How Does the Trace of a Matrix Relate to Its Components in Quantum Mechanics?

[Xyius]

Homework Statement

Suppose a 2x2 matrix X (not necessarily Hermitian, nor Unitary) is written as..
$$X=a_0+σ \cdot a$$
(In the book σ and a are both bold and are being dotted.)
Where $a_0$ and $a_{1,2,3}$ are numbers.

a.)How are $a_0$ and $a_k, (k=1,2,3)$ related to $tr(X)$ and $tr(σ_kX)$?
b.)Obtain $a_0$ and $a_k$ in terms of the matrix elements $X_{ij}$.

Homework Equations

$tr(X)$= The trace of X, meaning the sum of its diagonal components.
$tr(X)=\sum_{a'}\left\langle a'|X|a' \right\rangle$
Where the name a' represents base kets.

The Attempt at a Solution

I do not know where to start to be honest. My first question is how can a 2x2 matrix operator equal a number $a_0$ plus the dot product of two vectors? I know I must be misinterpreting this. Can anyone help?

 

[phyzguy]

The σ's are the 2x2 Pauli matrices, so what the problem means is :

X = a0*(2x2 Identity matrix) + ax*(2x2 Pauli matrix σx) + ay*(2x2 Pauli matrix σy)+ az*(2x2 Pauli matrix σz)

 

[Xyius]

The σ's are the 2x2 Pauli matrices, so what the problem means is :

X = a0*(2x2 Identity matrix) + ax*(2x2 Pauli matrix σx) + ay*(2x2 Pauli matrix σy)+ az*(2x2 Pauli matrix σz)

Ah that makes it MUCH more clear! So I basically plugged everything in and found the trace in each case and got..

$$tr(X)=2a_0$$
$$tr(σ_{1}X)=2a_1$$
$$tr(σ_{2}X)=2a_2$$
$$tr(σ_{3}X)=2a_3$$

Which I assume is what they are looking for for part A. Part B however is making me a bit confused. Do they want me to just solve the above expressions for a?

 

[vela]

No, you have some matrix X where
$$X = \begin{pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix} = a_0 + \vec{a}\cdot\vec{\sigma}$$The problem wants you to solve for a₀ and the a_k's in terms of the X_ij's. Start by writing down explicitly what $a_0 + \vec{a}\cdot\vec{\sigma}$ is equal to.

 

[Xyius]

Ohh okay so would this be on the right track?

$$a_k=\frac{1}{2}\left[ (σ_kX)_{11}+(σ_kX)_{22} \right]$$
Where k=0,1,2,3

EDIT: I got this by looking at the expressions I posted for part A and finding a common equation that suits all of them. I now see that you said to start by writing the original expression. I will try this.