# Commutator of function of operators

1. Aug 11, 2014

### kini.Amith

According to my teacher, for any two operators A and B, the commutators
[f(A),B]=[A,B]df(A)/dA
and [A,f(B)]=[A,B]df(B)/dB
He did not give any proof.
I can easily prove this for the particular cases
[f(x),p]=[x,p]df(x)/dx
and [x,pn]=[x,p]npn-1
But I don't see how the general formula is true. I also can't find it in any textbooks.

2. Aug 12, 2014

### bhobba

I am not sure that's true.

Expand f(A) in terms of a power series. Take the derivative of that to give df(A)/dA and then check if equal. I cant see how it is without further assumptions.

Thanks
Bill

Last edited: Aug 12, 2014
3. Aug 12, 2014

### kini.Amith

But he used this formula to prove eAeB=eA+B+[A,B]/2.
To be exact, he took [eAt,B]=[A,B]teAt in one step of the proof.

4. Aug 12, 2014

### bhobba

Expand them out and have a look at the first term - are they equal?

To be exact the first term on LHS doesn't have t in front - yet the first term on the RHS does.

Wait a minute - the first term contains I which always commutes so is zero. That might be the assumption he is making - if the first term of f(A) commutes with B it may hold. I will check at get back.

Yea - the assumption he is making is the power series expansion is in terms of scalars ie f(A) = a0 + a1A + a2A^2 ..... and a0, a1 etc are scalars.

I think if you insert and equate then the formula are true.

Thanks
Bill

Last edited: Aug 12, 2014
5. Aug 12, 2014

### Orodruin

Staff Emeritus
This is only true if
$$[A,[A,B]] = [B,[A,B]] = 0.$$
That is, the commutator of A and B must commute with A and B for the formula to be true. In the case of x and p, their commutator is a scalar and it therefore holds. However, there is no guarantee for arbitrary A and B. The relation
$$e^A e^B = e^{A+B + [A,B]/2}$$
is a special case of the Baker-Campbell-Hausdorff formula (see http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula) for when $[A,B]$ commutes with $A$ and $B$.

6. Aug 12, 2014

### Orodruin

Staff Emeritus
Just to expand a bit on the series expansion.

Expanding $f(a)$ gives
$$f(a) = \sum_{n=0}^\infty \frac{d^n f}{da^n} \frac{a^n}{n!} \equiv \sum_{n=0}^\infty k_n a^n$$
and thus
$$\frac{df(a)}{da} = \sum_{n=0}^\infty k_n n a^{n-1} = \sum_{n=1}^\infty n k_n a^{n-1}.$$
Now, in the same way
$$f(A) = \sum_{n=0}^\infty k_n A^n,$$
which implies
$$[f(A),B] = \sum_{n=0}^\infty k_n [A^n, B].$$
We can here make use of the commutator relation
$$[A^n, B] = \sum_{k = 1}^n A^{n-k} [A,B] A^{k-1}$$
to rewrite this as
$$[f(A),B] = \sum_{n=0}^\infty k_n \sum_{k = 1}^n A^{n-k} [A,B] A^{k-1}.$$
Now, if $[A,[A,B]] = 0$, then
$$A^{n-k} [A,B] A^{k-1} = [A,B] A^{n-1}$$
and
$$\sum_{k = 1}^n A^{n-k} [A,B] A^{k-1} = [A,B] A^{n-1} \sum_{k=1} ^n 1 = n [A,B] A^{n-1}.$$
The expression would then simplify to
$$[f(A),B] = [A,B] \sum_{n=0}^\infty k_n n A^{n-1} = [A,B] f'(A).$$
The last expression is of course also equal to $f'(A) [A,B]$, since $[A,B]$ commutes with $A$.

7. Aug 12, 2014

### kini.Amith

Thanks a ton.

8. Aug 12, 2014

### bhobba

Yea - me too.