Download e-book for kindle: Advanced quantum mechanics by Freeman J Dyson; David Derbes

By Freeman J Dyson; David Derbes

ISBN-10: 9812706224

ISBN-13: 9789812706225

ISBN-10: 9812706615

ISBN-13: 9789812706614

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Thus M gives the probability amplitude for arriving in state B after a long time when the system is known to be in state A to begin with. Hence M itself should be an invariant relativistically. e. consider the particles to go directly from the free-particle state A to the free-particle state B by applying the interaction operator once to state A. For electrons at reasonably high or relativistic velocities this will be a very good approximation (e 2 / v 1). Also we treat the electromagnetic interaction classically, just as in the O16 problem, taking the field produced by particle 1 according to classical Maxwell equations to act directly on particle 2.

Phys. 20 (1948) 367 Phys. Rev. 80 (1950) 440 J. Schwinger, Phys. Rev. 82 (1951) 914 It is relativistic all the way, and it is much simpler than the old methods. It is based directly on the Action Principle form of the classical theory which I have just given you, not the Hamiltonian form. In the quantum theory the φα are operators defined at each point of space-time as before. They satisfy the same field equations as before, and this is ensured if we assume that the Action Principle δI(Ω) = 0 1 I(Ω) = c 4 α α Ω L φ , φµ d x (171) holds for all variations δφα of the operators vanishing on the boundaries of Ω.

These quantum numbers are not enough to fix the state, because each value of j may correspond to two non-relativistic states with = j ± 21 . Therefore we need an additional operator which commutes with H, which will distinguish between states with σ parallel or antiparallel to J . The obvious choice is Q=σ·J But [H, σ] is non-zero and rather complicated. So it is better to try Q = βσ · J (73) which is the same in the non-relativistic limit. Then we have [H, Q] = [H, βσ · J] = [H, βσ] · J + βσ · [H, J ] But [H, J ] = 0; furthermore, since αk βσ = βσ αk k= and αk βσk = −βσk αk we get 3 [H, βσ] = −i c {(α · ∇) βσ − βσ (α · ∇)} = −2i c k=1 αk σk β ∇ k 19 The Dirac Theory Therefore 3 [H, βσ] · J = −2 c k=1 αk σk β ∇k (r × ∇)k − i c (α · ∇) β = −i c (α · ∇) β = H, 12 β because ∇·r×∇=0 α k σk = and 0 1 1 0 for all k’s.

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Advanced quantum mechanics by Freeman J Dyson; David Derbes


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