# Ladder operator commutator with arbitary function

Tags:
1. May 9, 2016

### AwesomeTrains

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

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.

2. Relevant equations
Product and chain rule for derivatives.
Commutator definition.

3. 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

2. May 9, 2016

### nrqed

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 $[a,(a^\dagger)^n]$.

3. May 9, 2016

### AwesomeTrains

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.

4. May 9, 2016

### AwesomeTrains

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

5. May 9, 2016

### nrqed

Are you sure about this??

6. May 9, 2016

### AwesomeTrains

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 $[a,(a^\dagger)^m]=\sum^{m-1}_{k=0}\binom{m-1}{k}(a^\dagger)^{m-k-1}[a,(a^\dagger)]a^k =(a^\dagger)^{m-1}\sum^{m-1}_{k=0}\binom{m-1}{k}=(a^\dagger)^{m-1}2^{m-1}$

7. May 10, 2016

### AwesomeTrains

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?

8. May 10, 2016

### blue_leaf77

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.

9. May 10, 2016

### AwesomeTrains

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.

10. May 10, 2016

### blue_leaf77

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

11. May 10, 2016

### AwesomeTrains

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?

12. May 10, 2016

### blue_leaf77

Are the operators in the commutator above in the right order?

Stop right there, how will you simplify $\frac{m}{m!}$?

13. May 10, 2016

### 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?

14. May 10, 2016

### blue_leaf77

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?

15. May 10, 2016

### nrqed

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.

16. May 10, 2016

### AwesomeTrains

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?

17. May 10, 2016

### AwesomeTrains

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

18. May 10, 2016

### blue_leaf77

Yes.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted