# Construction of Pauli Matrices

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?

• 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. Commented Jul 7, 2015 at 19:49
• 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. Commented Jul 7, 2015 at 19:52
• this might help web.uconn.edu/~ch351vc/pdfs/spin1.pdf same type of problem
– user81619
Commented Jul 7, 2015 at 20:20
• @Ultima You can prove that $\mathrm{tr}(\sigma_i)=0$ from the commutation relationships, so you don't need to assume that as well. Commented Jul 8, 2015 at 12:02

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.

• What is meant by lie bracket ? Commented Jul 8, 2015 at 13:44
• In this case, lie bracket refers to the commutator.
– Omry
Commented Jul 8, 2015 at 13:45
• @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"? Commented Jul 8, 2015 at 13:58
• 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. Commented Jul 9, 2015 at 8:07
• @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? Commented Jul 9, 2015 at 8:15

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 $$\mathbf{L}\times \mathbf{L}= i\mathbf{L} \tag{A-04}$$

by which

a) The allowable eigenvalues of the absolute value operator $$L^2\equiv L_1^2+L_2^2+L_3^2 \tag{A-05}$$ are
$$j\left(j+1\right)\;, \qquad j=0\:,\:\tfrac{1}{2}\:,\:1\:,\:\tfrac{3}{2}\:,\:2\:,\:\tfrac{5}{2}\:,\:\ldots \tag{A-06}$$ 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 $x_3$ $$m\;, \qquad m = -j\:,\:-j+1\:,\:\cdots \:,\:j-1\:,\:j \tag{A-07}$$ 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 : $$C=[A,B]=AB-BA \Longrightarrow TrC=Tr[A,B]=Tr(AB)-Tr(BA)=0 \tag{A-08}$$
(a4) since for any $\:j\:$ $$L_1^2+L_2^2+L_3^2=L^2=j\left(j+1\right)I \tag{A-09}$$ and there is no reason to distinguish or prefer between the components $\:L_1,L_2,L_3 \:$ we must have $$L_1^2=L_2^2=L_3^2=\dfrac{1}{3}L^2=\dfrac{j\left(j+1\right)}{3}I \tag{A-10}$$

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) $$L_1^2=L_2^2=L_3^2=\dfrac{1}{4}I \tag{B-01}$$ From this expression we define the equivalent set of matrices $\:\left(\sigma_{1},\sigma_{2},\sigma_{3}\right) \:$ by $$L_1=\dfrac{1}{2}\sigma_{1}\;, \quad L_2=\dfrac{1}{2}\sigma_{2}\;, \quad L_3=\dfrac{1}{2}\sigma_{3} \tag{B-02}$$ and the problem is to find 3 matrices $\:\left(\sigma_{1},\sigma_{2},\sigma_{3}\right) \:$ with properties

$$\sigma_{k}=\sigma_{k}^{\boldsymbol{*}}\;, \quad k=1,2,3 \qquad \text{(hermitian)} \tag{B-03a}$$ $$\sigma_{k}^{2}=I\;, \quad k=1,2,3 \qquad \text{(idempotent)} \tag{B-03b}$$ $$Tr\left(\sigma_{k}\right)=0\;, \quad k=1,2,3 \qquad \text{(traceless)} \tag{B-03c}$$ \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

$$W= \begin{bmatrix} &w_3&w_1-iw_2&\\ &w_1+iw_2&-w_3& \end{bmatrix} \in \mathbb{H} \tag{B-04}$$ 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 $\mathbb{R}^3$.

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

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 $$W^{2}=I \Longrightarrow \begin{bmatrix} &w_1^2+w_2^2+w_3^2&0&\\ &0&w_1^2+w_2^2+w_3^2& \end{bmatrix} =I \Longrightarrow \Vert\mathbf{w} \Vert^{2}=1 \tag{B-07}$$
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 $\mathbf{v}$
$$\mathbf{v}=(v_1,v_2,v_3)\in \mathbb{R}^3\;\longleftrightarrow \; V= \begin{bmatrix} &v_3&v_1-iv_2&\\ &v_1+iv_2&-v_3& \end{bmatrix} \in \mathbb{H} \tag{B-08}$$ 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 $$WV =(\mathbf{w}\circ \mathbf{v})I+i\begin{bmatrix} (\mathbf{w}\times \mathbf{v})_3&(\mathbf{w}\times \mathbf{v})_1-i(\mathbf{w}\times \mathbf{v})_2\\ (\mathbf{w}\times \mathbf{v})_1+i(\mathbf{w}\times \mathbf{v})_2&-(\mathbf{w}\times \mathbf{v})_3 \end{bmatrix} \tag{B-10}$$ where $(\mathbf{w}\circ \mathbf{v})$ is the inner (scalar) product of vectors $\mathbf{w},\mathbf{v}$ $$\mathbf{w}\circ \mathbf{v}=w_1v_1+w_2v_2+w_3v_3 \tag{B-11}$$ 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) $$VW =(\mathbf{w}\circ \mathbf{v})I-i\begin{bmatrix} (\mathbf{w}\times \mathbf{v})_3&(\mathbf{w}\times \mathbf{v})_1-i(\mathbf{w}\times \mathbf{v})_2\\ (\mathbf{w}\times \mathbf{v})_1+i(\mathbf{w}\times \mathbf{v})_2&-(\mathbf{w}\times \mathbf{v})_3 \end{bmatrix} \tag{B-13}$$ Subtracting from (B-10) we have for the commutator $[W,V]$

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

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 $[W,V]/{2i}$. $$\begin{bmatrix} &\mathbf{w}& \longleftrightarrow & W&\\ &\mathbf{v}& \longleftrightarrow & V& \end{bmatrix} \Longrightarrow \begin{bmatrix} &(\mathbf{w}\times \mathbf{v})& \longleftrightarrow & [WV-VW]/{2i}& \end{bmatrix} \tag{B-15}$$

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

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): $$L_1=\dfrac{1}{2}A_{1}\;, \quad L_2=\dfrac{1}{2}A_{2}\;, \quad L_3=\dfrac{1}{2}A_{3} \tag{B-17}$$

If as orthonormal right-handed basis $\:\left(\mathbf{a}_{1},\mathbf{a}_{2},\mathbf{a}_{3}\right)\:$ we choose the usual basis $$\mathbf{e}_{1}=\left(1,0,0\right),\quad \mathbf{e}_{2}=\left(0,1,0\right),\quad \mathbf{e}_{3}=\left(0,0,1\right) \tag{B-18}$$ 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}

• Whoa thats a very thorough answer. Looks like something from a textbook. Commented Jul 8, 2015 at 15:04

There are very good answers here already, however, for heuristic purposes, I will derive the Pauli matrices just from the 3 postulates presented, without any further assumptions, using only matrix multiplication and solving a few equations.

The 3 postulates presented by you can be written as:

$$(1)$$ $$\sigma_i$$ are $$2 \times 2$$ matrices;

$$(2)$$ $$\sigma^2_i=I$$;

$$(3)$$ $$\left [ \sigma_i,\sigma_j \right ]=2i\sigma_k,$$, where $$i≠j≠k$$ are indexes;

The general form of a $$2 \times 2$$ matrix is $$\sigma_i=\begin{bmatrix} a_i & b_i\\ c_i & d_i \end{bmatrix} \tag{4}$$

Applying $$(3)$$

$$\begin{bmatrix} a_i & b_i\\ c_i & d_i \end{bmatrix} \begin{bmatrix} a_j & b_j\\ c_j & d_j \end{bmatrix} -\begin{bmatrix} a_j & b_j\\ c_j & d_j \end{bmatrix} \begin{bmatrix} a_i & b_i\\ c_i & d_i \end{bmatrix} = 2i\begin{bmatrix} a_k & b_k\\ c_k & d_k \end{bmatrix}$$

Doing the math on the left-hand side and equating the terms occupying the same position at the right-hand side, we get a set of 4 equations, one of which is $$d_k=-a_k$$. Since, $$k$$ is a free index any of the particular indexes set there ($${1,2,3};{x,y,z}$$) must give the same result, meaning that the general form of the matrices is $$\sigma_i=\begin{bmatrix} a_i & b_i\\ c_i & -a_i \end{bmatrix} \tag{4'}$$ $$(4')$$ indicates that any $$2 \times 2$$ matrices satisfying condition $$(3)$$ must be traceless.

Squaring $$(4')$$ and applying condition $$(2)$$, we get $$\begin{bmatrix} a_i^2+b_ic_i & 0\\ 0 & a_i^2+b_ic_i \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ or $$a_i^2+b_ic_i=1 \tag{5}$$

If one imagines $$a,b$$ and $$c$$ as points in a 3D space, it is easy to check that once a particular choice for them is made, such a choice creates a basis vector into that particular direction, thus, every other choices that has a component in that direction, can be written as a linear combination of the former, consequently, individual choices do not reduce generality.

With this in mind, the solutions to $$(5)$$ are:

a) $$a=±1$$ (basis for all solutions that $$a$$ do not vanishes)

This choice demands every possible value for $$b$$ and $$c$$ to be in the $$(a,b)$$ or in the $$(a,c)$$ planes, meaning $$b=0,\,c=\lambda_1$$ or $$b=\lambda_2,\,c=0$$ with $$\lambda$$ any real or complex scalar. The 2 choices above give us 2 different matrices. However, it is easy to show (you can try by applying $$(3)$$ into those matrices) that any choice for $$\lambda_1$$ and $$\lambda_2$$ not simultaneously $$0$$ won't satisfy $$(3)$$, hence, for $$a=±1$$ there is only one possible solution, which is

$$\sigma_3=\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} \tag{6}$$ The solutions $$a=-1$$ are not needed as it will be a scalar multiple of $$(6)$$.

b) $$a=0$$ (solutions that vanishes the right-hand side of $$(5)$$’s sum)

$$bc=1 \tag{7}$$

This places the solutions into the $$(b,c)$$ plane. Considering that for the sake of generality $$a$$ and $$b$$ could be complex numbers, all possible points in the $$(b,c)$$ plane can just be written as linear combinations of a pure real and a pure imaginary basis vectors. The simplest real basis vector in the $$(b,c)$$ plane satisfying $$(7)$$ is $$b=1,\,c=1$$, that gives us

$$\sigma_1=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{8}$$

The pure imaginary basis vector to $$(7)$$ is $$b=-i,\,c=i$$ (or the scalar multiple of this $$b=i,\,c=-i$$, which is obviously the same choice).

$$\sigma_2=\begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix} \tag{9}$$

The 3 matrices $$(6),(8)$$ and $$(9)$$ are the Pauli matrices and they spans the set of all matrices that satisfy the proposed postulates. They form the full set of basis vector for the $$2 \times 2$$ matrix representation $$\sigma_i$$s.