Sakurai 1.15

[Bill Foster]

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

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.


3. 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

[csco]

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

[jdwood983]

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

[Bill Foster]

Yep. Looks like this one is done, too.

Danke.