So I came across this question that asked what is the probability of a hydrogen atom which is prepared in an initial state $\Psi (\vec{r},t)$ to be in the ground state $\psi_{100}(\vec{r}) =2exp(-r)Y_{0,0}(\theta, \phi)$ with $$\Psi (\vec{r},t=0) = 1/\sqrt{2}exp(-r/2)Y_{0,0}(\theta, \phi)$$ where $Y_{0,0}(\theta, \phi) = 1/\sqrt{4\pi}$. The answer is apparently found projecting the ground state onto the $\Psi (\vec{r},t=0)$: $$P = |\int{\psi_{100}^*}\Psi (\vec{r},t=0)dr^{3}|^2$$ What is the reasoning behind this?
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1$\begingroup$ This is how you determine the probability of any measurement in QM... $\endgroup$ – Aaron Stevens Apr 1 at 20:31
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Look for Measurement in quantum mechanics on Wikipedia. There you find that the probability of a measuring a state $|\psi_n\rangle$ with an initial given state $|\psi \rangle$ is given by \begin{align*} \operatorname{Pr}\left(\psi_{n}\right)=|\langle \psi_n | \psi\rangle|^{2} \end{align*} which is in a continuous space (which you have) given by the integral representation: \begin{align*} \operatorname{Pr}\left(\psi_{n}\right)=\int \mathrm d x \, \psi_n^*(x) \psi(x) \end{align*} Maybe this answers your question.
