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

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?

• 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 $\psi$ – K_inverse Jan 21 at 8:37
• You say the $\alpha$'s can be anything but surely they need to be normalised? – jacob1729 Jan 21 at 10:17
• @jacob1729, true, they need to be normalised – Mahathi Vempati Jan 21 at 10:18
• @MahathiVempati if $\alpha$'s are normalised then one (long winded) way of writing the number 1 is $1=\sum \alpha^*_w \delta_{wv}\alpha_v$. – jacob1729 Jan 21 at 11:31

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

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

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 $$B=0$$.