Annihilation creation operators in case of half H.oscillator

Click For Summary

Discussion Overview

The discussion revolves around the application of annihilation and creation operators in the context of a half harmonic oscillator, particularly in relation to the Ehrenfest theorem. Participants explore the validity of these operators, boundary conditions, and methods for calculating expectation values.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the annihilation and creation operators are valid for a half harmonic oscillator, noting a violation of the Ehrenfest theorem in their calculations.
  • Another participant asserts that the standard destruction and creation operators for a full harmonic oscillator do not exist in the space of states for a half harmonic oscillator, proposing alternative definitions.
  • Several participants discuss the boundary conditions applicable at ##x=0##, mentioning Dirichlet and Neumann conditions and their implications for the wave function.
  • A participant expresses confusion over a specific result involving the factor of "2" in their calculations and questions the applicability of the Ehrenfest theorem in this context.
  • Another participant suggests that the Ehrenfest theorem remains valid but may require a more nuanced approach than simply using creation and destruction operators.
  • There is mention of using Hermite polynomials for expectation value calculations and the potential use of the Heisenberg picture as a simpler method.
  • A participant expresses interest in sharing future ideas related to the half harmonic oscillator and inquires about the appropriateness of continuing in this thread versus starting a new one.

Areas of Agreement / Disagreement

Participants express differing views on the validity of annihilation and creation operators for the half harmonic oscillator, the interpretation of boundary conditions, and the application of the Ehrenfest theorem. The discussion remains unresolved with multiple competing perspectives present.

Contextual Notes

Participants note that the definitions and applications of operators may depend on the specific boundary conditions chosen, and there is uncertainty regarding the legitimacy of certain mathematical replacements in the context of the half harmonic oscillator.

Who May Find This Useful

This discussion may be of interest to those studying quantum mechanics, particularly in the context of harmonic oscillators, boundary conditions, and the Ehrenfest theorem.

SUVAM ROY
Messages
10
Reaction score
0
I'm trying to check if Ehrenfest theorem is satisfied for this wave function,
|Y>=(1/sqrt(2))*(|1>+|3>),
where |1> and |3> are the ground and 1st exited state wave functions of a half harmonic oscillator. When I'm calculating the expectation values of x and p using annihilation creation operators and trying to check the Ehrenfest theorem I'm getting
m*(time derivative of <x>)=2*<p>
which violates the Ehrenfest theorem. So, my question is whether annihilation creation operators are valid in case of a half harmonic oscillator.
 
Physics news on Phys.org
If ##a## and ##a^{\dagger}## are destruction and creation operators for the full harmonic oscillator, then the destruction and creation operators for the half harmonic oscillator are
$$A=aa$$
$$A^{\dagger}=a^{\dagger}a^{\dagger}$$
The operators ##a## and ##a^{\dagger}## are not defined (do not exist) in the space of states of the half harmonic oscillator.
 
  • Like
Likes   Reactions: SUVAM ROY
SUVAM ROY said:
a half harmonic oscillator.
What is this?
 
A. Neumaier said:
What is this?
Come on ! you can google it too !
 
BvU said:
Come on ! you can google it too !
There are two natural boundary conditions one can place at ##x=0##. It is odd that http://www.physicspages.com/2012/08/18/half-harmonic-oscillator/ only describes the odd half with Dirichlet boundary conditions; the even half with Neumann boundary conditions is not even mentioned.
 
Last edited:
  • Like
Likes   Reactions: dextercioby
Demystifier said:
If ##a## and ##a^{\dagger}## are destruction and creation operators for the full harmonic oscillator, then the destruction and creation operators for the half harmonic oscillator are
$$A=aa$$
$$A^{\dagger}=a^{\dagger}a^{\dagger}$$
The operators ##a## and ##a^{\dagger}## are not defined (do not exist) in the space of states of the half harmonic oscillator.
Thanks a lot.
 
A. Neumaier said:
There are two natural boundary conditions one can place at ##x=0##. It is odd that http://www.physicspages.com/2012/08/18/half-harmonic-oscillator/ only describes the odd half with Dirichlet boundary conditions; the even half with Neumann boundary conditions is not even mentioned.
This is the standard practice in QM (see e.g. a textbook treatment of infinite well potential) to require that ##\psi## (not ##\psi'##) vanishes at the points where the potential ##V## is infinite. The textbooks also give a physical reason for that. I has to do with the fact energy should be finite and that Schrödinger equation has ##\psi## (and ##\psi''##) but not ##\psi'##.

More generally, Neumann boundary conditions in physics usually correspond to ends that can oscillate freely. It should be intuitively clear that ends at infinite ##V## are far from being "free". Think, e.g., of a guitar string.
 
Last edited:
Demystifier said:
If ##a## and ##a^{\dagger}## are destruction and creation operators for the full harmonic oscillator, then the destruction and creation operators for the half harmonic oscillator are
$$A=aa$$
$$A^{\dagger}=a^{\dagger}a^{\dagger}$$
The operators ##a## and ##a^{\dagger}## are not defined (do not exist) in the space of states of the half harmonic oscillator.
Sorry, but I'm still getting the same $$2<p>$$result when I replaced ##a## and ##a^{\dagger}## by ##A## and ##A^{\dagger}## in expression for x and p and did the calculations. Why this "2" is coming in the expression? Is Ehrenfest theorem also not defined in this case?
 
SUVAM ROY said:
Sorry, but I'm still getting the same $$2<p>$$result when I replaced ##a## and ##a^{\dagger}## by ##A## and ##A^{\dagger}## in expression for x and p and did the calculations. Why this "2" is coming in the expression? Is Ehrenfest theorem also not defined in this case?
I did not say that such a replacement is legitimate. You should not express ##x## and ##p## in terms of creation/destruction operators. Ehrenfest theorem is still valid, but it is more subtle.
 
  • #10
Demystifier said:
I did not say that such a replacement is legitimate. You should not express ##x## and ##p## in terms of creation/destruction operators. Ehrenfest theorem is still valid, but it is more subtle.
So, I have to take the Hermite polynomials and calculate expectation values in integration method? Or, is there any other easier method to do this? So that I can verify Ehrenfest theorem.
 
  • #11
SUVAM ROY said:
So, I have to take the Hermite polynomials and calculate expectation values in integration method? Or, is there any other easier method to do this? So that I can verify Ehrenfest theorem.
The approach with the Heisenberg picture should be the easiest. The direct calculation with Hermite polynomials should work too, but make sure to first renormalize them because now they are defined only for x>0.
 
  • #12
Demystifier said:
The approach with the Heisenberg picture should be the easiest
I'll try that. Thanks again.
 
  • #13
Hello everyone :biggrin:
As I'm starting to study exactly the topic of the half oscillator for my PhD research, may I post my future replies/ideas/considerations here or would you rather have me start a brand new thread?

Thanks and have a nice day!
Frank

p.s.: the retro smileys! :headbang::oldbiggrin::oldmad::oldruck:
 
  • #14
Hm, I'd rather start a new thread.
 
  • Like
Likes   Reactions: jim mcnamara
  • #15
Please start a new thread.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
0
Views
820
  • · Replies 1 ·
Replies
1
Views
942
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
5K