Let $\mathcal{H} =\mathbb{C}^2, \mathcal{M}_1 = \mathbb{C}|0\rangle$ with $|\psi\rangle = \alpha |0\rangle + \beta|1\rangle$. Show $Pr(\mathcal{M_1}) = |\alpha|^2.$
We know that $\mathcal{M_1}$ is a subspace of the Hilbert space $\mathcal{H}$, and that $Pr(\mathcal{M_1}) = \langle \psi| Proj_\mathcal{M}|\psi\rangle.$
I'm confused on how to interpret the bra ket notation here and how to actually calculate $Pr(\mathcal{M_1})$ using the formula and information given.
Let any linear subspace $\mathcal M$ for a (for simplicity's sake: finite-dimensional) Hilbert space $\mathcal H$ with $\{u_1,\ldots,u_k\}$ being any orthonormal basis of $\mathcal M$ be given. Then the orthogonal projection onto $\mathcal M$ is given by $P_{\mathcal M}=\sum_{i=1}^k |u_i\rangle\langle u_i|$ so the probability of finding a pure state $|\psi\rangle$ in there is given by $$ \langle\psi|P_{\mathcal M}|\psi\rangle=\sum_{i=1}^k |\langle\psi|u_i\rangle|^2\ . $$ Side note: if the state of the system is not pure $|\psi\rangle\langle\psi|$ but a general mixed state $\rho$ (positive semi-definite, trace 1) then the expectation value generalizes to $\operatorname{tr}(\rho P_{\mathcal M})$---and if $\rho$ is pure then we get back $\langle\psi|P_{\mathcal M}|\psi\rangle$ as expected.