Commutation of 2 operators using braket notation?

[philip041]

How do you work out the commutator of two operators, A and B, which have been written in bra - ket notation?

alpha = a beta = b

A = 2|a><a| + |a><b| + 3|b><a|

B = |a><a| + 3|a><b| + 5|b><a| - 2|b><b|

The answer is a 4x4 matrix according to my lecturer...

Any help much appreciated!

I know the commutator is AB - BA but how do you do it for these big bra - ket eqns?

 

[Ben Niehoff]

Just write out AB - BA and apply the standard rules of algebra - expand and distribute. Just don't rearrange the order of terms that don't commute.

Are the two vectors |a> and |b> orthonormal? If so, then you can also use

<a|a> = <b|b> = 1

<a|b> = <b|a> = 0

If they are not orthonormal, then you cannot use the above reductions.

 

[philip041]

but how do you get a 4 x 4 matrix?

 

[Ben Niehoff]

B = |a><a| + 3|a><b| + 5|b><a| - 2|b><b|

This IS a matrix. Kets are column vectors and bras are row vectors.

But it looks like you should get a 2x2 matrix. Perhaps your professor misspoke?

Unless of course the kets |a> and |b> have additional internal structure that you haven't shown us.

 

[philip041]

ohh cheers

 

[Domnu]

If you have an n-state system, you could create a vector in R^n that corresponds to each (orthogonal) state. So for example, in this case, you could have that

<a| = (1 0)
<b| = (0 1)

And of course, the kets for each of the corresponding bras would just be column vectors which are transposes for each. You can just multiply everything out and it will work. I think this must be it =]