# I Kets and vectors

1. Jul 16, 2016

### dyn

Hi. I have read that when working in 3-D the following kets | x > and | p > are not vectors in 3-D. If that is correct what are they ? I know | ψ > is an abstract vector but I thought | x > and | p > would be 3-D vectors in the position and momentum representation ?
Thanks

2. Jul 16, 2016

### andrewkirk

They are vectors in an infinite-dimensional space. It is natural to feel that $|x\rangle$ should be in a 3d space because $x$, without the enclosing symbols, is a coordinate in 3D space. But $|x\rangle$ is not a coordinate or any other type of number. In the position representation it is the Dirac delta function $\delta_x$, which is an element of a space of functions, which - like most function spaces - is infinite-dimensional.

3. Aug 3, 2016

### t3rm1

<a is conjugate of a>
<a|b> kind of dot(a.conjugate , b)
b><b|a> projection into other direction like b*dot(b.conjugate , a)

v2=pol><pol|v> is equivalent with this pseudo code
amp=dot(v,pol)
v2.x = pol.x*amp
v2.y = pol.y*amp

conjugate of V is V.imaginary=-V.imaginary

4. Aug 3, 2016

### t3rm1

The Pauli matrices transform the 3d direction into 2d complex vector like spin.

5. Aug 3, 2016

### t3rm1

Why do we need conjugation?
because there is a minus sign in real part of multiplication of complex numbers.

complex operator *(complex c)
{
complex e;
e.real = this->real*c.real - this->img*c.img;
e.img = this->img*c.real + this->real*c.img;
return e;
}

6. Aug 3, 2016

### A. Neumaier

They are distributions, elements in the dual space of a Schwartz space of smooth functions. Only the labels are vectors in 3D.