Prove that if the expectation value of an operator in any state is 1, the operator is Identity

Question

I want to prove that if $ \langle \psi | A | \psi \rangle = 1$ for all $ \psi ,$ then $A=I .$

Let's write $A$ and $\psi$ in the same basis.

$$ \begin{alignat}{7} \left\langle \psi \middle| A \middle| \psi \right\rangle & ~=~ && \left( \sum_w \alpha_w^* \langle w | \right) \left(\sum_{pq} \gamma_{pq} |p\rangle \langle q| \right) \left(\sum_v \alpha_v | v \rangle \right) \\[5px] & ~=~ && \sum_{wvpq} \alpha_w^* \gamma_{pq} \alpha_v \langle w | p \rangle \langle q | v \rangle \\[5px] & ~=~ && \sum_{wv} \alpha_w^* \gamma_{wv} \alpha_v \end{alignat} $$

which is equal to $1 .$

We know that the $\alpha$'s can be anything, and we need to prove that the $\gamma$'s where $w = v$ are $1$ and when $w \neq v$ is $0 .$

How do I proceed?

Answer 1

We have the freedom to expand $A$ in its eigenbasis, then \begin{equation} A = \sum_{\alpha} \vert\alpha\rangle \langle \alpha\vert A\vert\alpha\rangle \langle \alpha\vert = \sum_{\alpha} \vert\alpha\rangle \langle \alpha\vert = 1, \end{equation} where in the second equality we used that by assumption $\langle \alpha\vert A\vert\alpha\rangle=1$.

Answer 2

Loewe's proof is very slick but assumes that $A$ can be diagonalized. Here is a proof with no such restriction.

First, setting $B=A-I$ the statements becomes

$$ \langle \psi, B \psi \rangle=0 ,\ \ \forall \parallel \psi \parallel=1 \Rightarrow B = 0. $$

By going to a basis, reasoning as the OP, one obtains

$$ \sum_{i,j} \psi_i^\ast \psi_j B_{i,j} =0 $$

from which one obtains that the hermitian part of $B$ must be zero. I.e., the statements holds if $B$ (and hence $A$) is hermitian. Loewe's proves it for the more general case for which $A$ can be diagonalized.

For full generality the trick is to realize that we also have information on the off-diagonal elements of $B$ via the polarization identity.

Define

$$ q(\psi) = \langle \psi, B\psi \rangle $$

then

$$ \langle \phi, B \psi \rangle = \frac{1}{4} \sum_{n=0}^3 i^n q(\phi + i^n \psi) $$

but by assumption $q(\xi)=0$ for all $\xi$, hence $\langle \phi, B \psi \rangle =0$ for all $\phi,\psi$, hence $B=0$.