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If $\lvert\psi\rangle$ denotes the state space corresponding to a qubit, then what is the meaning of the $$\int d\psi$$ where the integral is over whole state space of a qubit? How do I evaluate it? Just below equation (1) of this paper, the author says we can use this integral to normalize.

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    $\begingroup$ ...the meaning of integrating over the state space is generically to "sum over all possibilities". Without telling us what is being integrated, there can't be an actual answer to this. $\endgroup$ – ACuriousMind Jun 12 '15 at 12:44
  • $\begingroup$ @ACuriousMind I was reading arxiv.org/pdf/quant-ph/0205035v2.pdf In the beginning ( just below equation (1) )author says the above integral can be normalized. How is the integral equal to a constant I could not understand that. $\endgroup$ – sashas Jun 12 '15 at 12:47
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Integration over pure quantum states usually refers to the Haar measure, i.e., the unitarily invariant measure. Vaguely speaking, you assign the same volume (=weight in the integral) to any two set of states which are related by an arbitrary unitary rotation $U$. In the case of one qubit, this is equivalent to integrating over the Bloch sphere; i.e., we have that $$ \int d\psi\, f(\psi) = \int_0^{\pi}\sin\theta d\theta \int_0^{2\pi}d\phi \, f(\cos(\theta/2)\vert0\rangle + e^{i\phi}\sin(\theta/2)\vert1\rangle)\ . $$

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In general, the integral $$ V := \int \mathrm{d} \mu = \int 1 \mathrm{d}\mu$$ is the integration of the identity over the space the measure $\mu$ is defined on, and should be intuitively understood as the volume of the space with respect to the measure. (This is usually only finite for compact spaces.)

Normalizing the measure means sending $\mu \mapsto \frac{1}{V}\mu$, such that the integral over the whole space is now $1$.

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