1
$\begingroup$

In https://arxiv.org/abs/2302.13515 the authors discuss in page 23 the quantum geometric tensor, defined as $$\mathcal Q_{\mu\nu} = \langle\partial_\mu\Psi|(I-|\Psi\rangle\!\langle\Psi|)|\partial_\nu\Psi\rangle,$$ after the following intuitive argument:

Intuitively, the distance between states is the size of the change in the state. We need to take care however to ensure that our notion of distance is independent of the global phase, since physical states are defined by rays in the Hilbert space.

This quantum geometric tensor is then decomposed into symmetric and skew-symmetric components as $\mathcal Q=\mathcal F+i\Omega$, and $\mathcal F$ and $\Omega$ are argued to be the quantum Fisher information metric and the Berry curvature, respectively.

I don't quite understand how you go from looking for a global-phase-independent notion of distance, to the formula above the quantum geometric tensor. Is there a more rigorous way to derive this formula for $\mathcal Q$, and then possibly show its connection with Fisher information and Berry curvature?

$\endgroup$
6
  • $\begingroup$ Why do you say the global phase matters here? As long as it doesn't depend on $\lambda_\mu$ it should be fine $\endgroup$ Aug 8, 2023 at 21:19
  • $\begingroup$ The previous sentence in your source explicitly says that this quantity takes care of the global phase, no? $\endgroup$ Aug 8, 2023 at 21:24
  • $\begingroup$ @QuantumMechanic yes, that's what I meant. $\mathcal Q$ was said to be derived to have a quantity that does not depend on the global phase. I can see it doesn't, but the question is how to get to this formula from the "naive metric" which would be just the inner product of the derivatives. I guess one approach might be to work out the change of $\langle\partial_\mu\Psi|\partial_\nu\Psi\rangle$ wrt a gauge transformation $\Psi\to e^{i\alpha}\Psi$, which should somehow give rise to the term with $|\Psi\rangle\!\langle\Psi|$? I was wondering it there's also a more elegant approach though $\endgroup$
    – glS
    Aug 8, 2023 at 21:58
  • $\begingroup$ Oh I see. I don't think your expression depends on the global phase either! Could be wrong $\endgroup$ Aug 9, 2023 at 12:51
  • $\begingroup$ to be clear, when I mentioned dependence on global phase, I'm talking about the fact that if $\Psi\to e^{i\alpha}\Psi$ then $\delta(\partial_\mu\Psi)=i\partial_\mu\alpha\Psi$, and $\delta\langle\partial_\mu\Psi|\partial_\nu\Psi\rangle=i \partial_\nu\alpha\langle\partial_\mu\Psi|\Psi\rangle-i\partial_\mu\alpha\langle\Psi|\partial_\nu\Psi\rangle$. I think this dependence disappears introducing the correction factor. Another paper I found discussing this is arxiv.org/abs/1012.1337 $\endgroup$
    – glS
    Aug 9, 2023 at 16:51

1 Answer 1

1
$\begingroup$

One expands the pure state to leading order, as in the paper, with $$|\psi(\lambda_\mu+d\lambda_\mu)\rangle=|\psi(\lambda_\mu)\rangle+d\lambda_\mu|\partial_\mu \psi(\lambda_\mu)\rangle.$$ For multiple parameters $\lambda=(\lambda_1,\cdots)$, we can compare two of these states, one where $\lambda_\mu$ varies and one where $\lambda_\nu$ varies. I'll use the notation $\psi(\lambda_\mu+d\lambda_\mu)$ to imply that $\lambda_\mu$ varies while the rest of the parameters are held constant: $$|\langle\psi(\lambda_\nu+d\lambda_\nu) |\psi(\lambda_\mu+d\lambda_\mu)\rangle|^2=|1+d\lambda_\nu\langle \partial_\nu \psi|\psi\rangle+d\lambda_\mu\langle \psi|\partial_\mu\psi\rangle+d\lambda_\mu d\lambda_\nu\langle \partial_\nu \psi|\partial_\mu\psi\rangle|^2=1+2d\lambda_\mu d\lambda_\nu [\langle \partial_\nu \psi|\psi\rangle\langle \partial_\mu\psi|\psi\rangle+\Re(\langle \partial_\nu \psi|\partial_\mu\psi\rangle)]+d\lambda_\mu^2|\langle\psi|\partial_\mu\psi\rangle|^2+d\lambda_\nu^2|\langle\psi|\partial_\nu\psi\rangle|^2.$$ I ignored terms higher than order $\mathcal{O}(\lambda^2)$ and used $\langle \partial_\mu \psi|\psi\rangle=-\langle \psi|\partial_\mu\psi\rangle$ to get rid of the first order terms and collect some of the second order terms. I also assumed $d\lambda_\rho$ to be real.

We observe that the coefficient of $d\lambda_\mu d\lambda_\nu$ is proportional to the real part of $$Q_{\mu\nu}=\langle \partial_\nu \psi|\psi\rangle\langle \partial_\mu\psi|\psi\rangle+\langle \partial_\nu \psi|\partial_\mu\psi\rangle=\langle \partial_\nu \psi|(1-|\psi\rangle\langle \psi|)|\partial_\mu\psi\rangle,$$ which is the expression you are looking for.

The terms proportional to $d\lambda_\mu^2$ drop out when differentiating with respect to both $\lambda_\mu$ and $\lambda_\nu$ unless $\mu=\nu$, so we might want to add a delta function: $$Q_{\mu\nu}=\langle \partial_\nu \psi|[1+(\delta_{\mu\nu}-1)|\psi\rangle\langle \psi|]|\partial_\mu\psi\rangle.$$ The diagonal components look like your expected $\langle \partial_\nu \psi|\partial_\mu\psi\rangle$ and the off-diagonal components look like the expression in the paper.

$\endgroup$
3
  • $\begingroup$ thansk! I think there's a typo in the eq for $Q$, the minus sign is in the other direction. That aside, what's the idea here then? Looks like you're computing how the norm $\|\psi(\boldsymbol\lambda)\|^2$ changes with respect to different parameters, $\partial_\mu\partial_\nu \|\psi\|^2$ (I'm assuming that's what you meant with $\langle\psi(\lambda_\nu+d\lambda_\nu)|\psi(\lambda_\mu+d\lambda_\mu)\rangle$ etc). But if the parameters parametrise pure states, shouldn't this give zero? Otherwise I might be misinterpreting this:you're computing fidelity between states at different parameter values? $\endgroup$
    – glS
    Aug 9, 2023 at 16:46
  • $\begingroup$ @gIS the minus sign looks good to me, can you specify which part? I assume you say there is a minus sign missing even after realizing the complex conjugating $\langle \partial_\mu \psi|\psi\rangle$ picks up a minus sign? $\endgroup$ Aug 9, 2023 at 18:43
  • $\begingroup$ @glS I'm literally taking the previous equation in the paper, expressing what the state looks like for small changes in one parameter, and comparing the overlap of that changed state with the state resulting from changing another parameter. Sorry so yes, one would expect (and see that it is indeed true) that this expression gives unity when $\mu=\nu$. I am explicitly computing fidelity between states at different parameter values. That's what the quantity seems to be asking for - I cannot tell you much more about the why $\endgroup$ Aug 9, 2023 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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