# Dot products in commutators

Suppose $\hat r$ is an position operator, $\hat p$ is a momentum operator and $\vec c$ is a constant vector.

What does the commutator $[\hat p, \vec c\cdot\hat r]$ mean?

I see that you can expand the second term such that the commutator becomes $[\hat p, c_xr_x+c_yr_y+c_zr_z]$ but then one of the operators in the commutator is a "vector" whereas the other is a scalar? Perhaps I am interpreting this wrong.

What would the value of $[\hat p, \vec c\cdot\hat r]$ be? Given that $[x,p_x]=i\hbar$? where $x$ is a component of $\hat r$ and $p_x$ the corresponding component in $\hat p$.

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Yes, I think that the interpretation that you pose isn't right, if it were then the commutator would always be zero (since scalar multiplication of a vector is commutative) and so the definition wouldn't have much content. – Jonah Sinick Nov 1 '12 at 20:05

## 1 Answer

Using $$\left[\hat{p}_j, \hat{r}_i\right] = - i \hbar \delta_{i j},$$ $$\begin{eqnarray} \left[\hat{\bf p}, {\bf c} \bullet \hat{\bf r}\right] &=& \sum_{i,j=1}^3 \left[\hat{p}_i {\bf e}_i, c_j\hat{r}_j\right] = \sum_{i,j=1}^3 c_j \left[\hat{p}_i,\hat{r}_j\right] {\bf e}_i \end{eqnarray} = - i \hbar \sum_{i=1}^3 c_i {\bf e}_i = -i \hbar {\bf c}.$$

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