Plane wave limit of Gaussian packet

[RedX]

\psi (x)= \frac{1}{(\pi\Delta^2)^\frac{1}{4}} e^(\frac{i<p>(x-<x>)}{\hbar})e^(\frac{-(x-<x>)^2}{2\Delta^2}) as \Delta\rightarrow\infty should approach the plane wave \frac{1}{(2\pi\hbar)^\frac{1}{2}} e^(\frac{i<p>x}{\hbar}) up to a phase factor. I guess this happens by setting the e^(\frac{-(x-<x>)^2}{2\Delta^2}) term equal to 1. However, ignoring phase factor differences, the normalization factor out front still seems to be different. One is \frac{1}{(\pi\Delta^2)^\frac{1}{4}} while the other is \frac{1}{(2\pi\hbar)^\frac{1}{2}}. Am I not seeing something?

 

[Eye_in_the_Sky]

From where did you get that this ψ is a plane wave in the limit ∆ → ∞ ? This is clearly not the case. In that limit ψ(x) → 0 for all x. Moreover,

∫ |ψ(x)|² dx = 1 , for all values of ∆ ,

and thus, in particular, this is so also for the limit ∆ → ∞. A plane wave must have infinite norm, not norm equal to 1.

Furthermore, why are you looking in position space? It is most unintuitive to do so.

So ... let's look in momentum space. Write

[1] |ψ_ε> = ∫ f_ε(p - p_o) |p> dp .

This f_ε(p) is chosen in such a way that

[2] lim_{ε → 0+} f_ε(p - p_o) = δ(p - p_o) .

This means that lim_{ε → 0+} |ψ_ε> = |p_o>, a plane wave with momentum p_o.

In terms of a Gaussian, a suitable choice for f_ε(p) is

[3] f_ε(p - p_o) = (1/ε√π) exp{-(p - p_o)²/ε²} .
_______

Now, let's look at [1] in the position space of functions:

[4] ψ_ε(x) ≡ <x|ψ_ε> = ∫ f_ε(p - p_o) <x|p> dp ,

where, of course,

[5] <x|p> = 1/√(2πh_bar) exp{ipx/h_bar} .

The RHS of equation [4] is essentially a Fourier transform of f_ε(p - p_o). The precise result of [4], in light of [3] and [5], is

[6] ψ_ε(x) = 1/√(2πh_bar) exp{ip_ox/h_bar} ∙ exp{-x²ε²/4h_bar²} .

Clearly,

lim_{ε → 0+} ψ_ε(x) = 1/√(2πh_bar) exp{ip_ox/h_bar} ,

as required. Also,

∫ |ψ_ε(x)|² dx = const ∙ (1/ε) ,

so that in the limit ε → 0+, this quantity goes to ∞ , as required.

Equation [6] is a correct rendition of the ψ(x) you originally proposed. You can now redefine ε as you wish.

 

[R0man]

Your understanding is correct. As the width of the Gaussian packet, represented by the parameter \Delta, approaches infinity, the exponential term in the equation for \psi(x) becomes a constant, resulting in a plane wave solution. However, there is a slight difference in the normalization factor between the two equations. This is because the normalization factor for a plane wave solution is defined in terms of the momentum, p, while the normalization factor for a Gaussian packet is defined in terms of the position and width, <x> and \Delta. The difference in these factors can be accounted for by the uncertainty principle, which states that there is a trade-off between the precision in measuring position and momentum. Therefore, as the width of the Gaussian packet approaches infinity, the uncertainty in the position decreases, resulting in a larger normalization factor for the plane wave solution. This difference in normalization factors does not affect the overall behavior of the wave function, as the phase factor is the same for both equations.