Analytic Solutions of Control Mechanism in Single-Qubit Systems

Quantum control studies the evolution of systems whose Hamiltonian $H=H_{0}+H_{I}(t)$ is the sum of a field-free Hamiltonian $H_{0}$ and an interacting Hamiltonian $H_{I}(t)$ with zero diagonal which depends on the control field. The eigenstates of $H_{0}$ are denoted $\ket{i}$ with $H_{0}\ket{i}=E_{i}\ket{i}$ for $i\in\quantity{1,\dots,d}$, where $d$ is the dimension of the system. In the interaction picture, the time-dependent Schrödinger equation is given as:

$$i\hbar\derivative{t}U(t)=V(t)U(t)$$

(1)

where $V(t)\equiv\exp(iH_{0}t/\hbar)H_{I}(t)\exp(-iH_{0}t/\hbar)$. In the general case where $V(t)$ don’t commute at different times, this equation is solved by the time-ordered exponential $U(t)=\mathcal{T}\exp(-\frac{i}{\hbar}\int_{0}^{t}V(t)\differential{t})$. Expanding this exponential yields the perturbative expansion for $U(t)$, known as the Dyson series, which is given as follows:

	$\displaystyle U$	$\displaystyle(t)=1+\quantity(\frac{-i}{\hbar})\int_{0}^{t}V(t_{1})\differential{t_{1}}$
		$\displaystyle+\quantity(\frac{-i}{\hbar})^{2}\int_{0}^{t}\!\int_{0}^{t_{2}}V(t_{2})V(t_{1})\differential{t_{1}}\differential{t_{2}}$
		$\displaystyle+\quantity(\frac{-i}{\hbar})^{3}\int_{0}^{t}\!\int_{0}^{t_{3}}\!\int_{0}^{t_{2}}V(t_{3})V(t_{2})V(t_{1})\differential{t_{1}}\differential{t_{2}}\differential{t_{3}}$
		$\displaystyle+\ \cdots$		(2)

Noting that $\sum_{i=1}^{d}\outerproduct{i}{i}=1$, we insert a complete set of basis states in between every matrix product. Adopting the notation $U_{ba}=\matrixelement{b}{U(T)}{a}$ and $v_{mn}(t)=-\frac{i}{\hbar}\matrixelement{m}{V(t)}{n}=-\frac{i}{\hbar}\matrixelement{j}{H_{I}(t)}{i}e^{i(E_{j}-E_{i})t/\hbar}$ yields:

		$\displaystyle U_{ba}=\delta_{ab}+\int_{0}^{T}v_{ba}(t_{1})\differential{t_{1}}$
		$\displaystyle+\sum_{l=1}^{d}\int_{0}^{T}\!\!\!\int_{0}^{t_{2}}v_{bl}(t_{2})v_{la}(t_{1})\differential{t_{1}}\differential{t_{2}}$
		$\displaystyle+\sum_{m=1}^{d}\!\sum_{n=1}^{d}\int_{0}^{T}\!\!\!\int_{0}^{t_{3}}\!\!\!\int_{0}^{t_{2}}\!\!v_{bm}(t_{3})v_{mn}(t_{2})v_{na}(t_{1})\differential{t_{1}}\differential{t_{2}}\differential{t_{3}}$
		$\displaystyle+\cdots$

We can define mechanism in terms of pathways directly from this expansion. We can express the transition amplitude $U_{ba}$ as:

$$U_{ba}=\sum_{n=0}^{\infty}\sum_{l_{n-1}=1}^{d}\cdots\sum_{l_{1}=1}^{d}U_{ba}^{n(l_{1},\dots,l_{n-1})}$$

where $U^{0()}_{ba}=\delta_{ba}$ and:

	$\displaystyle U^{n(l_{1},\dots,l_{n-1})}_{ba}\hskip 30.0pt$
	$\displaystyle\equiv\int_{0}^{T}\int_{0}^{t_{n}}\cdots\int_{0}^{t_{2}}$	$\displaystyle v_{bl_{n-1}}(t_{n})v_{l_{n-1}l_{n-2}}(t_{n-1})\cdots$
		$\displaystyle v_{l_{1}a}(t_{1})\differential{t_{1}}\cdots\differential{t_{n-1}}\differential{t_{n}}$		(3)

Each $U^{n(l_{1},\dots,l_{n-1})}_{ba}$ is called a pathway amplitude for reasons which will soon become clear. The superscript $n$ denotes the order of the amplitude, while the $l_{i}$ are intermediate states inserted in the expansion. It should be noted that no two adjacent states in a pathway should be equal because $V(t)$ has zero diagonal. Because each pathway amplitude is defined by a unique initial state, sequence of intermediate states, and ending state, we say that $U^{n(l_{1},\dots,l_{n-1})}_{ba}$ is the amplitude of the $n$th-order perturbative pathway $\ket{a}\to\ket{l_{1}}\to\dots\to\ket{l_{n-1}}\to\ket{b}$. Pathways with large or small amplitudes have large and small contributions to the control dynamics, respectively, and different pathways constructively and destructively interfere to drive the control dynamics of the system.

Before we even consider the Hamiltonian, the simplicity of the two-level single-qubit system means we should first inspect the possible pathways in the system. Notably, for each order, there is at most one valid pathway from a given start state to a given end state. For example, from $\ket{0}$ to $\ket{1}$, there is one first-order pathway $\ket{0}\to\ket{1}$, no second-order pathway, one third-order pathway $\ket{0}\to\ket{1}\to\ket{0}\to\ket{1}$, no fourth-order pathway, one fifth-order pathway $\ket{0}\to\ket{1}\to\ket{0}\to\ket{1}\to\ket{0}\to\ket{1}$, and so on. $\ket{0}$ to $\ket{0}$ and $\ket{1}$ to $\ket{1}$ have only even pathways, while $\ket{0}$ to $\ket{1}$ and $\ket{1}$ to $\ket{0}$ have only odd pathways, so we call the former “even” transitions and the latter “odd” transitions.

We now turn to the Hamiltonian. A single qubit can be described by $H_{0}=\omega_{0}S_{z}$ and $H_{I}=\epsilon_{x}(t)S_{x}+\epsilon_{y}(t)S_{y}$ so:

	$\displaystyle H(t)$	$\displaystyle=\omega_{0}S_{z}+\epsilon_{x}(t)S_{x}+\epsilon_{y}(t)S_{y}$
	$\displaystyle V(t)$	$\displaystyle=e^{i\omega_{0}S_{z}t/\hbar}(\epsilon_{x}(t)S_{x}+\epsilon_{y}(t)S_{y})e^{-i\omega_{0}S_{z}t/\hbar}$

We can expand this by noting that the exponentials act as rotation operators on the spin matrices:

	$\displaystyle V(t)$	$\displaystyle=\phantom{}(\epsilon_{x}(t)\cos\frac{\omega_{0}t}{\hbar}+\epsilon_{y}(t)\sin\frac{\omega_{0}t}{\hbar})S_{x}$
		$\displaystyle\phantom{=}+(\epsilon_{y}(t)\cos\frac{\omega_{0}t}{\hbar}-\epsilon_{x}(t)\sin\frac{\omega_{0}t}{\hbar})S_{y}$

We can now define interaction fields $\tilde{\epsilon}_{x}(t),\tilde{\epsilon}_{y}(t)$ to obtain a much more convenient Hamiltonian:

	$\displaystyle\tilde{\epsilon}_{x}(t)$	$\displaystyle=\epsilon_{x}(t)\cos\frac{\omega_{0}t}{\hbar}+\epsilon_{y}(t)\sin\frac{\omega_{0}t}{\hbar}$
	$\displaystyle\tilde{\epsilon}_{y}(t)$	$\displaystyle=\epsilon_{y}(t)\cos\frac{\omega_{0}t}{\hbar}-\epsilon_{x}(t)\sin\frac{\omega_{0}t}{\hbar}$
	$\displaystyle V(t)$	$\displaystyle=\tilde{\epsilon}_{x}(t)S_{x}+\tilde{\epsilon}_{y}(t)S_{y}$

Other than $\omega_{0}=0$ and a time-dependent rotation of the fields, this is identical to our original Hamiltonian. We can thus forget about $\omega_{0}$ entirely and work exclusively in this rotating interaction frame; we will do so moving forward, setting $\omega_{0}=0$, and as a result we will drop the tilde on $\tilde{\epsilon}_{x}=\epsilon_{x},\tilde{\epsilon}_{y}=\epsilon_{y}$ and use $V$ synonymously with $H$.

With the field-free Hamiltonian eliminated and the powers of $S_{x},S_{y}$ well-known, we can now expand the mechanism of a single pulse which is constant in time and active over a finite time interval. With interval $0\leq t<T$, we’ll write the Hamiltonian as $H=\omega(S_{x}\cos\phi+S_{y}\sin\phi)\equiv\omega S_{\phi}$ in the interval and $H=0$ otherwise, so that in the interval $H\ket{0}=\frac{\hbar}{2}\omega e^{i\phi}\ket{1}$ and $H\ket{1}=\frac{\hbar}{2}\omega e^{-i\phi}\ket{0}$. If we consider the mechanism from $t=0$ from $t=T$, we can forget about the transience of the pulse entirely and consider only the nonzero Hamiltonian. For the purpose of the expansion, we’ll define $S_{\phi}=\frac{\hbar}{2}\sigma_{\phi}$ and note that $\sigma_{\phi}^{2}=1$, so from Eq. II.1 the Dyson series collapses to:

	$\displaystyle U(T)$	$\displaystyle=1+\quantity(\frac{-i}{\hbar})T\omega\frac{\hbar}{2}\sigma_{\phi}+\quantity(\frac{-i}{\hbar})^{2}\frac{1}{2!}T^{2}\omega^{2}\frac{\hbar^{2}}{4}$
		$\displaystyle+\quantity(\frac{-i}{\hbar})^{3}\frac{1}{3!}T^{3}\omega^{3}\frac{\hbar^{3}}{8}\sigma_{\phi}+\quantity(\frac{-i}{\hbar})^{4}\frac{1}{4!}T^{4}\omega^{4}\frac{\hbar^{4}}{16}+\ \cdots$

Since there’s (at most) one pathway per order, we can read the pathways directly from this expansion. Since we need to distinguish even and odd order $n$, we’ll introduce some new notation: $n_{e}$ and $n_{o}$ will indicate that $n$ is even or odd, respectively. Also, since $T$ and $\omega$ always show up together, we’ll define the pulse width $\tau=T\omega$. Finally, since there’s at most one pathway per order, we don’t need the parentheses in the amplitude superscript. Directly from the expansion above, we conclude that the pulse with width $\tau$ and phase $\phi$ has mechanism:

	$\displaystyle U_{00}^{n_{e}}(T)$	$\displaystyle=\frac{1}{n!}\pqty{\frac{-i}{2}}^{n}\tau^{n},$	$\displaystyle U_{00}^{n_{o}}(T)=0$
	$\displaystyle U^{n_{o}}_{10}(T)$	$\displaystyle=\frac{1}{n!}\pqty{\frac{-i}{2}}^{n}e^{i\phi}\tau^{n},$	$\displaystyle U^{n_{e}}_{10}(T)=0$
	$\displaystyle U^{n_{e}}_{11}(T)$	$\displaystyle=\frac{1}{n!}\pqty{\frac{-i}{2}}^{n}\tau^{n},$	$\displaystyle U^{n_{o}}_{11}(T)=0$
	$\displaystyle U^{n_{o}}_{01}(T)$	$\displaystyle=\frac{1}{n!}\pqty{\frac{-i}{2}}^{n}e^{-i\phi}\tau^{n},$	$\displaystyle U^{n_{e}}_{01}(T)=0$		(4)

We will use this expansion extensively when we turn to piecewise constant controls. It also allows us to extrapolate that for any pulse sequence, $U_{00}^{n}=\overline{U_{11}^{n}}$ and $U_{10}^{n}=-\overline{U_{01}^{n}}$; we will not prove this here, but it is straightforwardly shown from the concatenation expansion of Section II.3.

To expand our focus to piecewise constant pulses, we need to explore mechanism concatenation. Let’s assume that we have two pulses whose lengths sum to $T$, $H(0<t<t_{s})=H_{1}(t)$ and $H(t_{s}<t<T)=H_{2}(t)$. If $H_{1},H_{2}$ are each zero outside their time interval, we can write this Hamiltonian as $H=H_{1}+H_{2}$. Now we can expand the time-ordered exponential $U(T)=\mathcal{T}\exp\pqty{\frac{-i}{\hbar}\int_{0}^{T}(H_{1}(t)+H_{2}(t))\differential{t}}$:

	$\displaystyle U(T)=1$	$\displaystyle+\quantity(\frac{-i}{\hbar})\int_{0}^{T}(H_{1}(t_{1})+H_{2}(t_{1}))\differential{t_{1}}$
		$\displaystyle+\quantity(\frac{-i}{\hbar})^{2}\int_{0}^{T}\!\int_{0}^{t_{2}}(H_{1}(t_{2})+H_{2}(t_{2}))$
		$\displaystyle\phantom{+\quantity(\frac{-i}{\hbar})^{2}\int_{0}^{T}\!\int_{0}^{t_{2}}}\!\cdot(H_{1}(t_{1})+H_{2}(t_{1}))\differential{t_{1}}\differential{t_{2}}$
		$\displaystyle+\ \cdots$

Expanding each piece out by splitting the integrals on $t_{s}$, we see:

	$\displaystyle\int_{0}^{T}$	$\displaystyle(H_{1}(t_{1})+H_{2}(t_{1}))\differential{t_{1}}$
		$\displaystyle=\int_{0}^{t_{s}}H_{1}(t_{1})\differential{t_{1}}+\int_{t_{s}}^{T}H_{2}(t_{1})\differential{t_{1}}$

and:

	$\displaystyle\int_{0}^{t}\!\int_{0}^{t_{2}}$	$\displaystyle(H_{1}(t_{2})+H_{2}(t_{2}))(H_{1}(t_{1})+H_{2}(t_{1}))\differential{t_{1}}\differential{t_{2}}$
	$\displaystyle=$	$\displaystyle\int_{0}^{t_{s}}\!\int_{0}^{t_{2}}H_{1}(t_{2})H_{1}(t_{1})\differential{t_{1}}\differential{t_{2}}$
		$\displaystyle+\int_{t_{s}}^{t}H_{2}(t_{2})\differential{t_{2}}\int_{0}^{t_{s}}H_{1}(t_{1})\differential{t_{1}}$
		$\displaystyle+\int_{t_{s}}^{t}\int_{t_{s}}^{t_{2}}H_{2}(t_{2})H_{2}(t_{1})\differential{t_{1}}\differential{t_{2}}$

and so on. We can prove the validity of this expansion in general by noting that $H_{1}$ and $H_{2}$ commute at any given time and splitting the time-ordered exponential. The validity of this expansion reflects the fact that we can write each order of the concatenated Dyson series by expanding a product of the Dyson series of the two individual pulses. This allows us to write the mechanism of the concatenated pulse as a convolution of the mechanisms of the individual pulses as well: for example, if ${}^{1}U$ and ${}^{2}U$ denote pathway amplitudes from the first and second pulses respectively, the pathway amplitude $U^{3(23)}_{41}$ of $\ket{1}\to\ket{2}\to\ket{3}\to\ket{4}$ in a four-level system would be:

$\displaystyle U^{3(23)}_{41}={{}^{1}}U^{3(23)}_{41}+{{}^{1}}U^{2(2)}_{31}\,{{}^{2}}U^{1()}_{43}+{{}^{1}}U^{1()}_{21}\,{{}^{2}}U^{2(3)}_{42}+{{}^{2}}U^{3(23)}_{41}$

By repeating this concatenation process, we can describe the concatenation of any number of pulses. While pulse concatenation works in general, we generally calculate mechanism by encoding the Hamltonian rather than by direct computation Mitra and Rabitz (2003), so this process is usually not any more efficient than simply encoding the Hamiltonian for the concatenated system. We explore piecewise constant single-qubit controls in this paper because we can analytically solve each constant pulse, allowing us to write down and manipulate closed-form expressions for the concatenated amplitudes.

The two-pulse case is a special case because it lends itself to a neat final result which unfortunately does not generalize to higher numbers of pulses. We’ll start with $U_{10}^{N_{o}}$, noting the odd and even subscripts in the summations. Applying the techniques of Section II.3 to Eq. II.2:

	$\displaystyle U$	${}^{N_{o}}_{10}(T)$
		$\displaystyle=\quantity(\frac{-i}{2})^{N}\left(\sum_{n_{o}<N}\frac{1}{n!}e^{i\phi_{1}}\tau_{1}^{n}\frac{1}{(N-n)!}\tau_{2}^{N-n}\right.$
		$\displaystyle\qquad\qquad\quad\left.+\sum_{n_{e}<N}\frac{1}{n!}\tau_{1}^{n}\frac{1}{(N-n)!}e^{i\phi_{2}}\tau_{2}^{N-n}\right)$
		$\displaystyle=\quantity(\frac{-i}{2})^{N}\left(\frac{1}{N!}\sum_{n_{o}}\binom{N}{n}\tau_{1}^{n}\tau_{2}^{N-n}e^{i\phi_{1}}\right.$
		$\displaystyle\qquad\qquad\quad\left.+\frac{1}{N!}\sum_{n_{e}}\binom{N}{n}\tau_{1}^{n}\tau_{2}^{N-n}e^{i\phi_{2}}\right)$

We can deal with the odd and even sums by defining $\tau_{+}=\tau_{1}+\tau_{2}$ and $\tau_{-}=-\tau_{1}+\tau_{2}$ so that:

	$\displaystyle\tau_{+}^{N}$	$\displaystyle=(\tau_{1}+\tau_{2})^{N}=\sum_{n}\binom{N}{n}\tau_{1}^{n}\tau_{2}^{N-n}$
	$\displaystyle\tau_{-}^{N}$	$\displaystyle=(-\tau_{1}+\tau_{2})^{N}=\sum_{n}\binom{N}{n}(-\tau_{1})^{n}\tau_{2}^{N-n}$		(5)

These polynomials allow us to concisely describe the odd and even sums:

	$\displaystyle\sum_{n_{e}<N}\frac{1}{n!}\tau_{1}^{n}\frac{1}{(N-n)!}\tau_{2}^{N-n}$	$\displaystyle=\frac{1}{N!}\sum_{n_{e}}\binom{N}{n}\tau_{1}^{n}\tau_{2}^{N-n}$
		$\displaystyle=\frac{1}{2}\frac{1}{N!}(\tau_{+}^{N}+\tau_{-}^{N})$
	$\displaystyle\sum_{n_{o}<N}\frac{1}{n!}\tau_{1}^{n}\frac{1}{(N-n)!}\tau_{2}^{N-n}$	$\displaystyle=\frac{1}{N!}\sum_{n_{o}}\binom{N}{n}\tau_{1}^{n}\tau_{2}^{N-n}$
		$\displaystyle=\frac{1}{2}\frac{1}{N!}(\tau_{+}^{N}-\tau_{-}^{N})$

We can now proceed with the pathway amplitude:

	$\displaystyle U$	${}_{10}^{N_{o}}(T)$
		$\displaystyle=\!\pqty{\frac{-i}{2}}^{N}\!\!\pqty{e^{i\phi_{1}}\frac{1}{2}\frac{1}{N!}(\tau_{+}^{N}-\tau_{-}^{N})+\!e^{i\phi_{2}}\frac{1}{2}\frac{1}{N!}(\tau_{+}^{N}+\tau_{-}^{N})\!}$
		$\displaystyle=\!\pqty{\frac{-i}{2}}^{N}\!\frac{1}{N!}\frac{1}{2}\pqty{e^{i\phi_{1}}\tau_{+}^{N}-e^{i\phi_{1}}\tau_{-}^{N}+e^{i\phi_{2}}\tau_{+}^{N}+e^{i\phi_{2}}\tau_{-}^{N}}$
		$\displaystyle=\!\pqty{\frac{-i}{2}}^{N}\!\frac{1}{N!}\frac{1}{2}((e^{i\phi_{1}}+e^{i\phi_{2}})(\tau_{1}+\tau_{2})^{N}$
		$\displaystyle\qquad\qquad\qquad\quad+(-e^{i\phi_{1}}+e^{i\phi_{2}})(-\tau_{1}+\tau_{2})^{N})$

This form allows us to concisely evaluate the pathway amplitude without performing sums over $O(N)$ terms. For larger $M$, this will be our final form, but we can proceed to a more pleasing expression in the two-pulse case by defining $\phi_{+}=(\phi_{1}+\phi_{2})/2$ and $\phi_{-}=(\phi_{1}-\phi_{2})/2$:

	$\displaystyle U_{10}^{N_{o}}(T)=\!\pqty{\frac{-i}{2}}^{N}\!\frac{1}{N!}e^{i\phi_{+}}($	$\displaystyle(\tau_{1}+\tau_{2})^{N}\cos\phi_{-}$
	$\displaystyle+$	$\displaystyle(-\tau_{1}+\tau_{2})^{N}i\sin\phi_{-})$

A similar expansion for $U_{00}^{N_{e}}(T)$ gives:

	$\displaystyle U$	${}^{N_{e}}_{00}(T)$
		$\displaystyle=\quantity(\frac{-i}{2})^{N}\left(\sum_{n_{o}<N}\frac{1}{n!}e^{i\phi_{1}}\tau_{1}^{n}\frac{1}{(N-n)!}e^{-i\phi_{2}}\tau_{2}^{N-n}\right.$
		$\displaystyle\qquad\qquad\quad\left.+\sum_{n_{e}<N}\frac{1}{n!}\tau_{1}^{n}\frac{1}{(N-n)!}\tau_{2}^{N-n}\right)$
		$\displaystyle=\!\pqty{\frac{-i}{2}}^{N}\!\frac{1}{N!}\frac{1}{2}((1+e^{i(\phi_{1}-\phi_{2})})(\tau_{1}+\tau_{2})^{N}$
		$\displaystyle\qquad\qquad\qquad\quad+(1-e^{i(\phi_{1}-\phi_{2})})(\tau_{1}-\tau_{2})^{N})$
		$\displaystyle=\!\pqty{\frac{-i}{2}}^{N}\!\frac{1}{N!}e^{i\phi_{-}}((\tau_{1}+\tau_{2})^{N}\cos\phi_{-}$
		$\displaystyle\qquad\qquad\qquad\quad+(-\tau_{1}+\tau_{2})^{N}i\sin\phi_{-})$

These are the analytic solutions for the mechanism of the two-pulse case. Taking $\tau_{2}=0$ or $\phi_{1}=\phi_{2}$ reduces this to the one-pulse case and recovers Eq. II.2 as expected.

We will now proceed with the three-pulse case in largely the same way as the two-pulse case. Some choices of notation may seem odd at first, but they are chosen to be suggestive of the general case. We start by applying Section II.3 to Eq. II.2 twice:

	$\displaystyle U^{N_{o}}_{10}(T)\qquad$
	$\displaystyle=\pqty{\frac{-i}{2}}^{N}$	$\displaystyle\left(\sum_{n_{e}+m_{e}+l_{o}=N}\frac{1}{n!m!l!}\tau_{1}^{n}\tau_{2}^{m}\tau_{3}^{l}e^{i\phi_{3}}\right.$
	$\displaystyle+$	$\displaystyle\sum_{n_{o}+m_{e}+l_{e}=N}\frac{1}{n!m!l!}\tau_{1}^{n}\tau_{2}^{m}\tau_{3}^{l}e^{i\phi_{1}}$
	$\displaystyle+$	$\displaystyle\sum_{n_{e}+m_{o}+l_{e}=N}\frac{1}{n!m!l!}\tau_{1}^{n}\tau_{2}^{m}\tau_{3}^{l}e^{i\phi_{2}}$
	$\displaystyle+$	$\displaystyle\left.\sum_{n_{o}+m_{o}+l_{o}=N}\frac{1}{n!m!l!}\tau_{1}^{n}\tau_{2}^{m}\tau_{3}^{l}e^{i(\phi_{1}-\phi_{2}+\phi_{3})}\right)$
	$\displaystyle\equiv\!\pqty{\frac{-i}{2}}^{N}$	$\displaystyle\!\frac{1}{N!}(\Sigma_{\quantity{}}e^{i\phi_{3}}\!+\!\Sigma_{\quantity{1}}e^{i\phi_{1}}$
		$\displaystyle\quad+\Sigma_{\quantity{2}}e^{i\phi_{2}}\!+\!\Sigma_{\quantity{1,2}}e^{i(\phi_{1}\!-\!\phi_{2}\!+\!\phi_{3})})$		(6)

Where the sets in the $\Sigma$ subscripts describe which of the first two integers in the sum are odd. We once again deal with the odd and even sums by defining polynomials in $\tau$. Using the multinomial coefficients $\binom{N}{n,m,l}=\frac{N!}{n!m!l!}$ with $n+m+l=N$, we write:

	$\displaystyle\tau_{\Bqty{}}^{N}$	$\displaystyle\equiv(\tau_{1}+\tau_{2}+\tau_{3})^{N}$
		$\displaystyle=\sum_{n+m+l=N}\binom{N}{n,m,l}\tau_{1}^{n}\tau_{2}^{m}\tau_{3}^{l}$
	$\displaystyle\tau_{\Bqty{1}}^{N}$	$\displaystyle\equiv(-\tau_{1}+\tau_{2}+\tau_{3})^{N}$
		$\displaystyle=\sum_{n+m+l=N}\binom{N}{n,m,l}(-\tau_{1})^{n}\tau_{2}^{m}\tau_{3}^{l}$
	$\displaystyle\tau_{\Bqty{2}}^{N}$	$\displaystyle\equiv(\tau_{1}-\tau_{2}+\tau_{3})^{N}$
		$\displaystyle=\sum_{n+m+l=N}\binom{N}{n,m,l}\tau_{1}^{n}(-\tau_{2})^{m}\tau_{3}^{l}$
	$\displaystyle\tau_{\Bqty{1,2}}^{N}$	$\displaystyle\equiv(-\tau_{1}-\tau_{2}+\tau_{3})^{N}$
		$\displaystyle=\sum_{n+m+l=N}\binom{N}{n,m,l}(-\tau_{1})^{n}(-\tau_{2})^{m}\tau_{3}^{l}$		(7)

Where the sets in the $\tau$ subscripts describe which $\tau_{i}$ are negated. Such powers of $\tau$ sums can be written as a linear combination of $\Sigma$ sums as in Eq. III.1:

	$\displaystyle\tau_{\quantity{}}^{N_{o}}=(\tau_{1}+\tau_{2}+\tau_{3})^{N_{o}}$	$\displaystyle=\Sigma_{\quantity{}}+\Sigma_{\quantity{1}}+\Sigma_{\quantity{2}}+\Sigma_{\quantity{1,2}}$
	$\displaystyle\tau_{\quantity{1}}^{N_{o}}=(-\tau_{1}+\tau_{2}+\tau_{3})^{N_{o}}$	$\displaystyle=\Sigma_{\quantity{}}-\Sigma_{\quantity{1}}+\Sigma_{\quantity{2}}-\Sigma_{\quantity{1,2}}$
	$\displaystyle\tau_{\quantity{2}}^{N_{o}}=(\tau_{1}-\tau_{2}+\tau_{3})^{N_{o}}$	$\displaystyle=\Sigma_{\quantity{}}+\Sigma_{\quantity{1}}-\Sigma_{\quantity{2}}-\Sigma_{\quantity{1,2}}$
	$\displaystyle\tau_{\quantity{1,2}}^{N_{o}}=(-\tau_{1}-\tau_{2}+\tau_{3})^{N_{o}}$	$\displaystyle=\Sigma_{\quantity{}}-\Sigma_{\quantity{1}}-\Sigma_{\quantity{2}}+\Sigma_{\quantity{1,2}}$		(8)

This system is inverted as:

	$\displaystyle 4\Sigma_{\quantity{}}$	$\displaystyle=\tau^{N_{o}}_{\quantity{}}+\tau^{N_{o}}_{\quantity{1}}+\tau^{N_{o}}_{\quantity{2}}+\tau^{N_{o}}_{\quantity{1,2}}$
	$\displaystyle 4\Sigma_{\quantity{1}}$	$\displaystyle=\tau^{N_{o}}_{\quantity{}}-\tau^{N_{o}}_{\quantity{1}}+\tau^{N_{o}}_{\quantity{2}}-\tau^{N_{o}}_{\quantity{1,2}}$
	$\displaystyle 4\Sigma_{\quantity{2}}$	$\displaystyle=\tau^{N_{o}}_{\quantity{}}+\tau^{N_{o}}_{\quantity{1}}-\tau^{N_{o}}_{\quantity{2}}-\tau^{N_{o}}_{\quantity{1,2}}$
	$\displaystyle 4\Sigma_{\quantity{1,2}}$	$\displaystyle=\tau^{N_{o}}_{\quantity{}}-\tau^{N_{o}}_{\quantity{1}}-\tau^{N_{o}}_{\quantity{2}}+\tau^{N_{o}}_{\quantity{1,2}}$		(9)

So we can now write:

	$\displaystyle U^{N_{o}}_{10}(T)\qquad$
	$\displaystyle=\!\pqty{\frac{-i}{2}}^{N}\!\frac{1}{N!}$	$\displaystyle\hskip 1.5pt\frac{1}{4}(\tau^{N_{o}}_{\quantity{}}{}(e^{i\phi_{3}}+e^{i\phi_{1}}+e^{i\phi_{2}}+e^{i(\phi_{1}-\phi_{2}+\phi_{3})})$
		$\displaystyle+\tau^{N_{o}}_{\quantity{1}}{}(e^{i\phi_{3}}-e^{i\phi_{1}}+e^{i\phi_{2}}-e^{i(\phi_{1}-\phi_{2}+\phi_{3})})$
		$\displaystyle+\tau^{N_{o}}_{\quantity{2}}{}(e^{i\phi_{3}}+e^{i\phi_{1}}-e^{i\phi_{2}}-e^{i(\phi_{1}-\phi_{2}+\phi_{3})})$
		$\displaystyle+\tau^{N_{o}}_{\quantity{1,2}}{}(e^{i\phi_{3}}-e^{i\phi_{1}}-e^{i\phi_{2}}+e^{i(\phi_{1}-\phi_{2}+\phi_{3})}))$		(10)

This reflects the general form that we will see in the next section. The case of $U_{00}^{N_{e}}(T)$ is solved similarly, so we will not detail it here.

We will now describe a general procedure for calculating any pathway amplitude $U_{ba}^{N}$ for a piecewise-constant pulse on a single qubit. This procedure will assume that the parity of $N$ matches the parity of the transition (e.g. $U_{10}^{N}$ should have $N$ odd) since the amplitude is zero otherwise. We’ll break up the piecewise-constant pulse into $M$ constant pulses, where each pulse has width $\tau_{m}$ and phase $\phi_{m}$. To prepare, we will define the powerset $\mathcal{P}_{M-1}$ as the set of all subsets of $\Bqty{1,\dots,M-1}$; each $S\in\mathcal{P}_{M-1}$ is a set of integers $s_{i}$ with $1\leq s_{1}<\dots<s_{|S|}\leq M-1$.

1. 1.

   Mirroring Eq. III.2, for each $S\in\mathcal{P}_{M-1}$ we define a sum $\Sigma_{S}$ over sets over integers $\Bqty{n_{1},\dots,n_{M}}$ satisfying $\sum_{m}n_{m}=N$, where for $m<M$ we have $n_{m}$ is odd for $m\in S$ and even for $m\notin S$, and $n_{M}$ takes the necessary parity for $\sum_{m}n_{m}=N$ ($n_{M}$ is even if $|S|\equiv N\mod 2$, odd otherwise):

   $$\Sigma_{S}\equiv\sum_{\begin{subarray}{c}\sum_{m}n_{m}=N,\\ \text{parity defined by $S$}\end{subarray}}\binom{N}{n_{1},\dots,n_{M}}\prod_{k}\tau_{k}^{n_{k}}$$

   (11)

2. 2.

   For each $S$ we define a phase $\phi_{S}$. This phase is defined slightly differently depending on the parities of $N$ and $|S|$:

   $\displaystyle\phi_{S}\equiv\begin{cases}\displaystyle\sum_{0<k\leq|S|}(-1)^{k-1}\phi_{s_{k}}&\text{$|S|$ odd, $N$ odd}\\ \displaystyle\sum_{0<k\leq|S|}(-1)^{k-1}\phi_{s_{k}}+\phi_{M}&\text{$|S|$ even, $N$ odd}\\ \displaystyle\sum_{0<k\leq|S|}(-1)^{k-1}\phi_{s_{k}}-\phi_{M}&\text{$|S|$ odd, $N$ even}\\ \displaystyle\sum_{0<k\leq|S|}(-1)^{k-1}\phi_{s_{k}}&\text{$|S|$ even, $N$ even}\end{cases}$

   (12)

3. 3.

   With both definitions above, we can now write the pathway amplitude in a form mirroring Eq. III.2:

   $$U_{ba}^{N}=\pqty{\frac{-i}{2}}^{N}\frac{1}{N!}\sum_{S\in\mathcal{P}_{M-1}}\Sigma_{S}e^{i\phi_{S}}$$

   (13)

4. 4.

   Let $I_{S}(k)=1$ if $k\in S$ and $I_{S}(k)=0$ otherwise. Mirroring Eq. III.2, we define a $\tau_{S}$ for each $S$:

   $$\tau_{S}\equiv\sum_{0<k\leq M}(-1)^{I_{S}(k)}\tau_{k}$$

   (14)

5. 5.

   Mirroring Eq. III.2, each $\tau_{S}^{N}$ can be uniquely written as a linear combination of $\Sigma_{S}$:

   $$\tau_{S}^{N}=\sum_{S^{\prime}\in\mathcal{P}_{M-1}}(-1)^{|S\cap S^{\prime}|}\Sigma_{S^{\prime}}$$

   (15)

6. 6.

   Under the natural order ¹¹1This order can be derived by counting in binary, i.e. by placing $S$ before $S^{\prime}$ if $\sum_{0<k\leq|S|}2^{s_{k}}<\sum_{0<k\leq|S^{\prime}|}2^{s_{k}}$. $\Bqty{}$, $\Bqty{1}$, $\quantity{2}$, $\quantity{2,1}$, $\quantity{3}$, $\dots$, the matrix coefficients of Eq. 15 form a Hadamard matrix, so we can invert this directly mirroring Eq. III.2:

   $$\Sigma_{S}=\frac{1}{2^{M-1}}\sum_{S^{\prime}\in\mathcal{P}_{M-1}}(-1)^{|S\cap S^{\prime}|}\tau_{S^{\prime}}^{N}$$

   (16)

7. 7.

   Now, following Eq. III.2, we can write:

   		$\displaystyle U_{ba}^{N}=\pqty{\frac{-i}{2}}^{N}\frac{1}{N!}\sum_{S\in\mathcal{P}_{M-1}}\Sigma_{S}e^{i\phi_{S}}$
   		$\displaystyle=\pqty{\frac{-i}{2}}^{N}\frac{1}{N!}\frac{1}{2^{M-1}}\sum_{S\in\mathcal{P}_{M-1}}\sum_{S^{\prime}\in\mathcal{P}_{M-1}}(-1)^{|S\cap S^{\prime}|}\tau_{S^{\prime}}^{N}e^{i\phi_{S}}$
   		$\displaystyle=\pqty{\frac{-i}{2}}^{N}\frac{1}{N!}\frac{1}{2^{M-1}}\sum_{S\in\mathcal{P}_{M-1}}\!\tau_{S}^{N}\!\sum_{S^{\prime}\in\mathcal{P}_{M-1}}(-1)^{|S\cap S^{\prime}|}e^{i\phi_{S^{\prime}}}$		(17)

This procedure gives a general solution for piecewise constant pulses on a one-qubit system. Since Eq. 15 takes the form of a Hadamard matrix, Eq. 7 can be evaluated by a fast Walsh-Hadamard transform in $O(M2^{M})$ time. This is significant both because it does not depend on $N$ and because it does not involve numerical methods, whereas directly evaluating the Dyson terms of an $N$th-order pathway usually requires computationally expensive $N$-dimensional numerical integration over time.