Physlets run in a Java-enabled browser on the latest Windows & Mac operating systems.
If Physlets do not run, click here for help on updating Java and setting Java security.
Section 8.2: The Quantum-mechanical Free-particle Solution
Please wait for the animation to completely load.
In order to tackle the free-particle problem, we begin with the Schrödinger equation in one dimension with V(x) = 0,
[−(ħ2/2m)∂2Ψ(x,t)/∂x2 = iħ(∂/∂t)Ψ(x,t) . (8.2)
We can simplify the analysis somewhat by performing a separation of variables and therefore considering the time-independent Schrödinger equation:
−(ħ2/2m)(d2/dx2) ψ(x) = E ψ(x) , (8.3)
which we can rewrite as:
[(d2/dx2) + k2] ψ(x) = 0 , (8.4)
where k2 ≡ 2mE/ħ2. We find the solutions to the equation above are of the form ψ(x) = Aexp(ikx) where we allow k to take both positive and negative values.2 Unlike a bound-state problem, such as the infinite square well, there are no boundary conditions to restrict the k and therefore E values. However, each plane wave has a definite k value and therefore a definite momentum (and also a definite energy) since pψ(x) = −iħ(d/dx) ψ(x) = ħkψ(x), again with k taking on both positive and negative values (so that pψ(x) = ± ħ|k|ψ(x)). The time dependence is now straightforward from the Schrödinger equation:
iħ(∂/∂t) Ψ(x,t) = EΨ(x,t) , (8.6)
or by acting the time-evolution operator, UT(t) = e−iHt/ħ, on ψ(x, t). Both procedures yield ψ(x,t) = Aeikx − iEt/ħ (again k can take positive and negative values) and since E = p2/2m = ħ2k2/2m, we also have that
ψ(k > 0)(x,t) = Aexp(ikx − iħk2t/2m) or ψ(k < 0)(x,t) = Aexp(-i|k|x − iħk2t/2m), (8.7)
where ħk2/2m ≡ ω. These solutions describe both right-moving (k > 0) and left-moving (k < 0) plane waves. Recall that solutions to the classical wave equation are in the form of f(kx −/+ ωt) for a wave moving to the right (−) or left (+). These quantum-mechanical plane waves, however, are complex functions and can be written in the form f(±|k|x − ωt).
Restart. In the animation, ħ = 2m = 1. A right-moving plane wave is represented in terms of its amplitude and phase (as color) and also its real, cos(kx - ħk2t/2m), and imaginary, sin(kx − ħk2t/2m), parts.
What is the velocity of this wave? If this were a classical free particle with non-relativistic velocity, E = mv2/2 = p2/2m and vclassical = p/m as expected. But what about our solution? The velocity of our wave is ω/k which gives: ħk/2m = p/2m, half of the expected (classical) velocity! This velocity is the phase velocity. If instead we consider the group velocity,
vg = ∂ω/∂k, (8.8)
we find that
vg = ∂(ħk2/2m)/∂k = ħk/m, (8.9)
the expected (classical) velocity.
Consider the right-moving wave,
ψ(x,t) = Aexp(ikx − iħk2t/2m) ,
which has a definite momentum, p = ħk. We notice that the amplitude of the wave function, A, is a finite constant over all space. However, we also find that ∫ ψ*ψ dx = ∞ [integral from −∞ to +∞] even though ψ*ψ = |A|2 is finite. While the plane wave is a definite-momentum solution to the Schrödinger equation, it is not a localized solution. In this case then, we must discard, or somehow modify, these solutions if we wish a localized and normalized free-particle description.
2The most general solution to the differential equation is
ψ(x) = Aeikx + Be−ikx (8.5)
with k values positive.
3We can however, box normalize the wave function. In box normalization, we normalize the wave function such that over a finite region of space the wave function is normalized.