# Inner Product for Vectors in Complex Space

1. Oct 18, 2006

### Hacky

Very basic question but could someone briefly explain why the inner product for complex vector space involves the conjugate of the second vector. Of course if imaginary component is 0 then this reduces to dot product in real vector space. And I see that this definition makes sense to calculate "length" so that it is not a negative number. But is there another geometrical (using cosine?) or intuitively logical reason why the inner product is defined this way?

Thanks, Howard

2. Oct 18, 2006

### matt grime

There is a priori no reason why you have to use the conjugate at all - you are free to make whatever definitions you want and see what happens. Evidently not using the conjugate isn't very useful, or we wouldn't have this definition instead. Someone must have looked at the real dot product and decided that the useful properties are things like

it is always a real number greater than or equal to zero
x.x=0 implies x=0,

and so on are what we want to use. In particular, without the conjuagation, the vector (1,i) is orthogonal to itself in C^2.

3. Mar 6, 2008

### fakrudeen

If you don't use conjugate but simple symmetry it will produce a contradiction with the axiom X . X >= 0

assume X.Y = Y. X

X.X >=0

=> c1 X . c1 X >= 0 [where c1 is a complex number with an imaginary part]

=> c1 (c1 X . X) > = 0 by linearity

=> c1 (X . c1 X) > = 0 by symmetry

=> c1 . (c1 ( X. X) )>= 0 by linearity

=> (c1 . c1) . (X.X) > = 0 by associativity of complex numbers

to see this, if you take c1 = i [square root of -1]

=> -1 (X.X) >= 0

=> (X.X) <= 0

4. Mar 6, 2008

### OrderOfThings

An interesting property of a complex (hermitian) inner product is that it does not depend on the absolute phases of the complex vectors. You get

$$<e^{i\alpha}\mathbf{z}, e^{i\alpha}\mathbf{w}> \, = \, <\mathbf{z},\mathbf{w}>$$

5. Mar 9, 2008

### Peeter

One explaination for this and many other basic complex number properties, can be had, if one considers complex numbers to be a special representation of two dimensional vectors.

In a nutshell one can consider a complex number to be a vector with one of the unit vectors factored out, as in the following example:

$$r = x e_1 + y e_2 = e_1 (x + e_1 e_2 y) e_1 = e_1 Z$$

(one could left or right factor out either of the $$e_j$$ unit vectors resulting in similar "complex number" representations).

Now that may look a bit strange at first since we are taught that we can't multiply or divide vectors, but these can be considered perfectly well defined operations with respect to the geometric (clifford) vector product. In a nutshell, that product defines the square of a vector as a scalar with magnitude equal to its squared length. A consequence of this rule, plus associativity and linearity, is the product of two perpendicular vectors is an entity that changes sign with commutation. With these as rules for multiplication, squaring $$e_1 e_2$$ gives:

$$(e_1 e_2)(e_1 e_2) = e_1 (e_2 e_1) e_2 = - e_1 e_1 e_2 e_2 = -1$$

ie: a product $$e_1 e_2$$ operates as a unit imaginary. All the complex number operations (multiplication, conjgation, inversion, ...) have natural equivalents expressed in this fashion, including the inner product of your question. I was quite excited seeing this for the first time since the complex multiplication "rule" (like the cross product "rule" also better formulated as a geometric product) always seemed like something that had been pulled out of a magic hat as opposed to something that followed from a logical argument.

Now, this probably doesn't precisely answer your question, instead should point you down a path that can, with some study, provide a good answer if you want it. One possible starting place is lect1.pdf from the following:

http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraLectures.zip

6. Mar 25, 2008

### Eidos

Thanks alot Peeter! The explanation was very enlightening. That's also a fantastic link and exactly what Ive been looking for, for a while now. :D

7. Mar 25, 2008

### Peeter

You are welcome! The approach to vectors and geometry seems very worthwhile to me (I am studying in my spare time with the aim of understanding my old engineering texts on E&M better than I did in school ... wish I was back at studying physics with these tools instead of my software job;)

8. Mar 26, 2008

### Eidos

I'm also learning this in my spare time since I'd like to apply it to control theory (which is already happening). Im certain that this will be invaluable as a tool for people in control engineering.

A book I can recommend, just in terms of its scope, is "The Road to Reality" by Roger Penrose. Its quite a tome but well worth the money, since it covers all of the maths needed to understand the physics, in the first half of the book. Pretty much an entire physics course (from classical to post modern) written for the layman. Its great just as plain sense intro into some of the more advanced stuff out there. :)

9. Nov 6, 2010

### sagysrael

hello
i haven't read all the replies, but i'll try to answer the original question geometrically:
a norm is meant to define "length" of a vector. in the complex plane it's the distance from the origin.
now a vector can be defined by it's length and angle (polar system)
now a property of vector multiplication is: if x,y are vectors, then:
the vector (x*y) has length = length of x * length of y, and angle = angle of x + angle of y
now a vector has the inverse angle of its conjugate but the same length.
so when you multiply x with conj(x) you get a vector with angle 0 and length |x|^2. so basically this vector lies on the x-axis and is just a real number, so if you take its square root you'll get the original length of the vector x.

hope this helped
sagy

10. Dec 19, 2011

### dimension10

So that

$$\theta=\arccos\frac{\Re (\langle a,b \rangle)}{\left|\left| a \right|\right|\;\;\left|\left| b \right|\right|}$$

11. Dec 20, 2011

### micromass

Staff Emeritus
This post is 5 years old.