Exponential of creation/annihilation operators

[ansgar]

Hello!

I found on this webpage:

http://www-thphys.physics.ox.ac.uk/people/JohnCardy/qft/costate.pdf

page 1, on the bottom

that

$$e^{\phi^* a } f(a^{\dagger} , a ) = f(a^{\dagger} + \phi^*, a) e^{\phi^* a }$$

I have tried to prove this, writing both as taylor series, but the problem is to understand the "hint" :(

 

[strangerep]

I found on this webpage:

http://www-thphys.physics.ox.ac.uk/people/JohnCardy/qft/costate.pdf

page 1, on the bottom

that

$$e^{\phi^* a } f(a^{\dagger} , a ) = f(a^{\dagger} + \phi^*, a) e^{\phi^* a }$$

I have tried to prove this, writing both as taylor series, but the problem is to understand the "hint" :(

For the benefit of other readers, this pdf performs a quick derivation of the
coherent state path integral (in which a coherent state resolution of unity is used
between time slices rather than the usual resolutions using momentum and position
eigenstates).

The "hint" is

Use the fact that a acts like $$\partial/\partial a^\dagger$$, (...),

which refers to this trick:

$$[a, g(a^\dagger)] ~=~ \frac{\partial g(a^\dagger)}{\partial a^\dagger}$$

as may be shown by induction (perhaps modulo a sign and/or some
factors of i and $$\hbar$$, depending on one's conventions).

Start with $$g(a^\dagger) = a^\dagger$$, then progress to
$$g(a^\dagger) = (a^\dagger)^2$$, to see the pattern,
then prove it for $$g(a^\dagger) = (a^\dagger)^n$$ by induction.
Then you can extend the result to any well-behaved analytic function.)

HTH.

 

[ansgar]

ah yes now its clear! Thank you

For the benefit of other readers, this pdf performs a quick derivation of the
coherent state path integral (in which a coherent state resolution of unity is used
between time slices rather than the usual resolutions using momentum and position
eigenstates).

The "hint" is



which refers to this trick:

$$[a, g(a^\dagger)] ~=~ \frac{\partial g(a^\dagger)}{\partial a^\dagger}$$

as may be shown by induction (perhaps modulo a sign and/or some
factors of i and $$\hbar$$, depending on one's conventions).

Start with $$g(a^\dagger) = a^\dagger$$, then progress to
$$g(a^\dagger) = (a^\dagger)^2$$, to see the pattern,
then prove it for $$g(a^\dagger) = (a^\dagger)^n$$ by induction.
Then you can extend the result to any well-behaved analytic function.)

HTH.