Opposite momenta lead to opposite directions

When a particle decays into two fragments, the wavefunctions of the latter are spherical shells with expanding radii. In spite of this spherical symmetry, the two particles can be detected only in opposite directions. When particles at rest decay into two fragments, the latter have opposite momenta, p1 + p2 = 0. (1) Yet, the statistical distribution of each fragment is isotropic. The same holds for pairs of particles resulting from a collision described in the center-of-momentum frame. In classical physics, it is obvious that particles with opposite momenta will be found in opposite directions. The problem is whether this property still holds in quantum mechanics. Indeed, the operator p1 +p2 does not commute with q1 +q2 . Rather, there are uncertainty relations, ∆(p1 + p2)∆(q1 + q2) ≥ h̄, (2) for each one of the Cartesian components of these vectors. This equation says that if a system is prepared in such a way that (p1+p2) is sharp, then the mid-point between the two particles has a broad distribution. However, Eq. (2) does not restrict the angular alignment of the two particles. The purpose of the present article is to show that the operator equation (1) leads to an observable alignment of the detection points of the two particles. First, consider a simpler problem, namely the motion of a single free particle. Why does a typical wavepacket move along a straight line? Let us write the initial wavefunction as a Fourier integral, ψ(r, 0) = ∫ f(p) eip·r/h̄ dp. (3) After a time t, this wavefunction becomes ψ(r, t) = ∫ f(p) ei(p·r−Et)/h̄ dp, (4) where E = p/2m for nonrelativistic particles (only the nonrelativistic case is considered here). Let us write f(p) = |f(p)| e, (5) and let us assume that |f(p)| is peaked around p ' k, so that the main contribution to the integral in Eq. (4) comes from values of p in the vicinity of k. However, this is not the only condition on the parameters appearing in that integral. Its value is usually very small because of the rapid oscillations of the exponent. The integral will be appreciably different from zero only if the phase of the exponent is stationary, namely ∂S ∂p + r− ∂E ∂p t = 0, (6) where all the above expressions have to be evaluated for p = k. Recall that ∂E/∂p = v is the (classical) velocity of a particle with momentum p, and define r0 := ∂S/∂p (evaluated at p = k). We then have r = r0 + v t. (7)