# Sakurai 1.15

## Homework Statement

Let A and B be observables. Suppose the simultaneous eigenkets of A and B $$\left{|a_n,b_n\rangle\right}$$ form a complete orthonormal set of base kets. Can we always conclude that $$[A,B]=0$$ ? If “yes”, prove it. If “no”, give a counterexample.

## The Attempt at a Solution

One solution is given as follows:

$$\sum_m |a_m,b_m\rangle\langle a_m,b_m|=1$$
$$\sum_n |a_n,b_n\rangle\langle a_n,b_n|=1$$

$$\left[A,B\right]=AB-BA$$
$$=\sum_m |a_m,b_m\rangle\langle a_m,b_m|\left(AB-BA\right)\sum_n |a_n,b_n\rangle\langle a_n,b_n|$$
$$=\sum_n\sum_m |a_m,b_m\rangle\langle a_m,b_m|\left(AB-BA\right)|a_n,b_n\rangle\langle a_n,b_n|$$
$$=\sum_n\sum_m |a_m,b_m\rangle\langle a_m,b_m|\left(AB|a_n,b_n\rangle\langle a_n,b_n|-BA|a_n,b_n\rangle\langle a_n,b_n|\right)$$
$$=\sum_n\sum_m |a_m,b_m\rangle\langle a_m,b_m|\left(a_n b_n|a_n,b_n\rangle\langle a_n,b_n|-b_n a_n|a_n,b_n\rangle\langle a_n,b_n|\right)$$
$$=0$$

My question is this: how is it known that the following is true?

$$AB|a_n,b_n\rangle = a_n b_n|a_n,b_n\rangle$$
$$BA|a_n,b_n\rangle = b_n a_n|a_n,b_n\rangle$$

And since it is true, why can the following be an equally valid solution?

$$\left[A,B\right]=\left(AB-BA\right)\sum_n |a_n,b_n\rangle\langle a_n,b_n|$$
$$=\sum_n\left(AB-BA\right)|a_n,b_n\rangle\langle a_n,b_n|$$
$$=\sum_n\left(AB|a_n,b_n\rangle\langle a_n,b_n|-BA|a_n,b_n\rangle\langle a_n,b_n|\right)$$
$$=\sum_n\left(a_n b_n|a_n,b_n\rangle\langle a_n,b_n|-b_n a_n|a_n,b_n\rangle\langle a_n,b_n|\right)$$
$$=0$$

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By definition $$|a_n,b_n\rangle$$ are eigenkets of A and B with eigenvalues an and bn:

$$A|a_n,b_n\rangle = a_n|a_n,b_n\rangle$$
$$B|a_n,b_n\rangle = b_n|a_n,b_n\rangle$$

Using the previous equations and linearity of A and B you have:

$$BA|a_n,b_n\rangle = Ba_n|a_n,b_n\rangle = a_nB|a_n,b_n\rangle = a_nb_n|a_n,b_n\rangle$$
$$AB|a_n,b_n\rangle = Ab_n|a_n,b_n\rangle = b_nA|a_n,b_n\rangle = b_na_n|a_n,b_n\rangle$$

And since it is true, why can the following be an equally valid solution?

$$\left[A,B\right]=\left(AB-BA\right)\sum_n |a_n,b_n\rangle\langle a_n,b_n|$$
$$=\sum_n\left(AB-BA\right)|a_n,b_n\rangle\langle a_n,b_n|$$
$$=\sum_n\left(AB|a_n,b_n\rangle\langle a_n,b_n|-BA|a_n,b_n\rangle\langle a_n,b_n|\right)$$
$$=\sum_n\left(a_n b_n|a_n,b_n\rangle\langle a_n,b_n|-b_n a_n|a_n,b_n\rangle\langle a_n,b_n|\right)$$
$$=0$$

If $a_n$ and $b_n$ are both numbers, can't we say (inserting this between the last two lines above)

$$[A,B]=\sum_n\left(a_n b_n-b_na_n\right)|a_n,b_n\rangle\langle a_n,b_n|$$

that is, factor out $|a_n,b_n\rangle\langle a_n,b_n|$? Then from this, since they are numbers, $a_nb_n-b_na_n=a_nb_n-a_nb_n=0$

Yep. Looks like this one is done, too.

Danke.