Analytic function, creation and annihilation operators proof

[vst98]

Homework Statement

Show that

f(a^†a)a^† = a^†f(a^†a + 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 k_n(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.

 

[dextercioby]

So f is essentially a power series and you can then use induction for an arbitrary power of (a+ a)

 

[vst98]

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 ?

 

[vst98]

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/Z_G)*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.

 

[paranormal]

To solve this problem, we can use the fact that the creation and annihilation operators, a and a†, are related to each other through the commutation relation [a, a†] = 1. This means that we can use the commutation relation to manipulate the expression f(a†a)a† into a form that is easier to work with.

Starting with the given expression, we can use the commutation relation to write:

f(a†a)a† = a†f(aa†) = a†f(aa† - 1 + 1) = a†f([a, a†] + 1)

Now, since [a, a†] = 1, we can simplify this expression further to get:

f(a†a)a† = a†f(1 + 1) = a†f(2)

Finally, we can use the fact that f is an analytic function to expand this expression as a power series:

f(a†a)a† = a†Ʃn cn(2)n = Ʃn cn(2)n+1 = Ʃn cn(2n + 1)

Similarly, we can manipulate the expression a†f(a†a + 1) to get:

a†f(a†a + 1) = a†f([a, a†] + 1) = a†f(1 + 1) = a†Ʃn cn(2)n = Ʃn cn(2n + 1)

Therefore, we can see that both expressions are equal, and we have successfully proven the given statement.