Linearity of a Hilbert space function and complex numbers.

• Apr 30th 2011, 06:44 PM
topsquark
Linearity of a Hilbert space function and complex numbers.
I recently rediscovered a text written by one of my professors from Purdue. It has a section on vector spaces. Unfortunately the text clarifies as much as it obfuscates.

So here is my question. Say we have two Hilbert spaces S and S', and a function $\Omega : S \to S'$. $\Omega$ is linear if
$\Omega ( \phi + \psi ) = \Omega \phi +' \Omega \psi$

and
$\Omega ( a \psi ) = a \Omega \psi$

where all $\phi$ and $\psi$ in S and all complex numbers a.

Now a Hilbert space is a metric vector space, right? So $\phi$ and $\psi$ are vectors in S, + is the binary operation of addition as defined in S, and +' is the binary operation of addition in S'. But why is the scalar field defined as the complex numbers? Or is this another one of the text's "simplifications" for Physicists? (I can't think of a problem in Physics where the scalar field is not the complex numbers, but that doesn't mean it must be in general.)

-Dan

Edit: He also defines the metric in terms of complex numbers as well if that makes any difference:
$d: S \times S \to \mathbb{C}$

Now that I think of it, isn't a metric function defined in terms of a real number?
• Apr 30th 2011, 08:50 PM
Deveno
are you sure that d:S x S --> C? just saying, the complex field comes with a "natural metric" defined in terms of complex numbers, but which in fact is always real-valued:

d(z,w) = |z - w|. i am suspecting that instead of a metric, what is actually being defined is an inner product, which IS a function from SxS to C.

given an inner product, one can use the inner product to define a norm by |φ| = √(<φ|φ>) (or, in more standard mathematical notation, |φ| = √<φ,φ>).

a metric is then defined in terms of the norm.

Hilbert spaces can be over the reals or the complex field, but the field must be (topologically) complete. taking the field to be the complex field

is "more general" because most of the differences between the two cease to exist when the scalar is taken to be real. for example,

the complex inner product is sesquilinear, but the difference between that and bilinear disappears when the imaginary part is 0.

inner product spaces can only be defined over a field with an ordered subfield. this automatically forces the field to be of characteristic 0.

to define the norm as a square root, the field must be quadratically closed (every number needs to be a square).

Hilbert spaces require that Cauchy sequences of vectors converge, which means that the field has to be complete, and therefore has to contain

a complete ordered field, that is, the real numbers. so you have the demands of the algebra on one hand, and the demands of the topology

on the other. the complex field is particularly well-suited to these demands.

it is possible to have metric vector spaces over other fields.
• May 1st 2011, 03:32 AM
topsquark
Quote:

Originally Posted by Deveno
are you sure that d:S x S --> C? just saying, the complex field comes with a "natural metric" defined in terms of complex numbers, but which in fact is always real-valued:

d(z,w) = |z - w|. i am suspecting that instead of a metric, what is actually being defined is an inner product, which IS a function from SxS to C.

I'm sorry. I plead need for sleep. What my tired brain was trying to come up with was that the inner product is complex. I don't know why I got thinking that the metric was complex. I looked it up this morning and he defined it correctly.

As to the rest of it, more than I had anticipated for an answer, but as always interesting reading. :) Apparently I wasn't specific enough in my question. The origin of my question about the field is that an F vector space has, as scalars, a field F. Since scalar multiplication involves F then in the following
$\Omega (a \psi ) = a \Omega( \psi )$

a will be a member of F. Since this (as far as I know) can be defined on any F vector space, isn't F as the complex numbers only a specific case? Or is there something (perhaps the added extra structure of the inner product) that forces the field to be real or complex?

So you mentioned that, for a Hilbert space the field of a Hilbert space defined will be either real or complex. Is this in general? If so I missed that in the definition.

-Dan
• May 1st 2011, 05:41 AM
Deveno
it's true in general for inner product spaces (or pre-Hilbert spaces). it's possible to define the notion of an inner product space on quadratically closed fields (fields with all square roots), such as the algebraic numbers or the hyperreal numbers, but the positive-definite nature of an inner product limits the kinds of fields you can consider.

this is the property: for v ≠ 0, <v,v> > 0. for that to even make sense, you have to have a field in which ">" is defined. this needs only to be a subfield of the underlying field (in the case of a complex field <v,v> is always real, because you wind up taking (v1*)v1 + (v2*)v2 +....+ (vn*)vn, where vj* is the complex conjugate of vj, and (vj*)vj is real (some texts make the second term the conjugate, this is one of those arbitrary left-right sort of choices one finds in mathematics).

more generally, you can consider hermitian or quadratic forms, which may be positive-definite, semi-positive-definite, or simply non-degenerate. the nature of the underlying field greatly affects the theory (one can see that fields of characteristic 2 will be a special case, quadratic equations in a field of characteristic 2 are unlike such equations in other fields...there is no "completing the square").

one can also generalize inner product spaces to normed, or semi-normed spaces (every inner product induces a norm, but not necessarily vice-versa), again, certain inequalities that define a semi-norm require the field be of characteristic zero, and contain an ordered subfield, so one restricts one's attention to vector spaces over subfields of the complex numbers.

so, short answer: yes, it is the inner product structure that limits what we can choose as an underlying field. in particular, we want the cauchy-schwartz inequality (and thus the triangle inquality) to make sense. this is the basis of most analytic arguments involving continuity based on the metric induced by the inner product. in a vector space, we have vector sums, and "bounding" a function near some vector v, usually involves "bounding" some vector sum, and repeated use of the triangle inequality. inner product spaces (or more generally normed spaces) let us re-use the knowledge hard-gained in ordinary calculus in much more abstract settings. all that work involved in defining lub's and glb's doesn't go to waste ;)
• May 1st 2011, 10:36 AM
topsquark
Thank you. (Nod)

-Dan