By Gibbs G.W.

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Our choice implies the following wave equation iE i ψ˙ P = − ψP = − h ¯ h ¯ mc2 + ∂x ψP = i pψP , ¯ h We have also 1 2 p ψP . 38) from which we deduce i¯ hψ˙ P = mc2 ψP − ¯2 2 h ∂ ψP . e. deﬁning ψ ≡ eimc 2 t/¯ h ψP ∼ exp i h ¯ px − p2 t 2m . 38) and has the same probabilistic interpretation as ψP . Indeed both ρ and J are unchanged. The wave equation instead changes: 2 h 2 ¯ i¯ hψ˙ = − ∂ ψ ≡ Tψ. 41) This is the Schr¨ odinger equation for a free (non-relativistic) particle, in which the right-hand side has a natural interpretation in terms of the particle energy, which in the free case is only of kinetic type.

What is the velocity v cm of their center of mass frame? 46), it is easily found that vcm = 1/3 c, vcm = 2/3 c √ and |v cm | = 5/3 c. 31. A particle of mass M decays, while at rest, into three particles of equal mass m. What is the maximum and minimum possible energy for each of the outgoing particles? Answer: Let E1 , E2 , E3 and p1 , p2 , p3 be respectively the energies and the momenta of the three outgoing particles. We have to ﬁnd, for instance, the maximum and minimum value of E1 which are compatible with the constraints p1 +p2 +p3 = 0 and E1 + E2 + E3 = M c2 .

5 The Potential Barrier The most interesting physical situation is that in which particles are not free, but subject to forces corresponding to a potential energy V (x). 42) has to be used. Since the equation is linear, the study can be limited, without loss of generality, to solutions which are periodic in time, like: ψ(x, t) = e−iEt/¯h ψE (x) . 54) Indeed the general time dependent solution can always be decomposed in periodic components through a Fourier expansion, so that its knowledge is equivalent to that of ψE (x) plus the expansion coeﬃcients.