# Kets, vectors.

Dathascome
This might be a really stupid question but what is the signifigance of having a something like this

/r >

Which is supposed to be an r vector inside of a ket.

In the book I'm reading they use this notation when talking about the shrodinger equation, where r is the positon vector.

I'm sort of confused because at some point a big deal is made of making a distinction between kets/bras and the vectors (arrows) we all know and love. So the arrows are a certain subset of vecotrs in a vector space (R^3), and there's a relation between kets and vectors but I don't understand what it means to have a vector, like the position vector inside as a ket. Maybe I'm just thinking too much and making something out of nothing?

Last edited:

abercrombiems02
If what your talking about is like a little ^ right above the vector, it means that the vector is a unit vector. The elementary property of a unit vector is that its magnitude is 1 (or unity) thus the name "unit vector". I should also mention that unit vectors, themselves, are dimensionless. The physical quantity (or the scalar) that is associated with the unit vector, gives the vector its dimensions.

Last edited:
Dathascome
No it's not refering to a unit vector, just the position vector, normalized or not.

Last edited:
Staff Emeritus
Gold Member
A ket $| \alpha \rangle$ is nothing but a complex vector, or an element of the n-dimensional complex vector space $\mathbb {C}^n$.

Homework Helper
The ket $|\bold{r} \rangle$ is just a vector in your vector space. A ket IS a vector and a vector is a ket, same thing. But keep in mind they live in the state space or Hilbert space). Whatever you put inside the $| \rangle$ is just a label to distinguish between different kets. The label $\bold {r}$ is chosen because $| \bold{r} \rangle$ is the eigenket (or eigenvector) of the position operator $\vec R=\hat X \vec i + \hat Y \vec j +\hat Z \vec k$ corresponding to the eigenvalue $\bold r$ (yes, that's a vector, because $\vec R$ is a vector-operator. This vector DOES live in 3D geometric space).

If it helps, try to think of the 1D analog. The position operator $\hat X$ has eigenkets $|x \rangle$ corresponding to the eigenvalue x.

Last edited:
Dathascome
So then I have position vectors as the elements/components of the ket then right?

I think that the thing that confused me is that in the book I'm reading they talk about the matrix representation of kets/bras, where in they define the ket , /S>, as a nX1 column vector with it's elements/component given by the dot product <Pn/S> where /Pn> is the nth basis ket of /S>. Specifically I thought that this dot product should give me a scalar and not a vector, so it seemed sort of odd to me to have a ket with it's elements given by a vector, because I didn't see how you could get a vector out of that dot product.

Is this an understandable mistake or is there something that I'm truly missing?

Staff Emeritus
Gold Member
Is this an understandable mistake or is there something that I'm truly missing?

You've given "n" two different meanings.

The coordinate representation of |S> in the ordered basis {|Pk>} (k = 1 .. n) is the nx1 matrix whose k-th element is <Pk|S>

Dathascome
Right...my mistake...I'm at work and wasn't being careful As for my probelm though...?

Staff Emeritus