# Notation: Hamiltonian Simulation of Pauli Gates

Let $$\sigma^j_x$$ describe the following unitary over $$n$$ qubits: on the $$j$$-th qubit, it acts as the Pauli $$x$$ operator; instead, on any other qubit, it acts as the identity. A paper states now that

\begin{align} \exp(-it \sum_j \sigma^j_x) & = \bigotimes_j \begin{pmatrix} \cos(t) & -i \sin(t) \newline -i \sin(t) & \cos(t) \end{pmatrix} \end{align}

I assume that each element of $$\bigotimes$$ is meant to describe the unitary that acts as identity outside qubit $$j$$ and as the visualized $$2 \times 2$$ matrix on the $$j$$-th qubit itself. If this is correct, it seems that the overall term misses the factor $$e^{-it}$$ in front of $$\otimes$$. Is this an instance of the "global phase does not count" convention or did I miss something? Please help, at times the quantum notation is somewhat ambiguous.

Add-on. If there is no $$e^{-it}$$ missing, what is wrong with the following calculation?

In case of $$n = 1$$ qubits, we get \begin{align} \exp(-it \sigma_x^1) & = \begin{pmatrix} \cos(t) & -i \sin(t) \newline -i \sin(t) & \cos(t) \end{pmatrix} & \exp(-it \text{Id}^1) & = \begin{pmatrix} e^{-it} & 0 \newline 0 & e^{-it} \end{pmatrix} \end{align}

Consequently, for $$n = 2$$ qubits, we get:

\begin{align} \exp(-it \sigma_x^1) & = \begin{pmatrix} \cos(t) & -i \sin(t) & 0 & 0 \newline -i \sin(t) & \cos(t) & 0 & 0 \newline 0 & 0 & e^{-it} & 0 \newline 0 & 0 & 0 & e^{-it} \newline \end{pmatrix} \end{align}

\begin{align} \exp(-it \sigma_x^2) & = \begin{pmatrix} e^{-it} & 0 & 0 & 0 \newline 0 & e^{-it} & 0 & 0 \newline 0 & 0 & \cos(t) & -i \sin(t) \newline 0 & 0 & -i \sin(t) & \cos(t) \newline \end{pmatrix} \end{align}

Multiplying both matrices thus gives rise to the factor $$e^{-it}$$. The factor seems to vanish only if one replaces, outside the active qubit, the identity function with the zero function.

I think an example would be helpful here. Consider a 2-qubit system. $$\sigma_x^1=\sigma_x\otimes I, \quad \sigma_x^2=I\otimes \sigma_x.$$ Now, for any $$H$$ such that $$H^2=I$$, $$e^{-iHt}=\cos(t)I-i\sin(t)H$$. The $$\sigma_x^i$$ all obey this, so $$e^{-i\sigma_x^1 t}=\cos(t)I-i\sin(t)\sigma_x^1=\left(\begin{array}{cccc} \cot(t) & 0 & -i\sin(t) & 0 \\ 0 & \cos(t) & 0 & -i\sin(t) \\ -i\sin(t) & 0 & \cos(t) & 0 \\ 0 & -i\sin(t) & 0 & \cos(t) \end{array}\right)=(\cos(t)I-i\sin(t)\sigma_x)\otimes I.$$ Now, the second ingredient that you need is that for any two operators that commute, $$H_1H_2=H_2H_1$$, then $$e^{-i(H_1+H_2)t}=e^{-iH_1t}e^{-iH_2t}.$$ So, in the present case, you get $$(I\otimes (\cos(t)I-i\sin(t)\sigma_x))((\cos(t)I-i\sin(t)\sigma_x)\otimes I)$$ which is the same as $$(\cos(t)I-i\sin(t)\sigma_x)\otimes (\cos(t)I-i\sin(t)\sigma_x)$$

No - I don't believe there is an $$e^{-it}$$ in front of the $$\otimes$$.

$$\sigma_x^1 \rightarrow \sigma_x \oplus I$$

This is a so-called "direct sum", so when you take the exponential, you sensibly distributed the $$-it$$ to both qubits:

$$\exp(-it\sigma_x^1) \rightarrow e^{-it (\sigma_x \oplus I)} \rightarrow e^{(-it\sigma_x) ~\oplus~ (-itI)} \rightarrow e^{-it\sigma_x} \oplus e^{-itI}$$

The matrix representation you wrote would be correct if this were the correct interpretation. But the correct interpretation of $$\sigma_x^1$$ is:

$$\sigma_x^1 \equiv \sigma_x \otimes I$$

This is a direct product, not a direct sum. And that means the $$-it$$ does not distribute to each qubit:

$$\exp(-it\sigma_x^1) \equiv e^{-it(\sigma_x \otimes I)} = e^{(-it\sigma_x)\otimes I} = e^{-it\sigma_x} \otimes I$$

##### A somewhat thorough derivation for completeness:

The starting point $$\exp\left(-it\sum_j \sigma_x^j\right)$$ could be written as $$\exp\left[\sum_j (-it\sigma_x^j)\right]$$, or, most clearly of all, $$\prod_j \exp(-it\sigma_x^j)$$.

The action on qubit $$j$$ is therefore the single-qubit operator $$\exp(-it\sigma_x)$$, which has the matrix representation:

$$\exp(-it\sigma_x) = \left[ \array{ \cos t & -i \sin t \\ -i \sin t & \cos t } \right]$$

Understanding the notation $$\bigotimes_j$$ to implicitly insert identities on any qubit index not included in the set $$\{j\}$$, we reach the final identity:

$$\exp\left(-it\sum_j \sigma_x^j\right) = \bigotimes_j \left[ \array{ \cos t & -i \sin t \\ -i \sin t & \cos t } \right]$$

• Thanks Jecado. Unfortunately, I it is not clear from the reply why $e^{-it}$ vanishes. I expanded my question by a concrete calculation. It would be great if you (or anyone else) could help. Sep 13, 2023 at 7:49
• Aha - the edits to your question give me a better idea of where you might have gotten off. I've edited my answer so that its first half hopefully helps you reconcile why the matrix representation in @DaftWullie's answer is the correct one. Sep 13, 2023 at 15:27