# The free particle

1. Dec 4, 2012

### aaaa202

The free particle is the particle in no potential:

Hψ = Eψ

with V=0

you get waves travelling in + and - direction but they are not normalizeable. So for some reason my book comes up with the idea of constructing:

ψ(x,t) = ∫β(k)exp(i(kx-hk2/2m t)) dk

I have no idea where this came from, so far all the complete solutions to the SE I have seen have been discrete. What makes us suddenly go to continuous solutions?

I can see how the above resembles a fourier transform, but I am unsure what to make of it.

Can someone guide me through a more intuitive way of seeing the above. I have watched extra lectures in QM and know how the momentum space wave function is the fourier transform of the position space. But how does it exactly relate to the above? I mean all I can see is that for a free particle the solutions must be eigenstates to the momentum operator. But why is it that they are not normalizeable etc. etc.

2. Dec 4, 2012

### cattlecattle

Bound states are guaranteed to have quantized energy, but this is not true for free particle who can reach infinity. There are two kinds of normalizability in QM, the proper one will normalize to 1, but for ideally localized particle (or particle with definite momentum), the wave functions is only required to normalize to the Dirac delta function.

The proposed solution is indeed a fourier transform, because it's the time-dependent solution. The time-dependency being $e^{-iEt/\hbar}$ where $E=p^2/2m$, where $p=hk$ If you ignore this time term, the remaining term is exactly the inverse Fourier transform from p-basis to the x-basis.

3. Dec 5, 2012

### jmcelve

Hi aaaa202,

The "discreteness" of QM is actually manifest in potentials. It is not the case that "all energies are quantized" since it is boundary conditions that demand that certain eigenfunctions (and consequently certain energies) be solutions to the Schrodinger equation for particular potentials.

However, what we see here is that there is no potential. Since there are no boundary conditions on the solutions to the Schrodinger equation, the general solution will simply be superposition of plane waves. (Note I said a superposition of plane waves -- not just *a* plane wave.)

To obtain this solution, just solve the Schrodinger equation in the regular way. Here, your Hamiltonian will have no potential (V=0), so you just have the kinetic energy component and the time derivative. You'll have an equation on which you can use the separation of variables technique (separate ψ into a function of time and position and set both the x-dependent and t-dependent components to a constant).

However, we have obtained only *one* sollution for a particular value of momentum in this way. It turns out that *any* value of p can solve the Schrodinger equation for a free particle, and so the *most general* solution is a superposition of particular solutions (as I said above). That is why we use an integral. In the case where there is a potential (such as the infinite square well), the most general solution is a *sum* of solutions (the lowest ground state plus the first excited state plus..) because we have discrete states. Because the free particle lacks boundary conditions and consequently any value of p is allowable, we use an integral (analogous to the sum) from negative infinity to infinity with respect to k (which is related to p by p=hbar*k).

Now β(k) weights certain values of k within the integral. This is identical to the coefficients in a Fourier expansion where the coefficients correspond to the "amount" a certain particular basis function "contributes" to an arbitrary function. If that part doesn't quite make sense, think of it like this: Imagine you have a simple function f(x)=2*sin(x)+5*sin(2x). The Fourier series expansion of this is pretty simple, since it's 2*sin(x)+5*sin(2x) (the way this function decomposes into basis functions is obvious, but that is not true in general). But if we wrote it in summation form, we'd note that the generic coefficient cn is not the same for c1=2 as it is for c2=5. In the same way, β(k) weights certain values of k so that a generic function ψ(x,t) is weighted by certain particular solutions more than others (in general).

As far as why they aren't normalizable.. Well, just try to normalize them! What happens? The integral diverges! So clearly these solutions are not normalizable to 1. But we *can* normalize them to the Dirac delta.

I hope I've helped. Best,
jmcelve

Last edited: Dec 5, 2012
4. Dec 5, 2012

### chill_factor

only bound states have quantized energies. scattering states have continuous energy spectra. when you solve a PDE the general solution is in the form of a sum (see Mathematical Methods in Physical Sciences chapter 13); since the free particle can have any momentum it is better represented as an integral in momentum-space.

5. Dec 5, 2012

### Staff: Mentor

Because it's the only way you can get a normalizable free-particle wave function.

Correction: the definite-momentum states of the free particle are eigenstates of the momentum operator. They're not normalizable because they extend to infinity in both directions with constant amplitude.

6. Dec 5, 2012

### aaaa202

Thanks for the good answers all. Now 2 of you mentioned normalizing to the dirac delta. What exactly do you mean by that? I know the deltafunction acts as the orthonormality relation in the continous position basis:

<x'lx> = δ(x-x')

But where does that come in here?

7. Dec 5, 2012

### cattlecattle

For a momentum eigenstate |p>, its wave function in the x-basis is a plane wave:
$\langle x | p \rangle \rightarrow p(x)=e^{ipx/\hbar}$

If you attempt to impose the orthonormality of the p-basis, you'll find:
$\langle p'|p \rangle = \int \langle p' | x \rangle \langle x | p \rangle dx = \int \exp\left[i(p-p')x/\hbar\right]dx$

The latter integral is known to be $\delta(p-p')$ up to a constant factor.