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# 3.1: The Free Particle

#### Schrödinger's Equation for V(x) = 0

Not surprisingly, a particle free of the influence of a potential well is called a free particle. Of course, a free particle can encounter "bumps" in the potential energy (such as an unbound electron passing by a proton), and this greatly complicates the math compared to a flat $$V\left(x\right)=0$$ for all values of $$x$$, but there is much to be learned from the simplest case nonetheless.

Schrödinger's equation with zero potential in the position basis is:

$-\dfrac{\hbar^2}{2m} \dfrac{\partial^2}{\partial x^2} \psi\left(x,t\right) = i\hbar \dfrac{\partial}{\partial t} \psi \left(x,t\right)$

We return to our separation of variables trick (Equation 2.4.1) to look at a subset of possible quantum states. This has the effect of selecting a free particle with a single specific energy. The solution to the time portion of the separated wave function is the same as before (Equation 2.4.6), for which we will again choose $$A=1$$.

$T(t)=e^{-i\omega t}, \;\;\;\;\; \omega \equiv \dfrac{E}{\hbar}$

The solution to the spatial part is:

$-\dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} \psi_E\left(x\right) = E\; \psi_E \left(x\right) \;\;\; \Rightarrow \;\;\; \psi_E \left(x\right) = Ae^{ikx} , \;\;\;\;\; k \equiv \pm\dfrac{\sqrt{2mE}}{\hbar}$

Putting the two parts back together gives us the full wave function:

$\psi_E \left(x,t\right) = Ae^{ i\left(kx-\omega t\right)}$

Our separation of variables sorted out states of equal energy, and since the only form of energy here is momentum, the particle could have a momentum of either $$+\hbar k$$ or $$-\hbar k$$, and it would have the same energy. The fact that $$k$$ can either be positive or negative takes into account these two possibilities.

These single-energy separable solutions are called plane wave solutions, and while we know they are eigenstates of energy (because they are separable), they are also eigenstates of momentum, as is readily-shown by operating on $$\psi_E \left(x,t\right)$$ with the momentum operator. The eigenvalue of this state is $$\hbar k$$. These states are eigen functions of both energy and momentum because the energy is only kinetic ($$V\left(x\right)=0$$, and kinetic energy is a simple function of momentum.

#### Non-Plane-Wave Free Particles

It is actually quite useful to repeat this process in momentum space, for reasons that will become clear. In this basis, the hamiltonian operator is simply a function of momentum (no derivatives), giving this for Schrödinger's equation (note that we are going to continue to use functions of $$k$$, which means we replace $$p$$ with $$\hbar k$$):

$\dfrac{\hbar^2 k^2}{2m} \phi\left(k,t\right) = i\hbar \dfrac{\partial}{\partial t} \phi \left(k,t\right)$

Using separation of variables here gives a little different result than for the position basis:

$\dfrac{\hbar^2 k^2}{2m}\left[\phi\left(k\right)T\left(t\right)\right] = i\hbar \dfrac{\partial}{\partial t} \left[\phi\left(k\right)T\left(t\right)\right]$

When we divide both sides by $$\phi\left(k\right)T\left(t\right)$$, we see that the $$\phi\left(k\right)$$ portion just drops out entirely - no differential equation in $$k$$ remains as occurred in the position-basis case (though both sides still equal the constant that is the energy), while the time dependence is the same as before:

$\dfrac{\hbar^2 k^2}{2m} = E, \;\;\;\;\; T\left(t\right) = e^{-i\omega t} \;\;\;\;\; \omega \equiv \dfrac{E}{\hbar}$

If we are looking at a plane wave, then the part of the momentum-space wave function that describes the momentum must be a Dirac delta function (because there is only one momentum in the distribution). Indeed, if we do a fourier transform of the plane wave solution we found above, we get a delta function.

But not all free particles are plane waves – the general solution to Schrödinger's equation does not need to be separable. Such particles have a distribution of possible momenta (and therefore a distribution of possible energies). Each energy in this momentum state has a unique frequency $$\omega = \frac{E}{\hbar}$$, which means that the $$\omega$$ is a function of $$k$$. Specifically:

$\omega = \frac{1}{\hbar} E = \frac{1}{\hbar} \left(\dfrac{\hbar^2 k^2}{2m}\right) \;\;\; \Rightarrow \;\;\; \omega\left(k\right) = \dfrac{\hbar}{2m}k^2$

So in words, what we have here is this: A free particle that is not in an energy/momentum eigenstate is in a mix (superposition) of these eigenstates. Each component has its own momentum, and along with each momentum comes its own energy, and with each energy is own frequency (the rate at which the phasor rotates).

Notice that the relationship between the frequency and wave number is not the same as we have seen for other waves, such as like, which satisfies:

$\omega = 2\pi \;f = 2\pi \left(\dfrac{c}{\lambda}\right) = c \left(\dfrac{2\pi}{\lambda}\right) =c\;k \;\;\; \Rightarrow \;\;\; \omega\left(k\right)=c\;k$

So for our massive particle, the relationship between frequency and wave number is quadratic, while for light (a massless particle), it is linear. These functional forms of frequency in terms of wave number are called dispersion relations. It requires more detail than we will go into here to explain the origin of this name, but here is a taste:

A free particle that is a mixture of momentum eigenstates has several parts of its wave function (each eigenstate) all moving at different speeds. The speed for each constituent eigenstate is called the phase velocity of the that part of the wave function, and is simply the ratio of the frequency to the wave number:

$v_{phase} = \frac{\omega}{k} = \dfrac{\hbar k}{2m}$

But suppose we prepare a particle over and over again in one of these mixed states, and measure its position change after a fixed period of time over and over. We will get different results thanks to the uncertainty in momentum, but we can take an average. That is essentially the measured speed of the particle, and as it involves a group of momentum eigenstates, it is called the group velocity. This group velocity can be found from the dispersion relation by taking the derivative of the frequency with respect to the wave number and evaluating it at the average wave number $$k_o$$ for the group:

$v_{group} = \frac{d\omega}{dk}\Big|_{k_o} = \dfrac{\hbar k_o}{m}$

The result makes sense, as it is the average momentum divided by the mass. The word "dispersion" applies because the wave group has many parts that are all traveling at different phase velocities, which means that the group will spread over time (even as its average velocity remains fixed at the group velocity).