I learned that given that the eigenvalue equation is $$ \widehat{A}\left|u_{n}^{i}\right\rangle=\lambda_{n}\left|u_{n}^{i}\right\rangle $$ where $ i \in\{1,2, \ldots, g_n\} $, and that the state $ |\psi\rangle $ is not normalised, the probability of obtaining an eigenvalue $ \lambda_{n} $ is given by: $$ P\left(\lambda_{n}\right)=\frac{\sum_{i=1}^{g_{n}}\left|\left\langle u_{i}^{n} \mid \psi\right\rangle\right|^{2}}{\langle\psi \mid \psi\rangle} $$ Now, from a discrete case, if we shift to the case with a continuous eigenvalue equation given by: $$ \hat{A}\left|v_{\alpha}\right\rangle=\alpha\left|v_{\alpha}\right\rangle $$ where $\alpha$ is a continuous variable, then how can we naturally move on from the above probability expression to its continuous case wherein we happen to bring in the concept of probability density?
I tried to tackle this by first writing down an arbitrary ket in the state space as: $$ |\psi\rangle=\int c(\alpha)\left|v_{\alpha}\right\rangle d \alpha $$ where $c(\alpha)=\left\langle v_{\alpha} \mid \psi\right\rangle$ which is the wavefunction here.
Is it correct to use the same analogy for the discrete case and write down the probability of measuring an eigenvalue as given below? $$ P(\alpha)=|c(\alpha) d \alpha|^{2} $$ I somehow need to come to the result: $$ \frac{d P(\alpha)}{d \alpha}=\frac{|c(\alpha)|^{2}}{\langle\psi \mid \psi\rangle} $$
Any idea or a complete result will be appreciated. I just want a smooth transition from discrete case to a continuous case.
