    I am studying the Poincare group as I go along as well, the Lorentz group together with translations from a subset of the conformal group, so I suppose it is natural that I will touch upon it.
    Thanks.
     
  19. Jun 28, 2014 #18

    strangerep

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    I'll respond to the pieces of your earlier post separately (since some require more thought than others).

    That's the right idea, but not quite the correct way to express it. I need to explain a little more about representations...

    Mathematically, a representation (of a group, say) is a mapping from the abstract group elements to operators on a linear space. So, e.g., in the spin-1 representation, we have a mapping from the abstract elements of ##SO(3)## to 3x3 matrices acting on the usual 3D vector space. Those matrices are said to represent the abstract group elements, and the entire 3D vector space is said to be the "carrier space" for the representation -- meaning that there are operators (matrices) on the vector space that represent the group elements faithfully, including the multiplicative group properties, etc, etc.

    In the scalar (i.e., spin-0) case, the linear carrier space is 1-dimensional, hence all elements of ##SO(3)## map to 1x1 matrices. But since the determinant must be ##\pm 1##, those 1x1 matrices are just the numbers 1 and -1. (In the -1 case, we're actually dealing with "pseudo-scalars" -- meaning that they change sign under parity inversion, unlike ordinary scalars.)

    So your last sentence above should be changed to something like: In any scalar representation, the group elements are all represented trivially by the identity.
     
  20. Jun 28, 2014 #19

    strangerep

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    Again, that's heading in the right direction, but not quite correct in the detail. The trouble with your formulation is that the component ##\Phi^{+1/2}(x)## could have different x-dependence from ##\Phi^{-1/2}(x)##.

    It's better to think of it as a tensor product of spaces. Something like
    $$\{ |1/2\rangle \,,\, |-1/2\rangle\} ~\otimes~ L^2(R^3)$$I.e., the 2D Hilbert space ##C^2## tensored with the space of square-integrable functions on position space.

    No. The exponentials are operators, but the ket is a vector.

    As for ##\Phi## -- here there is a big difference between a classical field (where it's just a function), a QM function (which is a vector in a Hilbert space), and a quantum field (where it's an operator on a Fock space).

    The stuff that I referenced in Ballentine corresponds to the 2nd case (QM) where the ##\Phi## wavefunction is a vector, so we write the generic rotation transformation as ##\phi \to R \Phi##.

    But for the 3rd case (QFT), ##\Phi## is an operator so it transforms under a generic rotation as ##\Phi \to R \Phi R^{-1}##. So in section 4.2.1 of FMS, they're really dealing with the quantum field case, even though the untrained eye might think it's still the classical case.
     
  21. Jun 28, 2014 #20

    strangerep

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    I begin to dislike FMS more and more. Back on p98, in eq(4.18) they use the symbol ##L_{\mu\nu}## for orbital angular momentum. But here on p100 they silently change it to mean the total angular momentum. No word about the distinction between the abstract generator and specific representations of it. From what they've written, I don't think it is possible to go from their eq(4.25) to eq(4.26) in the way they sketch. I think they are just using their knowledge of the correct result to fudge an explanation. My advice is to put that book aside for now.

    The correct underlying idea is as follows. The difference between "total angular momentum of a field about a point ##a_0##" and "total angular momentum of a field about a point ##a_1##" does indeed equal an orbital-like term ##-b_\mu P_\nu + b_\nu P_\mu##, where ##b = a_1 - a_0##. To see this, can you access a copy of the "Gravitation" bible by Misner, Thorne & Wheeler? See Box 5.6 on pp157-159. Then they show how to go from the spin part to the total expression. This is the best compact (classical) treatment of the relationships between total, spin and orbital angular momenta that I've found. (It's also good for debunking the oft-quoted phrase that "spin is a quantum concept". It's not, of course, as MTW shows. Only the quantized values for angular momentum arise from the quantum framework.)

    I also think you should (as a high priority) get a copy of Greiner's "Field Quantization" (if you haven't already) and study ch2 thoroughly -- which really does deal with classical field theory. They use Poisson brackets to represent abstract group commutators, which is the correct way to do things in the classical case.
     
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