So, I've read that $|0\rangle$ does not mean the zero vector, that is just represented by $0$. So what does $|0\rangle$ and $\langle 0|$ "equal" in standard matrix notation? In particular, are there some general rules I can follow for the translation of bra-ket notation into standard matrix notation? I come up against problems like $\langle 0|0\rangle$ and I understand what is happening intuitively, but I have no idea how to practically carry it out (for what it's worth, I'm used to normal matrix notation and can solve problems in it).
Thanks!
EDIT:
Perhaps I didn't make my question clear. I know that $\langle 0|0\rangle$ represents an inner product or dot product, that $\langle 0|$ represents a row vector, and that $|0\rangle$ represents a column vector; this is what I meant by understanding it intuitively. However, given the problem $\langle 0|0\rangle$ I don't know what the answer is, because I don't know how to work it out; I'm used to standard matrix notation.
Best Answer
The numbers in notations like $|n\rangle$ are the analogues of indices in matrix notation. That is, $|0\rangle=e_0$, $|1\rangle=e_1$, etc., where $e_n$ is the vector which has a $1$ in the $n$th position and $0$ in the other entries. Unfortunately, this notation is unspecific about the dimension of the base space. For qubits in quantum computers, the dimension is $2$, so we only have $|0\rangle=e_0=(1,0)$ and $|1\rangle=e_1=(0,1)$. It is also common to have a countable infinity of basis vectors, so we get $|n\rangle$ for each $n\in\Bbb N$. In quantum mechanics one also deals with this notation for larger dimensional spaces; for example we may have $|x\rangle$ for each $x\in\Bbb R^3$ (the position basis), which is a vector space of uncountable dimension $|\Bbb R^3|=2^{\aleph_0}$.
In any case, these vectors are usually enumerating a basis of some kind, and the details beyond that depend on the context.
The notation $\langle 0|0\rangle$ is written in linear algebra notation as $e_0^Te_0$, which is a $1\times 1$ matrix whose value can be identified with the dot product $e_0\cdot e_0$. Provided that the vector is normalized, this will always be $1$. So a general answer is $\langle m|n\rangle=0$ if $m\ne n$, and $\langle n|n\rangle=1$, which expresses that the vectors $(|n\rangle)_{n\in\Bbb N}$ are an orthonormal basis for the space.
For some general rules, then, we have $|n\rangle=e_n$ and $\langle n|=e_n^T$ (or $e_n^\dagger$ in complex vector spaces), where we understand the first as a $d\times 1$ matrix so that the second is $1\times d$, where $d$ is the dimension of the space. Then the inner product is $\langle m|n\rangle=e_m^Te_n=e_m\cdot e_n$, and the outer product is $|n\rangle\langle m|=e_ne_m^T$, which is a $d\times d$ matrix with a single $1$ at the index $(n,m)$. Note that these notations are also used for arbitrary vectors; for example we might write $|\psi\rangle=v$ for some vector $v$, and then $\langle\psi|=v^T$, $\langle\psi|\psi\rangle=\|v\|^2$, and $|\psi\rangle\langle\psi|$ is the projection matrix in the direction of $v$.