# Problem with vectors and matrices.

1. Feb 6, 2016

### Dewgale

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$\sigma_1 = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$
$$\sigma_2 = \begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix}$$
$$\sigma_3 = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$$
3. 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!!

2. Feb 7, 2016

### 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.

3. Feb 7, 2016

### Dewgale

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?

4. Feb 7, 2016

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

5. Feb 8, 2016

### 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.