How can you describe position of a particle? Need three numbers. What three numbers? Depend on my location. So particle P knows three numbers, that depend on where we're standing.

How? If I move by (column) vector [\del x, \del y, \del z], P's numbers change by [x,y,z]\mapsto [x+\del x, y+\del y, z+\del z].

This is called an

*action*: there's an \R^3 of translations, which "act" on the \R^3 of possible states of the particle.

Now let's be mathematicians, and generalize this. But let's start wih a very simple case: let's say that P only has _one_ number, but that that number depends on where we're standing when we measure it. I.e. we have a fn \Phi: locations \to numbers.

Ok, now let's say that we move by some vector \vec{\del} = [\del x,\del y,\del z]. How does \Phi change? Well, by some function:

\Phi(location + \vec{\del}) = \Phi(loc) + f(loc,\vec{\del})

So, laws of nature seem to not depend on location. Of course, \Phi depends explicitly on location. So let's demand the next best thing: let's say that \Phi is symmetrical enough that f does not depend on location:

\Phi(location + \vec{\del}) = \Phi(loc) + f(\vec{\del})

Well, what if we move and move again?

\Phi(loc) + f(\vec\del_1 + \vec\del_2) = \Phi(loc + \vec\del_1 + \vec\del_2) = \Phi(loc + \vec\del_1) + f(\vec\del_2) = \Phi(loc) + f(\vec\del_1) + f(\vec\del_2).

So f is linear! (One student correctly asked about multiplication by constants; our functions will satisfy some smoothness condition (for \R, suffices to by cont's; for \C, suffices to be \C-differentiable), and then this \Z-linearity is enough to guarantee linearity.)

By linear algebra, a linear function f:\R^3 \to \R is exactly a row vector <f_x, f_y, f_z> s.t.

<f_x, f_y, f_z> [\del_x, \del_y, \del_z] = f(\vec\del)

(I'm writing <> for row vectors, [] for columns. This then is the usual matrix product.) Why? You can read off f_x, for instance, by stepping in one unit in the x-direction, and seeing how much \Phi changes; I'll leave it as an exercise to prove that the three numbers f_x,f_y,f_z exactly give you your function.

In QM, we call the vector <f_x, f_y, f_z> the

*momentum*of the particle.

Let me step back and say what we've done. CM is problematic in a really important way: if a particle is over there,how can I measure it, if everything is local? So, what we've said, is that no, the "particle" is _everywhere_, and remembers a number, which we can only measure _here_.

Now, what if particle 1 turns into particle 2? Say 2 also knows a number \Phi_2, so \Phi_1(loc)\to\Phi_2(loc) as 1\to 2.

And let's say that this is as simple as possible: in particular,

\Phi_2(loc) = \Phi_1(loc) + \varphi

where \varphi does not depend on location --- i.e. is a constant.

Then

\Phi_2(loc) + f_2(\del) = \Phi_2(loc+\del) = \Phi_1(loc+\del) + \varphi = \Phi_1(loc) + f_1(\del) + \varphi

So f_1 = f_2! Conservation of momentum!

Now, we're mathematicians. Let's generalize.

Does it matter that \Phi: \to \R? What if, for instance, \Phi: locations \to S = circle? Then f(\del) is an angle change. We still demand the same linearity: f(\del_1+\del_2) = f(\del_1) + f(\del_2). By taking \del very small, we can assure that f does not go all the way around the circle, in which case it still acts like a number, and it turns out that f is still given exactly by a triple of numbers. I'll leave it as an exercise, for those who know some calculus, to make this rigorous.

By the same argument, of course, this triple, which still deserves the name momentum, is conserved.

But, other things can act on S. See, only \R (and its products, like \R^3) can act on \R, because \R is too long for something compact to act on it. But actions on S can also include S actions! So let's just imagine that, if in addition to moving in three-space, we could move in some circular direction. Then "location" includes an angle-valued component as well as the three large components: We can move by some \del(angle) = \theta, and then

\Phi \to \Phi + c(\theta)

Just as before, c should be linear:

c(\theta_1 + \theta_2) = c(\theta_1) + c(\theta_2)

But also, \theta is an angle, so \theta = \theta + 2\pi, so

c(\theta + 2\pi) = c(\theta)

And c(\theta) is also an angle, so

c(\theta) = c(\theta) + 2\pi.

So, exercise: the only linear maps satisfying this are

c(\theta) = k\theta, for k\in\Z, i.e. for k an integer!

Intuition: this is just topology! How can you wrap a circle around a circle? (Negative numbers correspond to going the wrong direction.)

This number k is called the

*charge*of the particle! Charge is conserved exactly as it was for momentum, and this explains why charge is quantized.

Or, rather, it explains it _if_ space is (3 dimsensions) \times circle. Can it be? Well.... yes, if circle is very small, so small that atoms are bigger than size of circle.

This is called the

*Kaluza-Klein picture*: (gravity in \R^3\times S) = (gravity in \R^3) \tensor (EM in \R^3).

BTW, in QM, physicists prefer everything to be multiplicative, not additive. So, fine: we can use \Psi(loc) = e^{\Phi(loc)}, turning all our "plus"es into "times"es. But we want \Phi to take values in a circle. In \C, we can do this via

\Psi(loc) = e^{i \Phi(loc)}

because e^{i2\pi} = 1. Then

\Psi(loc + \del + \theta) = e^{i\Phi(loc)} e^{i p.\del} e^{i k \theta}

Where p is the (row) vector of momentum, so p.\del is the dot-product, and k is the electric charge. The nice thing about using complex-valued \Psi is it lets us add them, etc., do QM, still w/ this rule. And assuming moving by (\del,\theta) is linear and that space really does have these symmetries, we get conservation of momentum, charge, etc.

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