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We will next demonstrate the impossibility of processes of the form (32313). 

Immediately after the first collision, in the frame of 1, 2 is moving away from 1 and 3 is moving toward 1. Let the velocity of 3 be w and the velocity of 2 be v exp( iα ) . We view the next collision, (2), in the frame of particle 1. After the collision (2), 2 needs to have its velocity shifted by a phase of π so that it changes from an outgoing particle to an ingoing particle. The collision matrix for collision (2) is: 

Before the second collision, the velocities of 2 and 3 are ( v exp( iα ) , w ) $^{T}$. If we apply the above matrix to this vector, we find the velocities after the second collision: 

<img src='content_image/363692.jpg'>

<img src='content_image/363691.jpg'>

As mentioned, we need to have the velocity of 2 shifted by a phase of π . This requires that the term 2 bw sin( θ/ 2) exp[ i ( θ + π ) / 2] / ( a + b ) have opposite phase from v exp( iα ) and be larger in magnitude than v . For the phase of the correction term to be right, we need θ = 2 α + π . Plugging this into Eq. A11 gives the velocities as: 

and for the velocity of particle 2 to be turned around, we need: 

<img src='content_image/363690.jpg'>

<img src='content_image/363689.jpg'>

We will analyze the third collision by working in the frame of 1 before the collision and by working in the frame of 2 after the collision. That way, before the collision, 3 is moving away from 1 and 2 is moving toward 1 and after the collision, 1 is moving away from 2 and 3 is moving toward 2. At the moment of the third collision, 2 replaces 1 at the origin. Thus before the collision, 3 is moving away from the origin and after the collision it is moving toward the origin. So we need the collision to change the phase of particle 3 by π . The matrix for collision (3) in which we work in the frame of 1 before the collision and the frame of 2 after the collision is: 

<img src='content_image/363688.jpg'>

