Why does the inner produce in Hilbert space use fg* and not f*g?

[nomadreid]

The inner product is supposed to give the probability amplitude for state u turning into state v. Taking a little mini-universe of two dimensions in the complex plane if I rotate vector u by multiplying it by scalar a to get vector v, then I end up that a = u*v/(||u||.||v||); given that the lengths are irrelevant for quantum physics, the numerator is taken as the inner product at a point (with the dot product being its real part). But the inner product is defined as the integral of fg*, not f*g. Why?

Or did I calculate that wrong?

 

[mathman]

What is the meaning of fg*? f*g is the usual notation for product. The only caveat is that if the functions are complex, you need to use the complex conjugate of g.

 

[Matterwave]

<f|g> is an integral f*g isn't it...where are you getting fg*?

 

[nomadreid]

First, Matterwave: I am taking this from the standard definition for L^2 spaces; for example, in http://en.wikipedia.org/wiki/Lp_space, where it is the second term, not the first, which has the complex conjugate in the definition of the inner product.
Second, mathman: I was using the * to indicate the complex conjugate, not multiplication. So you repeat in words what I used this notation for. Yes, a better notation is with the line over it, $\int$f$\bar{g}$d$\mu$, so in the rest of this discussion I will use that.
So, back to my original question: to simplify, I will stick to unit vectors. if I have vectors $\vec{u}$ and $\vec{v}$,and a complex scalar a representing the transformation from $\vec{u}$ to $\vec{v}$, that is,a$\vec{u}$ = $\vec{v}$ , then being a bit sloppy and just using complex numbers for the vectors, a=v/u = $\bar{u}$v/u$\bar{u}$ =$\bar{u}$v. So I end up with $\bar{u}$v, not the desired u$\bar{v}$. What am I doing wrong?

 

[mathman]

First, Matterwave: I am taking this from the standard definition for L^2 spaces; for example, in http://en.wikipedia.org/wiki/Lp_space, where it is the second term, not the first, which has the complex conjugate in the definition of the inner product.
Second, mathman: I was using the * to indicate the complex conjugate, not multiplication. So you repeat in words what I used this notation for. Yes, a better notation is with the line over it, $\int$f$\bar{g}$d$\mu$, so in the rest of this discussion I will use that.
So, back to my original question: to simplify, I will stick to unit vectors. if I have vectors $\vec{u}$ and $\vec{v}$,and a complex scalar a representing the transformation from $\vec{u}$ to $\vec{v}$, that is,a$\vec{u}$ = $\vec{v}$ , then being a bit sloppy and just using complex numbers for the vectors, a=v/u = $\bar{u}$v/u$\bar{u}$ =$\bar{u}$v. So I end up with $\bar{u}$v, not the desired u$\bar{v}$. What am I doing wrong?

Your last two lines are a little confusing. At times you have arrows over the symbol, others have a bar (conjugate), and some have nothing. Finally, what is u/v? If these are vectors, it doesn't make sense. Just to clarify, by definition the inner product (u,v) is always the integral of a product, where the term on the right is the complex conjugate.

 

[nomadreid]

$\vec{}$When I have an arrow over the symbol, it is to indicate that I am regarding this as a vector. The letter alone is the complex number associated with this vector (when centered at the origin) (so u/v is the result of the corresponding division). The line over the symbol is the complex conjugate of the symbol. When I said that I was being a little sloppy, I just meant that I was skipping a couple of steps, not that the calculations are sloppy. Sorry about the notational confusion; I hope that clears it up.

Yes, I am aware of the fact that one kind of inner product is the integral of a product with the term on the right being the complex conjugate. (However, this is not the definition of an inner product, and there are other variations.) I think the basis of my problem may be the confusion between the notations (leaving out all the arrows etc.) <x,y> and <x|y>, and the fact that some places define the linearity in the first argument, and some in the second argument.

 

[mathman]

The letter alone is the complex number associated with this vector (when centered at the origin) (so u/v is the result of the corresponding division).

What is the complex number?. The only scalar associated with a vector is its magnitude, a non-negative real number.

 

[nomadreid]

In my first post I alluded to the fact that I was taking the example of the two-dimensional vector space (in which the inner product is just the dot product) formed by the set of complex numbers and the appropriate structures, over the field of complex numbers for my scalars. An explicit treatment starts with the equivalence classes of the vectors and the corresponding operations on these equivalence classes. For example, for any two complex numbers y and z, we can define the vector as the equivalence class of all ordered pairs of complex numbers (a,b) under the equivalence relation "(a,b)~(y,z) iff z-y = b-a". A typical representative of such a class is then (0, z-y). This vector can then be identified by this representative, which in its turn can can then be identified with the complex number w=z-y, taking care to distinguish when we are using the complex number w as that which we identify with a vector, and when we are using it as a scalar. Usually this is clear from the context. Addition and multiplication are then correspondingly defined on these classes (which can be shown to be sets). [This sort of thing is implicitly done in the real plane in every secondary mathematics and physics course when the vectors are represented by arrows. The isomorphism from the real to the complex plane then completes the intuition.] Every vector can be associated with a complex number. Now, with complex numbers being used both as scalars and as representatives of vectors, one must be careful with operations: as an example, if y and z are vectors, then y/z is meaningless. However, I can define a function t on the vector space to the complex numbers (as the universe of the underlying field) roughly as t(z) = "the scalar with the same complex number as is used to represent the vector z". Then, it is not hard to show that if I have a scalar k such that, for vectors y and z, y = kz, then k = t(y)/t(z). This boils down in practice to finding k by using the complex number representing y and dividing it by the complex number representing z. When one is careful, one can dispense with the notation such as [v] for the equivalence class, special symbols for the operations for operations between vectors or between vectors and scalars, t(z), and so forth, working with the complex numbers and keeping these distinctions in mind. So, in sum, by "the letter alone" I meant this t(z).

 

[mathman]

For example, for any two complex numbers y and z, we can define the vector as the equivalence class of all ordered pairs of complex numbers (a,b) under the equivalence relation "(a,b)~(y,z) iff z-y = b-a". A typical representative of such a class is then (0, z-y).

What is the vector? Your description doesn't look like anything I would call a vector. It is an equivalence class of number pairs.

 

[Fredrik]

Most math books define inner products and semi-inner products to be linear in the first variable and antilinear in the second. Most physics books define them to be linear in the second variable and antilinear in the first. So most physics books will define $$\langle f,g\rangle=\int f^*g d\mu=\int f(x)^*g(x)dx.$$ I don't understand your arguments. I think I understand what you're doing, but not why you're doing it. In particular, I don't understand what the definition of an inner product has to do with the question of what number u must be multiplied with to get the result v. (The answer is clearly v/u, or to put that differently, u*v/|u|²). I also don't understand why you want to define "vectors" as equivalence classes of pairs of complex numbers. ℂ is already a (1-dimensional) vector space over ℂ, so its members are vector and scalars at the same time. It looks like what you're aiming for is an affine space, not a vector space, but I don't think affine spaces will help you understand inner product spaces or semi-inner product spaces.

 

[nomadreid]

First, mathman: The set of equivalence sets of pairs of complex numbers with the operation of addition appropriately defined over the field of complex numbers with the operation of a scalar times an equivalence class appropriately defined satisfy the axioms for vectors (Closure, Associativity and commutativity of addition of vectors, identity and inverse elements for addition, distributivity of scalar multiplication over addition of vectors, and so forth). Indeed, this corresponds to the intuition of vectors as those things represented by arrows on the plane.

Secondly, Frederick. Thanks very much, I think you have put your finger on my difficulty with your characterization of who put the antilinearity first, and who second. I was trying to illustrate (not show: I was not trying to define vectors by my space, but just to take an example of a vector space) why it would make sense in physics to define the inner products with the linearity in the second variable (using the idea of the inner product as connected with converting one state to another) -- which you pointed out that it does, but I thought that it defined it with the linearity in the first variable because I was consulting a math book!

You’re also right that I would not want to take this into affine spaces, as this would not be relevant to my question.

Thanks to both of you.

 

[vanhees71]

Physicists don't use the mathematicians definition with the first argument of the scalar product linear and the second antilinear because this destroys Dirac's great bra-ket notation. I know that mathematicians don't buy such arguments, but once I got them when I took the functional analysis lecture as my math subject for the physics diploma examination. When it came to Hilbert space, I did the problems first in bra-ket notation, where most of them where just simple formal manipulations. Then I translated it to the mathematicians clumsy notation. That was way faster than using the mathematician's notation from the beginning.

The same holds for Ricci calculus with Einstein's summation convention. Mathematicians usually laugh about it and write everything "coordinate free". However, for pure calculational purposes the Ricci formalism is way more easy and straight forward to use than the coordinate and index free notation.