# Given a unitary $U_p:|0\rangle\to\sum_\omega\sqrt{P(\omega)}|\omega\rangle$, what does $|0\rangle$ represent exactly?

Consider a random variable $$X$$ on a probability space $$(\Omega, 2^\Omega, P)$$. Let $$H_\Omega$$ be a Hilbert space with basis states $${| \omega \rangle}_{\omega \in \Omega}$$, and fix a unitary $$U_P$$ acting on $$H_{\Omega}$$ such that $$U_p:| 0 \rangle \rightarrow \sum_{\omega \in \Omega}\sqrt{P(\omega)}| \omega \rangle$$,

assuming $$0 \in \Omega$$. We define a quantum experiment as the process of applying the unitary $$U_P$$ or its inverse $$U_P^{-1}$$ on any state in $$H_\Omega$$.

My question is what $$| 0 \rangle$$ exactly represents. Does it mean that we fix some basis state in the Hilbert space $$H_{\Omega}$$ from which we generate a superposition of all basis states. Is the choice of this basis state $$| 0 \rangle$$ arbitary? Could it be done for every $$| \omega \rangle \in H_{\Omega}$$ ?

• It is just initial state. Of course, depending on number of elements in $\Omega$ (lets say $N$) the initial state will be composed of $\log_2(N)$ qubits in state $|0\rangle$. This initial state is one of basis states. Commented Jun 14, 2023 at 20:56
• @MartinVesely: Is this some kind of basis state that represent for example a low energy level? So the initial state is given by $| 0 \rangle = \sum_{\omega \in \Omega } a_{\omega} |\omega \rangle$ Commented Jun 14, 2023 at 21:03
• I think that you can assing zero bit string $|0...0\rangle$ to any member of $\Omega$. Technically, this state does not have to be lowest energy state. It depends on your definition of a qubit in particular technology. However, according to Di Vincenzo criteria (en.wikipedia.org/wiki/DiVincenzo%27s_criteria), you must have one precisely defined state which is assumed to be an initial state of a quantum computer. Note that almost any quantum algorithm assumes that initially, the computer is in state $|0...0\rangle$. Commented Jun 15, 2023 at 17:39
• Thank you very much:) Commented Jun 17, 2023 at 17:38