I Qbit pure vs mixed state space

[gentzen]

I'm still confused about the meaning of a writing like this: $$\ket{0}$$
Is it an actual vector in qbit's 2-dimensional Hilbert space

Yes, it is.

or is it the "column vector" of the coordinates/components in a given Hilbert space basis ?

If you want a density matrix, then it is easiest to see it as a column vector in a specific basis.

Otherwise, you should imagine a density operator. This is not difficult either, it is just a different picture. In that case, you "imagine" how it applies/operates on a given vector. For example, if you apply the operator $\ket{\psi}\bra{\psi}$ to the vector $\ket{x}$, you get $\ket{\psi}\bra{\psi}\ \ket{x}=\ket{\psi}\ (\langle\psi|x\rangle)=(\langle\psi|x\rangle)\ \ket{\psi}$.

 

[cianfa72]

For example, if you apply the operator $\ket{\psi}\bra{\psi}$ to the vector $\ket{x}$, you get $\ket{\psi}\bra{\psi}\ \ket{x}=\ket{\psi}\ (\langle\psi|x\rangle)=(\langle\psi|x\rangle)\ \ket{\psi}$.

Ah yes, as I said before in this thread $\ket{\psi}\bra{\psi}$ should be actually a dyadic tensor product.

 

[gentzen]

Sorry, then $\ket{\psi}\bra{\psi}$ should be actually the (dyadic) product tensor of the ket $\ket{\psi}$ times the dual-vector (bra) $\bra{\psi}$ ?

Edit: it is a dyad according to Wikipedia (tensor of order two and rank 1).

If the "dyad picture" works nice for you, then you can think like that. However, the typical contexts where the "dyad picture" arises is in formulas from classical electrodynamics that are written with explicit scalar products (and vector products). In this context, the "dyad picture" is a convenient way to generalize the behavior of scalar products, such that a "tensor like" value can be put into a scalar product, and you get out a vector, instead of a scalar.

 

[PeterDonis]

Is it an actual vector in qbit's 2-dimensional Hilbert space or is it the "column vector" of the coordinates/components in a given Hilbert space basis ?

You seem to be confused about what the thing you are calling a "column vector" actually is. It is a representation of the vector itself in a particular basis. The vector itself is an element of a vector space, which is an abstract space that exists and has certain properties independent of any particular choice of representation/basis. If you choose a basis, then you can write the vector as a column vector in that basis.

A ket like $\ket{0}$ can represent either the abstract vector itself, or the column vector in a specific basis. It depends on how you want to view the label $0$ that you put on the ket. Is that label referring to the particular abstract vector that you have in mind? Or is it referring to the column vector that represents that abstract vector in the particular basis in which that vector is an eigenvector whose eigenvalue corresponds to the label $0$? It could be either; the only person who knows which is the person who wrote down the ket, which in this case is you.

 

[cianfa72]

Or is it referring to the column vector that represents that abstract vector in the particular basis in which that vector is an eigenvector whose eigenvalue corresponds to the label $0$?

A vector is an eigenvector for an operator regardless of the picked basis. Therefore I believe what you meant is that if $0$ was understood as the label for an eigenvalue, then ket $\ket{0}$ would be the "column vector" representing the associated eigenvector in the basis in which an element is that eigenvector. That means it could be for example in the form $$\ket{0}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

 

[PeterDonis]

A vector is an eigenvector for an operator regardless of the picked basis.

Yes, but it is one of the basis vectors in only one basis.

it could be for example in the form $$\ket{0}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Yes, it would be a basis vector in this basis, but not in any other basis.