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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.

Qmechanic
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Pratham Hullamballi
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    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 Sep 28 '22 at 09:42
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    @lcv, if possible, could you answer this question using the non-trivial functional analysis you are referring to? It would be very helpful. – Pratham Hullamballi Oct 10 '22 at 13:57
  • I can try to give it a go when I find some time. – lcv Oct 10 '22 at 19:17
  • related https://physics.stackexchange.com/questions/98455/conceptual-difficulty-in-understanding-continuous-vector-space – user35952 Dec 08 '22 at 10:41

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You are nearly on the right track. I will give a simple explanation without any mathematical rigour.

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 $$\frac{dP(\alpha)}{d\alpha} = \frac{|c(\alpha)|^2}{\langle\psi|\psi\rangle}.$$

Thomas Fritsch
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