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Suppose I have a quantum Hamiltonian $H(\{\theta_{k}\}, \{n^{\theta}_{k}\}; \lambda)$, where each $\theta_{k}$ and $n_{k}$ are conjugate variables satisfying a commutation relations $[\theta_{k}, n^{\theta}_{k'}] = i \delta_{k,k'}$, while $\lambda$ is taken as a classical parameter (let's assume, for now, that it's time-independent). In practice $\lambda$ may be "grouped" with different terms in the Hamiltonian, but one could define transformations of the form $U=\exp(-i \lambda n^{\theta}_{k})$ to "change the grouping". For example, let us consider the following $H=U + T$, where $$U=- E_J \cos(\theta) + E_L(\theta - \lambda)^2 $$ and $$T=E_C (n^{\theta})^2 $$ So clearly, $\lambda$ is "grouped" with the term with prefactor $E_L$. Now let's define $$U=\exp(-i \lambda n^{\theta}))$$ which lets us write a transformed Hamiltonian*** $$H' = U^{\dagger} H U$$ (assuming $\lambda$ is time independent) that results in $$H'= E_C (n^{\theta})^2 - E_J \cos(\theta + \lambda) + E_L(\theta)^2 $$ Clearly now the "grouping" of $\lambda$ is with the terms with prefactor $E_J$.

We should expect the physics of those two descriptions to be the same. Namely if $|k>$ and $|k'>$ are the eigenvectors of the non-primed and primes Hamiltonians, then $<k|H|m> = <k'|H'|m'>$, and the same would be true for any observable operator A and A'. Namely $<k|A|m> = <k'|A'|m'>$.

... now my problem comes with the following. It seems it's very common to assume that $\lambda$ can be written as $$\lambda = \lambda_0 + \delta \lambda$$ where $\lambda_0$ is typically assumed static and $\delta \lambda$ as small (usually time-dependent) correction. Then one can write an approximate Hamiltonian, using a Taylor expansion, as $$H \approx H(\lambda=\lambda_0) + \frac{\partial H }{\partial \lambda}|_{\lambda_0} \delta \lambda $$ The second term can then treated as a perturbation, with $\delta \lambda$ regarded as a "noise term". Then (for example) using Fermi's Golden Rule, one can calculate decay rates between eigenstates of $H(\lambda=\lambda_{0})$, which depend on matrix elements $<k| \frac{\partial H }{\partial \lambda} |m>$.

The problem for me comes with the fact that it would seem like the answer (say the rates) that I would get actually depends on how I have "grouped" $\lambda$. This can be shown for the example above, but also in general it can be shown that $$<k| \frac{\partial H }{\partial \lambda} |m> \ne <k'| \frac{\partial H' }{\partial \lambda}|m'>$$ if $k\ne m$. So this would tell me that one of the grouping of $\lambda$ is "more special" than the other, but of course this cannot be, since I know that I can go between the groupings of $\lambda$ with a simple unitary transformation.

So my question is really: what am I missing in this discussion? I guess, the Taylor expansion is only an approximation to the effective Hamiltonian, but even if so, then which particular grouping is a better reflection of the "reality", and will lead to more accurate results (for example when calculating relaxation rates)?

For those interested, this technique of looking at small perturbative Taylor expansions, is used (for example) when looking at effects of noise (due to flux for example) in superconducting circuits.

Thanks!

***: I realize that in order to arrive at $H'$, I assumed that $\lambda$ is time independent, but later in the Taylor expansion, I take $\lambda \delta$ to potentially have time-dependence, as is customary in these calculations. I think, however, that the question is still valid, even If I were to assume $\delta \lambda$ is small and time-independent.

A couple of specific references, where this Taylor expansion is done:

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I suspect your confusion comes from the fact that the unitary transformation $U$ you are applying depends of $\lambda$.

When you compute transition rates, you are looking at a decomposition of your state vector as:

$$\left| \psi \right\rangle =: \sum_k \psi_k(t) e^{-ikt} \left|k\right\rangle$$

where $\left|k\right\rangle$ are the eigenvectors of $H(\lambda_o)$. The transition rates govern the time-evolution of the $\psi_k(t)$:

$$\frac{d\psi_k}{dt}(t) = -i \sum_l \left\langle k \middle| \delta H \middle| l \right\rangle \psi_l(t) e^{i(k-l)t}$$

and $\delta H = H(\lambda) - H(\lambda_o)$ is the difference between the Hamiltonian $H(\lambda)$ under which $\left|\psi\right\rangle$ evolves and the unperturbed Hamiltonian $H(\lambda_o)$ whose eigenvectors are the $\left|k\right\rangle$.

So, if we want to use $\delta H' = H'(\lambda) - H'(\lambda_o)$ instead, we need to understand the exact meaning of the transition rates we would be computing. Now, the perturbed Hamiltonian under which $\left|\psi\right\rangle'$ evolves is $H'(\lambda)$, so $\left|\psi\right\rangle'$ must be $U^{\dagger}(\lambda) \left|\psi\right\rangle$ (assuming that $\lambda$ is time-independent...), while the basis vectors $\left|k\right\rangle'$ are eigenvectors of $H'(\lambda_o)$, ie $\left|k\right\rangle' = U^{\dagger}(\lambda_o) \left|k\right\rangle$. In other words, we are looking at:

$$\left| \psi \right\rangle' = \sum_l \psi_l(t) e^{-ilt} U^{\dagger}(\lambda) \left|l\right\rangle\\ = \sum_k \left(\sum_l \psi_l(t) e^{i(k-l)t} \left\langle k \middle| U(\lambda_o) U^{\dagger}(\lambda) \middle|l\right\rangle \right) e^{-ikt} \left|k\right\rangle'\\ =: \sum_k \psi'_k(t) e^{-ikt} \left|k\right\rangle'$$

Oops, $\psi'_k(t) \neq \psi_k(t)$! That's why the transition rates that govern their time evolution are different! In fact, if we go through the calculation, say at order 1 in $\delta\lambda$, the transition rates will differ exactly in the right way to account for the different definition of $\psi_k(t)$ vs $\psi'_k(t)$ (expanded at order 1 as well).

Of course, which answer is the "right" one depends, at the end of the day, of what exactly you are measuring, ie whether your experiment follows the evolution of $\psi'_k(t)$ or of $\psi_k(t)$. To illustrate this, let's consider two simple experimental protocols:

  1. The initial decomposition of $\left|\psi\right\rangle$ on the basis $\left|k\right\rangle$ is measured at $t=0^-$, then, from $t=0^+$ to $t=\tau^-$, the classical perturbation $\delta\lambda$ is turned on, so that, during this time $\left|\psi\right\rangle$ evolves under $H(\lambda)$, and, finally, the decomposition of $\left|\psi\right\rangle$ on the basis $\left|k\right\rangle$ is measured again at $t=\tau^+$. In this case, the experience measures the evolution of $\psi_k(t)$.

  2. Same protocol, but the details of "turning on" $\delta\lambda$ now have the side effect of rotating the state vector by $U(\lambda)U^{\dagger}(\lambda_o)$, so that $$\left|\psi(0^+)\right\rangle = U(\lambda)U^{\dagger}(\lambda_o) \left|\psi(0^-)\right\rangle$$ and $$\left|\psi(\tau^+)\right\rangle = U(\lambda_o)U^{\dagger}(\lambda) \left|\psi(\tau^-)\right\rangle$$ Equivalently, $$\left|\psi(0^+)\right\rangle' = \sum_k \psi'_k(0) \left|k\right\rangle'$$ with the $\psi'_k(0)$ coinciding with the coefficients of $\left|\psi\right\rangle$ on the basis $\left|k\right\rangle$ at $t=0^-$, and similarly at $t=\tau^{\pm}$. Thus, in this case, the experience measures the evolution of $\psi'_k(t)$.

TL;DR:

  • If we consider only a single value of $\lambda$, the choice to apply $U(\lambda)$ or not is completely arbitrary: it's just a unitary change of basis after all, no physical outcome can ever depend of the basis in which we do the calculations.

  • But if the experiment at hand compares the evolution between different values of $\lambda$, this comparison will be affected by $\lambda$-dependent changes of basis, and we need to analyze precisely eg the dynamics of how precisely $\lambda$ is turned on to get the correct answer (as the example above show, we need to use the $\lambda$-dependent basis in which the "turning-on" is "transparent", ie only modify the Hamiltonian during the time $\delta\lambda$ is on, without affecting the state vector at the times it is turned on/off).

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  • $\begingroup$ thank you for your answer... i will get back to you shortly. $\endgroup$ – heythereitsme May 11 '17 at 16:34
  • $\begingroup$ Finally a chance to reply. Regarding your first sentence; I don't think I'm confused about $U$ depending on $\lambda$ - I'm ok with this, and in fact used this fact to show that $<k| \frac{\partial H }{\partial \lambda} |m> \ne <k'| \frac{\partial H' }{\partial \lambda}|m'>$... $\endgroup$ – heythereitsme May 12 '17 at 21:33
  • $\begingroup$ My "confusion" (trouble?) lies more with the fact that in papers I know, the grouping of $\lambda$ is usually somewhat arbitrary (because after all the dynamics are grouping-independent), but perhaps this whole discussion shows is that when taking the derivative to calculate the (approximate) noise-coupling operator (i.e. $I_{\rm noise} = \partial H/\partial \lambda$), the grouping really does matter... $\endgroup$ – heythereitsme May 12 '17 at 21:34
  • $\begingroup$ So maybe what one needs instead, is to start with some particular $\lambda$ grouping that corresponds to a "lab frame" - then obtain $I_{\rm noise}$ within that frame (by an argument that this is the "right" frame that defines $I_{\rm noise}$ in the lab), then, since any "regrouping" of $\lambda$ is done by some unitary operator U, that operator also needs to be applied to $I_{\rm noise}$ in order to get consistent results... does this sound reasonable to you? thanks again for taking the time to read and answer this post. $\endgroup$ – heythereitsme May 12 '17 at 21:34
  • $\begingroup$ I'm not familiar with your specific application domain, but the way I understand it, at the general level, is as follow: $\endgroup$ – Luzanne May 12 '17 at 21:45

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