1
$\begingroup$

Any unitary operation $U \in U(N)$ can be represented as $U = e^{iA}$, where $A$ is an arbitrary $N \times N$ Hermitian matrix. In the case of a time-independent Hamiltonian $H$, we have $A = -itH$ (thus $U = e^{-iHt}$), which corresponds to a unitary operator generated by $H$ for an evolution time $t$. For the time-dependent case, $A$ becomes more complex and is expressed in terms of a time-ordered exponential.

My question is about the matrix element of $A$, denoted as $A_{pq}$. Assuming I can always find a Hamiltonian $H$ (or $H(t)$) that can generate any unitary operator $U \in U(N)$, does this imply that each matrix element $A_{pq}$ of $A$ must possess a unique functional dependence on $H$ (or $H(t)$)?

The origin of my question lies in this Science report paper about quantum optimal control landscapes. Particularly, the question is essentially related to the transition from Eq.(2) to (3) there. The main argument they present is as follows: First, consider a time-dependent Hamiltonian $H(t)$ that depends on a control field $C(t)$. The unitary operator $U$ is generated by this Hamiltonian, indicating that $U$ is also a functional of $C(t)$. The cost function, which is the functional derivative of the transition probability from state $\lvert i \rangle$ to $\lvert f \rangle$ (i.e., $\lvert \langle i \rvert U \lvert f \rangle \rvert^2 = \lvert U_{if} \rvert^2$) with respect to $C(t)$, is denoted as \begin{equation} \frac{\delta \lvert U_{if}\rvert^2}{\delta C(t)}. \end{equation} The paper then makes use of the fact that $U = e^{iA}$ where $A$ is an arbitrary Hermitian matrix, and therefore, $U_{if} = \langle i \rvert U \rvert f \rangle = \langle i \rvert e^{iA} \lvert f \rangle$, leading to the expression

\begin{equation} \frac{\delta \lvert U_{if} \rvert^2}{\delta C(t)} = \sum_{p,q} \frac{\partial \lvert U_{if} \rvert^2}{\partial A_{pq}} \frac{\delta A_{pq}}{\delta C(t)}. \end{equation}

The argument they then rely on is that each matrix element $A_{pq}[C(t)]$ of $A$ must be independently addressable, implying that each $A_{pq}[C(t)]$ should have a unique functional dependence on $C(t)$ based on the "controllability assumption" they introduced earlier. Therefore, they claimed that the set of functions $\{\delta A_{pq}/\delta C(t)\}$ are linearly independent for all $p,q,$ and $t$. As I understood from Eq.(1), the controllability assumption essentially states that there exists at least one control field $C(t)$ such that $\lvert U_{if} \rvert^2 = 1$ corresponds to $\frac{\delta \lvert U_{if} \rvert^2}{\delta C(t)} = 0$ for all time $t \in [0, T]$. However, I don't fully understand how such arguments put together to get such conclusion. I'd be greatly appreciate it if anyone could provide any insight and explanation. I know my question tends to compress a lot of things all at once, so please feel free to leave any question or comment for any clarification.

$\endgroup$
6
  • $\begingroup$ Hello! Sorry, I find the question very unclear. What do you want to know, in simple terms? Which is the point that is puzzling you? $\endgroup$
    – Quillo
    Commented Jul 23, 2023 at 8:04
  • $\begingroup$ @Quillo hey, thanks for the comment. Denote an arbitrary unitary $U = e^{iA}$, where $A$ is an arbitrary Hermitian matrix. $U$ is generated by Hamiltonian $H(t)$, which depends on some control function $C(t)$. For example, $H(t) = C(t)H$ where $H$ is some Hermitian matrix. Then, I am curious to know under what condition each matrix element $A_{pq}$ of $A$ can be independently addressable, i.e. a set of functions $\{ \delta A_{pq}/ \delta C(t) \}$ are linearly independent for all $p,q$. $\endgroup$
    – Jon Megan
    Commented Jul 23, 2023 at 22:54
  • 1
    $\begingroup$ @Quillo It seems like in the paper, such condition is when there always exists $C(t)$ such that $U[C(t)]$ can produce any target state $\lvert f \rangle$. It roughly makes sense to me because if so, then every $N^2$ matrix elements of $A$ can be generated independently by $C(t)$. However, I want to see it more rigorously. Also, if so, then doesn't it also imply that every $N^2$ matrix elements of $U$ can also be addressed independently by $C(t)$? I'm not sure about the relationship between the matrix element of $A$ or $U$ and the quantum state that it can generate/reach. $\endgroup$
    – Jon Megan
    Commented Jul 23, 2023 at 23:06
  • $\begingroup$ I think your question is mostly about scrutinising a (sloppy but well-motivated) physical argument related to an integral of a matrix and then about its imaginary exponential. It might be better answered in Maths.SE under Lie groups. $\endgroup$ Commented Jul 24, 2023 at 7:57
  • $\begingroup$ @JonMegan I totally agree with your concerns, the paper is very unclear and it seems that this point is not discussed at all! I posted a very incomplete answer, I hope it could be somehow useful to other interested people. $\endgroup$
    – Quillo
    Commented Jul 24, 2023 at 11:59

1 Answer 1

2
$\begingroup$

Note: This is NOT a satisfactory answer, as I feel that I am missing some possibly important point! However, the question is delicate and technical and I hope that this starting point could help someone else to provide a better answer (or an interested reader to study the problem). For an introduction to quantum control, see Quantum control Landscapes (from the same authors of the paper discussed in this post) or the links here.

The analysis in Quantum Optimally Controlled Transition Landscapes is restricted to quantum systems with a finite number $N$ of possible states (the Hilbert space is just a finite-dimensional vector space). The fundamental assumption is that the entire $N$-dimensional Hilbert space is just the "controllable" portion of a possibly larger system that also contains states that are not addressed by the control.

Quantum control - In this context, "quantum control" means that we want to maximize the probability $P_{if}=|U_{if}|^2$ to transition from an initial state $|i\rangle$ to a chosen final state $|f\rangle$, where $U_{if}=\langle i|U|f \rangle \in U(N)$ is a $N\times N$ unitary matrix. The time evolution operator $U=U[C(t)]$ is a functional of the control function $C(t)$: typically one adds to the Hamiltonian $H_0$ an external field that depends on $C(t)$. The goal is to maximize the chance to transition to a desired state $|f\rangle$ by choosing the best $C(t)$. In other terms, the whole point is to maximize $P_{if}[C]$ with respect to $C(t)$ by considering the functional derivative $\delta P_{if}[C]/\delta C(t)$. In most cases, $A= \int_0^T dt (H_0+C(t) f)$, where $f$ is a suitable operator that models experimental control methodology (i.e. a "Dirac-delta" version of more general kernel operators, see equation 2.1c here). More generally you should have a T-ordered exp.

Notation - The paper proceeds by saying that it is more convenient to introduce the generator $A$, $U=\exp(iA)$ with $A^\dagger = A$. Therefore, $A$ is an Hermitian $N\times N$ matrix. Since we have $N$ independent states, call $|q\rangle$ a basis of our Hilbert space ($q=1...N$), so that $|f \rangle = \sum_{q=1}^N|q \rangle \langle q|f \rangle $ and similarly for $|i \rangle$. Hence, $A_{pq} = \langle p|A|q \rangle$ is the matrix element on this basis ($p$ is just another index to label the $N$ basis states of the Hilbert space).

Controllability - We can write $\delta P_{if}[C]/\delta C(t)$ in terms of $A$ via simple calculation, as described in the question:

$$ \dfrac{P_{if}[C]}{\delta C(t)} = \sum_{p,q} \frac{\partial \lvert U_{if} \rvert^2}{\partial A_{pq}} \frac{\delta A_{pq}}{\delta C(t)} $$

This equation in terms of $A$ turns out to be convenient and useful if the only way of having $\dfrac{P_{if}[C]}{\delta C(t)} =0$ is that $\frac{\partial \lvert U_{if} \rvert^2}{\partial A_{pq}}=0$ for all $p,q$!

Therefore, it seems to me that there is nothing particularly deep: the authors just claim that it is possible to make all the (not well motivated IMO) assumptions they need to guarantee this fact, e.g. "the set of functions $\delta A_{pq}/\delta C(t)$ for $p,q=1...N$ are all linearly independent (whatever this means)" and $A=\sum_{q,p}|p\rangle A_{pq}[C(t)]\langle q|$ is an element of a Hermitian operator, at each time $t$ (namely, time dependence does not spoil the good properties of $A$).

$\endgroup$
3
  • 1
    $\begingroup$ Thanks you for your comment. I am not sure if I understand the first paragraph properly. What do you mean by "𝑁-dimensional Hilbert space is just the "controllable" portion of a possibly larger system"? $\endgroup$
    – Jon Megan
    Commented Jul 24, 2023 at 17:21
  • $\begingroup$ I also believe that this assumption is not well-explained. I have thought about it deeper and now it at least makes sense that every matrix element of $A$ should be independent with respect to $C(t)$. This is because we produce every $2^N$ computational basis states: $\lvert 00 \cdots 00 \rangle, \dotsc, \lvert 11 \cdots 11\rangle$. Since this is a (sub)set of basis for any $U$, it kind of makes sense that every element matrix of $U$ can be independently addressable w.r.t $C(t)$, and so thus $A$ ($A$ is hermitian, so matrix element of $A$ must be directly related to $U = e^{iA}$). $\endgroup$
    – Jon Megan
    Commented Jul 24, 2023 at 17:26
  • 1
    $\begingroup$ @JonMegan it means that typically the Hilbert space is much larger and, possibly, uncontrollable (it is infinite-dimensional). However, there may be a finite number $N$ of (orthogonal) states that can be reached ("prepared") via experimental techniques. Therefore, the analysis is restricted to this subspace (or to spin systems that are naturally finite dimensional.. or an ensemble of Q-bits). $\endgroup$
    – Quillo
    Commented Jul 24, 2023 at 17:31