A Wave Packet that Moves!

And now let’s add time dependence...

The wave packet amplitude function Ψ(x)

We start with the same wave function we had last week, but written to include time evolution.  The initial width is (Subscript[d, 0]) and wave-number (Subscript[k, 0]).
(See also Liboff pg 160).

$Assumptions={ℏ>0,m>0,d0>0,Element[{k0,ℏ,m,d0,x,t},Reals]}
Ψ=Sqrt[((d0 (E^(-(((d0 k0))^2))))/(Sqrt[π] ((d0^2)+I t ℏ/m)))] (E^(-((((x- I (d0^2) k0))^2)/(2((d0^2)+I t ℏ/m )))))

Let’s check the normalization...

Integrate[Abs[Ψ]^2/.{t->0},{x,-Infinity,Infinity}]

Mathematica needs some help.

ComplexExpand[-Re[(d0^2) (k0^2)]-Re[((((-I (d0^2) k0+x))^2)/(d0^2))]]

(∫_(-∞)^∞)((E^(-((x^2)/(d0^2))))/((Sqrt[π]) Abs[d0]))\[DifferentialD]x

So, the normalization is ok.  Is this a solution to the Schrödinger equation?

FullSimplify[ (∂_t) Ψ==((I ℏ)/(2m))(∂_x) (∂_x) Ψ]

Yes!  Finally, a particle that moves!

Manipulate[
Plot[{Re[Ψ],Abs[Ψ]}/.{d0->10^logD,k0->(Subscript[k, 0]),t->tNow,ℏ->1,m->1},{x,-5,20},PlotRange->{-1,1},Filling->Axis,PlotLabel->"Ψ(x,t)"],{{logD,0},-1,1,0.1,Appearance->"Labeled"},{{(Subscript[k, 0]),4},-5,10,0.5,Appearance->"Labeled"},{{tNow,0},0,3,0.02,Appearance->"Labeled"}]

Momentum Space

The Fourier transform looks much like the wave-packet from last week, but with a new time-dependent phase factor.

(OverTilde[Ψ])=1/Sqrt[2π]Integrate[Ψ Exp[-I k x],{ x,-Infinity,Infinity}]

What do the expectation values look like?

Momentum Statistics

Mathematica needs some help with the math, but the answers are the same as before.

(OverBar[p])=Integrate[Ψ\[Conjugate](-I ℏ (∂_x) Ψ),{x,-Infinity,Infinity}]
Simplify[(OverBar[p])/.t->0]

Could <p> be a function of time?  Why not?

(OverBar[p2])=Integrate[Ψ\[Conjugate](-I ℏ (∂_x)(-I ℏ (∂_x) Ψ)),{x,-Infinity,Infinity}]
Simplify[(OverBar[p2])/.t->0]

Δp=FullSimplify[Sqrt[(OverBar[p2])-((OverBar[p])^2)]/.t->0]

Energy Statistics

(OverBar[E])=Integrate[Ψ\[Conjugate](I ℏ (∂_t) Ψ),{x,-Infinity,Infinity}]
Simplify[(OverBar[E])/.t->0]

hmmm... why isn’t that zero when (Subscript[k, 0])=0?  If the particle has no momentum, why does it have energy?
(The striking similarity to (OverBar[p2]) is not an accident.)

Position Statistics

(OverBar[x])=Integrate[Ψ\[Conjugate]x Ψ,{x,-Infinity,Infinity}]

(OverBar[x])=Simplify[((k0 t ℏ m )/(m^2))]

(OverBar[x2])=Integrate[Ψ\[Conjugate](x^2) Ψ,{x,-Infinity,Infinity}]

(OverBar[x2])=Simplify[((m((d0^4) (m^2)+(1+2 (d0^2) (k0^2)) (t^2) (ℏ^2)))/(2 (d0^2) (m^3)))]

Δx=FullSimplify[Sqrt[(OverBar[x2])-((OverBar[x])^2)]]
FullSimplify[Δx==(d0/( (Sqrt[2])))(Sqrt[1+(((((t ℏ)/( (d0^2) m))))^2)])]

And the uncertainty product?

FullSimplify[Δx Δp]
FullSimplify[Δx Δp==(ℏ/( 2)) (Sqrt[1+(((((t ℏ)/( (d0^2) m))))^2)])]

... grows with time!