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,

[−(*ħ*^{2}/2*m*)∂^{2}Ψ(*x*,*t*)/∂*x*^{2} = *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:

−(*ħ*^{2}/2*m*)(*d*^{2}/*dx*^{2}) ψ(*x*) = *E* ψ(*x*) , (8.3)

which we can rewrite as:

[(*d*^{2}/*dx*^{2}) + *k*^{2}] ψ(*x*) = 0 , (8.4)

where *k*^{2 }≡ 2*mE*/*ħ*^{2}. We find the solutions to the equation above are of the form ψ(*x*) = *A*exp(*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 *p*ψ(x) = −*iħ*(*d*/*dx*) ψ(*x*) = *ħk*ψ(*x*), again with *k* taking on both positive and negative values (so that *p*ψ(*x*) = ± *ħ*|*k*|ψ(*x*)). The time dependence is now straightforward from the Schrödinger equation:

*iħ*(∂/∂*t*) Ψ(*x,t*) = *E*Ψ(*x,t*) , (8.6)

or by acting the time-evolution operator, *U*_{T}(*t*) = e^{−iHt/ħ}, on ψ(*x, t*). Both procedures yield ψ(*x,t*) = *A*e^{ikx − }^{iEt/ħ} (again *k* can take positive and negative values) and since *E* = *p*^{2}/2*m* = *ħ*^{2}*k*^{2}/2*m*, we also have that

ψ_{(k > 0)}(*x,t*) = *A*exp(*i**kx* − *iħk*^{2}*t*/2*m*) or ψ_{(k < 0)}(*x,t*) = *A*exp(-*i*|*k*|*x* − *iħk*^{2}*t*/2*m*), (8.7)

where *ħk*^{2}/2*m* ≡ 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* −/+ ω*t*) 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 * − ω*t*).

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*-

*ħk*

^{2}

*t*/2

*m*), and imaginary, sin(

*kx*−

*ħk*

^{2}

*t*/2

*m*), parts.

What is the velocity of this wave? If this were a classical free particle with non-relativistic velocity, *E* = *m*v^{2}/2 = *p*^{2}/2*m* and v_{classical }= *p*/*m* as expected. But what about our solution? The *velocity* of our wave is ω/k which gives: *ħk*/2*m* = *p*/2*m*, half of the expected (classical) velocity! This velocity is the *phase velocity*. If instead we consider the *group velocity*,

v_{g} = ∂ω/∂*k*, (8.8)

)we find that

v_{g }= ∂(*ħk*^{2}/2*m*)/∂*k* = *ħk*/*m*, (8.9)

the expected (classical) velocity.

Consider the right-moving wave,

ψ(*x,t*) = *A*exp(*i**kx* − *iħk*^{2}*t*/2*m*) ,

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.

^{2}The most general solution to the differential equation is

ψ(*x*) = *A*e^{ikx} + *B*e^{−ikx} (8.5)

with *k* values positive.

^{3}We 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.

« previous

next »