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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.

ReplyOptionsWe will use the following representations of the Pauli operators:

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

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

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

ReplyOptions