# About the first postulate

1. Sep 6, 2009

### snoopies622

I see how a vector in an infinite-dimensional space can be thought of as a function of one continuous variable in that both are a list of numbers (complex numbers in the case of quantum mechanics) that is infinitely long. But since wave functions in the position basis are functions of three dimensions (three continuous variables), how is all that information represented with only one vector?

Thanks.

(Edit) Afterthought: since momentum in three-dimensional space has three components, I guess this applies to the momentum basis too.

2. Sep 6, 2009

### atyy

Cheating: In 2D, discretize each space dimension into n pieces, then a region of space will have nXn indices, and the wavefunction is a an nXn matrix To make them in a row, just take all the first row of the matrix, then string the second row after it, then the third row ....

More correctly, a vectors of a vector space are just things that obey some formal rules like http://mathworld.wolfram.com/VectorSpace.html . In their notation X,Y,Z are vectors, but you will see that functions obey those rules just as well, and so are vectors by that definition. The one thing to note is that a scalar product is not part of the vector space definition, and is considered an additional structure. For functions, the scalar product is the overlap integral of the two functions (for complex functions, there's complex conjugation somewhere too, but the basic idea is the same).

3. Sep 6, 2009

### atyy

Oh yes, the more correct thing I said above is actually still cheating. If you hunt around these forums, you can see what George Jones says about rigged Hilbert spaces or something very very correct.

4. Sep 7, 2009

### snoopies622

Thanks atyy. I forgot that there was a more general definition of vectors and vector spaces of which the familiar $R^n$ Euclidean kind was only one class. (And I took two linear algebra classes in college! Of course that was a some time ago..)

Wow, that's neat!