While reading Chapter 9.2.1 Trace distance in "Quantum Computation and Quantum Information," I encountered a question. What is the vector of Pauli matrices referring to?

$$ \vec{\sigma} = (\sigma_x,\sigma_y,\sigma_z) $$

Also, why does matrix $(\vec{r}-\vec{s}) \cdot \vec{\sigma}$ have an eigenvalues $\pm|\vec{r}-\vec{s}|$?

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1 Answer 1


The $\vec{\sigma}$ is a shorthand notation. It literally represents a vector of Pauli matrices. It is useful to denote things algebraically like this for conciseness. We usually use it to represent a linear combination of the Pauli matrices by taking a vector dot product of $\vec{\sigma}$ with a vector of scalars, which are essentially the coefficients of the linear combination. For example, see equations $(7.1)$ to $(7.5)$.

$$ \vec{\sigma} = \begin{bmatrix}\sigma_x \\ \sigma_y \\ \sigma_z \end{bmatrix} = \begin{bmatrix} \begin{bmatrix} 0 &1 \\ 1 & 0 \end{bmatrix} \\ \begin{bmatrix} 0 &-i \\ i & 0 \end{bmatrix}\\ \begin{bmatrix} 1 &0 \\ 0 & -1 \end{bmatrix} \end{bmatrix} \,.\tag{1} $$

Now, let $\vec{r}$ be

$$\vec{r} = \begin{bmatrix}r_1\\r_2 \\r_3 \end{bmatrix}\,.\tag{2}$$


$$\vec{s} = \begin{bmatrix}s_1\\s_2 \\s_3 \end{bmatrix}\,.\tag{3}$$

Now, we can calculate $\vec{r} - \vec{s}\,.$

$$\vec{r} - \vec{s} = \begin{bmatrix}r_1 - s_1\\r_2 - s_2 \\r_3 - s_3 \end{bmatrix}\,.\tag{4}$$

For the sake of simplicity, say

$$r_1 - s_1= a\,,\tag{5.1}$$ $$r_2 - s_2 = b\,,\tag{5.2}$$ $$r_3 - s_3 = c\,.\tag{5.3}$$


$$\vec{r} - \vec{s} = \begin{bmatrix}a\\b \\c \end{bmatrix}\,.\tag{6}$$

Now, we can calculate the matrix for $(\vec{r} - \vec{s})\cdot \vec{\sigma}\,.$

$$\begin{align} (\vec{r} - \vec{s}) \cdot \vec{\sigma} &= \begin{bmatrix}a\\b \\c \end{bmatrix} \cdot \begin{bmatrix}\sigma_x \\ \sigma_y \\ \sigma_z \end{bmatrix}\,, \tag{7.1}\\ &= a\sigma_x + b\sigma_y + c \sigma_z\,,\tag{7.2}\\ &= a \begin{bmatrix} 0 &1 \\ 1 & 0 \end{bmatrix} + b \begin{bmatrix} 0 &-i \\ i & 0 \end{bmatrix} + c \begin{bmatrix} 1 &0 \\ 0 & -1 \end{bmatrix}\tag{7.3}\,,\\ &= \begin{bmatrix} 0 &a \\ a & 0 \end{bmatrix} + \begin{bmatrix} 0 &-ib \\ ib & 0 \end{bmatrix} + \begin{bmatrix} c &0 \\ 0 & -c \end{bmatrix}\,,\tag{7.4}\\ &= \begin{bmatrix}c & a-ib \\a+ib & -c \end{bmatrix}\,.\tag{7.5} \end{align}$$

Now, if you calculate the eigenvalues $(\lambda)$ of this matrix, you will find,

$$ \begin{align} \lambda &= \pm \sqrt{a^2 + b^2 + c^2}\tag{8.1}\,,\\ &= \pm \sqrt{(r_1 - s_1)^2 + (r_2 - s_2)^2 + (r_3 - s_3)^2}\,,\tag{8.2}\\ &= \pm |\vec{r} - \vec{s}|\tag{8.3}\,. \end{align} $$

  • $\begingroup$ Thank you very much! I initially thought this was related to the eigenvalues of Pauli matrices being +1 and -1. So, may I ask if this trace distance calculation can be extended to 2-qubit or N-qubit systems? That is, decomposing the density matrix of a quantum state into the Pauli basis, and the trace distance between the two quantum states is the difference in coefficients in the Pauli basis. $\endgroup$ Jan 10 at 7:18
  • $\begingroup$ In higher dimensions, you don't have Bloch vector anymore. You have a coherence vector. The geometry of "Bloch sphere" is also much more complicated in higher dimensions. It is no longer a high-dimensional sphere. So, we generally use the density matrices and calculate trace distance, which you can read more about on wikipeadia. $\endgroup$
    – FDGod
    Jan 10 at 7:23
  • $\begingroup$ Although there is no Bloch vector, can't we still decompose the quantum state onto different Pauli bases? Can the Bloch sphere be regarded as a special case of decomposing the quantum state into the Pauli basis? $\endgroup$ Jan 10 at 7:30
  • 1
    $\begingroup$ Yes, we can decompose the quantum state in terms of Pauli bases, but that doesn't do anything. The Bloch vector is special because the components are real. In a Hilbert space of $d$ dimensions, the density matrix becomes a $d \times d$ matrix, and the trace constraints remove one matrix element. Since the dimensionality of the geometric representation is, therefore, $d^2 − 1$ for a $d$-level system. Read more about it here and here. $\endgroup$
    – FDGod
    Jan 10 at 7:39
  • 1
    $\begingroup$ Here is another good reference for the shape of the set of quantum states in general $d$. $\endgroup$ Jan 10 at 8:16

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