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Section 8.5: Towards a Wave Packet Solution
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) eipx/ħ dp [integral from −∞ to +∞] (8.11)
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 +∞] (8.12)
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) = (α1/2/π1/4) 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.}