The particle receives an instantaneous momentum p at t = 0 and its velocity changes to p/m instantaneously. The duration of the impulse is, however, too short for the wave function to change. Hence, in the view of a frame K' moving with the particle, the latter is still in the ground state of the harmonic oscillator \(\psi_0\) (x). But in the view of a stationary frame K, it is in the state \(\psi_0\) (x) exp (-ipz/h). We may reasonably treat the position of the particle as constant during the process, so that at the end of the impulse the coordinate of the particle is the same for both K and K'. Hence the initial wave function in K is
\(\psi'_0= \psi_0(x) \exp(-px/h)\)
Thus, the probability that the particle remains in its ground state after the impulse is
