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Considering our failure with using only one solution to the Schrödinger equation for the free-particle problem (the lack of localization and normalization), 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) eipx/ħ dp [integral from −∞ to +∞] (8.8)
which is called a Fourier transform. The Fourier transform adds a continuum of plane wave solutions, eipx/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 +∞] (8.9)
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) = (α1/2/π1/4) exp[−α2(p − p0)2/2]. (8.10)
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. (8.10) in Eq. (8.8) 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). (8.11)
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.
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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.}