In quantum computation, a common operation performed between two quantum states is the tensor product, which allows us to create a new and higher-dimensional state from two lower-dimensional states. The tensor product is usually denoted by the symbol $\otimes$. So, if $\lvert \psi\rangle = \alpha \lvert 0\rangle + \beta \lvert 1\rangle \in \mathbb{C}^n$ and $\lvert \phi\rangle = \gamma \lvert 0\rangle + \delta \lvert 1\rangle \in \mathbb{C}^n$ are the states of two qubits, then $\lvert \psi\rangle \otimes \lvert \phi\rangle \in \mathbb{C}^{n^2}$ is their tensor product.

I have just come across the notation $\lvert \mathbf{x}, 0\rangle$ while reading section $1$ of this paper. What does it mean? I understood $\mathbf{x}$ is a (classical) bit string of length $n$ (and I think this the reason it's in bold).

I am aware of the fact that the tensor product of two vectors $\lvert \psi\rangle$ and $\lvert \phi\rangle$ can be shortened as follows $\lvert \psi\rangle \lvert \phi\rangle$. So, I don't think $\lvert \mathbf{x}, 0\rangle$ is also a shorthand for $\lvert \mathbf{x} \rangle \otimes\lvert 0\rangle$.


1 Answer 1


Yes, $|\mathbf{x},0\rangle$ is a shorthand for $|\mathbf x\rangle\otimes |0\rangle$.

Note that $|\mathbf x\rangle$ itself, with $\mathbf x = x_1x_2\dots x_N$ a bit string, is just a shorthand for $$ |\mathbf x\rangle \equiv |x_1\rangle \otimes |x_2\rangle \otimes\cdots |x_N\rangle\ . $$


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