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Michael Weiss: Seeing Double

Michael Weiss: Seeing Double

OK, John gave us the diagram:
 \    /
  \  /  <------ forwards lightcone (w >= 0)
  /  \  <------ backwards lightcone (w <= 0)
 /    \
where pure waves exp(i(kx - wt)) live. Let me draw in the w and k axes:
 \  |  /
  \ | /            
  / | \
 /  |  \    
("Rule, Britannia! Britannia rules the waves" - which I guess she did when Maxwell was alive...)

The diagram stretches out to infinity in all directions, of course, but let's say this piece just goes from w=-1 to w=+1, and k=-1 to k=+1. At the four corners we *could* put the four functions:

exp(i(-x -t))        exp(i(x - t))
exp(i(-x +t))        exp(i(x + t))

which span a 4-complex-dimensional subspace of the space of *all* complex-valued functions on R. (Nice thing about this post - I can just talk about complex vector spaces, and fuggedabout inner-products, measures, and so forth for the nonce.)

Here I'm using the *wrong* complex structure - the one where to compute i times f(x), you just multiply using ordinary complex arithmetic. With that definition, I'm sure our four functions are linearly independent.

With John's *clever* complex structure, I get a little confused. Do we really have four independent vectors, or am I seeing double?

Let me think about the *real-valued* functions first. I've been working with these four till now, two that travel rightward, two leftward:

leftward             rightward
---------            --------
sin(-x -t)           sin(x - t)
cos(-x -t)           cos(x - t)

Since these are all real-valued, we can form a real vector space with them, no sweat. 4-dimensional as a real vector, natch.

Where do these go on my "Rule Britannia" diagram? Well, = that's a bit ambiguous. Does cos(x - t) has positive frequency, or should I think of it as cos(-x + t) with negative frequency? Both and neither!

Blame Euler, who noticed that cos(x - t) is the average of *two* of our exp waves, the positive frequency exp(i(x - t)) and the negative frequency exp(i(-x +t)). We have the problem deciding whether sin(x - t) has positive or negative frequency.

Anyway, following John's instructions, we now define the operation of J on sin(x - t) by the simple rule: "roll back 1/4 period". This gives cos(x - t), either way you look at it. Likewise J applied to cos(x - t) gives -sin(x - t) (as it must, if we're gonna have J^2 = -1).

And that turns our 4-dimensional *real* vector space into a 2-dimensional *complex* vector space! After all, if you know what r times v means, for any real number r and vector v, *and* you know what i times v means, then you *ought* to know what (r+is) times v means--- or, to be crystal clear, (r+Js) times v. Our two rightward waves now become linearly dependent; one's just J times the other.

How about our exp waves? Hmm, Euler always said, "shut up and calculate". (Or something like that.) Let's try this:

   exp( i(x - t)) = cos(x - t) + J sin(x - t) = 2 cos(x - t)

   exp(-i(x - t)) = cos(x - t) - J sin (x - t) = 0

Seems a little odd. But OK, I guess. I suppose I'm trying to find a complex-linear map from:

the space of complex-valued functions on R, with the ordinary unclever boring "i" complex structure
the space of real-valued functions on R, with the clever interesting "J" complex structure
and I've found one! And it maps the 4-complex-dimensional space spanned by the four corners of the British Empire, down to the 2-complex-dimensional space spanned by our trig waves.

This map has a couple of nice properties. First, say f belongs to the first space, but just happens to be real-valued. Then the map leaves f alone - it just maps f to itself. Next, the *negative* frequency waves just get annihilated! How's *that* for the power of positive thinking!

(Hmm, you think I can throw away my glasses?)

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