# What is a general scalar and what a (complex conjugate)

I've been reading something about Quantum Mechanics where they introduce the maths slightly more rigorously. They talk about vector spaces and an inner product which yields a scalar. Moreover complex conjugation appears.

Of course I know about complex numbers, but is there a more general framework which defines more generally what a scalar means and also what conjugation (and inner product) means? Maybe some more general algebra which also satisfies some minimum axioms?

In the context of vector spaces, a scalar is a member of the underlying field.

An inner product is a special kind of bilinear form on a vector space over the reals or complexes. Inner products satisfy conjugate symmetry, which over the reals is just plain symmetry.

• I guess I have to look up more closely what field and vector space means. Can scalars be something else than reals or complex numbers? – Gerenuk Feb 15 '12 at 12:37
• – Paul Slevin Feb 15 '12 at 12:41
• I'll go through these, but I'm not an expert to see the general picture confidently enough and not make premature judgements. So can scalars in vector spaces be something different than real numbers or complex numbers? – Gerenuk Feb 15 '12 at 14:58
• @Gerenuk, sure. The reals are a vector space over the rationals and so are polynomials with rational coefficients and the set $\mathbb Q(\sqrt 2)=\{ a + b \sqrt 2 : a,b \in\mathbb Q\} \subset \mathbb R$. – lhf Feb 15 '12 at 15:03
• Thanks! Hmm, not sure if I understand it right. I basically wonder if the result of the inner product of two vectors can be something different than a real number or a complex number. You are saying they can? And if such algebra exists, how would you define the general conjugation neccessary for the definition or inner product? – Gerenuk Feb 15 '12 at 15:25

To nit-pick on your question, scalars could be anything that you can define a field with. As long as you also define addition, subtraction, commutative multiplication, and division without zero, your scalars might be rabbits, or anything. Of course, this is completely equivalent to choosing numbers instead.

The concept of a "number" extends to hypercomplex numbers like quaternions or octonions. However, their multiplication is not commutative, and hence they do not form a field. A vector space requires an underlying field. Multiplication of integer, rational, real, or complex numbers satisfy commutativity. This, among other requirements, makes them candidates for fields.

Refer to the Wikipedia article on fields to get an impression of more possibilities, most notably finite fields.

In the context of modules (a generalized notion of vector spaces), a scalar is not required to form a field, but only a ring instead, i.e., division need not be defined. This includes quaternions as scalars, in this context (!). Considering the fact that you delve in quantum mechanics, you might want to extend these notions to octonions as well. Many arithmetic and algebraic lemmata also apply to these.

• For completion, some (as far as I know, French) authors develop the theory of vector fields in the generality of division rings (not necessairly commutative). – user228113 May 18 '15 at 11:37
• Yes, a division ring (or skew field) is a special case of a ring,, namely one with division. The ring of quaternions is an example of a non-commutative division ring. – Douba May 18 '15 at 13:26
• I actually thought I had deleted that comment... Well, I'm sorry, you have a point. – user228113 May 18 '15 at 13:37