# Is there a reason why operators are often written as exponentials in quantum optics?

I have noticed in many quantum optics papers that operators are written as exponentials. Is there a reason for this beyond style or convention? For example, is it physically significant or more amenable to calculation? If the latter, what specific theorems make it better?

As an example, in [1], they write the controlled phase flip gate as a matrix exponential $e^{i\pi P}$ where $P$ is a projection operator. The exponential of a projection operator has a simple Taylor expansion \begin{align} e^{i\alpha \hat{P}_{+}} &=I+(i\alpha) \hat{P}_{+} + \frac{(i\alpha)^{2}}{2}\hat{P}_{+}+\cdots \\& =I+\hat{P}_{+}(i\alpha + \frac{(i\alpha)^2}{2}+\cdots) \\&=I+\hat{P}_{+}(e^{i\alpha}-1) \end{align} which, from my understanding, you would generally need to perform first in order to use the operator for calculations. So why not just write it in the expanded form to begin with?

1. Scalable Photonic Quantum Computation through Cavity-Assisted Interactions. L.-M. Duan and H. J. Kimble. Phys. Rev. Lett. 92, 127902 (2004), arXiv:quant-ph/0309187.
• I don't know if this is the primary reason here, but operators in exponential form have the benefit of being all bounded, which makes analysis much simpler on them. Commented Mar 17, 2018 at 16:43

I'm not familiar with quantum optics, but from a more general mathematical perspective, I'd propose the following: if the exponential expression is more compact than other alternatives, then motivated by notational parsimony, perhaps a more natural question would be "when can we get away with using the exponential form?" This is analogous to preferring to write $\sin\theta$ and using its algebraic properties to avoid expanding it in a power series to perform computations.

There are lots of facts about operator exponentials, modulo some mathematical subtleties that physicists tend to ignore, that make this possible in many contexts:

1. Let $O$ be a linear operator, and let $|\lambda\rangle$ be an eigenvector of $O$ corresponding to eigenvalue $\lambda$, then $|\lambda\rangle$ is an eigenvector of $e^O$ with corresponding eigenvalue $e^\lambda$:

\begin{align} e^{O}|\lambda\rangle &= \sum_{k=0}^\infty \frac{O^k}{k!}|\lambda\rangle = \sum_{k=0}^\infty \frac{\lambda^k}{k!}|\lambda\rangle = e^\lambda|\lambda\rangle \end{align}

In fact more generally, under the hypotheses of the observation, if $f$ is any analytic function, then $|\lambda\rangle$ will be an eigenvector of $f(O)$ with eigenvalue $f(\lambda)$.

2. Basic algebraic properties of the operator exponential:

3. The BCH formula and its cousins, especially this lemma about adjoints.

• What did you mean by BCH's cousins?
– 0x90
Commented Mar 18, 2018 at 13:59
• @ox90 Any and all formulas that are closely related to it. Commented Mar 18, 2018 at 18:08

If two operators are equal, $A=B$, then there is nothing wrong with using either form for the same object. With that in mind, the choice of form obeys a wide array of reasons depending on the context and what one wants to express, though generally speaking there is a large premium on notational compactness and conceptual clarity.

So, along that vein, here are some relevant points to consider:

• In almost all cases, the exponential form is more compact, and it uses less ink. That, by itself, already puts it a step up from any alternative formulation.
• Exponentials of anti-hermitian operators are immediately recognizable as unitary. This is definitely at play in the example you mention: the form $\hat U=\exp(i\alpha \hat P)$ is obviously unitary, where on the other hand $\hat U = 1+\hat P(e^{i\alpha}-1)$ can be seen to be unitary through an easy but non-obvious calculation (i.e. you can't do it in your head without completely losing the thread of what the paper was talking about).
• Imaginary exponentials offer more conceptual clarity in that they are immediately recognizable as rotation operators. As an example, if $\hat \sigma$ is an involution (i.e. $\hat \sigma^2=1$, including all Pauli matrices) then $$e^{i\theta \hat \sigma} = \cos(\theta) + i \hat\sigma\sin(\theta),$$ but the former makes the rotation (say, on the Bloch sphere, around the eigen-axis of $\hat \sigma$) evident, while the latter is much more obscure.

In that regard, operator exponentials act rather like special functions, which, as Michael Berry memorably put it, "enshrine sets of recognizable and communicable patterns and so constitute a common currency". By expressing an operator as an exponential, you're not just giving a specification of what operator you're talking about; you're also making a meaningful statement about how you're thinking about that operator.

• Broken-out forms are not necessarily easier to calculate and conceptualize, and they will often be harder (or harder to deal with in the conceptual frame the text is operating on). This is particularly the case if you're already operating on an eigenbasis of the operator $\hat A$ being exponentiated: if you're already working on a basis of eigenstates $\hat A|a\rangle = a|a\rangle$, then the exponential $$e^{i\theta \hat A}|a\rangle = e^{i\theta a}|a\rangle$$ is just the basis-vector-dependent number $e^{i\theta a}$.

This is at play in the controlled-phase-flip unitary $$U_\mathrm{CPF} = e^{i\pi |0⟩⟨0| \otimes |h⟩⟨h|}$$ in the example you mention: this is easily analyzed as $1$ if the control qubit is on$|1⟩$, and as the operator $e^{i\pi|h⟩⟨h|}$ if the control qubit is on $|1⟩$; moreover, it is also directly seen as the fase-flip $-1=e^{i\pi}$ for the target qubit on the state $|h⟩$ and as unity on its orthogonal complement $|\bar h⟩$.

• +1: Great answer, especially because of the awesome Berry quote that I will now be using whenever I want to explain that concept to students. Commented Mar 18, 2018 at 18:11
• @josh, Yeah, that's an awesome paper in so many respects. Commented Mar 18, 2018 at 19:42