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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?

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    $\begingroup$ Can we prove it in the opposite direction (i.e. contra-positive)? if $A \neq I$, then $\langle \psi | A | \psi \rangle \neq 1$ for some $\psi$ $\endgroup$ – K_inverse Jan 21 at 8:37
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    $\begingroup$ You say the $\alpha$'s can be anything but surely they need to be normalised? $\endgroup$ – jacob1729 Jan 21 at 10:17
  • $\begingroup$ @jacob1729, true, they need to be normalised $\endgroup$ – Mahathi Vempati Jan 21 at 10:18
  • $\begingroup$ @MahathiVempati if $\alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=\sum \alpha^*_w \delta_{wv}\alpha_v$. $\endgroup$ – jacob1729 Jan 21 at 11:31
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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 $\langle \alpha\vert A\vert\alpha\rangle=1$.

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  • $\begingroup$ Thank you! Is it possible to continue my line of proof, though? $\endgroup$ – Mahathi Vempati Jan 21 at 9:12
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    $\begingroup$ This is slick but assumes that $A$ can be diagonalized.. $\endgroup$ – lcv Jan 21 at 19:19
  • $\begingroup$ @MahathiVempati I sort of addressed your comment. $\endgroup$ – lcv Jan 22 at 9:03
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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 $B=0$.

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