# Transition from discrete case to a continuous case with regards to the Born's Rule

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.

• A proper answer to this question requires some non trivial functional analysis and treatment of infinite dimensional linear spaces. To start, if $\alpha$ belongs to the continuum spectrum then it has no eigenvector. In any case the Born rule (which is what you are after) can be written with projectors in a way which is valid for all cases.
– lcv
Commented Sep 28, 2022 at 9:42
• @lcv, if possible, could you answer this question using the non-trivial functional analysis you are referring to? It would be very helpful. Commented Oct 10, 2022 at 13:57
• I can try to give it a go when I find some time.
– lcv
Commented Oct 10, 2022 at 19:17
• Commented Dec 8, 2022 at 10:41

When the system is in state $$|\psi\rangle = \int c(\alpha)|v_\alpha\rangle d\alpha$$ then, when measuring $$\hat{A}$$, the probability of obtaining a value in the range $$[\alpha, \alpha + d\alpha]$$ is $$dP(\alpha) = \frac{|c(\alpha)|^2 d\alpha}{\langle\psi|\psi\rangle}.$$ Note that the probability $$dP(\alpha)$$ needs to be a differential (when $$d\alpha\to 0$$ you also have $$dP(\alpha)\to 0$$). Therefore it is an equation between infinitesimal quantities. Dividing this equation by $$d\alpha$$ you get a probability density $$\frac{dP(\alpha)}{d\alpha} = \frac{|c(\alpha)|^2}{\langle\psi|\psi\rangle}.$$