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How can we construct the Pauli matrices starting from $$\sigma_i=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$$ by using the conditions $$\sigma^2_i=1,$$$$\left [ \sigma_x,\sigma_y \right ]=2i\sigma_z,$$ and so on?

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  • $\begingroup$ Have you attempted this yourself? Notice that all of these conditions produce a set of equations relating the $\{a,b,c,d\}$. Also, not sure if it is given, by $\text{Tr}(\sigma_i) = 0$ as well. $\endgroup$ – Ultima Jul 7 '15 at 19:49
  • $\begingroup$ The problem is I don't know how to start from ?, I found the matrix elements of pauli matrix by using inner product relation and I am completely lost in it. It would be great if anyone could give me any hint to start from. $\endgroup$ – Roshan Shrestha Jul 7 '15 at 19:52
  • $\begingroup$ this might help web.uconn.edu/~ch351vc/pdfs/spin1.pdf same type of problem $\endgroup$ – user81619 Jul 7 '15 at 20:20
  • $\begingroup$ @Ultima You can prove that $\mathrm{tr}(\sigma_i)=0$ from the commutation relationships, so you don't need to assume that as well. $\endgroup$ – WetSavannaAnimal Jul 8 '15 at 12:02
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Since each $\sigma_i$ is a scalar multiple of a Lie bracket of other finite matrices, each $\sigma_i$ must be traceless. So straight away we know:

$$\sigma_i=\left(\begin{array}{cc}a&b\\c&-a\end{array}\right)\tag{1}$$

and $\sigma_i^2=\mathrm{id}$ then yields $a^2 + b\,c=1$.

The eigenvalues of any matrix of the form in (1) with $a^2 + b\,c=1$ are $\pm\sqrt{a^2+b\,c} = \pm1$. Therefore, for any set of matrices we find fulfilling all the given relationships, we can do a similarity transformation on the whole set and thus (1) diagonalize any member of the set we choose whilst (2) keeping all the required relationships intact. Exercise: Prove that the given relationships (Lie brackets and $\sigma_i^2=\mathrm{id}$) are indeed invariant under any similarity transformation.

Thus, without loss of generalness, we can always choose one of the set to be:

$$\sigma_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\tag{2}$$

So now work out the Lie bracket of $\sigma_z$ and $\sigma_x = \left(\begin{array}{cc}a_x&b_x\\c_x&-a_x\end{array}\right)$: result must be $2\,i\,\sigma_y$ and so we get:

$$\sigma_y = \left(\begin{array}{cc}0 & -i\,b_x \\i\,c_x & 0 \\\end{array}\right)\tag{3}$$

But given $\sigma_y^2=1$ we get $b_x\,c_x=1$ whence $a_x=0$ (since $a_x^2 + b_x\,c_x=1$). So our remaining two matrices are of the forms:

$$\sigma_x = \left(\begin{array}{cc}0 & b_x \\\frac{1}{b_x} & 0 \\\end{array}\right)$$ $$\sigma_y = \left(\begin{array}{cc}0 & -i\,b_x \\\frac{i}{b_x} & 0\\\end{array}\right)\tag{4}$$

and the remaining commutation relationships then give you the unknown constant $b_x$.

Once you have found $b_x$, we know from our comments above that any set of matrices fulfilling the required commutation relationships and $\sigma_i^2=\mathrm{id}$ is gotten from this particular set (the "standard" Pauli matrices) by a similarity transformation.

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  • $\begingroup$ What is meant by lie bracket ? $\endgroup$ – Roshan Shrestha Jul 8 '15 at 13:44
  • $\begingroup$ In this case, lie bracket refers to the commutator. $\endgroup$ – Omry Jul 8 '15 at 13:45
  • $\begingroup$ @RoshanShrestha I worked out the commutator (Lie bracket) $[\sigma_z,\,\sigma_x]$ for the assumed form of $\sigma_x$ defined by $a_x,\,b_x,\,c_x$ in the answer. Is this "an example"? $\endgroup$ – WetSavannaAnimal Jul 8 '15 at 13:58
  • $\begingroup$ Can we also use $\sigma_x$ for generality. Can't we find all the components of Pauli matrices without using any one of them as a generality. $\endgroup$ – Roshan Shrestha Jul 9 '15 at 8:07
  • $\begingroup$ @RoshanShrestha I'm only sticking to convention: it's customary to call $\sigma_z$ the diagonal one. I'm simply saying you can make a similarity transformation on any set of three matrices fulfilling the given relationships so that any one of them is equal to the diagonal one $\mathrm{diag}(1,\,-1)$. You could call this one $\sigma_x$ if you liked and you'd get a valid set. However, this latter set, like any other set that fulfills the given relationship, can still be derived from the standard set by a similarity transformation. Is this what you mean? $\endgroup$ – WetSavannaAnimal Jul 9 '15 at 8:15
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SECTION A : Angular Momentum in Quantum Mechanics - The minimum non-zero case $\:j=1/2\:$

The angular momentum in Quantum Mechanics has the following properties :


THEOREM :

The components of the dimensionless orbital angular momentum of a particle $\:\mathbf{L}=\mathbf{J}/\hbar =\left(\mathbf{r}\times\mathbf{p}\right)/ \hbar $ satisfy the commutation relations \begin{eqnarray} \left[L_1,L_2\right]& =& L_1L_2-L_2L_1 \ =\ i L_3 \tag{A-01}\\ \left[L_2,L_3\right]&=& L_2L_3-L_3L_2 \ =\ i L_1 \tag{A-02}\\ \left[L_3,L_1\right]&=& L_3L_1-L_1L_3 \ =\ i L_2 \tag{A-03} \end{eqnarray} or in one stroke \begin{equation} \mathbf{L}\times \mathbf{L}= i\mathbf{L} \tag{A-04} \end{equation}

by which

a) The allowable eigenvalues of the absolute value operator \begin{equation} L^2\equiv L_1^2+L_2^2+L_3^2 \tag{A-05} \end{equation} are
\begin{equation} j\left(j+1\right)\;, \qquad j=0\:,\:\tfrac{1}{2}\:,\:1\:,\:\tfrac{3}{2}\:,\:2\:,\:\tfrac{5}{2}\:,\:\ldots \tag{A-06} \end{equation} b) The eigenvalue $\ j\left(j+1\right)\ $ has $\ (2j+1)$-multiplicity to which correspond the $\ (2j+1)\ $ possible eigenvalues of the component $L_3$ across an arbitrary axis $x_3$ \begin{equation} m\;, \qquad m = -j\:,\:-j+1\:,\:\cdots \:,\:j-1\:,\:j \tag{A-07} \end{equation} c) A complete set of common eigenfunctions of $L^2$ and $L_3$ is enumerated by the pair $(j,m)$.

(Note : it has been proved that the orbital angular momentum has only integer values of $j$ while the half-integer values are due exclusively to spin. Moreover, spin can take integer values too).


From this begins a whole story to find the matrix representations of $\:L,L_1,L_2,L_3 \:$ for any value of the aforementioned $\:j\:$. These matrices must satisfy the following conditions for any $\:j\:$ :

(a1) they must be square matrices, say $\:n\times n\:$.

(a2) they must be hermitian, since they represent observables. A measurement of any of them gives a value between its real eigenvalues.

(a3) they must be traceless (the trace of a square matrix is the sum of its diagonal elements). This results from the commutation relations (A-01,02,03) and the property that the trace of the product of two square matrices is independent of their order : \begin{equation} C=[A,B]=AB-BA \Longrightarrow TrC=Tr[A,B]=Tr(AB)-Tr(BA)=0 \tag{A-08} \end{equation}
(a4) since for any $\:j\:$ \begin{equation} L_1^2+L_2^2+L_3^2=L^2=j\left(j+1\right)I \tag{A-09} \end{equation} and there is no reason to distinguish or prefer between the components $\:L_1,L_2,L_3 \:$ we must have \begin{equation} L_1^2=L_2^2=L_3^2=\dfrac{1}{3}L^2=\dfrac{j\left(j+1\right)}{3}I \tag{A-10} \end{equation}

SECTION B : The case $\:j=1/2\:$ and the Pauli matrices $\:\boldsymbol{\sigma}=\left(\sigma_{1},\sigma_{2},\sigma_{3}\right)\:$

Suppose now that we must find a matrix representation of $\:L,L_1,L_2,L_3 \:$ for the special case $\:j=1/2\:$. Note that this case concerns only spin of a particle, since the orbital angular momentum takes only integer values of $\:j\:$. In our case from (A-10) \begin{equation} L_1^2=L_2^2=L_3^2=\dfrac{1}{4}I \tag{B-01} \end{equation} From this expression we define the equivalent set of matrices $\:\left(\sigma_{1},\sigma_{2},\sigma_{3}\right) \:$ by \begin{equation} L_1=\dfrac{1}{2}\sigma_{1}\;, \quad L_2=\dfrac{1}{2}\sigma_{2}\;, \quad L_3=\dfrac{1}{2}\sigma_{3} \tag{B-02} \end{equation} and the problem is to find 3 matrices $\:\left(\sigma_{1},\sigma_{2},\sigma_{3}\right) \:$ with properties

\begin{equation} \sigma_{k}=\sigma_{k}^{\boldsymbol{*}}\;, \quad k=1,2,3 \qquad \text{(hermitian)} \tag{B-03a} \end{equation} \begin{equation} \sigma_{k}^{2}=I\;, \quad k=1,2,3 \qquad \text{(idempotent)} \tag{B-03b} \end{equation} \begin{equation} Tr\left(\sigma_{k}\right)=0\;, \quad k=1,2,3 \qquad \text{(traceless)} \tag{B-03c} \end{equation} \begin{align} & \left[\sigma_{1},\sigma_{2}\right]= \sigma_{1}\sigma_{2}- \sigma_{2}\sigma_{1}=2i\sigma_{3} \tag{B-03d.1}\\ & \left[\sigma_{2},\sigma_{3}\right]=\sigma_{2}\sigma_{3}- \sigma_{3}\sigma_{2}=2i\sigma_{1} \tag{B-03d.2}\\ & \left[\sigma_{3},\sigma_{1}\right]=\sigma_{3}\sigma_{1}- \sigma_{1}\sigma_{3}=2i\sigma_{2} \tag{B-03d.3} \end{align}

We begin with matrices of the minimum dimensions. Matrices $\:1\times 1\:$, that is scalars, can't obey the commutation relations. So we proceed with matrices $\:2\times 2\:$. Note that if it's impossible to find matrices $\:2\times 2\:$ that satisfy above properties we proceed to $\:3\times 3\:$ matrices and so on.

So, let a $\:2\times 2\:$ matrix $\:W\:$ that satisfies at first properties (B-03a), (B-03c) that is hermitian and traceless. It's not difficult to see that the general form of such a matrix is

\begin{equation} W= \begin{bmatrix} &w_3&w_1-iw_2&\\ &w_1+iw_2&-w_3& \end{bmatrix} \in \mathbb{H} \tag{B-04} \end{equation} where $w_1,w_2,w_3$ are real parameters and $\mathbb{H}$ the linear space of hermitian traceless $2\times2$ matrices. So, there exists a bijection (one-to-one and onto correspondence) between $\mathbb{H}$ and $\mathbb{R}^3$.

\begin{equation} \mathbf{w}=(w_1,w_2,w_3)\in \mathbb{R}^3\;\longleftrightarrow \; W= \begin{bmatrix} &w_3&w_1-iw_2&\\ &w_1+iw_2&-w_3& \end{bmatrix} \in \mathbb{H} \tag{B-05} \end{equation}

Since in $\mathbb{R}^3$ any set of 3 linearly independent vectors $\:\mathbf{a}_{1},\mathbf{a}_{2},\mathbf{a}_{3}\:$ constitute a basis, then the following correspondent set of the 3 hermitian traceless $2\times2$ matrices is a basis for the space $\mathbb{H}$ :

\begin{eqnarray} \mathbf{a}_1 &=&(a_{11},a_{21},a_{31}) \longleftrightarrow A_1= \begin{bmatrix} a_{31}&a_{11}-ia_{21}\\ a_{11}+ia_{21}&-a_{31} \end{bmatrix} \tag{B-06a}\\ \mathbf{a}_2 &=&(a_{12},a_{22},a_{32}) \longleftrightarrow A_2= \begin{bmatrix} a_{32}&a_{12}-ia_{22}\\ a_{12}+ia_{22}&-a_{32} \end{bmatrix} \tag{B-06b}\\ \mathbf{a}_3 &=&(a_{13},a_{23},a_{33}) \longleftrightarrow A_3= \begin{bmatrix} \ a_{33}&a_{13}-ia_{23}\\ a_{13}+ia_{23}&-a_{33} \end{bmatrix} \tag{B-06c} \end{eqnarray}

Now, for the general hermitian traceless $2\times2$ matrix $\:W\:$ of equation (B-04) to obey property (B-03b) we must have \begin{equation} W^{2}=I \Longrightarrow \begin{bmatrix} &w_1^2+w_2^2+w_3^2&0&\\ &0&w_1^2+w_2^2+w_3^2& \end{bmatrix} =I \Longrightarrow \Vert\mathbf{w} \Vert^{2}=1 \tag{B-07} \end{equation}
which means that the set of $2\times2$ hermitian traceless idempotent matrices, a subset of $\mathbb{H}$, is in one-to-one and onto correspondence with the surface of the unit sphere of $\mathbb{R}^3$.

So, if the 3 linear independent vectors $\:\mathbf{a}_{1},\mathbf{a}_{2},\mathbf{a}_{3}\:$ are of unit norm $\:\Vert\mathbf{a}_{1} \Vert=\Vert\mathbf{a}_{2}\Vert=\Vert\mathbf{a}_{3}\Vert=1\:$, then the correspondent matrices $\:A_{1},A_{2},A_{3}\:$ of (B-06) have all the properties (B-03a,b,c ).

It remains now to find the necessary properties that the set of 3 matrices $\:A_{1},A_{2},A_{3}\:$ must have in order to satisfy the commutation relations (B-03d). We'll see in the following that these commutation relations impose orthogonality relations between the 3 linear independent unit vectors $\:\mathbf{a}_{1},\mathbf{a}_{2},\mathbf{a}_{3}\:$ and later on how the above analysis gives the standard form of the Pauli matrices.

Now, let another matrix $V\in \mathbb{H}$ generated by a vector $\mathbf{v}$
\begin{equation} \mathbf{v}=(v_1,v_2,v_3)\in \mathbb{R}^3\;\longleftrightarrow \; V= \begin{bmatrix} &v_3&v_1-iv_2&\\ &v_1+iv_2&-v_3& \end{bmatrix} \in \mathbb{H} \tag{B-08} \end{equation} and $\:W \in \mathbb{H}\:$ the matrix of (B-05) generated by the vector $\mathbf{w}$. The vectors $\mathbf{w},\mathbf{v}$ have not special properties, like unit norm or normal to each other. Then it's easy to see that \begin{eqnarray} WV &=& \begin{bmatrix} &w_3&w_1-iw_2&\\ &w_1+iw_2&-w_3& \end{bmatrix} \begin{bmatrix} &v_3&v_1-iv_2&\\ &v_1+iv_2&-v_3& \end{bmatrix} \nonumber\\ &=& \begin{bmatrix} (\mathbf{w}\circ \mathbf{v})+i(\mathbf{w}\times \mathbf{v})_3&\ &i(\mathbf{w}\times \mathbf{v})_1+(\mathbf{w}\times \mathbf{v})_2\\ i(\mathbf{w}\times \mathbf{v})_1-(\mathbf{w}\times \mathbf{v})_2&\ &(\mathbf{w}\circ \mathbf{v})-i(\mathbf{w}\times \mathbf{v})_3 \end{bmatrix} \tag{B-09} \end{eqnarray} or \begin{equation} WV =(\mathbf{w}\circ \mathbf{v})I+i\begin{bmatrix} (\mathbf{w}\times \mathbf{v})_3&(\mathbf{w}\times \mathbf{v})_1-i(\mathbf{w}\times \mathbf{v})_2\\ (\mathbf{w}\times \mathbf{v})_1+i(\mathbf{w}\times \mathbf{v})_2&-(\mathbf{w}\times \mathbf{v})_3 \end{bmatrix} \tag{B-10} \end{equation} where $(\mathbf{w}\circ \mathbf{v})$ is the inner (scalar) product of vectors $\mathbf{w},\mathbf{v}$ \begin{equation} \mathbf{w}\circ \mathbf{v}=w_1v_1+w_2v_2+w_3v_3 \tag{B-11} \end{equation} and $(\mathbf{w}\times \mathbf{v})_\jmath$ are the components of the outer (vector) product between vectors $\mathbf{w},\mathbf{v}$ \begin{eqnarray} (\mathbf{w}\times \mathbf{v})_1&=& w_2v_3-w_3v_2 \nonumber\\ (\mathbf{w}\times \mathbf{v})_2&=& w_3v_1-w_1v_3 \nonumber\\ (\mathbf{w}\times \mathbf{v})_3&=& w_1v_2-w_2v_1 \tag{B-12} \end{eqnarray} Interchanging $W \leftrightarrow V$ and $\mathbf{w}\leftrightarrow \mathbf{v}$ in equation (B-10) \begin{equation} VW =(\mathbf{w}\circ \mathbf{v})I-i\begin{bmatrix} (\mathbf{w}\times \mathbf{v})_3&(\mathbf{w}\times \mathbf{v})_1-i(\mathbf{w}\times \mathbf{v})_2\\ (\mathbf{w}\times \mathbf{v})_1+i(\mathbf{w}\times \mathbf{v})_2&-(\mathbf{w}\times \mathbf{v})_3 \end{bmatrix} \tag{B-13} \end{equation} Subtracting from (B-10) we have for the commutator $[W,V]$

\begin{equation} [W,V]=WV-VW =2i\begin{bmatrix} (\mathbf{w}\times \mathbf{v})_3&(\mathbf{w}\times \mathbf{v})_1-i(\mathbf{w}\times \mathbf{v})_2\\ (\mathbf{w}\times \mathbf{v})_1+i(\mathbf{w}\times \mathbf{v})_2&-(\mathbf{w}\times \mathbf{v})_3 \end{bmatrix} \tag{B-14} \end{equation}

So, if the vectors $\mathbf{w},\mathbf{v}\in \mathbb{R}^3$ generate the hermitian traceless $2 \times 2$ matrices $W,V \in \mathbb{H}$ respectively, then the outer product $\mathbf{w}\times \mathbf{v}$ generates $[W,V]/{2i}$. \begin{equation} \begin{bmatrix} &\mathbf{w}& \longleftrightarrow & W&\\ &\mathbf{v}& \longleftrightarrow & V& \end{bmatrix} \Longrightarrow \begin{bmatrix} &(\mathbf{w}\times \mathbf{v})& \longleftrightarrow & [WV-VW]/{2i}& \end{bmatrix} \tag{B-15} \end{equation}

Also, adding equations(B-10) ,(B-13) we have for the anti-commutator $\left\{W,V\right\}$ \begin{equation} \left\{W,V\right\}=WV+VW =2(\mathbf{w}\circ \mathbf{v})I \\ \tag{B-16} \end{equation}

Equation (B-14) yields the final solution to our problem :

Any set of 3 matrices $\:A_{1},A_{2},A_ {3}\:$ generated as in equations (B-06) from an orthonormal right-handed basis $\:\left(\mathbf{a}_{1},\mathbf{a}_{2},\mathbf{a}_{3}\right)\:$ satisfies all conditions (B-03) and so it can be considered as a representation of the components of dimensionless (spin) angular momentum for $\:j=1/2\:$ in the sense of equations (B-02): \begin{equation} L_1=\dfrac{1}{2}A_{1}\;, \quad L_2=\dfrac{1}{2}A_{2}\;, \quad L_3=\dfrac{1}{2}A_{3} \tag{B-17} \end{equation}

If as orthonormal right-handed basis $\:\left(\mathbf{a}_{1},\mathbf{a}_{2},\mathbf{a}_{3}\right)\:$ we choose the usual basis \begin{equation} \mathbf{e}_{1}=\left(1,0,0\right),\quad \mathbf{e}_{2}=\left(0,1,0\right),\quad \mathbf{e}_{3}=\left(0,0,1\right) \tag{B-18} \end{equation} respectively then we have in place of the matrices $\:A_{1},A_{2},A_ {3}\:$ the well-known Pauli Matrices $\:\sigma_{1},\sigma_{2},\sigma_ {3}\:$ respectively :

\begin{eqnarray} \mathbf{e}_1 &=&(1,0,0)\qquad \longleftrightarrow \qquad \sigma_1= \begin{bmatrix} &0&1&\\ &1&0& \end{bmatrix} \tag{B-19a}\\ \mathbf{e}_2 &=&(0,1,0)\qquad \longleftrightarrow \qquad \sigma_2= \begin{bmatrix} &0&-i\\ &i&0 \end{bmatrix} \tag{B-19b}\\ \mathbf{e}_3 &=&(0,0,1)\qquad \longleftrightarrow \qquad \sigma_3= \begin{bmatrix} &1&0\\ &0&-1 \end{bmatrix} \tag{B-19c} \end{eqnarray}

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  • $\begingroup$ Whoa thats a very thorough answer. Looks like something from a textbook. $\endgroup$ – Horus Jul 8 '15 at 15:04

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