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Lorentz invariance When we say theory is Lorentz invariant we mean it is invariant under proper orthochronous subgroup only (those that can be obtained by compounding ILTs) based on S-2 Lorentz transformation (linear, homogeneous change of coordinates): that preserves the interval Transformations that take us out of proper orthochronous subgroup are parity and time reversal: : orthochronous but improper All Lorentz transformations form a group: product of 2 LT is another LT nonorthochronous and improper identity transformation: inverse: A quantum field theory doesn’t have to be invariant under P or T. for inverse can be used to prove: 20 Infinitesimal Lorentz transformation: 22 How do operators and quantum fields transform? Lorentz transformation (proper, orthochronous) is represented by a unitary operator that must obey the composition rule: thus there are 6 independent ILTs: 3 rotations and 3 boosts not all LT can be obtained by compounding ILTs! infinitesimal transformation can be written as: +1 proper are hermitian operators = generators of the Lorentz group -1 improper from proper LTs form a subgroup of Lorentz group; ILTs are proper! using and expanding both sides, keeping only linear terms in Another subgroup - orthochronous LTs, since we get: are arbitrary ILTs are orthochronous! general rule: each vector index undergoes its own Lorentz transformation! 21 23 Finally, let’s look at transformation of a quantum scalar field: using and expanding to linear order in Recall time evolution in Heisenberg picture: we get: this is generalized to: These comm. relations specify the Lie algebra of the Lorentz group. P x = P µ xµ = P · x − Ht We can identify components of the angular momentum and boost operators: x is just a label we can write the same formula for x-a: e+iP a/! e−iP x/! φ(0)e+iP x/! e−iP a/! = φ(x − a) we define space-time translation operator: and find: and obtain: 24 in a similar way for the energy-momentum four vector P µ = (H/c, P i ) we find: using and expanding to linear order in 26 Similarly: we get: Derivatives carry vector indices: or in components: in addition: is Lorentz invariant Comm. relations for J, K, P, H form the Lie algebra of the Poincare group. 25 27 Canonical quantization of scalar fields Relativistic generalization based on S-3 Hamiltonian for free relativistic particles: Hamiltonian for free nonrelativistic particles: spin zero, but can be either bosons or fermions Furier transform: a(x) = ! d3 p eip·x ã(p) ! 3/2 (2π) d3 x (2π)3 we get: Is this theory Lorentz invariant? eip·x = δ 3 (p) Let’s prove it from a different direction, direction that we will use for any quantum field theory from now: start from a Lorentz invariant lagrangian or action derive equation of motion (for scalar fields it is K.-G. equation) can go back to x using: ! find solutions of equation of motion 3 d p ip·x e = δ 3 (x) (2π)3 show the Hamiltonian is the same as the one above 28 Canonical quantization of scalar fields 30 A theory is described by an action: where (Anti)commutation relations: is the lagrangian. Equations of motion should be local, and so where is the lagrangian density. Thus: [A, B]∓ = AB ∓ BA Vacuum is annihilated by is Lorentz invariant: : is a state of momentum is eigenstate of , eigenstate of For the action to be invariant we need: with with energy eigenvalue: the lagrangian density must be a Lorentz scalar! 29 31 Any polynomial of a scalar field is a Lorentz scalar and so are products of derivatives with all indices contracted. arbitrary constant Let’s consider: real solutions: ! = 1, c = 1 and let’s find the equation of motion, Euler-Lagrange equation: k −→ −k (we find eq. of motion from variation of an action: making an infinitesimal variation in and requiring the variation of the action to vanish) thus we get: integration by parts, and δφ(x) = 0 at infinity in any direction (including time) is arbitrary function of x and so the equation of motion is Klein-Gordon equation (such a k µ is said to be on the mass shell) 32 Solutions of the Klein-Gordon equation: 34 Finally let’s choose one classical solution is a plane wave: so that is Lorentz invariant: manifestly invariant under orthochronous Lorentz transformations on the other hand is arbitrary real wave vector and The general classical solution of K-G equation: sum over zeros of g, in our case the only zero is k 0 = ω where and are arbitrary functions of , and is a function of |k| (introduced for later convenience) for any the differential it is convenient to take the Lorentz invariant differential is: if we tried to interpret as a quantum wave function, the second term would represent contributions with negative energy to the wave function! 33 is Lorentz invariant for which 35 In our case: Finally we have a real classical solution of the K.-G. equation: , where again: For later use we can express , in terms of : Inserting where and we will call Note, we get: . is time independent. 36 38 Constructing the hamiltonian: Recall, in classical mechanics, starting with lagrangian as a function of coordinates and their time derivatives we define conjugate momenta and the hamiltonian is then given as: ! In field theory: d3 x ip·x e = δ 3 (p) (2π)3 hamiltonian density and the hamiltonian is given as: 37 39 Summary: From classical to quantum (canonical quantization): “coordinates” and “momenta” are promoted to operators satisfying canonical commutation relations: is equivalent to: operators are taken at equal times in the Heisenberg picture for: We have rederived the hamiltonian of free relativistic bosons by quantization of a scalar field whose equation of motion is the KleinGordon equation (starting with manifestly Lorentz invariant lagrangian). does not work for fermions, anticommutators lead to trivial hamiltonian! 40 We have derived the classical hamiltonian: We kept ordering of a’s unchanged, so that we can easily generalize it to quantum theory where classical functions will become operators that may not commute. The hamiltonian of the quantum theory: (2π)3 δ 3 (0) = V see the formula for delta function is the total zero point energy per unit volume we are free to choose: the ground state has zero energy eigenvalue. 41 42