Ladder operator commutator with arbitary function

In summary: 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]=[A,B]C+B[A,C], \thickspace \thickspace [a,(a^\dagger)^3]=[a,a^\dagger(a^\dagger)^2]\\=[a,a^\dagger](a^\dagger)^2+a^\dagger[a,a^\dagger](a^\dagger)^
  • #1
AwesomeTrains
116
3
Hey there!
1. Homework Statement

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

Homework Equations


Product and chain rule for derivatives.
Commutator definition.

The Attempt at a Solution


[itex][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)[/itex] <-This is where I'm stuck :(

Can this be simplified further? Is this even correct so far?
Kind regards
Alex
 
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  • #2
AwesomeTrains said:
Hey there!

Homework Statement


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

Homework Equations


Product and chain rule for derivatives.
Commutator definition.

The Attempt at a Solution


[itex][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)[/itex] <-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]##.
 
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  • #3
nrqed said:
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 [itex][a,(a^\dagger)^n][/itex]. I'll try and do it your way, thanks :). I'll come back if I get stuck again.
 
  • #4
How does this look? I calculated [itex][a,(a^\dagger)^m]=2^{m-1}a^{m-1}[/itex].
Then I did the taylor expansion [itex][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}[/itex]
 
  • #5
AwesomeTrains said:
How does this look? I calculated [itex][a,(a^\dagger)^m]=2^{m-1}a^{m-1}[/itex].
Are you sure about this??
 
  • #6
nrqed said:
Are you sure about this??
No, I'm not :) How is this?: [itex][a,(a^\dagger)^m]=2^{m-1}(a^\dagger)^{m-1}[/itex]? I wrote out the commutator for the first few m and found that [itex][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}[/itex]
 
  • #7
We simplified this commutator in the lecture today and got this: [itex][a,(a^\dagger)^n]=n(a^\dagger)^{n-1}[/itex] Can you see where my mistake in the previous post is?
 
  • #8
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
blue_leaf77 said:
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 [itex]n=3[/itex] using a commutation identity: [itex][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[/itex]
This is how our prof did: [itex][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[/itex] which generalizes to [itex][a,(a^\dagger)^n]=n (a^\dagger)^{n-1}[/itex]
Why do I get some other result, can't I use the identity [itex][A,BC]=B[A,C]+[A,B]C[/itex]? Or is it something else that I'm doing wrong.
 
  • #10
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
blue_leaf77 said:
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 :) [itex][a,a^\dagger (a^\dagger)^2]=a^\dagger[a,(a^\dagger)^2]+[a,a^\dagger](a^\dagger)^2[/itex]
Well then using the correct result I get this for the original question:
[itex][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}[/itex] Doesn't seem right though. Can I even simplify that series?
 
  • #12
AwesomeTrains said:
$$\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}[(a^\dagger)^m,a]$$
Are the operators in the commutator above in the right order?

AwesomeTrains said:
$$\sum^\infty_{m=0}\frac{f^{(m)}(0)}{m!}m (a^\dagger)^{m-1}$$
Stop right there, how will you simplify ##\frac{m}{m!}##?
 
  • #13
[itex][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}[/itex] Can I return this to the original taylor approximation so that I can write [itex]f(a^\dagger)[/itex] instead of the series?
 
  • #14
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?
 
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  • #15
AwesomeTrains said:
[itex][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}[/itex] Can I return this to the original taylor approximation so that I can write [itex]f(a^\dagger)[/itex] 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.
 
  • #16
Well it can't start at 0 since [itex](-1)![/itex] isn't defined. The next reasonable index is 1. Ah, it's the derivative of the function at [itex]a^\dagger[/itex]. What I get is the shifted series. Am I right?
 
  • #17
[itex][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)[/itex]
 
  • #18
AwesomeTrains said:
Ah, it's the derivative of the function at a†
Yes.
 

1. What is a ladder operator commutator with arbitrary function?

A ladder operator commutator with arbitrary function is a mathematical tool used in quantum mechanics to describe the behavior of a quantum system. It is a combination of two operators, known as ladder operators, and an arbitrary function, which is used to represent physical quantities in the system.

2. How does a ladder operator commutator with arbitrary function work?

The commutator of a ladder operator with an arbitrary function is calculated by taking the difference between the product of the two operators and the product of the two operators in reverse order. This commutator is used to determine the change in physical quantities when a system is acted upon by the operators.

3. What is the significance of a ladder operator commutator with arbitrary function in quantum mechanics?

In quantum mechanics, the ladder operator commutator with arbitrary function is an important tool for understanding the behavior of quantum systems. It allows us to calculate the changes in physical quantities and make predictions about the behavior of the system under different conditions.

4. Is a ladder operator commutator with arbitrary function always used in quantum mechanics?

No, a ladder operator commutator with arbitrary function is not always used in quantum mechanics. It is most commonly used in the study of quantum harmonic oscillators, but may also be applied to other systems depending on the specific problem being investigated.

5. How can a ladder operator commutator with arbitrary function be applied in practical scientific research?

A ladder operator commutator with arbitrary function can be used in practical scientific research to model and analyze the behavior of quantum systems, such as molecules or subatomic particles. It can also be used to make predictions about the behavior of these systems under different conditions, which can then be tested through experiments.

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