# What is the significance of a complex commutator?

1. Mar 12, 2012

### Darkmisc

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

If A and B are two operators such that
[A,B] = λ , where l is a complex number, and
if μis a second complex number, show that:
exp[μ(A + B)] = exp(μA)exp(μB)exp(- λμ^2/ 2).

2. Relevant equations

3. The attempt at a solution

I'm stuck on where to begin. I know that the result [x, p_x] = ih_bar indicates that x and p_x relate by the uncertainty principle. I'm not sure if this conclusion applies to complex commutators in general.

Also, I don't understand the significance of the exponential function. Do I lose anything by expressing that information as

μ(A+B)=(-λμ^2)/2 ?

Any ideas?

Thanks

2. Mar 12, 2012

### turin

I can't figure out why the imaginaryness of λ and μ is important (at least formally). Exponentiating an operator means nothing more than the infinite power series of the exponent (Taylor series). I think that it's actually technically defined this way. So, for example:

exp(μA) = 1 + μA + (1/2)μ2A2 + (1/6)μ3A3 ...

So, just expand each exponential, multiply the r.h.s. out, and then collect terms of the same order in μ. If you can stand it, you might work all the way up to order μ4, but you will begin to see the point even at order μ2.

CAUTION: WHEN YOU EXPAND OPERATOR EXPRESSIONS, THE COMMUTATIVE LAW FOR MULTIPLICATION MAY NOT APPLY ...

Where did you get that?

Last edited: Mar 13, 2012
3. Mar 13, 2012

### Darkmisc

oops, I messed up that expression. I meant to equate the indices.

4. Mar 13, 2012

### turin

There are operators A and B, and then there are parameters λ and μ; what indices?

5. Mar 15, 2012

### sgnl03

Could this be done easier using BCH formula due to non-commuting? Expansion doesn't seem to get anywhere.

edit: I guess I don't mean putting it through the entire thing, just the result for exp(A +B) = exp(A)exp(B)exp([A,B]/2), but then again I'm also clueless.

6. Mar 15, 2012

### turin

Sure, it would be MUCH easier using the BCH formula (almost trivial). I was assuming that the problem is an excercise to demonstrate a nontrivial exponentiation of operators, and if the point was to use the BCH formula, then they would have at least made the commutator more complicated (like proportional to A or B). Of course, this is just an assumption, and since the problem statement doesn't disallow it, BCH formula is probably the way to show something like this in practice.

I don't know what you mean. Expansion must get somewhere, and it is a good excercise to formally match terms according to an order parameter (μ in this case.) Of course, then you must also make use of
BA = AB - [A,B]
etc.

This is correct. You know that you can stop there because the commutator is a scalar. In effect, that IS "the entire thing", and anyway "the entire thing" apparently doesn't have (what a would call) a closed form (at least it doesn't look very elegant in general).

7. Mar 18, 2012

### Darkmisc

Cheers, Thanks.

Last edited: Mar 18, 2012