[Q]Analytic function of operator A

[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.

[clem]

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

[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

[clem]

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

[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.

[Hurkyl]

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

[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.

[clem]

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.

It is wrong.