Problem with vectors and matrices.

[Dewgale]

Homework Statement

Calculate $(\vec a \cdot \vec \sigma)^2$, $(\vec a \cdot \vec \sigma)^3$, and $(\vec a \cdot \vec \sigma)^4$, where $\vec a$ is a 3D-vector and $\vec \sigma$ is a 3D-vector formed from the $\sigma_i$ vectors.

Homework Equations

$$\sigma_1 = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$$
$$\sigma_2 = \begin{bmatrix}
0 & -i\\
i & 0
\end{bmatrix}$$
$$\sigma_3 = \begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}$$

The Attempt at a Solution

This makes very little sense to me, since there are no $\sigma_i$ vectors, just matrices. My main thought was to take the determinant of each matrix and set it as a component, such that
$$\vec \sigma_i = <-1,-1,-1>$$

Then $\vec a \cdot \vec \sigma_i$ is $-(a_1 + a_2 + a_3)$.
$(\vec a \cdot \vec \sigma_i)^2$ is $(a_1 + a_2 + a_3)^2$,
$(\vec a \cdot \vec \sigma_i)^3$ is $-(a_1 + a_2 + a_3)^3$ and
$(\vec a \cdot \vec \sigma_i)^4$ is $(a_1 + a_2 + a_3)^4$.

I have no idea, however, whether this is the right approach. Some guidance on this would be nice, thank you!

 

[blue_leaf77]

A notation like $\mathbf{a}\cdot\mathbf{\sigma}$ is just a way of saying $a_1\sigma_1 + a_2\sigma_2 + a_3\sigma_3$. $\mathbf{a}$ is known to be a 3D vector, therefore its components are just numbers.

 

[Dewgale]

A notation like $\mathbf{a}\cdot\mathbf{\sigma}$ is just a way of saying $a_1\sigma_1 + a_2\sigma_2 + a_3\sigma_3$. $\mathbf{a}$ is known to be a 3D vector, therefore its components are just numbers.

Yes, I know that. Thank you though. I'm confused though as to how to form a 3D vector from three 2x2 matrices. Is there a prescribed method or do I just need to do something along the lines of what I did?

 

[blue_leaf77]

The sigma vector $\mathbf{\sigma}$ is just one example of vector operators, another example would be the orbital angular momentum operator $\mathbf{L}$. Operators in quantum mechanics (as well as in linear algebra in general) need not always be represented by a matrix. The matrix representation is especially helpful when you are working in the basis formed by the eigenstates of the operator being represented as a matrix. If, on the other hand, you are given a problem in which you have to operate $L_z$ on $Y_{lm}(\theta,\phi)$ ($|l,m\rangle$ in position basis), you will then resort to the position form of $L_z$, which is equal to $-i\hbar\frac{\partial}{\partial\phi}$, instead of its matrix form.

I'm confused though as to how to form a 3D vector from three 2x2 matrices.

Especially for spin operators, there is no position representation for them. Therefore, the most commonly used representation for these operators are the matrix form.

 

[haruspex]

I think what's throwing you is the last word in the problem statement. If you change it from 'vectors' to 'matrices' does it make sense for you?
$a_i\sigma_i$ is just a scalar multiplied by a 2x2 matrix, giving another 2x2 matrix. The dot product $\vec a.\vec \sigma$ is a sum of these, so is another 2x2 matrix. This can be raised to integer powers.