Fusilli, tagliatelle o fettuccine electrons?


In my last entry on Hamilton’s method for systems with constraints I left some unfinished work: The general derivation that I gave is not completely correct; although the method does work, as one can see clearly in the example. As soon as I have some time I will add the explanations.

Now I want to add some comments on certain “exotic” quantum states that keep on appearing in the literature over the last years giving rise to a certatin amount of surprise. These states are basically attached to photons, but very similar arguments are presumably valid for particles satisfying the Schrödinger equation, as the paraxial approximation of optics is  formally equivalent to a Schrödinger equation.

These states are known as Laguerre-Gaussian, Hermite-Gaussian, etc. Basically it consists in the existence of quantum states representing possible evolution modes for particles freely  propagating along one direction and with an orbital angular momentum packaged within the beam, identifyable with an orbital motion in the direction perpendicular to the one of free motion. Thus, the z component, say, is a free-propagating Gaussian packet, while the x and y components (the factor corresponding to the wave function, that is), are  represented by, e.g., a Laguerre or Hermite polynomial. This is paradoxical because we have been taught that free evolution always leads to dispersive or spreading wave packets, and yet these states seem to be carrying along a confined, orbital motion in their evolution.

The reason that we find this paradoxical is that free propagation is incorrectly accounted for in quantum mechanics books. I will try to explain this. The usual procedure of axiomatically  building quantum mechanics is based on Von Neumann, and it tells us that there are complete sets of compatible observables. Once these compatible observables expanding any state  and containing all the possible statistical information on the state are identified, they are instrumental to represent the evolution of an arbitrary quantum state. In the case of linear  momentum, there are three compatible operators (commuting in pairs), \left(\hbar/i\right)\partial/\partial x\overset{{\scriptstyle \textrm{def}}}{=}P_{x}\left(\hbar/i\right)\partial/\partial y\overset{{\scriptstyle \textrm{def}}}{=}P_{y}, \left(\hbar/i\right)\partial/\partial z\overset{{\scriptstyle \textrm{def}}} {=}P_{z}.

This assumes that, in a laboratory, it is feasible to guarantee for a free particle to be an eigenstate of P_{z}, and P_{x}, and P_{y}. That’s not the way it works: First, it has to be an eigenstate of P_{z}, where z is the filtering direction chosen, but not necessarily of P_{y} and P_{x}. In fact. diaphragms and collimators, slits, etc., that are used are essentially physical obstacles conditioning in position coordinates x and y, and consequently representable by “obstacle potential functions” V_{i}\left(x,y\right), so they will have produced a certain \left(x,y\right) profile that neither is a free-propagating packet nor does it have to be.

Thus the natural thing to say is that a free particle with a selected linear momentum is an eigenstate of P_{z}, where z is the chosen direction of filtering,  and that in principle we haven’t the slightest idea what it is in P_{x} and P_{y}. People generally do not see this, because in general they consider that the distinction between filtering or non-filtering state preparations are a nicety or unnecessary erudition, when actually it is absolutely crucial. Let us go directly to the relaxation of the  statement “we haven’t the slightest idea”, as the truth is we can postulate how the state in p_{x} and p_{y} looks like in a very natural way.

If after a certain relaxation time the particle is indeed free, we can posit that it must be an eigenstate of the free Hamiltonian (kinetic energy),

H=-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)

We add now the demand that it be an eigenstate of the filtered linear momentum in the chosen direction. We have,

-\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)\psi=\frac{p_{z}^{2}}{2m}\psi

\frac{\hbar}{i}\frac{\partial}{\partial z}\psi=p_{z}\psi

This amounts to,

\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)\psi=0

\frac{\hbar}{i}\frac{\partial}{\partial z}\psi=p_{z}\psi

 

Thus the state that represents a particle free-propagating with momentum determined in p_{z}, perhaps not the most general one, but definitely general enough, is a function that is harmonic in x, y multiplied by an eigenstate of P_{z}.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: