From: Robin Chapman Subject: Re: Commutators Date: Sun, 30 Apr 2000 20:42:38 GMT Newsgroups: sci.math Summary: Algebra of vector fields on a manifold In article <390C8A3B.4BE7828@brfree.com.br>, brunobg@geocities.com wrote: > Gravitation (Misner et al) defines commutators this way: > > [u, v] = [d_u, d_v ] = d_u d_v - d_v d_u > > Where u, v are vectors, Not quite. One needs to distinguish between vectors (elements of the tangent space at a point of a manifold) and vector fields (sections of the tangent bundle over an open subset of the manifold). This defintion is valid for vector fields. > and d_u is a directional derivative in the > direction of u; this definition is based on the viewpoint that > a vector u can be defined as d_u (isomorphism between vectors and > directional derivatives). > Now, the questions. First, why there's a isomorphism? Locally a manifold is isomorphic to R^n. Consider an open subset U of R^n. One proves that each derivation from C^infinity(U) to itself (an R-linear map satisfying the Leibniz identity) has the form D = f_1 d/dx_1 + ... + f_n d/dx_n where the f_j are in C^infinity(U). If you consider what D does at a point u; you see that Dg(u) is the directional dervvative of g in the direction (f_1(u), ..., f_n(u)). > Second, it asks to show that > > [u, v] = (u^b v^a_{,b} - v^b u^a_{,b}) E_a > > where E_a are the basis vectors (and not components), is valid for any > coordinate basis. I've showed it, but I'm not satisfied with my proof > (I think I forced it). How can you prove it? A general guideline would be > enough. Your proof is probably fine! Write u as a differential operator u_1 d/dx_1 + ... + u_n d/dx_n etc. and just work it out. > It also says that for any basis { E_a }, one defines commutation > coefficients: > > [E_b, E_d] = c_bd^a E_a > > And that they are zero for coordinate basis. Is it zero because the > derivatives vanish? E_b, I suppose will correspond to d/dx_b for the coordinate basis. This is just "equality of mixed partials". > What is the *meaning* of c? ? Isn't the above equation is meaning? > Later it's described its > connection to "covariant connection coefficients"; what role exactly does > it play there? I'm lost here. You need a connection before you have connection coefficients. > I've usually seen these c.c.c. as Christoffel symbols > (without the c); in fact, the development of covariant derivatives I've > read never had commutation coefficients; where do they come from? > Last, but not least, consider a spherical coordinate system, with > E_r = dP/dr and E_\theta = 1/r dP/d\theta. I calculated c_r\theta^theta and > found it to be 1/r^2, which is wrong (it should be -1/r). Can you please > show me why? Let's write t for theta and compute [E_r, E_t]. Let f be any smooth function on the plane minus the origin. By definition [E_r, E_t](f) = E_r(E_t f) - E_t(E_r f) = E_r( (1/r) df/dt) - E_t( df/dr) = (-1/r^2) df/dt +(1/r) d^2f/dr dt - (1/r) d^2f/dt dr = (-1/r^2) df/dt = (-1/r) E_t(f). Hence [E_r, E_t] = (-1/r) E_t so c_{rt}^t = -1/r like the book sez. -- Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html "`The twenty-first century didn't begin until a minute past midnight January first 2001.'" John Brunner, _Stand on Zanzibar_ (1968) Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: Matthew Donald Subject: Re: Quantisation of orbital angular momentum. Date: Tue, 2 May 2000 03:50:26 GMT Newsgroups: sci.physics.research Summary: angular momentum and commutators Gerard Westendorp wrote > Angular momentum seems to come in lumps of h_bar/2. > But you can do a change in origin, and get: > L -> L + (r0 x p0) > Where p0 is the momentum of the center of mass, and > r0 is the change in origin. > How is the quantisation property preserved, or is it? Standard textbook treatments of angular momentum show that for any vector operator J with commutation relations given by [J_1, J_2] = i hbar J_3 and its cyclic permutations J^2/hbar^2 is quantized in the form j(j + 1) and each component of J/hbar is quantized in corresponding half-integer units. However, in general, the eigenfunctions of J_1 are not also eigenfunctions of J_2 or of J_3. For orbital angular momentum L = r x p, we deduce the required commutation relations from [ r, p ] = i hbar, and then we note that invariance under 2 pi rotations implies that the quantization of L should be in integer units. Under a shift of co-ordinates, the commutation relations [ r, p ] = i hbar and hence [L_1, L_2] = i hbar L_3 and its cyclic permutations are unchanged. These commuation relations imply that angular momentum is still quantized in the new co-ordinates. However, the eigenfunctions in the new co-ordinates will not, in general, also be eigenfunctions in the original co-ordinates. The physically relevant co-ordinates are the centre of mass co-ordinates for systems with a rotation invariant interaction. Only for those co-ordinates can the angular momentum eigenfunctions be used in combination with radial eigenfunctions to express the energy eigenfunctions. Matthew Donald (matthew.donald@phy.cam.ac.uk) web site: http://www.poco.phy.cam.ac.uk/~mjd1014 ``a many-minds interpretation of quantum theory'' *************************************************** ============================================================================== From: Matthew Donald Subject: Re: Quantisation of orbital angular momentum. Date: 4 May 2000 21:38:06 GMT Newsgroups: sci.physics.research I wrote explaining that after a change of co-ordinates angular momentum will still be quantized because the basic commutation relation [ r, p ] = i hbar is preserved. I went on > However, the eigenfunctions in the new co-ordinates will not, in > general, also be eigenfunctions in the original co-ordinates. Gerard Westendorp replied > If I understand it correctly: > A specific wave function may have non-integer orbital am. > But if you try to measure am. you express the wave function as a > superposition of eigenfunctions of the L-operator. Each > eigenfunction has an integer eigenvalue (multiplied by h_bar). > It is only the outcome of actual measurements that are quantised. In the present case, we have two angular momentum operators (to be entirely specific, consider an atomic system). L = r x p is the angular momentum in the physically important co-ordinates of an electron relative to the center of mass. L' = r' x p is angular momentum in some spurious co-ordinates defined by r' = r + a where a is some fixed c-number-valued vector. Both L and L' are quantized along any chosen direction in the mathematical sense that both have eigenvalues defined by pairs of integers (j, m) with j = 0, 1, 2, . . . and m = -j, -j+1, . . . , j-1, j. This follows from [r, p] = [r', p] = i hbar. In this mathematical sense, it is not correct to say that ``it is only the outcome of actual measurements that are quantised''. ``Quantization'' can be (and is) used to refer both to the mathematical property of the discreteness of the spectrum of some self-adjoint operators and to the physical consequences of that property. Eigenfunctions of L are also eigenfunctions of energy. So these eigenfunctions are physically important states. For example, if the system is in an eigenfunction of L with respect to some axis with (j, m) = (1, 1), then it may spontaneously emit a photon and end up in the ground state with (j, m) = (0, 0). Or perhaps you could make a physical choice of direction by placing the system in a strong magnetic field. Then the physical light spectrum of the system could allow you to observe directly the splitting of the energy spectrum corresponding to different values of m. Both of these examples refer to eigenfunctions of L rather than L' because photons interact with atoms as a small perturbation on the atomic energy eigenstates. The L' eigenfunctions are mathematically just as good as the L eigenfunctions but they are physically irrelevant. Decoherence theory suggests that (for significant values of a) they are physically unstable: Because of collisions between atoms, if the state of the relative co-ordinate of an atom is originally an eigenfunction of L', it will rapidly change to a density matrix of the form of a sum of the much more stable, physically localized, energy eigenfunctions (eigenfunctions of L). If you could measure L' then of course it would be physically quantized. But how would you do it? Matthew Donald (matthew.donald@phy.cam.ac.uk) web site: http://www.poco.phy.cam.ac.uk/~mjd1014 ``a many-minds interpretation of quantum theory'' ***************************************************