# Commutator question

1. Jan 12, 2013

### LagrangeEuler

If $\hat{n}=\hat{a}^+\hat{a}$ is number operator and $$\hat{a}^+$$,$$\hat{a}$$ are Bose operators. Is there then some formula for
$$[f(\hat{n}),\hat{a}]$$
$$[f(\hat{n}),\hat{a}^+]$$

2. Jan 13, 2013

### Bill_K

f(N)a = af(N+I)
f(N)a = af(N-I)

So for example if f(x) = x3,
N3a = a(N3 + 3N2 + 3N + I)

3. Jan 13, 2013

### LagrangeEuler

Thanks for the answer. How you see that?

$$[f(\hat{n}),\hat{a}^+]=f(\hat{n})\hat{a}^+-\hat{a}^+f(\hat{n})=\hat{a}^+f(\hat{n}+1)-\hat{a}^+f(\hat{n})=\hat{a}^+(f(\hat{n}+1)-f(\hat{n}))$$
For $f(\hat{n})=\hat{n}$ its ok. Because.
$$[\hat{n},\hat{a}^+]=\hat{a}^+$$

4. Jan 13, 2013

### andrien

For every function which can be written in terms of a series expansion,it will work.It is an assumption.

5. Jan 13, 2013

### Bill_K

Take as basis |n>, the usual eigenstates of the number operator.

N|n> = n|n> and for any function f, f(N)|n> = f(n)|n>.

Then f(N) a|n> = f(N) √(n+1)|n+1> = √(n+1) f(N) |n+1> = √(n+1)f(n+1) |n+1>.
Also a f(N+1) |n> = a f(n+1)|n> = f(n+1) √(n+1) |n+1>.

so these two expressions are equal. Does not depend on expressing f as a power series.

6. Jan 14, 2013

### LagrangeEuler

Don't understand this step
$$f(\hat{n})|n+1\rangle=f(n+1)|n+1\rangle$$

7. Jan 14, 2013

### andrien

when it is not a power series then it is hard to verify the validity
f(N)|n> = f(n)|n>
see dirac'principal of quantum mechanics' section 11.

8. Jan 14, 2013

### LagrangeEuler

Could I say that $\hat{a}^2=(\hat{a}^+)^2=0$?

9. Jan 14, 2013

### Bill_K

The state |n+1> is the eigenstate of N with eigenvalue n+1. Applying N to one of its eigenstates just multiplies it by the eigenvalue, namely n+1 in this case: N|n+1> = (n+1) |n+1>. Also N2|n+1> = (n+1)2|n+1> etc. And in general f(N)|n+1> = f(n+1)|n+1>. Contrary to what andrien says, this is not an "assumption", does not depend on a power series expansion, need "verification", or anything like that. It's just the way eigenstates work.
No, a is the raising operator. Applied to an eigenstate |n> it gives you back a multiple of the next eigenstate. |n+1>. Applying it twice raises twice and gives a multiple of |n+2>. Likewise a2 lowers by two steps.

10. Jan 15, 2013

### andrien

f(N)|n+1> = f(n+1)|n+1>,this may be an obvious thing.But it is not very obvious to me at least when it is a function.Since N operates and becomes the eigenvalue,all N2 or any power will do that.But if there is any function like Sin2N,it is not very obvious that it will become Sin2n,until unless one resorts to a series form.However I don't know the general proof,but it is ok.

11. Jan 15, 2013

### CompuChip

If not by series expansion, could you suggest how to define a function of an operator (e.g. sin(N))?

12. Jan 15, 2013

### strangerep

Elaborating a little on Bill_K's "It's just the way eigenstates work"....

The extension from power series to continuous functions is the subject of more sophisticated versions of the spectral theorem. Cf. Kreyszig, theorems 9.9-1 and 9.10-1.

Wikipedia http://en.wikipedia.org/wiki/Spectral_theorem explains how, for an ordinary self-adjoint operator $A$ with spectrum $\sigma(A)$, one can express $A$ as
$$\def\<{\langle} \def\>{\rangle} A ~=~ \int_{\sigma(A)} \lambda dE_\lambda ~=~ \int_{\sigma(A)} \lambda |\lambda\>\<\lambda| ~.$$
Unfortunately, Wiki doesn't explain extensively that the extension to a continuous real-valued function $f$ then takes the form
$$f(A) ~=~ \int_{m-0}^M f(\lambda) dE_\lambda ~,$$
where the integral is understood in the sense of uniform operator convergence -- cf. http://en.wikipedia.org/wiki/Operator_topology , and $f$ is defined on $[m,M]$, where
$$m ~=~ \inf_{\|x\|=1} \<A\psi, \psi\> ~,~~~~~ M ~=~ \sup_{\|x\|=1} \<A\psi, \psi\> ~.$$
(For details, one must read Kreyszig, or some other book on functional analysis.)

Anyway..., although there are still some caveats on which functions can be used with which operators, for physics purposes it's a very class.

Additional warning: there are various operator topologies (i.e., various senses in which operators can be considered equal), so one must take care when applying the more sophisticated versions of the spectral theorem.