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.


2 Answers 2


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 $\otimes$.

In your question, you seem to have made the following interpretation:

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

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:

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

  • 1
    $\begingroup$ 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. $\endgroup$
    – user20374
    Sep 13, 2023 at 7:49
  • $\begingroup$ 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. $\endgroup$
    – jecado
    Sep 13, 2023 at 15:27

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