States on finite dimensional Hilbert spaces In my quantum theory lecture we talked about states on finite dimensional Hilbert spaces and had the following statement:
Let $\mathcal{H}$ be a finite dimensional complex Hilbert space, $\mathcal{L}(\mathcal{H})$ the set of bounded linear operators, $\omega \colon \mathcal{L}(\mathcal{H}) \to \mathbb{C}$ a state and $H\in \mathcal{L}(\mathcal{H})$ be self-adjoint. Then $\omega$ is a ground state of $H$, i.e.
$$ \omega(H)\leq \tilde{\omega}(H)\quad \text{for all states } \tilde{\omega} \text{ on } \mathcal{L}(\mathcal{H})$$
if and only if
$$ \omega(A^*[H,A])\geq 0\quad \text{for all }A\in\mathcal{L}(\mathcal{H}).$$
I tried to compute both sides and get to the other, but it absolutely didn't work... I have no idea how to see that this statement is true. I would be greatful for tips or help!
 A: We start by expanding
$$
\omega(A^*[H,A])=\omega(A^*H A-A^*AH).
$$ We use the usual linear operation $\omega(B)=\mathrm{Tr}(\rho_{\omega}B)$ where $\rho_\omega$ is the state represented by $\omega$. Since this operation is linear, we have that
$$
\omega(A^*[H,A])=\omega(A^*H A)-\omega(A^*AH)=\mathrm{Tr}(A\rho_{\omega}A^* H)-\mathrm{Tr}(H\rho_\omega A^*A)\ge 0\, ,
$$ where we have used the circular property of the trace. The operator $A^*A$ will be positive for any $A$, and the combination $A\rho_\omega A^*$ will be proportional to another state $\rho_{\tilde{\omega}}$ in the same Hilbert space. This yields a promising inequality
$$
\alpha\mathrm{Tr}(\rho_{\tilde{\omega}}H)\geq \mathrm{Tr}(H \rho_{\omega}A^*A),\qquad \alpha\equiv \mathrm{Tr}(A\rho_\omega A^*)=\mathrm{Tr}(A^*A\rho_\omega).
$$ Equivalently,
$$
\tilde{\omega}(H)=\mathrm{Tr}(\rho_{\tilde{\omega}}H)\geq \frac{\mathrm{Tr}(H \rho_{\omega}A^*A)}{\mathrm{Tr}(\rho_\omega A^*A)}.
$$
The final step is to prove that
$$
\frac{\mathrm{Tr}(H \rho_{\omega}M)}{\mathrm{Tr}(\rho_\omega M)}\geq \mathrm{Tr}(H \rho_{\omega})=\omega(H)
$$ for all positive operators $M$, which I can leave to you.
A: Every pure state $\psi\in\cal H$ gives rise to a linear operator $\omega:\cal O\mapsto \omega(\cal O)$ by the familiar $\cal O\mapsto\langle \cal O\psi,\psi\rangle\,.$ For two arbitrary pure states $\psi$ and $\tilde\psi\,,$ there always exists a unitary operator $A$ such that $\tilde\psi=A\psi\,.$ Therefore, restricted to pure states $\omega,\tilde\omega\,,$ the first statement becomes
$$
\langle H\psi,\psi\rangle\le\langle HA\psi,A\psi\rangle\mbox{ for all unitary }A\,.
$$
Since the LHS equals $\langle AH\psi,A\psi\rangle$ it is easy to see that the statment is equivalent to
$$
\langle [H,A]\psi,A\psi\rangle\ge 0\mbox{ for all unitary }A\,,
$$
resp. to
$$
\langle A^*[H,A]\psi,\psi\rangle\ge 0\mbox{ for all unitary }A\,.
$$
Perhaps your theorem is stronger than that but I hope the tips are good enough.
