Exponential of creation/annihilation operators

  • Thread starter ansgar
  • Start date
505
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Hello!

I found on this webpage:

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

page 1, on the bottom

that

[tex] e^{\phi^* a } f(a^{\dagger} , a ) = f(a^{\dagger} + \phi^*, a) e^{\phi^* a }[/tex]

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

strangerep

Science Advisor
2,982
771
I found on this webpage:

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

page 1, on the bottom

that

[tex] e^{\phi^* a } f(a^{\dagger} , a ) = f(a^{\dagger} + \phi^*, a) e^{\phi^* a }[/tex]

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 [tex]\partial/\partial a^\dagger[/tex], (....),
which refers to this trick:

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

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

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

HTH.
 
505
0
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:

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

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

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

HTH.
 

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