If there is anything to be salvaged from this quantum card game, it will necessarily involve some issues with dimensions, spaces, bases, and functions.
There are, it seems to me, a few central problems.
1) What a vector space or subspace is (what conditions must be met to make something such a space)
2) What "infinite-dimensional" means and how it relates to dimensional spaces
3) What eigenvectors and eigenvalues are.
The 2nd is the easiest because it narrows (infinitely, actually) what we have to deal with..
Wave functions can be thought of as vectors because the field we are building the vectors out of is the square integrable functions.
So, how do you specify a function over an infinite space by example?
The answer is easy. You don't do that. Because Hilbert space isn't a function over an infinite space, nor do we think of vectors as built out of squared integrable functions, but rather that these constitute an infinite number of functions
on or
over a space. They're a
metric imposing mathematical structure extending Euclidean space.
The right question is, "what is the field of square integrable functions?" The answer:
Take any function
f. If
f satisfies the following:
then
f qualifies as a square integrable function. Basically, it means that a function defined anywhere on the number line s.t. it is itself squared qualifies. To qualify as a Hilbert space, we need only ensure there is an inner product defined for any two function
f and
g and the we have the necessary metric or measure
mu:
All fascinating stuff, and a very scary was of saying that Hilbert space is similar to the space we seem to move around in (3D Euclidean space), only it has a few extra properties and it is
infinite dimensional.
Let's say we really were working in a 52-dimensional space. How much larger is that space than a 4-dimensional space? Well, you can get pretty technical here or just realize that both spaces are infinitely dense and extend infinitely in all directions. The main difference is just the number of "directions" in which the space can extend infinitely.
The point, however, as that a 52-dimensional space
is not an infinite dimensional space. So any worry we might have about functions dealing with infinite dimensional spaces is needless.
So now we are back in good ol' 52-dimensional space. Specifically, a linear finite Hilbert space. What do we do with it? Well, if we're interested at understanding how to approach this space using vectors, we figure out what vectors in this space would look like. In
R2, a point is defined by two coordinates (x, y). The same is true here, only we have 52 coordinates. A vector in
R2 corresponds to some point (x, y) but it has a direction, or represents an increment. The differences in physics are usually pretty clear, but in mathematics in general it is rather arbitrary. The temperature of a room a some time
t is a point in 1D space. Same with the temperature of the room at some time
t + 1 hour. Both are points. However, the increment or change from the first to the second has a direction and magnitude (the temperature change is a certain amount in a certain direction). That gives us a vector.
Once again, it is pretty much exactly the same in 52D. A vector is nothing more than a list of 52 entries that correspond to a point in 52D space, but from some direction to that point. Some vectors, however, are more important than others. Because if I take the linear combination of certain 52-D vectors (for each vector, I can multiply it by a scalar, including 0), I can "hit" any point anywhere in the entire infinitely extending and infinitely dense 52D space. And if, by removing any one of these vectors, I can no longer do this, then I have basis vectors for that space.
The only real problem left is this german Eigen-whats it. THE equation in linear algebra is
Ax=b.
A is a matrix. However, a matrix is not just a group of numbers but a
function (more than that actually). Specifically, we call it a linear transformation that takes the vector
x and transforms it into
b. Both vectors, in 52D space, have 52 entries. The matrix
A must, therefore, have as many columns as the vector does rows (52). The equation we need to get both an eigenvalue and eigenvector is A
x=
λx. The reason this is a lot more important than it looks is because the linear transformation takes the vector
x and maps it onto some multiple of itself. The eigenvalue is the scalar that can do what the matrix
A does, and it corresponds to the eigenvector that is mapped to a multiple of itself. Things do get a little more complicated when we have an eigenvalue
i or -
i, because for every coordinate we have both a "real" and an "imaginary" or complex part, but as this doesn't actually change anything for a quantum shuffle, we don't have to worry about calculating something that can't be calculated.