# What does $|x⟩|0⟩$ actually mean in bra-ket notation?

Consider the following quote from Wikipedia's page on Shor's algorithm:

Initialize the registers to

$Q^{-1/2} \sum_{x=0}^{Q-1} \left|x\right\rangle \left|0\right\rangle$

where $x$ runs from 0 to $Q-1$. This initial state is a superposition of $Q$ states.

I am having a hard time getting what $\left|x\right\rangle \left|0\right\rangle$ actually is. Is it outer product of ket $x$ and ket $0$? Or what does it actually mean?

Juxtaposing two kets is, except on rare occasions, a shorthand for the tensor product (try also this article for a gentler introduction) of the two vectors, which would be written in full as $|x\rangle\otimes|0\rangle$.
• In this particular context it maybe worth mentioning that $| x \rangle$ is itself a tensor product of qubit kets corresponding to the binary representation of $x$. – Slaviks Mar 9 '15 at 18:06