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In Ran Cheng's review of the quantum geometric tensor, eq. (11) gives the tensor as: $$ Q_{\mu\nu}=\sum_{n\neq 0}\frac{\langle\phi_0|\partial_\mu H|\phi_n\rangle\langle\phi_n|\partial_\nu H|\phi_0\rangle}{(E_0-E_n)^2}. $$ Per eq. (6), we see that the quantum metric may be defined as: $$ g_{\mu\nu}=\text{Re }Q_{\mu\nu}. $$ I now want to calculate the distance between two arbitrary quantum states of a two-level system parameterized by two variables $(k_x,k_y)$ using eq. (9): $$ |\langle\phi(\lambda_F)|\phi(\lambda_I)\rangle|=1-\frac{1}{2}\int_{\lambda_I}^{\lambda_F} g_{\mu\nu}(\lambda)d\lambda^\mu d\lambda^\nu. $$

Could someone verify that the steps I'm taking are correct? The results of my numerical simulations seem suspicious.

  1. Find the two eigenvectors of the 2-level Hamiltonian (not resulting in the nice standard eigenvectors discussed in Cheng's work).
  2. Use these to calculate $g_{k_x k_x},g_{k_x k_y},g_{k_y k_x}$ and $g_{k_y k_y}$.
  3. To perform the path integral, I first parameterize some path in $k$-space (for example using $k_x (t) = \sin(t), k_y(t) = \cos(t)$).
  4. Plug these $k_x(t),k_y(t)$ into $g_{\mu\nu}$ found in step 2.
  5. Using motivation from eq. (21) of the review, numerically integrate the following, for absolute value of some $x$ indicated by $|x|$: $$ |\langle\phi(\lambda_F)|\phi(\lambda_I)\rangle|=1-\frac{1}{2}\int_{\lambda_I}^{\lambda_F} \sqrt{ \left|g_{k_x k_x} \left(\frac{dk_x}{dt}\right)^2\right|+ \left|g_{k_x k_y}\left(\frac{dk_x}{dt}\right)\left(\frac{dk_y}{dt}\right)\right|+ \left|g_{k_y k_x}\left(\frac{dk_y}{dt}\right)\left(\frac{dk_x}{dt}\right)\right|+ \left|g_{k_y k_y}\left(\frac{dk_y}{dt}\right)^2\right| } dt. $$ I am supposed to get an answer between $0$ and $1$, but I am not sure I incorporated/defined the metric correctly. Any advice?

UPDATE: As requested, here is (Mathematica) code that I used:

(* Some Hamiltonian H[kx,ky] with eigenvectors |n[kx,ky]>,|p[kx,ky]> \
and eigenvalues Energy[kx,ky]*)
m = 0.1;
sx = {{0, 1}, {1, 0}}; sy = {{0, -I}, {I, 0}}; sz = {{1, 0}, {0, -1}};
H[kx_, ky_] = m sz + kx sz + ky sy;
p[kx_, ky_] = 
  Eigenvectors[H[kx, ky]][[1]]/Norm[Eigenvectors[H[kx, ky]][[1]]];
n[kx_, ky_] = 
  Eigenvectors[H[kx, ky]][[2]]/Norm[Eigenvectors[H[kx, ky]][[2]]];
Energy[kx_, ky_] = Eigenvalues[H[kx, ky]];
(* Components of Quantum geometric tensor *)
Qxx[kx_, ky_] = ((Conjugate[n[kx, ky]] . D[H[kx, ky], kx] . 
       p[kx, ky]) (Conjugate[p[kx, ky]] . D[H[kx, ky], kx] . 
       n[kx, ky]))/(Energy[kx, ky][[2]] - Energy[kx, ky][[1]])^2;
Qxy[kx_, ky_] = ((Conjugate[n[kx, ky]] . D[H[kx, ky], kx] . 
       p[kx, ky]) (Conjugate[p[kx, ky]] . D[H[kx, ky], ky] . 
       n[kx, ky]))/(Energy[kx, ky][[2]] - Energy[kx, ky][[1]])^2;
Qyx[kx_, ky_] = ((Conjugate[n[kx, ky]] . D[H[kx, ky], ky] . 
       p[kx, ky]) (Conjugate[p[kx, ky]] . D[H[kx, ky], kx] . 
       n[kx, ky]))/(Energy[kx, ky][[2]] - Energy[kx, ky][[1]])^2;
Qyy[kx_, ky_] = ((Conjugate[n[kx, ky]] . D[H[kx, ky], ky] . 
       p[kx, ky]) (Conjugate[p[kx, ky]] . D[H[kx, ky], ky] . 
       n[kx, ky]))/(Energy[kx, ky][[2]] - Energy[kx, ky][[1]])^2;
(* For integral, parameterize by g the k-space path (qx[g],qy[g]) of \
radius r, center (kx0,ky0). *)
r = 1; kx0 = 0; ky0 = 0;
qx[g_] = kx0 + r Cos[g];
qy[g_] = ky0 + r Sin[g];
(* Derivatives of qx[g],qy[g] for integral metric*)
dqx[g_] = D[qx[g], g]; dqy[g_] = D[qy[g], g];
(* Components of metric tensor*)
gxx[kx_, ky_] = Re[Qxx[kx, ky]];
gxy[kx_, ky_] = Re[Qxy[kx, ky]];
gyx[kx_, ky_] = Re[Qyx[kx, ky]];
gyy[kx_, ky_] = Re[Qyy[kx, ky]];
(* Starting and ending points of g, for integration limits. If \
gend=2Pi, I expect the distance to = 0, since they are the same point \
as the parameterized path is then that of a closed circle. *)
gstart = 0;
gend = 3/4 Pi;
(* Quantum distance: I am not sure whether my definition of the \
metric is correct. *)
distance = 
 1 - (1/2) NIntegrate[
Sqrt[Abs[gxx[qx[g], qy[g]] dqx[g] dqx[g]] + 
  Abs[gxy[qx[g], qy[g]] dqx[g] dqy[g]] + 
  Abs[gyx[qx[g], qy[g]] dqy[g] dqx[g]] + 
  Abs[gyy[qx[g], qy[g]] dqy[g] dqy[g]]], {g, gstart, gend}]
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  • 2
    $\begingroup$ +1. Can you show us the MATLAB code you used and some example numbers (input and output)? I can try to see if I get the same output with the same input, or something else. $\endgroup$ Nov 29, 2021 at 4:52
  • $\begingroup$ @NikeDattani thanks. Unfortunately, I used Mathematica for the numerical integration. But I updated my post with my code. $\endgroup$ Nov 29, 2021 at 5:13
  • 1
    $\begingroup$ Maybe change the tolerance in NIntegrate? $\endgroup$ Nov 29, 2021 at 7:52
  • $\begingroup$ @NikeDattani thank you, do you know whether my definition/implementation of the metric tensor inside the integral is correct? I tried other variants as well. $\endgroup$ Nov 29, 2021 at 16:36

1 Answer 1

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The issue was the absolute values, and misunderstanding the notation. Per eq. (2) of this paper, we have: $$ ds^2 =1-|\langle\phi(\lambda_F)|\phi(\lambda_I)\rangle| =\sum_{\mu\nu}g_{\mu\nu}(\lambda)d\lambda^\mu d\lambda^\nu $$

Then, to get the quantum distance, all I needed to do was get $\int ds$:

NIntegrate[Sqrt[gxx[g] + gxy[g] + gyx[g] + gyy[g]], {g, gstart, gend}],

where I re-defined the functions gxx, etc for convenience:

gxx[g_] = gxx[qx[g], qy[g]] dqx[g] dqx[g];
gxy[g_] = gxy[qx[g], qy[g]] dqx[g] dqy[g];
gyx[g_] = gyx[qx[g], qy[g]] dqy[g] dqx[g];
gyy[g_] = gyy[qx[g], qy[g]] dqy[g] dqy[g];

after re-defining each component in terms of parameter g alone:

gxx[g_]=gxx[qx[g],qy[g]]
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