Operator acting on ket state n

Glogower ladder operator, and it should act on a |n> state as follows:$$\left(\hat{N}+\mathbf{1}\right)^{-1 / 2} \alpha|n\rangle =\left(1-\frac{1}{2} \hat{N}+\frac{3}{8} \hat{N}^{2}-\cdots\right)\left(\sqrt{n}|n-1\rangle\right)=\frac{\sqrt{n}}{\sqrt{n-1}}|n-1\rangle$$In summary, the conversation is about how the operator (N+1)^-1/2 α acts on a |n> state of the harmonic oscillator,
  • #1
Jean-Mathys du bois
2
0
TL;DR Summary
how does the operator (N+1)^-1/2 α act on a |n> state of harmonic osciliator? N is the number operator N|n>=n|n> and α anihilation operator
I tried playing with the number's operator eigenvalues equation but couldn't get anywhere, can s/b help me out?
 
Physics news on Phys.org
  • #2
If $$\hat{N}\left|n\right\rangle =n\left|n\right\rangle $$
then, by the very definition of function of operators, we have that
$$f(\hat{N})\left|n\right\rangle =f(n)\left|n\right\rangle $$
 
  • Like
Likes DrClaude and vanhees71
  • #3
Jean-Mathys du bois said:
Summary:: how does the operator (N+1)^-1/2 α act on a |n> state of harmonic osciliator? N is the number operator N|n>=n|n> and α anihilation operator

I tried playing with the number's operator eigenvalues equation but couldn't get anywhere, can s/b help me out?
To make sense of a non-polynomial function of an operator, you can interpret it as a Taylor series:
$$(1 + N)^{-1/2} = 1 - \frac 1 2 N + \frac 3 8 N^2 \dots $$
 
  • #4
The operator (N+1)^-1/2 α , i think is called Susskind
 

Similar threads

Replies
16
Views
1K
Replies
11
Views
1K
Replies
21
Views
2K
Replies
8
Views
1K
  • Quantum Physics
Replies
7
Views
526
Replies
0
Views
150
  • Quantum Physics
Replies
1
Views
828
  • Quantum Physics
Replies
10
Views
2K
  • Quantum Physics
Replies
2
Views
973
Replies
7
Views
2K
Back
Top