Ex5 7

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 5.7}$

Show that the Pauli operators form a basis for all linear operators on a two dimensional space.

Easy Simultaneous Equations

Hephaestis 14 Jan 2017 00:26

Let M be an arbitrary 2x2 matrix, with entries A and B in the top row and entries C and D in the second row.

Set M = aI + bX + cY + dZ, where I, X, Y, and Z are the Pauli operators and a, b, c, and d are any scalars.

Then,
A = a + d. B = b + c.
C = b - c. D = a - d.

Solving simultaneously the equations involving a and d, we get a = (A+D)/2 and d = (A-D)/2.
Similarly solving for b and c, we get b = (B+C)/2 and c = (B-C)/2.

This shows that any 2x2 matrix is a linear combination of the 4 Pauli matrices. Since there are 4 degrees of freedom in choosing the 4 entries of M, the basis must have 4 elements.

rshaffer 19 Feb 2019 16:30

We will use the following representations of the Pauli operators:

(1)

\begin{align} \hat{I} = \hat{\sigma}_I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \qquad \hat{X} = \hat{\sigma}_X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \qquad \hat{Y} = \hat{\sigma}_Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \qquad \hat{Z} = \hat{\sigma}_Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{align}

We are asked to show that the Pauli operators form a basis for all linear operators on a two-dimensional space. This means that, given any linear operator $\hat{A}$, we want to show that we can express it as a linear combination of the Pauli operators:

(2)

\begin{align} \hat{A} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = b_{I} \hat{\sigma}_I + b_{X} \hat{\sigma}_X + b_{Y} \hat{\sigma}_Y + b_{Z} \hat{\sigma}_Z = \begin{pmatrix} b_{I} + b_{Z} & b_{X} - i b_{Y} \\ b_{X} + i b_{Y} & b_{I} - b_{Z} \end{pmatrix} \end{align}

for some (complex) coefficients $b_{I}$, $b_{X}$, $b_{Y}$, and $b_{Z}$. Clearly, this relation requires that we satisfy the system of linear equations:

(3)

\begin{align} a_{11} = b_I + b_Z \qquad a_{12} = b_X - i b_Y \qquad a_{21} = b_X + i b_Y \qquad a_{22} = b_I - b_Z \end{align}

Solving for $b_{I}$, $b_{X}$, $b_{Y}$, and $b_{Z}$, we find:

(4)

\begin{align} b_I = \tfrac{1}{2}(a_{11} + a_{22}) \qquad b_X = \tfrac{1}{2}(a_{12} + a_{21}) \qquad b_Y = \tfrac{i}{2}(a_{12} - a_{21}) \qquad b_Z = \tfrac{1}{2}(a_{11} - a_{22}) \end{align}

Therefore, we can write any arbitrary linear operator $\hat{A}$ as a linear combination of the Pauli operators as follows:

(5)

\begin{align} \hat{A} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \tfrac{1}{2}(a_{11} + a_{22}) \hat{\sigma}_I + \tfrac{1}{2}(a_{12} + a_{21}) \hat{\sigma}_X + \tfrac{i}{2}(a_{12} - a_{21}) \hat{\sigma}_Y + \tfrac{1}{2}(a_{11} - a_{22}) \hat{\sigma}_Z \end{align}

This confirms that the Pauli operators form a basis for all such linear operators on a two-dimensional space.