Sakurai 1.15

  • #1
338
0

Homework Statement



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


The Attempt at a Solution



One solution is given as follows:

[tex]\sum_m |a_m,b_m\rangle\langle a_m,b_m|=1[/tex]
[tex]\sum_n |a_n,b_n\rangle\langle a_n,b_n|=1[/tex]

[tex]\left[A,B\right]=AB-BA[/tex]
[tex]=\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|[/tex]
[tex]=\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|[/tex]
[tex]=\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)[/tex]
[tex]=\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)[/tex]
[tex]=0[/tex]

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

[tex]AB|a_n,b_n\rangle = a_n b_n|a_n,b_n\rangle[/tex]
[tex]BA|a_n,b_n\rangle = b_n a_n|a_n,b_n\rangle[/tex]

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

[tex]\left[A,B\right]=\left(AB-BA\right)\sum_n |a_n,b_n\rangle\langle a_n,b_n|[/tex]
[tex]=\sum_n\left(AB-BA\right)|a_n,b_n\rangle\langle a_n,b_n|[/tex]
[tex]=\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)[/tex]
[tex]=\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)[/tex]
[tex]=0[/tex]
 
Last edited:

Answers and Replies

  • #2
14
0
By definition [tex]|a_n,b_n\rangle[/tex] are eigenkets of A and B with eigenvalues an and bn:

[tex]A|a_n,b_n\rangle = a_n|a_n,b_n\rangle[/tex]
[tex]B|a_n,b_n\rangle = b_n|a_n,b_n\rangle[/tex]

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

[tex]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[/tex]
[tex]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[/tex]
 
  • #3
383
0
And since it is true, why can the following be an equally valid solution?

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

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

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

that is, factor out [itex]|a_n,b_n\rangle\langle a_n,b_n|[/itex]? Then from this, since they are numbers, [itex]a_nb_n-b_na_n=a_nb_n-a_nb_n=0[/itex]
 
  • #4
338
0
Yep. Looks like this one is done, too.

Danke.
 

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