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I was wondering if anyone knew how the author got to equation 12 on page four of this paper, I will attempt to explain the situation below.

Given $C$, a target operator, we wish to create an approximation starting with $\rho_0$, and acting sequentially with propagators over equal times steps $\delta t =T/N$, where $T$ is the total time- these propagators are given by: $$U_j = \exp \left\{-i\Delta t \left( \mathcal{H}_o + \ \sum^m_{k=1} u_k(j) \mathcal{H}_k \right) \right\} \tag{1} \, .$$ The $\mathcal{H}_k$ is simply a control Hamiltonian, and $u_k(j)$ describes the amplitude of the $k^{th}$ component at time step $j$. We thus define a performance function, that we wish to maximize, given by:

$$\Phi_o =\langle U^{\dagger}_{j+1}... U^{\dagger}_{N}CU_N...U_{j+1}|U_{j}... U_1CU_1^{\dagger}...U^{\dagger}_{j}\rangle=\langle\lambda_j|\rho_j\rangle \, . \tag{2}$$ This is simply the (trace) inner product of the backwards propagated target operator, with the forward propagated starting operator. Taking the change in $U_j$ to first order in $\delta u_k(j)$, and with $\Delta t$ small we have approximately (see the paper for the exact result): $$\delta U_j =-i\Delta t \delta u_k(j)\mathcal{H}_kU_j \, . \tag{3}$$ Finally the paper claims that the following is true, however I am at a loss as to how to proceed: $$\frac{\delta\Phi_o}{\delta u_k(j)}=-\langle\lambda_j|i\Delta t[\mathcal{H}_k,\rho_j]\rangle \, . \tag{4}$$ In particular, the appearance of the commutator is completely baffling to me- but I'm fairly certain I just am not sufficiently well versed in algebra. Any suggestions?

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  • $\begingroup$ Since the only thing you're saying about $\bar{H}$ is that it's approximately the same thing as $H$, why not just write $H$ everywhere and simplify your (and our) life? Don't worry about focusing too much on what the paper writes, but instead focusing on understanding the specific point that doesn't make sense to you. $\endgroup$
    – DanielSank
    Jun 27, 2015 at 0:29
  • $\begingroup$ Point taken, amending now. (And thanks for cleaning it up) $\endgroup$
    – user83928
    Jun 27, 2015 at 0:30
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    $\begingroup$ I think it would be good if you could include some more context. Not everyone knows what GRAPE is and the equations kind of don't make sense without knowing what's going on. It's considered much better on this site to include all relevant material from any references (because links rot, and not everyone can access them). The point is to be self contained. I'm harping on making this question good because I'm interested. $\endgroup$
    – DanielSank
    Jun 27, 2015 at 0:32
  • $\begingroup$ Alright, let me try to fix it up more. $\endgroup$
    – user83928
    Jun 27, 2015 at 0:33
  • $\begingroup$ What have you tried? Usually in life the best thing to do is to start from the definitions and proceed slowly and methodically. You have all the definitions in front of you. If you try to do the problem yourself what answer do you get? $\endgroup$
    – DanielSank
    Jun 29, 2015 at 7:10

1 Answer 1

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Alas, it was very simple! $$\delta U_j =-i\Delta t \delta u_k(j)\mathcal{H}_kU_j \, . \tag{1}$$

$$\frac{\delta U_j}{\delta u_k(j)} =-i\Delta t \mathcal{H}_kU_j \, . \tag{2}$$

Now, looking at $\Phi_o$, we see the only terms that actually have a $u_k(j)$ dependence are $U_j$ and $U_j^{\dagger}$, hence we have the following via the product rule: $$\frac{\delta\Phi_o}{\delta u_k(j)} =\langle U^{\dagger}_{j+1}... U^{\dagger}_{N}CU_N...U_{j+1}|\frac{\delta U_{j}}{\delta u_k(j)}... U_1CU_1^{\dagger}...U^{\dagger}_{j}\rangle + \langle U^{\dagger}_{j+1}... U^{\dagger}_{N}CU_N...U_{j+1}|U_{j}... U_1CU_1^{\dagger}...\frac{\delta U^{\dagger}_{j}}{\delta u_k(j)}\rangle\, . \tag{3}$$ From equation two, and the adjoint of equation two, we see that this simply reduces to the following: $$-\langle\lambda_j|i\Delta t(\mathcal{H}_k\rho_j-\rho_j\mathcal{H}_k)\rangle=-\langle\lambda_j|i\Delta t[\mathcal{H}_k,\rho_j]\rangle \, . \tag{4}$$

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    $\begingroup$ Glad you figured it out. Way to get your own bounty :-) $\endgroup$
    – DanielSank
    Jun 30, 2015 at 0:39

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