Sorry, when I see you talking about topology and group reps I keep forgetting you're a mathematician, not a physicist. Please Register or Log in to view the hidden image! That's not an insult, I was a mathematician when I learnt this stuff but I went down the mathematics of theoretical physics path. P, T and C are the 3 discrete symmetries of field theory. C isn't relevent here (it swaps particles and antiparticles) but P and T are. Given a 4 vector (t,x,y,z) we have : Spacial parity P : P(t,x,y,z) = (t,-x,-y,-z) Time parity T : T(t,x,y,z) = (-t,x,y,z) So PT(t,x,y,z) = (-t,-x,-y,-z) For operators you have \(A \to OAO^{-1}\), so you get that \(\Lambda \to \Lambda ' = P\Lambda P^{-1}\) and so on. Hence you get that under these parity operators you end up doing things like going from the physicists Lorentz group with \(\{\Lambda^{0}_{0} \geq 1\} \; \cup \; \{ |\Lambda| = 1\}\) to the three other options : T : \(\{\Lambda^{0}_{0} \leq -1\} \; \cup \; \{ |\Lambda| = 1\}\) P : \(\{\Lambda^{0}_{0} \geq 1\} \; \cup \; \{ |\Lambda| = -1\}\) PT :\(\{\Lambda^{0}_{0} \leq -1\} \; \cup \; \{ |\Lambda| = -1\}\) I think that's what they do. Whatever it is precisely, you can see that the three P, T and PT map to the other 3 parts of the full Lorentz group from the physicist's Lorentz group. You need the part connected to the identity, since you must be able to consider varying your transformations smoothly down to the identity, because you can rotate your coordinates by \(\theta\) and/or boost your description by a velocity v and obviously you must be able to vary them smoothly down to 0.
Physicists have been saying that for 120 years near enough. Please Register or Log in to view the hidden image!
Alpha, et al.: Thank you all again, but I wonder if we're not in danger of losing the plot here. Let's take it slow; I re-iterate the problem immediately to hand. We found the "native" basis for the algebra su(2) to be, in some sense, twice as big as it needs to be. So we factored by 1/2. This, on exponentiation, generated a group with t = t + 4 pi symmetry, which I was happy to allow described particles with 1/2-integer spin, i.e. fermions. We also agreed on the gluons - these present no real problems (!) But the problem before me is this: I have "used up my quota" of SU(2), recalling this group is all det = 1 2x2 Hermitian transformations (in the sense of Lie). So where do I find the bosons that mediate the weak force? So, I have two propositions: either I am being excessively naive in thinking that Lie theory can provide at least a framework for quantum theory, or maybe this may work....... Given SU(2) as above, i.e. the matrices that describe t = t + 4 pi symmetry, is it possible that the product SU(2) X U(1) induces a t = t + 2 pi symmetry? I haven't had time to do the calculation, in fact, to be truthful, I'm not sure I know how. But, bearing in mind that U(1) is the circle group, it may drop out like that. This might create a problem for you guys; if this can be done, then, as I was told that the basis for u(1) describes the photon, then I would have to lump the B, W and Z bosons together with the photon, which is surely not right? Finally let me say this: I have absolutely no pride on this matter; I haven't the slightest fear of being made a fool of, either by myself or by anyone else. So - bring it on!
Hey, hey, chaps. You obviously have a history, which is of no interest here. So, leave it out. But first, without going into detail, by doing a quick calculation, I found that, provided only that \(SU(2) \times U(1)\) is a valid construction, and provided only that there is an algebra (not proven!) \(su_2 \oplus u_1\), then the Jacobi identity is satisfied and so is skew symmetry (it's pretty ugly, so I shan't show it). But look. I am taking Mrs QH to Paris tomorrow for a few days (how sweet am I?). So, I may see you all soon
Just to prove that there's an xkcd cartoon for everything... Please Register or Log in to view the hidden image!
I've been in this situation in respect to the material in this thread : Please Register or Log in to view the hidden image!
I have to come clean; I have been a complete moron over SU(2). My only excuse (other than blaming Ben!) is that I was taught to think of the Lie groups as purely abstract constructions; specifically, the group SU(2) is the set of all unitary 2x2 matrices with det = 1. This is nice, as the group properties follow instantly from the property of unitarity. No doubt physicists regarded this as a form of intellectual onanism, and quite naturally want to know what Lie groups do rather that what they are. So let's see this. (sorry for the lecture!) The "physical" action of a Lie group is referred to as its realization. But there need not be a unique realization for any Lie group, in fact we know there isn't. Where a realization is the group of transformations on some vector space, one calls it a representation. And once again, representations need not be unique. Let's see this too. Recalling that SU(2) is a double covering for SO(3), let's define a 2:1 surjection \(f:SU(2) \to SO(3)\) by \(f(h_1) = f(h_2) = g \in SO(3) \). Suppose now there is a representation \(R: SO(3) \to V\). Then this induces a representation \(S: SU(2) \to V\) such that, for all h in SU(2), \(S(h) = R(f(h))\), hence, by the above, \( S(h_1) = S(h_2)\) for all representations with this property. This implies that, for some symmetry \(t = t+2\pi n\) described by SO(3), there is a corresponding symmetry \(t = t + 4 \pi n\) described by SU(2). We had agreed to call this this "half integer spin", had we not? But I may have another representation \(T: SU(2) \to W\) where \( T(h_1) \ne T(h_2)\), which seems to imply that there is another symmetry described by SU(2) (umm, I dunno how to write it down!!) which is totally unrelated to the symmetry described by SO(3), and which I am content to call "integer spin". Incidentally. Mrs. QH wants you to know that any calculations taking place in her chest must, of necessity, by rather brief. I wonder what she means??
Yes, it's convenient to consider the action that SU(2) has on say a doublet, in such a way as to preserve the inner product. Consider a Hilbert space \(\mathcal{H}\) of states \(|\phi\rangle\) with operators \(\mathcal{O} : \mathcal{H} \to \mathcal{H}\). All physical observables are defined by the expectation of a Hermitian operator as \(\langle \mathcal{O} \rangle = \langle \phi | \mathcal{O} | \phi \rangle \). A gauge transformation is, essentially, a way of counting the redundancy in this expectation value, how many choices of \(|\phi\rangle\) do we have which give the same s \(\langle \mathcal{O} \rangle \)? Well under a gauge transform we have that \(|\phi\rangle \to |\phi '\rangle = A|\phi\rangle\) and \(\mathcal{O} \to A \mathcal{O} A^{\dag}\) (this is needed to keep the operator hermitian, rather than say \(\mathcal{O} \to A \mathcal{O} A^{-1}\). Therefore we have that \( \langle \phi | \mathcal{O} | \phi \rangle = \langle \phi |A^{\dag} A \mathcal{O} A^{\dag} A | \phi \rangle\) so \(A^{\dag}A = \mathbb{I}\). Thus we end up working with U(n) or SU(n) in general (there's also Sp(n) and USp(n) but they mostly come up in string theory). We can imagine the specific action of SU(2) on say a doublet \(|\phi\rangle = \left| \begin{array}{c} \phi_{+} \\ \phi_{-} \end{array} \right\rangle\) by A in SU(2) \(|\phi\rangle \to |\phi ' \rangle \left| \begin{array}{c} \alpha \phi_{+} + \beta \phi_{-} \\ -\bar{\beta}\phi_{+}+\bar{\alpha}\phi_{-} \end{array} \right\rangle\) where \(|\alpha|^{2}+|\beta|^{2} = det(A)\langle \phi |\phi \rangle\). If you have that \(\langle \phi |\phi \rangle = 1 = det(A)\) (ie dealing with SU(2) rather than U(2) and normalised states) then you will see that \(\langle \phi '|\phi ' \rangle = 1\) and \(\langle \mathcal{O} \rangle \) is preserved. Spiffy. Yes, the rep is important. As I gave in a previous example, the 3x3 rep of SU(2) leads you to consider a spin system with spins 1, 0 and -1, rather than +1/2 and -1/2. 4x4 rep gives you +3/2. +1/2, -1/2 and -3/2. The pattern is obvious. This gives you the physical description of multi-fermion states. With it you can predict things like meson decays/interactions into simpler particles! Not perfectly, when dealing with stuff like isospin which is an approximate symmetry, but enough to get you into the ballpark. On a more complicated level you have different reps of the same group. For instance, SU(3) has the adjoint rep with dimenion 8 (there's 8 bosons) but the fundamental rep of dimension 3 (there's 3 colours). It explains why 3 colours have 8 'charge carriers' when you might naively expect there to be 9 (since the gluons are things like 'red-antigreen' and 'green-antiblue'). On a physical level you cannot get a 9th because you'd have the ability to construct a colour singlet and we'd see colour in day to day interactions. But it's confined. Group theory and representations explains why colour is a symmetry but not a 'visible' one!
Alpha: I thank you for that. I am still ploughing through it, but have a couple of questions. Right. What little I know of Hilbert spaces is that they are (infinite dimensional) vector spaces whose elements are the square integrable functions (in the sense of Lebesgue) on a closed interval in \(\mathbb{C}\). You want these "elements", these functions, to be states? I am not sure that I follow And I don't know what the "expectation" of an Hermitian operator means OK, that's a definition, fair play. Another reason to dislike the Dirac notation! It seems to suggest that the operator \(\langle \mathcal{O} \rangle\) is a vector that is its own dual. Well, very loosely speaking, I suppose this is a definition of Hermiticity provided only that I am prepared to regard this operator as a vector. Hmm. Why not, I ask myself? (no answer) But we know, by the definition of Hermiticity, that \(A^{\dag} = A\). So how does this follow? So we seem to have the following construction. There is an Hilbert space \(\mathcal{H}\), and an operator \(\mathcal{O}: \mathcal{H} \to \mathcal{H}\). Moreover we have another operator \(A\), an hermitian operator, acting on \(\mathcal{O}\) such that, for \(\phi \in \mathcal{H}\), something like \(A(\mathcal{O}(\phi)) = A^{\dag}(\mathcal{O}(\phi))\in \mathcal{H}\), which implies that the operator \(\mathcal{O}\) is an element of some vector space, whose transformations \(A,\;A^{\dag}\) can in in some sense "descend" to transformations on \(\mathcal{H}\). Weird enough. Leave it with me for a day or so. Thanks again for your trouble, sorry to be so dense.
Look at it as an emergent condition. An unlighted matchstick contains no fire within the matchhead. When struck the emergent fire appears.:shrug:
Yes, elements in the vector space are states of the quantum field and the inner product gives us a way of defining expectations. For instance \(\langle \psi | \phi \rangle\) is the amount of overlap between the state psi and the state phi. In terms of basis vectors and a continous space of states, so functions of x, you'd get \(\langle x | x' \rangle = \delta(x-x')\). http://www.hep.phys.soton.ac.uk/~g.j.weatherill/lecturenotes/II/pqm.pdf The first few chapters cover the Dirac notation nicely. Not really. If anything, it allows you to construct it as a matrix. Suppose you have a countable set of basis vectors |a>, then you can construct the matrix representation of an operator O as \(\mathcal{O}_{ab} = \langle a | \mathcal{O} |b\rangle\). No, \(\mathcal{O}\) is Hermitian, not A. A is the gauge freedom, whose properties we derive from the transformation behaviour of \(\mathcal{O}\). Expectations of operators should be real, since they are basically physical quantities. You can either use the "Hermitian matrices have real eigenvalues" linear algebra result or you can do the following method : \(\langle \phi |\mathcal{O}| \phi \rangle\) is real. Therefore \( \langle \phi |\mathcal{O}| \phi \rangle = (\langle \phi |\mathcal{O}| \phi \rangle)^{\dag} = \langle \phi |\mathcal{O}^{\dag}| \phi \rangle\) so \(\mathcal{O}^{\dag} = \mathcal{O}\). This remains when you chuck in gauge choice too, since you need \(A\mathcal{O}A^{\dag}\) to be hermitian, so \(A\mathcal{O}A^{\dag} = (A\mathcal{O}A^{\dag})^{\dag} = A\mathcal{O}^{\dag}A^{\dag}\) so \(\mathcal{O} = \mathcal{O}^{\dag}\). As you can see, the \(A\) and \(A^{\dag}\) terms are tagged on in such a way not to upset the hermitian conditions of the operator and also they preserve the inner product by cancelling with the transformation applied on the states.
There's no neutrinos within the nucleus. It's just a slew of quarks and gluons being exchanged between them. It's not defined like that. It's just defined as the interactions due to an SU(3) gauge field. The strong force has the interesting property that it gets weaker as you increase energies of interactions. The weak and EM forces get stronger. At GUT levels (\(10^{16}\) GeV) they end up being the same. I don't understand what you mean as '1'. 1 what? Actually it's done none in relation to strings and cosmic strings. We don't know the mass of the neutrinos. Besides, there's 3 of them and they have different masses, just like the electron, muon and tau have different masses. In all probability, given current experimental date, they have masses a lot less than 1 eV. Also, they are nothing to do with the force which moves electrons up and down orbitals. That's the electromagnetic force. Photons are emitted and absorbed when electrons move down or up orbital levels. The energy levels involved are around the 0.1~20eV level but the fact the neutrino masses are perhaps in the lower end of that range is just a coincidence. Neutrinos and electrons don't directly couple. The photons are massless but they respond to the curvature of space-time since they move along null geodesics. Just because the photon and graviton interact doesn't mean they were thought to be the same. Infact, they cannot be because they have different properties. The photon has spin 1, the graviton has spin 2. This is complete nonsense. The photon isn't a scalar field, it's a vector field. The graviton isn't a scalar field, it's a rank 2 tensor field. The photon can only interact, directly, with particles with EM charge. Infact, that's practically tautological! The graviton's interactions are a little open to questions. Supergravity is the study of supersymmetric gravitational theories which are weak field limits of gravity. I know, I do it. Practically nothing you said was right. Or even close to right. 5 minutes reading Wikipedia would have told you that.
OK, first this. Alpha provided a link to an online text (plus there is always Wikipedia et al) but I find it more entertaining to be educated here. I hope nobody thinks I am wasting bandwidth or their time thereby? Silly me, I did know that the product \(\langle x|H|x \rangle \) defines the expectation value for the Hermitian operator H. So all would be well if I had a clearer understanding of what is meant by a "state". Let's leave that a mo. Fair enough, except I don't fully understand the term "gauge", let alone "gauge freedom". Ben, bless him, did try to explain gauge invariance, but I was too dense to follow him. Wanna have a crack? Yes, I am familiar with both these proofs and their inverses. So your task, should you choose to accept, is to explain what a gauge is, what gauge freedom is and what a gauge choice is. Couched in language that morons can understand!
Um, why is it me? A gauge is just a 'standard' measurement - actually the term is borrowed I believe, from the width, or gauge of rail tracks. Anyways, I found an article by one Gerard t'Hooft, about gauge theories and fundamental forces (SA Jun 1980). He says the four forces now all have "non-Abelian gauge theories with local symmetry". He says Maxwell's theory of electromagnetic fields, was the first gauge theory with local symmetry. Local symmetry is what makes the theory invariant, with respect to a local transformation, or the appearance of a force. He maintains that the "addition" of a force is what makes a gauge theory invariant - maybe you can work out why...? You understand how the theories are about fields - scalar and vector. And how photons are the quantum of exchange, between charged particles. QM can represent particles as fields - a wave packet with a finite extension, or as a pointlike particle. The "choice" is to do with which symmetry to use locally, to define the field (say, the electric field), and how a transform leaves a local symmetry. A trivial example is how changing an electric potential globally leaves local electric potentials unchanged. Gauge is another word for "scale", I guess. And that's probably made gauge theories even more confusing.
I thank you (though I am pretty sure that photons are not charged!). Anyway, after much beard tugging, I wonder if the following might be an intuitional first floundering: Suppose that \(\psi\) is a field, any field. Then let me (sloppily) write \(\psi_p\) for the value of this field at the point p. Now suppose I require the value to be real, then I must have \(\psi_p \overline{\psi}_p = \alpha\) at p. Let me further suppose that this value be unchanged by evaluating at some other point, say q. So what guarantee do I have that this must be so? Well, Ben showed how! Let the transition function \(\psi_p \to \psi_q\) be given by \(\psi_q =e^{i \beta}\psi_p\), whereby the evaluation \(\psi_q \overline{\psi}_q = \alpha '\) is \(e^{i \beta}\psi_p \overline{e^{i \beta} \psi_p} = \psi_q \overline{\psi}_q = \alpha '\). Yet \(\psi_q \overline{\psi}_q = [e^{i \beta}e^{-i \beta}]\psi_p \overline{\psi}_p = \psi_p \overline{\psi}_p \Rightarrow \alpha ' = \alpha\) This "transition function" I may call a gauge transformation????
A gauge is a redundancy between a physical state and a mathematical description. It is not quite the same as a symmetry of the system. For instance, in electromagnetism we describe the physical 'Field Strength' \(F_{ab}\) by an electromagnetic potential \(A_{a}\) via \(F_{ab} = \partial_{a}A_{b} - \partial_{b}A_{a}\). Given A, you know F. But given F, which is what what we use to construct the Lagrangian of what we measure, can we find A? No, we can only find A up to some 'gauge'. For instance \(A_{a}\) and \(A_{a} + \partial_{a}\Lambda\) for a scalar \(\Lambda(x)\) both give the same F, since partial derivatives commute. So we have a gauge choice. A global symmetry is \(\Lambda\) is constant. This is trivially a symmetry since we don't actually change A, but if \(\Lambda = \Lambda(x)\) then it's upgraded to 'local' and you change A in a non-trivial way but F is always unchanged (for reference, if you're working over a non-trivial manifold where you have a non-zero connection then you upgrade your partial derivatives to covariant ones and everything still holds). A changes the mathematical description but not the physics. You have to consider the 'size' of this extra choice when you do integrations because you might overcount particular states due to gauge freedom. Particular examples of this in field theory leads to 'gauge fixing'. I'm not sure how to put it into a short, succinct, explaination. PM me your email and I'll send you a pdf version of an excellent book on all of this which directly addresses the relationship between sections of principle bundles, Lie groups and transition functions. You basically end up with something like : Given a potential A_i and a transition function t_ij then the A_j is (i,j are not space-time indices but labels for my U_i coverings) \(A_{j} = t_{ij}^{-1}A_{i}t_{ij} + t_{ij}^{-1}dt_{ij}\) For the U(1) bundle (ie the EM tensor setup) you have that \(t_{ij} = e^{i\theta_{ij}}\) so this reduces to \(A_{j} = A_{i} + i\partial \theta_{ij}(x)\) Which is precisely as I said it was. If the fibre is a non-abelian group then it doesn't factor nicely but it does leave the following definition of the physical quantity \(F_{ab}F^{ab}\) invariant with F \(F_{ab} = \partial_{a}A_{b} - \partial_{b}A_{a} + \big[ A_{a},A_{b} \big]\) Here \(A_{a}\) is Lie algebra values, so \(A_{a} = A_{a}^{i}T^{i}\) with \(T^{i}\) my generators (it's true for U(1) but it's only got 1 generator and the commutator is zero so it's generally not written in this manner) so it becomes \(F_{ab}^{i} = \partial_{a}A_{b}^{i} - \partial_{b}A_{a}^{i} + f_{ab}^{i}\), where \(f^{i}_{ab}\) are my Lie algebra's structure constants.
More from Mr t'Hooft: He reckons photons are charged, and QED is an Abelian theory, because phase rotations in that field are commutative. FYI: That description of the U(1) group and EM bundle corresponds to the Ae^st surface (where s is complex frequency), and transforms in the time-domain (electronic circuits with a "transfer function", which is dependent on the geometry of passive components in a network). The EM field is ubiquitous; electronics is just another way to study tensor calculus and Hamiltonians (but they don't generally tell you that's what it is).
