Please wait for the animation to completely load.

Considering our failure (the lack of localization and normalization) with using only one solution to the Schrödinger equation (a single time-dependent energy eigenfunction) for the free-particle problem, what about a superposition of plane wave solutions which you have explored in Sections 8.3 and 8.4?   Restart.  While these constructions approach a localized solution, there are always copies of the localized solution created. Instead of a sum of individual solutions, consider an integral,

Ψ(x) = 1/(2πħ)^1/2 ∫ Φ(p) e^ipx/ħ dp [integral from −∞ to +∞] (8.11)

which is called a Fourier transform. The Fourier transform adds a continuum of plane wave solutions, e^ipx/ħ, weighted by a function of momentum, φ(p). This function of momentum is called the momentum-space wave function since it plays the same role in momentum space as ψ(x) does in position space. The momentum-space wave function, φ(p), is itself the inverse Fourier transform of ψ(x) and is given by:

Φ(p) = 1/(2πħ)^1/2 ∫ Ψ(x) e^−ipx/ħ dp [integral from −∞ to +∞] (8.12)

Now, we seek to understand the generic wave function as defined by the Fourier transform in the first equation by substituting a reasonable function for φ(p) and calculating the position-space wave function. Consider a normalized Gaussian distribution in momentum centered on a momentum, p₀, such that

Φ(p) =  (α^1/2/π^1/4) exp[−α²(p − p₀)²/2]. (8.13)

Note that |Φ(p)|² goes to 1/e of its maximum value when p = p₀ ± 1/α. Therefore 1/α tells us something about the spread of the momentum-space wave function. This momentum-space wave function is shown in the bottom panel of the animation.  In the animation, ħ = 2m = 1.

To find the position-space wave function, we must use Eq. (8.13) in Eq. (8.11) and evaluate the resulting integral. When we do this Gaussian integral, we get:⁴

Ψ(x) =  [π^−1/4 (αħ)^−1/2] exp(ip₀x/ħ − x²/2α²ħ²). (8.14)

Look at the animation to see how the position-space wave function is related to the original momentum-space wave function.  The bottom panel shows momentum space and the top panel shows position space. Vary p₀ and α and see what happens. As p₀ gets larger and positive, the momentum-space wave function shifts to the right and is centered on the new value of p₀. The position-space wave function now has bands of color which represent the exp(ip₀x/ħ) factor in the wave function. As α increases, the momentum-space wave function narrows and the position-space wave function widens (which is a result of the Heisenberg uncertainty principle).

Our packet has almost all of the right features we want in a packet that simulates a particle. However, it does not have a time dependence and it does not allow us to shift the initial position, x₀, of the packet to any value we like. We will add these features next.

⁴Since the momentum-space wave function was normalized, so is the resulting position-space wave function. In general, due to the relationship between ψ(x) and φ(p) as expressed in the first and second equation, we have that: ∫ |ψ(x)|² dx = ∫ |φ(p)|² dp [integrals from −∞ to +∞], and hence if one is normalized, so is the other.}