# Creation/Anhilation Operator Exponential Commutator Relation

1. Jan 18, 2015

### teroenza

1. The problem statement, all variables and given/known data
Given that the function f can be expanded in a power series of $a$ and $a^\dagger$, show that:

$[a,f(a,a^{\dagger})]=\frac{\partial f }{\partial a^\dagger}$

and that

$[a,e^{-\alpha a^\dagger a}] = (e^{-\alpha}-1)e^{-\alpha a^{\dagger} a}a$

3. The attempt at a solution
I've tied using simple algebra, but got nowhere. Writing the exponential operators as infinite sums did not seems to be leading to an equality either. I am not sure how to correctly order the result of the derivative resulting from the first equation. I.e. $e^{-\alpha a^{\dagger} a}(-\alpha a)$ or $(-\alpha a)e^{-\alpha a^{\dagger} a}$.

2. Jan 19, 2015

Perhaps check first that the formula holds for simple monomials of the form $f=a^m (a^\dagger)^n .$

3. Jan 19, 2015

### teroenza

Ok. I think part of my problem is that I'm not clear on how to series expand a multivariable function. From
http://en.wikipedia.org/wiki/Power_series#Power_series_in_several_variables

I think (if I'm reading the notation right) that:
$f(a,a^{\dagger})=\sum_{i=0}^{\infty} \sum_{j=0}^{\infty}c_i c_ja^i (a^{\dagger})^j$.

Before looking at it in terms of series, I had tried to show that:
$[a,a^m(a^{\dagger})^n] = a^{m+1}(a^{\dagger})^n-a^m(a^{\dagger})^na = \frac{\mathrm{d} f}{\mathrm{d} a^{\dagger}}=a^mn(a^{\dagger})^{n-1}$

Last edited: Jan 19, 2015
4. Jan 19, 2015

You must be careful. The Wiki entry is for power series with commuting variables, while $a$ and $a^\dagger$ do not commute. You need to fix the ordering. Putting first powers of $a$, then powers of $a^\dagger$ is called anti-normal ordering. Remember the rule:

$$[A,BC]=B[A,C]+[A,B]C$$

Then calculate

$$[a,a^\dagger ]=1$$
$$[a,(a^\dagger)^2]=a^\dagger [a,a^\dagger]+[a,a^\dagger]a^\dagger=2a^\dagger$$
$$[a,(a^\dagger)^3]=a^\dagger[a,(a^\dagger)^2]+[a,a^\dagger](a^\dagger)^2=3(a^\dagger)^2$$
....
$$[a,(a^\dagger)^n]=n(a^\dagger)^{n-1}.$$
$$[a,a^m(a^\dagger)^n]=a^m[a,(a^\dagger)^n]+[a,a^m](a^\dagger)^n=a^m[a,(a^\dagger)^n]=na^m(a^\dagger)^{n-1}.$$

5. Jan 20, 2015

### teroenza

Ok. So I need to pick an ordering and stick with it. I'm not sure how to make the jump from the monomial to the exponential. Perhaps thinking of the commutator as $[a,(e^{- a^{\dagger}a})^{\alpha}]$? But that seems just as bad, because I would need to know $[a,e^{- a^{\dagger}a}]$

6. Jan 20, 2015

Look at the formula you need to prove: $[a,e^{-\alpha a^\dagger a}] = (e^{-\alpha}-1)e^{-\alpha a^{\dagger} a}a$

Expand the commutator on the LHS, expand the parenthesis on the RHS. Simplify. Multiply both sides by $e^{\alpha a^{\dagger} a}$ form the left. You will get a simpler formula that can be proven in several ways.

Last edited: Jan 20, 2015
7. Jan 20, 2015

### teroenza

Since the usual way to expand the LHS commutator would mix the ordering, the only way to expand the commutator would be to use what I proved before and use the derivative.

$e^{-\alpha a^\dagger a}(-\alpha a) = e^{-\alpha(1+aa^\dagger)}a - e^{-\alpha a a^\dagger}$

8. Jan 21, 2015

I will use $t$ instead of $\alpha$, because $\alpha$ is too similar to $a$.
You messed up. $LHS=[a,e^{-t a^\dagger a}]=ae^{-t a^\dagger a}-e^{-t a^\dagger a}a$.
$RHS=e^{-t}e^{-t a^{\dagger} a}a-e^{-t a^{\dagger} a}a$. Therefore you need to show that $ae^{-t a^\dagger a}=e^{-t}e^{-t a^{\dagger} a}a$. Multiply from the left by $e^{t a^\dagger a}$ taking into account the fact that $e^{-t}$ is a c-number, thus commutes with any operator. You need to prove that
$$e^{t a^\dagger a}ae^{-t a^\dagger a}=e^{-t}\,a$$
Define $f(t)=e^{t a^\dagger a}ae^{-t a^\dagger a}$. This is an operator valued function of a real variable $t$. Verify that $f(0)=a$. Now calculate the derivative with respect to $t$. You can use the chain rule, but you must pay attention to the order of noncommuting operators. You get
$$f'(t)=e^{t a^\dagger a}(a^\dagger a) a e^{-t a^\dagger a} -e^{t a^\dagger a}a(a^\dagger a) e^{-t a^\dagger a}=e^{t a^\dagger a}[a^\dagger a, a] e^{-t a^\dagger a}$$
The commutator you already know, and it is $-a$. Therefore $f(t)$ satisfies the differential equation $$f'(t)=-f(t).$$
Check that the RHS satisfies the same differential equation. Check the initial values are the same. From the uniqueness of solutions of first order linear differential equations you have that LHS=RHS for all $t$.

Last edited: Jan 21, 2015
9. Jan 21, 2015

### teroenza

Thank you so much for your help.