Consider the Heisenberg picture of Quantum Mechanics. For a two state system we have the Pauli matrices evolving according to the relation $$\sigma_i(t)=U^+\sigma_i(0)U$$ where $U=e^{-iHt/\hbar}$ and $i=x,y,z.$

But in a particular research paper I saw the evolution written as $$\vec\sigma(t)=U^+\vec\sigma(0)U$$ where $\vec\sigma=(\sigma_x,\sigma_y,\sigma_z)$. Thinking about the matrix representation of this equation, it is a 2x2 matrix $U^+$ multiplied by a 3x1 matrix $\vec\sigma$ multiplied by a 2x2 matrix $U$ which is obviously wrong. Where am I making mistakes in interpreting the above equation?

  • $\begingroup$ the second expression is just a funny way to write the first one (which is the one that is right) $\endgroup$ Apr 28, 2016 at 8:45
  • $\begingroup$ @AccidentalFourierTransform drive.google.com/open?id=0B-f2WgH6tfE-UGtZZUNmdWNwM1U and drive.google.com/open?id=0B-f2WgH6tfE-bXFWdVRRSTVRY0k. This is the context in which I saw the above equation. Specifically, equation (3) in the second link. $\endgroup$ Apr 28, 2016 at 8:57
  • 1
    $\begingroup$ $\vec{\sigma}$ is a $2\times 2$ matrix, $\sigma_x \vec{e}_x + \sigma_y \vec{e}_y + \sigma_z \vec{e}_z$. $\endgroup$
    – auxsvr
    Apr 28, 2016 at 9:03
  • $\begingroup$ @auxsvr towards the end of the paragraph in the second link they are talking about $\sigma_{ij}$..what does the i and j stand for in this? $\endgroup$ Apr 28, 2016 at 9:06
  • $\begingroup$ $\sigma_{ij}$ is the $i,j$-th element of the matrix $\vec{\sigma}$. A common convention is to interpret $i$ as the row and $j$ as the column. $\endgroup$
    – auxsvr
    Apr 28, 2016 at 9:11

2 Answers 2


Take your first relation for the 3 Pauli matrices individually:




Now you define a "vector" for notational convenience like the OP says in the question. I will choose to rewrite it as a column vector to visually show the relation of the above 3 relations: \begin{equation} \vec\sigma= \begin{pmatrix} \sigma_x\\ \sigma_y\\ \sigma_z \end{pmatrix} \end{equation}

Now we can write the above 3 notations as: $$\vec\sigma(t)=U^\dagger\vec\sigma(0)U$$

What we have really done here is define a new product that acts on the vectors defined for notational convenience. It acts as follows:

\begin{equation} U \begin{pmatrix} \sigma_x\\ \sigma_y\\ \sigma_z \end{pmatrix} := \begin{pmatrix} U \sigma_x\\ U \sigma_y\\ U \sigma_z \end{pmatrix} \end{equation}

Note this is a definition for notational convenience and there is nothing physical about it.

  • $\begingroup$ This notation can be used even if the coefficient matrices are linearly dependent! $\endgroup$
    – auxsvr
    Apr 28, 2016 at 9:09
  • $\begingroup$ @auxsvr thanks good point! i edited my answer. i guess i was anticipating that we could define other products $\endgroup$ Apr 28, 2016 at 9:11

If $\mathbf{x}=\left(x_{1},x_{2},x_{3}\right)$ is a 3-vector rotated to $\mathbf{x}^{\prime}=\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)$ then this rotation is expressed via special unitary matrices $U \in SU\left(2\right)$ as follows :

\begin{equation} \mathbf{X}^{\prime}\equiv \begin{bmatrix} x^{'}_3&x^{'}_1-ix^{'}_2\\ x^{'}_1+ix^{'}_2&-x^{'}_3 \end{bmatrix} =U \begin{bmatrix} x_3&x_1-ix_2\\ x_1+ix_2&-x_3 \end{bmatrix} U^{*} =U\mathbf{X}U^{*} \tag{01} \end{equation} So, I think that we must consider $\boldsymbol{\sigma}=\left(\sigma_{1},\sigma_{2},\sigma_{3}\right)$ typically as a 3-vector and interpret equation as follows :

\begin{equation} \begin{split} \mathbf{\Sigma}\left(t\right)& \equiv \begin{bmatrix} \sigma_{3}\left(t\right) & \sigma_{1}\left(t\right)-i\sigma_{2}\left(t\right)\\ \sigma_{1}\left(t\right)+i\sigma_{2}\left(t\right) &-\sigma_{3}\left(t\right) \end{bmatrix} =U \begin{bmatrix} \sigma_{3}\left(0\right) & \sigma_{1}\left(0\right)-i\sigma_{2}\left(0\right)\\ \sigma_{1}\left(0\right)+i\sigma_{2}\left(t\right) &-\sigma_{3}\left(0\right) \end{bmatrix} U^{*}\\ &=U\mathbf{\Sigma}\left(0\right)U^{*} \end{split} \end{equation}

\begin{equation} ------------------------------ \tag{02} \end{equation}

Note that if $\mathbb{A}$ is the $3\times3$ rotation matrix from which the special unitary matrix $U$ or $-U$ is created then typically :

\begin{equation} \boldsymbol{\sigma}\left(t\right)\equiv \begin{bmatrix} \sigma_{1}\left(t\right)\\ \sigma_{2}\left(t\right)\\ \sigma_{3}\left(t\right) \end{bmatrix} =\mathbb{A} \begin{bmatrix} \sigma_{1}\left(0\right)\\ \sigma_{2}\left(0\right)\\ \sigma_{3}\left(0\right) \end{bmatrix} =\mathbb{A}\boldsymbol{\sigma}\left(0\right) \tag{03} \end{equation}

  • $\begingroup$ Why does the elements of X have that particular form? Like why is the second element in first row $x_1-ix_2$? $\endgroup$ Apr 28, 2016 at 9:53
  • $\begingroup$ @Rajath Krishna R : May be you must enjoy with the details to pass from a rotation equation expressed with real 3-vectors and real orthonormal $3\times3$ rotation matrix, like equation (03) in my answer, to the corresponding equation with $2\times2$ complex matrices, like equation (01). $\endgroup$
    – Frobenius
    Apr 28, 2016 at 10:16
  • $\begingroup$ The paper mixes two conventions and probably defines $\vec{\sigma}(t) = \sigma_1(t) \vec{e}_1 + \sigma_2(t) \vec{e}_2 + \sigma_3(t) \vec{e}_3$, where the vector space is defined by the Pauli matrices, i.e. $\vec{e}_1 = \sigma_x$, etc. $\endgroup$
    – auxsvr
    Apr 28, 2016 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.