Analytic function, creation and annihilation operators proof

  • Thread starter vst98
  • Start date
  • #1
51
0

Homework Statement


Show that

f(aa)a = af(aa + 1)

Where f is any analytic function and a and a† satisfy commutation relation [a, a] = 1.

The Attempt at a Solution


I have used [a, a†] = aa†-a†a=1 to write the expression like

f(a†a)a†= a†f(aa†)

but I don't know what to do next.

I know that analytic function can be written like f(x)=Ʃ kn(x-x0)2 and that it is infinitely differentiable, but I don't see how can I sucesssfuly
apply this, or there some other trick here.
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,023
576
So f is essentially a power series and you can then use induction for an arbitrary power of (a+ a)
 
  • #3
51
0
I think now I understand,
so on the left hand side for f(a†a) I will have powers of a†a
like a†a + a†aa†a ...
and on the right hand side for f(aa†) i will have powers of aa†
like aa† + aa†aa† ...

but since I have to multiply these series by a† from the opposite sides
(a†a + a†aa†a ... ) /a† = a†aa† + a†aa†aa† ...
a†\ ( aa† + aa†aa† ... ) = a†aa† + a†aa†aa† ...

they turn out to be the same, right ?
 
  • #4
51
0
also I have a question which builds on this one (so I will stay in this thread),
A bosonic one level system can be described by the Hamiltonian
H= εa†a,

The expectation value of n = a†a is defined as n(ε) = <n> = tr(ρa†a) where

ρ=(1/ZG)*Exp[-β(ε-μ)a†a] , tr(ρ)=1

is the grand canonical density matrix.
Use f(a†a)a† = a†f(a†a + 1) to show that the form

n(ε) = 1/(Exp[β(ε-μ)]

of the Bose-Einstein distribution directly follows from the bosonic commutation
relation for a and a†.

I don't see where to apply my result from the first part, to trace function maybe ?
But that does not have appropriate form.
 

Related Threads on Analytic function, creation and annihilation operators proof

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
5
Views
7K
Replies
5
Views
1K
Replies
2
Views
3K
Replies
0
Views
531
Replies
5
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
1K
Replies
3
Views
209
Replies
1
Views
2K
Top