Exponentials and logarithms of Pauli matrices

Exponentiating Hamiltonian terms

Let \(A\) and \(B\) be one qubit unitaries (usually, Pauli matrices), acting on different qubits. They will therefore satisfy \(A^{2}=B^{2}=\mathbb{1}\), \(AB=BA\). Then, defining \(Q=\mathbb{1}+A+B+AB\), we have \(Q^{2}=4Q\). More generally, if \(A_{1},A_{2},...,A_{m}\) are \(m\) commuting unitaries, then the operator \(Q\) defined as the sum of all their possible products, \(Q=\prod_{k=1}^{m}(1+A_{k}),\) satisfies \(Q^{2}=2^{m}Q\), as is easily seen from the fact that \(A_{i}Q=Q\) for any \(i\). It follows that \(e^{i\theta Q}=1+\frac{Q}{2^{m}}(e^{i\theta 2^{m}}-1)\).

Let \(P\) be a projector operator, that is, an operator satisfying \(P^{2}=P\). Then its exponential is given by \(e^{i\theta P}=\mathbb{1}+P(e^{i\theta}-1)\), or, for \(\theta=\pi\), \(e^{i\pi P}=1-2P\).

Operator relation Exponential
\(Q^{2}=\lambda Q\) \(e^{i\theta Q}=1+\frac{Q}{\lambda}(e^{i\theta\lambda}-1)\)
\(P^{2}=P\) \(e^{i\theta P}=1+P(e^{i\theta}-1)\)
\(P^{2}=P\) \(e^{i\pi P}=1-2P\)
\(A^{2}=1\) \(e^{i\frac{\pi}{2}(1-A)}=A\)
\(A^{2}=1\) \(e^{i\theta A}=\cos\theta+i\sin\theta A\)

Using the above relations we see that \(e^{i\frac{\pi}{2}(1-\sigma_{i})}=ie^{-i\frac{\pi}{2}\sigma_{i}}=\sigma_{i}\), for all Pauli matrices \(\sigma_{i}\). More generally for any operator \(A\) such that \(A^{2}=\mathbb{1}\) we have \(\exp\left[i\frac{\pi}{2}(1-A)\right]=A\). Another example is provided by the Heisenberg interaction term: \(H_{12}=X_{1}X_{2}+Y_{1}Y_{2}+Z_{1}Z_{2}\). It can be checked that \(\left[(1-H_{12})/4\right]^{2}=(1-H_{12})/4\), so that \(\exp\left[i\pi\left(\frac{1-H_{12}}{4}\right)\right]=\operatorname{SWAP}\).

Hamiltonians exponentiating to the same unitary

Let now \(A_{i}\) denote a 1-qubit operator (e.g. a Pauli matrix) acting on the \(i\)-th qubit, and such that \(A_{i}^{2}=1\). We therefore know that for every \(i\) we have \(\exp\left[i\pi\left(\frac{1-A_{i}}{2}\right)\right]=A_{i}\). For any \(i,j\) we also have \((A_{i}A_{j})^{2}=1\), so that any such pair can similarly be written as an exponential: \(\exp\left[i\pi\left(\frac{1-A_{i}A_{j}}{2}\right)\right]=A_{i}A_{j}\). The same reasoning can clearly be extended to the product of any number of 1-qubit matrices. On the other hand, by definition, \(A_{i}\) and \(A_{j}\) act on different qubits for \(i\neq j\), therefore they commute, and the generator of their product must equal the sum of their individual generators, leading to the alternative expression:

\begin{equation} A_{i}A_{j}=\exp\left[i\pi\left(\frac{1-A_{i}A_{j}}{2}\right)\right]=\exp\left[i\pi\left(\frac{2-A_{i}-A_{j}}{2}\right)\right].\nonumber \\ \end{equation}

More generally, for the product of \(n\) 1-qubit operators, one has the following:

\begin{equation} A_{1}\cdots A_{n}=\exp\left[i\pi\left(\frac{1-A_{1}\cdots A_{n}}{2}\right)\right]=\exp\left[i\pi\left(\frac{n-(A_{1}+...+A_{n})}{2}\right)\right].\nonumber \\ \end{equation}

However, notably, these two exponential expressions are only equal for \(\theta=\pi\), in the sense that in general \(\exp\left[i\theta\left(\frac{1-A_{1}\cdots A_{n}}{2}\right)\right]\neq\exp\left[i\theta\left(\frac{n-(A_{1}+...+A_{n})}{2}\right)\right]\). This can be seen explicitly by writing the closed expressions for these exponentials, which turn out to be (denoting for brevity \(c_{\theta}=\cos(\theta/2)\) and \(s_{\theta}=\sin(\theta/2)\)):

\begin{equation} \exp\left[i\theta\left(\frac{1-A_{1}\cdots A_{n}}{2}\right)\right]=e^{i\theta/2}\left[c_{\theta}-is_{\theta}(A_{1}\cdots A_{n})\right],\nonumber \\ \end{equation} \begin{equation} \exp\left[i\theta\left(\frac{n-(A_{1}+...+A_{n})}{2}\right)\right]=e^{in\theta/2}\prod_{k=1}^{n}(c_{\theta}-is_{\theta}A_{k}).\nonumber \\ \end{equation}

Hamitonians with projectors

Let \(P\) and \(Q\) be two different projectors, \(P^{2}=P\), \(Q^{2}=Q\), \(PQ=QP=0\). Then, for any pair of operators \(A\) and \(B\) commuting with \(P\) (and therefore with \(Q\)), we have

\begin{equation} \exp[i\theta(AP+BQ)]=e^{i\theta AP}e^{i\theta BQ}=1+P\left(e^{i\theta A}-1\right)+Q\left(e^{i\theta B}-1\right).\nonumber \\ \end{equation}

Moreover, if \(P+Q=1\), then \(\exp[i\theta(AP+BQ)]=e^{i\theta A}P+e^{i\theta B}Q\).

As a corollary, if \(P\) and \(Q\) are orthogonal projectors such that \(P+Q=1\), and \(A\) and \(B\) are operators that commute with \(P\) and \(Q\) and such that \(A^{2}=B^{2}=1\), then

\begin{equation} AP+BQ=\exp\left\{i\frac{\pi}{2}\left[(A-1)P+(B-1)Q\right]\right\}=\exp\left\{i\frac{\pi}{2}\left[AP+BQ-1\right]\right\}.\nonumber \\ \end{equation}

In the particular case in which \(A=1\), we get \(e^{i\theta AP}=Q+Pe^{i\theta A}\)