## Monday, August 26, 2024

### Vectors+ 10 Complex Hilbert Space: UNIZOR.COM - Math+ & Problems - Vectors

Notes to a video lecture on http://www.unizor.com

Vectors+ 10
Complex Hilbert Spaces

As we know, a Hilbert space consists of
(a) an abstract vector space V with operation of addition
v1,v2V: v1+v2V
(b) a scalar space S of numbers (we consider only two cases: S=ℝ - a set of all real numbers or S=ℂ - a set of all complex numbers) with operation of multiplication of a scalar from this set S by a vector from vector space V
λS and ∀vV: λ·vV
(c) an operation of a scalar product of two vectors from V denoted as ⟨v1,v2⟩, resulting in a scalar from S
v1,v2V: ⟨v1,v2⟩∈S

Operations of addition of two vectors, resulting in a vector, multiplication of a vector by a scalar, resulting in a vector, and scalar product of two vectors, resulting in a scalar, must satisfy certain set of axioms presented in lecture Vectors 08 of this part of a course, including commutative, associative and distributive laws.

In all the previously considered cases our scalar space S was a set of all real numbers .
Now we will consider a set of all complex numbers as a scalar space S and examine the validity of our axioms as applied to a simple vector space.
By all accounts this should not present any problems since arithmetic operations with complex numbers are similar to those with real numbers.
There is, however, one complication.

One of the axioms of a scalar product was:
For any vector a from a vector space V, which is not a null-vector, its scalar product with itself is a positive real number
aV, a≠0: ⟨a,a > 0
This axiom is needed to introduce a concept of a length (or a magnitude, or a norm) of any vector as
||a|| = √⟨a,a⟩

Consider now a simple case of a vector space V - a set of all complex numbers with a scalar space S being a set of complex numbers as well.
We will use the word vector as an element of a vector space V and the word scalar as an element of a scalar space S, but both spaces are sets of all complex numbers .

Addition of two vectors (complex numbers) and multiplication of a vector (a complex number) by a scalar (also a complex number) are defined, as regular operations with complex numbers.
Scalar product of two vectors (two complex numbers) is defined as their regular operation of multiplication.
Let's check how valid our definition of a length of a vector is in this case with all the axioms previously introduced for a set of all real numbers being the scalar space S.

Vector a from V (a complex number) can be expressed as a1+i·a2, where a1 and a2 are also vectors from V, but belong to a subset of only real numbers, while i is an imaginary unit −1, an element of a scalar space S (also a set of all complex numbers).

Let's calculate a scalar product of vector a with itself using the old laws of operations previously defined
a1+i·a2,a1+i·a2 =
[since a1 and a2 are real numbers]
= a1²+2
i·a1·a2+·a2² =
= a1²+2
i·a1·a2−a2²

The problem is, this is not a positive number since it contains an imaginary part.
So, the set of axioms we introduced before for a scalar space being a set of all real numbers is contradictory in case of complex numbers as a scalar space for a simple case above.

To overcome this problem, we have to revise our axioms in such a way that they will hold for examples like above for complex scalars, while being held as well for real numbers as scalars, since real numbers are a subset of complex ones.

There are two small changes in the axioms for a scalar product that we introduce to solve this problem.
Let a and b be two vectors and λ - a complex number as a scalar.
(1) λ·a,b = λ·a,b = a,λ·b
(2) ⟨a,b = b,a
where horizontal bar above a complex number means complex conjugate number, that is a number x−i·y for a number x+i·y and a number x+i·y for a number x−i·y.
Incidentally, a conjugate to imaginary unit i (that is, 0+i·1) is −i (that is, 0−i·1).

First of all, if a complex number x+i·y is, actually, a real one (that is, if y=0), its conjugate number is the same as the original. So, in case of a scalar space being a set of all real numbers these axioms are exactly the same as the old previously accepted ones.

Secondly, for complex scalars these modified axioms solve the problem in a simple case of V=ℂ and S=ℂ mentioned before for a vector a=a1+i·a2, where a1 and a2 are from a subset of only real numbers.

Let's calculate a scalar product using the modified rules.
a1+i·a2,a1+i·a2 =
[using distributive law]
=
a1,a1 + a1,i·a2 +
+
i·a2,a1 + i·a2,i·a2 =
= a1² +
a1,i·a2 +
+
i·a2,a1 + a2·−i·i·a2 =
[since −i²=1 and addition is commutative]
= a1² + a2² +
a1,i·a2 + i·a2,a1 =
[using a newly modified rule]
λ·a·b⟩ = ⟨λ·a,b⟩ = ⟨a,λ·b⟩]
= a1² + a2² +
i·a1,a2 + i·a2,a1 =
= a1² + a2² +
(−i)·
a1,a2 + i·a2,a1 =
[since a1 and a2 are vectors from a subset of real numbers, their scalar product is commutative, and the last two terms cancel each other]
= a1² + a2²

which is a positive real number for any non-zero complex number.

This expression fully corresponds to a concept of an absolute value of a complex number a1+i·a2 and to a concept of a length of a vector in two-dimensional Euclidean space with abscissa a1 and ordinate a2 used as graphical representation of a complex number a1+i·a2.

Moreover, similar calculations can be applied to N-vector of complex numbers as a vector space with a set of all complex numbers as a scalar space.
Indeed, consider for brevity a case of N=2 since the more general case of any N is totally analogous.
The element of our two-dimensional vector space is a pair {a,b}, where both components are complex numbers.
Its scalar product with itself is defined as
⟨{a,b},{a,b}⟩ = (a·a) + (b·b)
and each term on the right of this equation, as we determined, is a positive real number.

As we see, a small modification of the axioms of scalar product in case of complex scalars solves the problem and, at the same time, does not change anything we knew about scalar product with real scalars.

As we did before, we can say that in case of a complex scalar space a scalar product of a non-null vector by itself is positive and the square root of it is its length or magnitude or norm
||a||² = a,a

Problem A
Prove the Cauchy-Schwarz-Bunyakovsky inequality for an abstract vector space V and a complex scalar space S=ℂ
a,bV: |a·b|² ≤ a·a·b·b
where absolute value of any complex number Z=X+i·Y is defined as
|Z|² = X² + Y² = Z·Z

Proof A
Consider any non-zero scalar (complex number) x and non-negative (by axiom) scalar product of vector a+xb by itself
0 ≤ a+x·b,a+x·b
Use distributive properties to open all parenthesis
0a,a+a,x·b+x·b,a+x·b,x·b
Using the modified commutative rules introduced in this lecture, we can transform individual members of this expression as follows
a,x·b = x·a,b
x·b,a = x·b,a
x·b,x·b = x·b,b

Now our inequality looks like
0 x·b,b+x·a,b+x·b,a+a,a

This inequality is true for any complex number x.
If vector b equals a null-vector, the inequality is held because b,b=0 and ⟨a,b=0 (see Problem A of lecture Vectors 08 of this part of a course).
Assume, vector b is not a null-vector, and let's see how our inequality looks for specific value of x:
x = −a,b/b,b

Let's evaluate each member of the inequality above.
x·b,b = a,b·a,b/b,b
x·a,b = −a,b·a,b/b,b
x·a·b = −a,b·a,b/b,b

Putting these values into our inequality and cancelling plus and minus of the same numbers, we obtain
0 ≤ a,b·a,b/b,b + a,a
Multiplying by positive ⟨b,b⟩ and separating members into different sides of an inequality, we obtain
a,b·a,ba,a·b,b
or, since |Z|²=Z·Z for any complex Z,
|a,b|² ≤ a,a·b,b
End of proof.