# Ladder operator commutator with arbitary function

• AwesomeTrains

#### AwesomeTrains

Hey there!
1. Homework Statement

I've been given the operators
$a=\sqrt\frac{mw}{2\hbar}x+i\frac{p}{\sqrt{2m\hbar w}}$ and $a^\dagger=\sqrt\frac{mw}{2\hbar}x-i\frac{p}{\sqrt{2m\hbar w}}$ without the constants and definition of the momentum operator:
$a=x+\partial_x$ and $a^\dagger=x-\partial_x$ with $\partial_x=\frac{\partial}{\partial x}$
I have to calculate $[a,f(a^\dagger)]$, with f some arbitrary function.

## Homework Equations

Product and chain rule for derivatives.
Commutator definition.

## The Attempt at a Solution

$[a,f(a^\dagger)]=(x+\partial_x)f(x-\partial_x)-f(x-\partial_x)(x+\partial_x)\\ =xf(x-\partial_x)+\partial_x f(x-\partial_x) - f(x-\partial_x)x- f(x-\partial_x)\partial_x\\ =xf(x-\partial_x)+(\partial_x f)(1-{\partial_x}^2)+f(x-\partial_x)\partial_x- f(x-\partial_x)x-f(x-\partial_x)\partial_x\\ =[x,f(a^\dagger)]+(\partial_x f)(1-{\partial_x}^2)$ <-This is where I'm stuck :(

Can this be simplified further? Is this even correct so far?
Kind regards
Alex

Hey there!

## Homework Statement

I've been given the operators
$a=\sqrt\frac{mw}{2\hbar}x+i\frac{p}{\sqrt{2m\hbar w}}$ and $a^\dagger=\sqrt\frac{mw}{2\hbar}x-i\frac{p}{\sqrt{2m\hbar w}}$ without the constants and definition of the momentum operator:
$a=x+\partial_x$ and $a^\dagger=x-\partial_x$ with $\partial_x=\frac{\partial}{\partial x}$
I have to calculate $[a,f(a^\dagger)]$, with f some arbitrary function.

## Homework Equations

Product and chain rule for derivatives.
Commutator definition.

## The Attempt at a Solution

$[a,f(a^\dagger)]=(x+\partial_x)f(x-\partial_x)-f(x-\partial_x)(x+\partial_x)\\ =xf(x-\partial_x)+\partial_x f(x-\partial_x) - f(x-\partial_x)x- f(x-\partial_x)\partial_x\\ =xf(x-\partial_x)+(\partial_x f)(1-{\partial_x}^2)+f(x-\partial_x)\partial_x- f(x-\partial_x)x-f(x-\partial_x)\partial_x\\ =[x,f(a^\dagger)]+(\partial_x f)(1-{\partial_x}^2)$ <-This is where I'm stuck :(

Can this be simplified further? Is this even correct so far?
Kind regards
Alex
Are you instructed to use the expressions of ##a## and ##a^\dagger## in terms of x and ##\partial_x## or was this your idea? Because I would be tempted to simply write ##f(a^\dagger)## as a Taylor series and then use the commutator ##[a,(a^\dagger)^n]##.

AwesomeTrains
Are you instructed to use the expressions of ##a## and ##a^\dagger## in terms of x and ##\partial_x## or was this your idea? Because I would be tempted to simply write ##f(a^\dagger)## as a Taylor series and then use the commutator ##[a,(a^\dagger)^n]##.

That was my idea. Sorry for putting it at the first point. It would make sense to do it that way. In the problem before on this assignment I calculated $[a,(a^\dagger)^n]$. I'll try and do it your way, thanks :). I'll come back if I get stuck again.

How does this look? I calculated $[a,(a^\dagger)^m]=2^{m-1}a^{m-1}$.
Then I did the taylor expansion $[a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[(a^\dagger)^m,a]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}2^{m-1}a^{m-1}$

How does this look? I calculated $[a,(a^\dagger)^m]=2^{m-1}a^{m-1}$.

No, I'm not :) How is this?: $[a,(a^\dagger)^m]=2^{m-1}(a^\dagger)^{m-1}$? I wrote out the commutator for the first few m and found that $[a,(a^\dagger)^m]=\sum^{m-1}_{k=0}\binom{m-1}{k}(a^\dagger)^{m-k-1}[a,(a^\dagger)]a^k =(a^\dagger)^{m-1}\sum^{m-1}_{k=0}\binom{m-1}{k}=(a^\dagger)^{m-1}2^{m-1}$

We simplified this commutator in the lecture today and got this: $[a,(a^\dagger)^n]=n(a^\dagger)^{n-1}$ Can you see where my mistake in the previous post is?

There is a simpler way though to calculate ##[a,(a^\dagger)^n]## using induction. You can start by calculating this commutator for ##n=0,1,2##. Then infer a pattern for this three values of ##n##'s. Next step is to make this pattern as a hypothesis and proceed through the standard induction procedure.

There is a simpler way though to calculate ##[a,(a^\dagger)^n]## using induction. You can start by calculating this commutator for ##n=0,1,2##. Then infer a pattern for this three values of ##n##'s. Next step is to make this pattern as a hypothesis and proceed through the standard induction procedure.
That's how we did in the lecture. I don't understand though why I get a different result. I did like this, for $n=3$ using a commutation identity: $[a,(a^\dagger)^3]=[a,(a^\dagger)^2]a^\dagger+a^\dagger[a,(a^\dagger)^2]\\=([a,a^\dagger]a^\dagger+a^\dagger[a,a^\dagger])a^\dagger+a^\dagger([a,a^\dagger]a^\dagger+a^\dagger[a,a^\dagger])\\=(a^\dagger+a^\dagger)a^\dagger+a^\dagger(a^\dagger+a^\dagger)=2(a^\dagger)^2+2(a^\dagger)^2=4(a^\dagger)^2$
This is how our prof did: $[a,(a^\dagger)^3]=a(a^\dagger)^3-(a^\dagger)^3a=(1+aa^\dagger)(a^\dagger)^2-(a^\dagger)^3a\\=(a^\dagger)^2+a^\dagger a a^\dagger a^\dagger-(a^\dagger)^3a\\=(a^\dagger)^2+a^\dagger(1+a^\dagger a)a^\dagger - (a^\dagger)^3a\\=(a^\dagger)^2+(a^\dagger)^2+(a^\dagger)^2aa^\dagger-(a^\dagger)^3a\\=2(a^\dagger)^2+(a^\dagger)^2(1+a^\dagger a)-(a^\dagger)^3a\\=3(a^\dagger)^2+(a^\dagger)^3a-(a^\dagger)^3a\\=3(a^\dagger)^2$ which generalizes to $[a,(a^\dagger)^n]=n (a^\dagger)^{n-1}$
Why do I get some other result, can't I use the identity $[A,BC]=B[A,C]+[A,B]C$? Or is it something else that I'm doing wrong.

You made mistake in the first line already. There you have ##[a,(a^\dagger)^3] = [a,a^\dagger(a^\dagger)^2]##. Denoting ##A = a##, ##B = a^\dagger##, and ##C = (a^\dagger)^2##, how do they transform according to the identity ##
[A,BC]=B[A,C]+[A,B]C##?

You made mistake in the first line already. There you have ##[a,(a^\dagger)^3] = [a,a^\dagger(a^\dagger)^2]##. Denoting ##A = a##, ##B = a^\dagger##, and ##C = (a^\dagger)^2##, how do they transform according to the identity ##
[A,BC]=B[A,C]+[A,B]C##?
Oh that's silly of me it has been bugging me the whole day, thank you :) $[a,a^\dagger (a^\dagger)^2]=a^\dagger[a,(a^\dagger)^2]+[a,a^\dagger](a^\dagger)^2$
Well then using the correct result I get this for the original question:
$[a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[(a^\dagger)^m,a]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}\\=\sum^\infty_{m=1}\frac{f^{(m)}(0)}{(m)!} (a^\dagger)^{m-1}\\=(a^\dagger)^{-1}\sum^\infty_{m=0}\frac{f^{(m)}(0)}{(m)!} (a^\dagger)^{m}-f(0)(a^\dagger)^{-1}\\=(a^\dagger)^{-1} f(a^\dagger)-f(0)(a^\dagger)^{-1}$ Doesn't seem right though. Can I even simplify that series?

$$\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[(a^\dagger)^m,a]$$
Are the operators in the commutator above in the right order?

$$\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}$$
Stop right there, how will you simplify ##\frac{m}{m!}##?

$[a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[a,(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{(m-1)!} (a^\dagger)^{m-1}$ Can I return this to the original taylor approximation so that I can write $f(a^\dagger)$ instead of the series?

Does the summation index really start from zero in the last expression?
If you compare the last expression (after fixing the starting summation index) with the Taylor expansion of ##f(a^\dagger)##, which is
$$\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m,$$
what do you think about the relation of them?

AwesomeTrains
$[a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[a,(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{(m-1)!} (a^\dagger)^{m-1}$ Can I return this to the original taylor approximation so that I can write $f(a^\dagger)$ instead of the series?
Yes, you should reexpress it in terms of ##f(a^\dagger)## but it is not the function itself you get, right? It is something else...And the end result is quite interesting.

Well it can't start at 0 since $(-1)!$ isn't defined. The next reasonable index is 1. Ah, it's the derivative of the function at $a^\dagger$. What I get is the shifted series. Am I right?

$[a,f(a^\dagger)]=[a,\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[a,(a^\dagger)^m]=\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}=\sum^\infty_{m=1}\frac{f^{(m)}(0)}{(m-1)!} (a^\dagger)^{m-1}=\sum^\infty_{m=1}\frac{f^{(m)}(0)}{(m-1)!} (a^\dagger)^{m-1}=\sum^\infty_{m=0}\frac{f^{(m+1)}(0)}{(m)!} (a^\dagger)^{m}=\sum^\infty_{m=0}\frac{f'^{(m)}(0)}{(m)!} (a^\dagger)^{m}=f'(a^\dagger)$

Ah, it's the derivative of the function at a†
Yes.