1
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Jul 2, 2021 at 16:29

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Mechanics thanks a lot! I did not have the idea of using this identity $\omega(A)=\text{Tr}(\rho_\omega A)$ $\endgroup$
    – uzizi_1
    Jul 2, 2021 at 6:12
  • $\begingroup$ I am sorry, i tried to see why your last inequality is true.. but i don't $\endgroup$
    – uzizi_1
    Jul 2, 2021 at 9:22
  • $\begingroup$ 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. $\endgroup$ Jul 2, 2021 at 15:12
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thank you very much! maybe its a stupid question, but what does LHS stand for? $\endgroup$
    – uzizi_1
    Jul 1, 2021 at 16:02
  • $\begingroup$ @uzizi_1 LHS= left hand side $\endgroup$ Jul 1, 2021 at 17:01
  • $\begingroup$ oh okay it definitely was a stupid question.. thanks a lot! $\endgroup$
    – uzizi_1
    Jul 1, 2021 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.