here is a derivation of equation 2.28 in the book

We’ll do that by differentiating with respect to time and show that this iszero.So, let A be proportional to the RHS of Equation 1:A =and let us compute dAdt .∫ [ ψ(x, t) ∂ψ∗ (x, t)∂x− ψ ∗ (x, t)2.1 puase, take a deep breath ...∫dA[dt = ∂ψ(x, t) ∂ψ ∗ (x, t)+ ψ(x, t) ∂2 ψ ∗ (x, t)∂t ∂x∂x∂tThe integrand can be written[∂ ∂ψ ∗ (x, t)ψ(x, t) +∂x ∂xwhich can, in turn, be written[∂ ∂ψ ∗ (x, t)ψ(x, t) +∂x ∂x]∂ψ(x, t)ψ ∗ (x, t) −2∂x]∂ψ(x, t)dx∂x]− ∂ψ∗ (x, t) ∂ψ(x, t)− ψ ∗ (x, t) ∂2 ψ(x, t)dx∂t ∂x∂x∂t[ ∂ψ ∗ (x, t)∂t∂ψ(x, t)∂x]∂ψ(x, t)ψ ∗ (x, t) − 2 ∂ [ψ ∗ (x, t)∂x∂t]+ ∂2 ψ(x, t)ψ ∗ (x, t)∂x∂t]∂ψ(x, t)∂xSo, as usual, the left part goes away when we integrate by parts, leavingdAdt = −2 ∂ ∫ [ ]ψ ∗ ∂ψ(x, t)(x, t) dx (2)∂t∂xNow, remember we’re trying to show that this is zero, since that willgive us uniform motion of the wavepacket. So, what do we know is certainlyindependent of time? For a free particle it is true that the normalization isindependent of time: ∫ψ ∗ (x, t)ψ(x, t)dx = 1.Somehow we need to introduce this normalization. Let’s try:∫∂[ ]ψ ∗ ∂ψ(x, t)(x, t) dx = ∂ ∫∂t∂x ∂t∂∂x |ψ(x, t)|2 dx− ∂ ∂t∫ [ ]ψ(x, t) ∂ψ∗ (x, t)dx∂x(3)3

The first term on the right side of the equal sign will be zero once weintegrate by parts. Then, the two terms on either side of the equal sign willbe complex conjugates of one another. Calling these integrals B and B ∗ wethen have:∂∂t [B + B∗ ] = 0.While not true in general, it is true for a free particle that this implies thatB itself is independent of time. HencedAdt = 0.Putting everything together, we have shown that the time derivative ofthe center of the wavepacket 〈x〉 was a constant in time. This means that3 caveat〈x〉 t = 〈x〉 0 + v 0 tLater, we will take a more general approach to this whole question by lookingat probability currents and the time derivative of the wavefunction normalization.That will be in Chapter 3 (around page 50). But for now, we areonly speaking of a free particle.4 the spreading of the wavepacketThe book denotes the variance of a wavepacket by ∆x 2 t . You will showin an exercise that variance of the wavepacket for a free particle increasesquadratically with time. I.e.,(∆x t ) 2 = (∆x 0 ) 2 + ξ 1 t + ∆v 2 t 2The sign of ξ can be positive or negative. Hence at any given time (∆x t )could be greater or less than ∆x 0 ). But, at late times the quadratic termdominates and since (∆v) 2 is positive, the wavepacket must spread accordingto∆x t ∝ ∆v|t|4