Understanding the Significance of Kets and Vectors in the Schrodinger Equation

[Dathascome]

This might be a really stupid question but what is the significance 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?

 

[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.

 

[Dathascome]

No it's not referring to a unit vector, just the position vector, normalized or not.

 

[Gokul43201]

A ket | \alpha \rangle is nothing but a complex vector, or an element of the n-dimensional complex vector space \mathbb {C}^n.

 

[Galileo]

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.

 

[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?

 

[Hurkyl]

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...?

 

[Hurkyl]

How many <Pk|S>'s are there?

BTW, they're components of the coordinate representation of the ket, not the ket itself. (The phrase "components of the ket" might not even make sense!)

 

[Dathascome]

Sometimes I find these little details to be the hardest part of learning this sort of stuff. I know that they are important and saying the wrong thing can lead to thinking the wrong thing, but the terminology and phrasing is hard to get down sometimes.