# 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!

• Adding this comment up here: $A^* [H,A]$ is not self-adjoint in general, even for pure states, so we should not expect $\omega(A^* [H,A])$ to be real in general, so the second inequality is suspect. Jul 2, 2021 at 16:29

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.

• Mechanics thanks a lot! I did not have the idea of using this identity $\omega(A)=\text{Tr}(\rho_\omega A)$ Jul 2, 2021 at 6:12
• I am sorry, i tried to see why your last inequality is true.. but i don't Jul 2, 2021 at 9:22
• To be honest, I'm not sure... $A^* HA$ is self-adjoint, but $A^*A H$ is not necessarily self-adjoint, so the inequality with complex numbers does not make much sense. My final inequality becomes an equality in the case where $\rho_\omega$ is an eigenstate of $H$, but in other circumstances I don't know what $\omega(G)\geq 0$ means when $G$ is not self-adjoint and so $\omega(G)$ is complex. Jul 2, 2021 at 15:12

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.

• Thank you very much! maybe its a stupid question, but what does LHS stand for? Jul 1, 2021 at 16:02
• @uzizi_1 LHS= left hand side Jul 1, 2021 at 17:01
• oh okay it definitely was a stupid question.. thanks a lot! Jul 1, 2021 at 17:26