States on finite dimensional Hilbert spaces

Question

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!

Answer 1

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.

Answer 2

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.