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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?

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  • $\begingroup$ 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
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    $\begingroup$ $\vec{\sigma}$ 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

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Take your first relation for the 3 Pauli matrices individually:

$$\sigma_1(t)=U^\dagger\sigma_1(0)U$$

$$\sigma_1(t)=U^\dagger\sigma_2(0)U$$

$$\sigma_3(t)=U^\dagger\sigma_3(0)U$$

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.

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  • $\begingroup$ 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
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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}

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  • $\begingroup$ 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

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