# On Born's rule and Cauchy-Schwarz inequality

Citing Born's rule:

If an observable corresponding to a self-adjoint operator $${\textstyle A}$$ with discrete spectrum is measured in a system with normalized wave function $${\textstyle |\psi \rangle }$$, then

• the measured result will be one of the eigenvalues $$\lambda }$$ of $$A}$$, and
• the probability of measuring a given eigenvalue $$\lambda _{i}}$$ will equal $$\langle \psi |P_{i}|\psi \rangle }$$, where $$P_{i}}$$ is the projection onto the eigenspace of $$A}$$ corresponding to $$\lambda _{i}}$$.

"Proof". If we assume $$\scr H$$ finite-dimensional, then every linear self-adjoint operator is compact, and hence Spectral Theorem applies:

1. $$\displaystyle A =\sum_{\sigma_p(A)} \lambda_i P_i$$
2. The eigenvectors $$\{\psi_i\}_{i \in \mathbb N}$$ of $$A$$, each corresponding to a different eigenvalue $$\lambda_i$$, form a base for $$\scr H$$.

Using 2., we can represent each element $$\psi \in \scr H$$ in the following way: $$\displaystyle \psi = \sum_i \langle \psi_i, \psi \rangle\psi_i$$. Copenhagen interpretation imposes $$\| \psi \| ^2 = 1$$. It follows then, according to Parseval's identity, $$\| \psi \| ^2 = \displaystyle \sum_i |\langle \psi_i, \psi \rangle|^2$$, where each real number $$|\langle \psi_i, \psi \rangle|^2$$ has to be interpreted as the probability of obtaining the result $$\lambda_i$$ from a measure of $$| \psi \rangle$$.

Now, using 1. $$\psi_i = P_i \psi$$, and remembering that $$P$$ is a positive operator, we obtain $$|\langle \psi_i, \psi \rangle| = \langle \psi_i, \psi \rangle = \langle \psi | P_i | \psi \rangle$$.

There are two major problems:

1. It's the square of $$| \langle \psi_i, \psi \rangle |$$ that should matters.

2. Since $$P$$ is also idempotent and self-adjoint, one can manipulate further $$\langle \psi | P_i | \psi \rangle = \langle \psi | P_iP_i | \psi \rangle = \langle \psi | P_i^{\dagger}P_i | \psi \rangle = \| P_i | \psi \rangle \|^2$$... but that means

$$| \langle \psi_i, \psi \rangle | = \|\psi_i\|^2$$

While following Cauchy-Schwarz inequality: $$|\langle \psi_i ,\psi \rangle| \leq \|\psi_i\|$$.

I don't know, there's something missing...

• $P_i|\psi\rangle \neq |\psi_i\rangle$. Instead $P_i|\psi\rangle = |\psi_i\rangle \langle\psi_i|\psi\rangle$. Jul 8, 2021 at 10:22
• I don't think what you have there is proof of the Born rule. You say, > It follows then, according to Parseval's identity, $\| \psi \| ^2 = \displaystyle \sum_i |\langle \psi_i, \psi \rangle|^2$, where each real number $|\langle \psi_i, \psi \rangle|^2$ has to be interpreted as the probability of obtaining the result $\lambda_i$ from a measure of $| \psi \rangle$. I don't see how that follows at all. As far as I know, deriving the Born rule from the other postulates is something that's not clear how to do. Copenhagen interpretation considers it to be an axiom. Jul 8, 2021 at 11:43

Let's start from

$$\psi = \sum\limits_i (\psi_i,\psi) \, \psi_i \quad .$$

Applying $$P_j$$ yields

$$P_j \, \psi = \sum\limits_i (\psi_i,\psi)\, \underbrace{P_j\, \psi_i}_{= \delta_{ij} \, \psi_j} = (\psi_j,\psi)\, \psi_j \quad$$

and we thus find

$$(\psi,P_j\, \psi) = |(\psi_j,\psi)|^2 \quad .$$

Similarly to what you've done, we calculate:

$$(\psi,P_j\, \psi) = ||P_j\,\psi||^2 = \underbrace{|(\psi_j,\psi)|^2}_{\leq 1} \,||\psi_j||^2 \leq ||\psi_j||^2 \quad .$$

From Cauchy-Schwarz we'd have obtained the following result:

$$(\psi,P_j\,\psi) \leq ||\psi|| \, ||P_j\,\psi|| =|(\psi_j,\psi)| \, ||\psi_j|| \quad .$$ All in all, there is no contradiction.

• Yes my error, as mentioned by @flippiefanus, was the identity $P_i |\psi\rangle = |\psi_i \rangle$. Jul 8, 2021 at 10:29
• @ric.san Yes, correctly. As I also commented (I deleted it since I wrote an answer anyway) this was immediately clear because, as you said, $P_i$ is positive semi-definite, while the expansion coefficients of the wave function are in general complex. Jul 8, 2021 at 10:31