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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 $| \omega \rangle \in H_{\Omega}$ ?

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    $\begingroup$ 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. $\endgroup$ Commented Jun 14, 2023 at 20:56
  • $\begingroup$ @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 $ $\endgroup$
    – Simon
    Commented Jun 14, 2023 at 21:03
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    $\begingroup$ 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 $|0...0\rangle$. $\endgroup$ Commented Jun 15, 2023 at 17:39
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    $\begingroup$ Thank you very much:) $\endgroup$
    – Simon
    Commented Jun 17, 2023 at 17:38

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