If $A$ is self-adjoint and nonnegative then linear functional is nonnegative

Question

Assume $\dim(\mathcal{H})=\infty$. Let $$V=\{B+cI|B\in\mathcal{K}(\mathcal{H}), c\in\mathbb{C}\}$$ where $\mathcal{K}(\mathcal{H})$ is the closure of the space of finite-rank operators with respect to the operator norm on $\mathcal{B}(\mathcal{H})$ (the space of bounded operators on $\mathcal{H}$). Define $\Phi:V\to\mathbb{C}$ by $$\Phi(B+cI)=c.$$ It can be shown that $|\Phi(A)\mid \le\|A\|$ for all $A\in V$. By the Hahn-Banach theorem, there exists a linear functional $\Psi_1:\mathcal{B}(\mathcal{H})\to\mathbb{C}$ such that $\Psi_1=\Phi$ on $V$ and such that $|\Psi_1(A)|\le\|A\|$ for all $A\in\mathcal{B}(\mathcal{H})$. Let $\Psi_2:\mathcal{B}(\mathcal{H})\to\mathbb{C}$ be defined by $$\Psi_2(A)=\frac12(\Psi_1(A)+\overline{\Psi_1(A^*)}).$$ Show that

1. $\Psi_2(I)=1$
2. $\Psi_2(A)\text{ is real whenever } A\text{ is self-adjoint}$
3. $\Psi(A)\ge0\text{ whenever } A\text{ is self-adjoint and non-negative}$
4. $\text{There does not exist any density matrix }\rho\text{ such that }\Psi_2(A)=\text{trace}(\rho A)\text{ for all }A\in\mathcal{B}(\mathcal{H}).$

This problem shows that there exists a linear functional which satisfies properties (1)-(3) but which is not a family of expectation values (because if it were, it wouldn't satisfy (4)). A density matrix is an operator $\rho\in\mathcal{B}(\mathcal{H})$ such that $\rho$ is self-adjoint, non-negative, and $\text{trace}(\rho)=1$.

Proof (1) Since $0\in\mathcal{K}(\mathcal{H})$, we have $I=0+I\in V$ and so $\Psi_1(I)=\Phi(I)=\Phi(0+I)=1$. Thus, $$\Psi_2(I)=\frac12(\Psi_1(I)+\overline{\Psi_1(I)})=1.$$ (2) Suppose $A$ is self-adjoint. That is, $A^*=A$. Then $$\Psi_2(A)=\frac12(\Psi_1(A)+\overline{\Psi_1(A^*)})=\frac12(\Psi_1(A)+\overline{\Psi_1(A)}).$$ So $\Psi_2(A)=\overline{\Psi_2(A)}$ and therefore $\Psi_2(A)\in\mathbb{R}$. $\square$

What I am stuck on is proving properties (3) and (4). In part (3), non-negativity of $A$ means $\langle\psi, A\psi\rangle\ge0$ for all $\psi\in\mathcal{H}$.

Answer 1

The key fact is the following: given a C$^*$-algebra $A$ (in particular, you can take $A=B(H)$) and linear functional $f:A\to \mathbb C$, the following statements are equivalent:

• $f\geq0$ (as in your condition 3)

• $f(1)=\|f\|$.

The above equivalence is very well-known; here is a proof.

Since $\Psi_1(A)\leq \|A\|$ for all $A$, we have that $\|\Psi_1\|\leq1$. But $\Psi_1(1)=1$, so $\|\Psi_1\|=1=\Psi_1(1)$. It follows from the equivalence above that $\Psi_1\geq0$. As $\Psi_2=\Psi_1$ on self-adjoint operators, we also get that $\Psi_2\geq0$.

And 4 is completely straightforward: you have by construction that $\Psi_2(A)=0$ for all compact $A$. Then you would get $$ \operatorname{Tr}(\rho^2)=\operatorname{Tr}(\rho\rho^*)=\Psi_2(\rho^*)=0 $$ and so $\rho=0$, a contradiction.