Finding orthonormal simultaneous eigenkets of two operator matrices

[voxel]

Homework Statement

Consider a thee-dimensional ket space. If a certain set of orthonormal kets - say, |1>, |2>, and |3> - are used as the base kets, the operators A and B are represented by
A =
a 0 0
0 -a 0
0 0 -a

B =
b 0 0
0 0 -ib
0 ib 0

with a and b both real.
a) Obviously A exhibits a degenerate spectrum. Does B also exhibit a degenerate spectrum?
b) Show that A and B commute.
c) Find a new set of orthonormal kets which are simultaneous eigenkets of both A and B. Specify the eigenvalues of A and B for each of the three eigenkets. Does your specification of eigenvalues completely characterize each eigenket?

Homework Equations

[A,B] = AB - BA = 0 if A & B commute

The Attempt at a Solution

a) B has eigenvalues b,b,-b. So yes, B is degenerate.
b) I have no problem showing that A & B commute.
c) I know how to find eigenkets (eigenvectors) of a matrix using the matrix eigenvalues, but I do not know how to go about finding eigenkets that are simultaneous eigenkets of both A and B?
I tried finding eigenvalues of the matrix AB, which come out to ab,ab,-ab (also degenerate) but can only construct 2 eigenkets, and when I construct a third orthonormal eigenket using the cross product |1>\otimes|2> ) it does not give A|3> = a|3> or B|3> = b|3>.

Any hints or suggestions would be greatly appreciated!

edit:
I understand that the eigenkets should satisfy A|a'b'> = a'|a'b'> & B|a'b'> = b'|a'b'>, I just don't know how to go about finding |a'b'>.

 

[gabbagabbahey]

If A and B commute, theeigenkets of either will also automatically be eigenkets of the other...I suspect you have already proven this in class, or in your text?

 

[voxel]

Thanks for the reply gabbagabbahey,
Yes, I found that in the textbook. I ended up leaving the eigenkets in a generalized form, then taking a cross product and choosing 3 kets that satisfied the generalized forms for all the eigenkets.

 

[marcelnv]

I know that when two operators A and B commute, any eigenket of A is also an eigenket of B.
As such, i think(Not certain though) that it suffices in this case to find eigenkets of either A or B and then normalize the vectors. What do you guys think?

 

[vela]

That's not quite accurate because of degeneracy. For example, for a free particle, H=p²/2m. The Hamiltonian and the momentum operator p commute, but the state \vert p \rangle + \vert -p \rangle is an eigenstate of H, but it's not an eigenstate of p.

 

[marcelnv]

Oh, that's true.
In this case, what condition should we place on the statement i did mention above?

i.e if A and B commute, then any eigenket of A is an eigenket of B.