# [Q]Analytic function of operator A

1. Nov 4, 2008

### good_phy

Hi, 5.27 problem in liboff says that if $g(A)f(\varphi) = g(a)f(\varphi),where A\varphi = a\varphi$

I tried to solve this problem with tylar expansion.

$f(\varphi) = f(0) + f^{'}(0)\varphi + \frac{f^{''}(0)}{2!}\varphi^2 + \frac{f^{(3)}(0)}{3!}\varphi^3 + ...$

$g(A) = g(0) + g^{'}(0)A + \frac{g^{''}(0)}{2!}A^2 + \frac{g^{(3)}(0)}{3!}A^3 + ...$

But when i applied g(A) to f($\varphi$) i can not get right hand side of equality

written above because i don't know how can i deal with unseen term such as

$A\varphi^2$or $A^2\varphi$ or etc

please assist to me.

2. Nov 4, 2008

### clem

Don't expand f.
Show that A^n f=a^n f, which is easy to show by repeated use of Af=af.

3. Nov 5, 2008

### good_phy

Ya I now your way, but How can i deal with $A\varphi^n$ showed in product

of $f(\varphi)$ expansion and g(A) expansion

4. Nov 5, 2008

### clem

As you show, expanding f is useless, because you don't know $$A\phi^2$$.

5. Nov 5, 2008

### cks

I think in this way, in Dirac notation

$$A(\|phi>)^2=(A\|phi>)|phi>=(a\|phi>)|phi>=a|phi>^2$$
Similarly,
$$A^2\|phi>=A(A\|phi>)=A(a\|phi>)=a(A|phi>)=a^2|phi>$$

The second equation is straightforward.
The first equation sounds a bit weird to me. I feel like I maybe wrong.

6. Nov 5, 2008

### Hurkyl

Staff Emeritus
Just checking -- what do you think $f''(0)$ and $\varphi^2$ mean?

7. Nov 5, 2008

### cks

Like Hurkyl pointed out,

one single operator acts on $$\psi^2$$ , acts on two wavefunction, sounds weird.I don't know how to explain it.

8. Nov 5, 2008

It is wrong.