On Born's rule and Cauchy-Schwarz inequality

Question

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

"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...

Answer 1

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.