# Wave packet

Some researchers have been able to produce a wave packet that is somewhat similar to a “chopped wave plane”. Namely that a wave packet is just a slice from an infinitely long single frequency plane wave. We will model this at $t=0$ as:

$psi(x,t=0)=left{begin{matrix}0&x<-a\A*e^{ipx/hbar}&-1leq xleq a\0&x>aend{matrix}right.$

a) Find $A$

b) Sketch the real part of $psi(x,t-0)$

c) Determine the momentum space wave function $phi(p)$

d) Sketch the plot of $|phi(p)|^2$

e) if the width $a$ were decreased how would that change your plot in part d. Explain

f) Write the integral expression for $<p>$ for the wave packet to the point where this integral could be evaluated, but do not evaluate the integral. Qualitatively argue what the evaluation of <p> will yield.

g) Find the expectation value of the position at $t=0$. Is it the same as the sketch in b)? Will the expectation value change with time?

a)

$1=A^2*∫_{-a}^a ψ^* ψdx=A^2*∫_{-a}^a d x=2aA^2$ that is $A=sqrt{1/2a}$

b)

c)

$ϕ(p)=∫_{-a}^a ψ(x)*exp(-i p x/hbar)d x=A∫_{-a}^a exp[i(p_0-p)/hbar *x] d x$

$ϕ(k)=A∫_{-a}^a exp [i(k_0-k)x]*d x=A/(i(k_0-k))*exp[i(k_0-k)x |_{-a}^a=2$

$=A/(k_0-k)*sin[(k_0-k)a]$

d) The graph is above taking $a=1$ and $k_0=1$

e) If the width of the wave packet is decreased then the amplitude of the above graph (probability is momentum space) will decrease. Quantum explanation is the following: decreasing a, increases the amplitude of the wave $A=sqrt{1/2a}$ and this increases the expectation value of the position. From the Heisenberg principle the momentum of the particle need to decrease.

f)

$<p>=∫_{-a}^a ψ^* (-ihbar*frac{d}{d x})ψdx=$

$=(1/2a)*(-ihbar )*(i k_0 )*∫_{-a}^a e^{-i k_0 x}*e^{i k_0 x}*d x=(hbar k_0)/2a*∫_{-a}^a d x$

The explanation is simple: the expectation value of the momentum is a constant value

$<p>=hbar k_0=p_0$

g)

$<x>=∫_{-a}^a ψ^* xψ*d x=(1/2a)*∫_{-a}^a x*d x=0$

At $t=0$ the wave packet is centered on $x =0$ so the figure from b) and the above answer agree.

Basically the particle being a wave packet it needs to obey the classical law for its position

$<x>=<v>*t=(<p>)/m*t=p_0/m*t$