I am reading Cohen-Tannoudji's Quantum Mechanics Vol. 1. In problem 11 from Chapter II there are two given operators defined in a space generated by $\{\left|u_1\right>,\left|u_2\right>,\left|u_3\right>\}$:

$H=h\omega\left(\begin{matrix}1&0&0\\0&-1&0\\0&0&-1\end{matrix}\right) \qquad B=b\left(\begin{matrix}1&0&0\\0&0&1\\0&1&0\end{matrix}\right)$

The problem is to show that these commute. It is easy to show it with matrix multiplication. But, my question comes up because of the solution given in the book, it says the next:

"$\left|u_1\right>$ is an eigenvector common to H and B. We therefore have, obviously $HB\left|u_1\right>=BH\left|u_1\right>.$ We see, then, that for $H$ and $B$ to commute, it is suficient that the restrictions of these operators to the subspace $\mathscr{E}_2$, spanned by $\left|u_2\right>$ and $\left|u_3\right>$, commute. Now, in this subspace, the matrix representing H is equal to $-h\omega I$ (where $I$ is the $2\times2$ unit matrix), which commutes with all $2\times 2$ matrices. $H$ and $B$ therefore commute."

I need an explanation of this answer. What is the restriction of an operator and why is it sufficient to show that $H$ and $B$ commute if their restrictions in space $\mathscr{E}_2$ commute?


1 Answer 1


The space generated by $|u_1 \rangle, \, |u_2 \rangle$ and $|u_3 \rangle$ is $R^3$ itself, since they are linearly independent. Consider a subspace of $R^3$, like $R^2$ for example. We can inquire about how an operator defined in the whole space acts solely on this subspace, and for that we need only to ignore everything it does outside of our subspace. In your case we want to forget about what the operator does in the subspace generated by $|u_1 \rangle $, which we know that is an eigenvector of both $H $ and $B $, so we focus in the subspace complementary to it, which is the plane generated by $ |u_2 \rangle$ and $|u_3 \rangle$. We are therefore restricting the action of $H $ and $B $ to a subspace. On this subspace we see, by forgeting about the first line and column of $H $ and $B $, that $H $ commutes with $B$. We therefore arrive at the conclusion that $H$ and $B$ commute on a plane and the one dimensional subspace complementary to it is generated by a common eigenvector: therefore they commute in the whole $R^3$.


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