Why don't creation and destruction operators conmute?

  • Context: Graduate 
  • Thread starter Thread starter carllacan
  • Start date Start date
  • Tags Tags
    Creation Operators
Click For Summary

Discussion Overview

The discussion revolves around the non-commutation of creation and destruction operators in quantum mechanics, specifically addressing the physical implications and interpretations of this mathematical property. Participants explore the symmetry of these operations, their relationship to physical observables, and the implications for particle states.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the physical meaning behind the non-commutation of creation (a+) and destruction (a-) operators, suggesting that they should cancel each other out due to their symmetric nature.
  • Another participant argues that a+ and a- are not Hermitian and therefore do not represent physical measurable observables, contrasting them with the number operator, which is a physical observable.
  • A different viewpoint emphasizes that the operations of creation and destruction are not symmetric, providing an example involving the vacuum state where destruction followed by creation is physically impossible.
  • Participants discuss the mathematical implications of the operators, noting that the results of applying them in different orders can lead to different states, particularly highlighting the importance of coefficients in the transformations.
  • One participant shares a link to external material that discusses deeper theoretical concepts related to creation and destruction operators, including a combinatorial interpretation of their non-commutation.

Areas of Agreement / Disagreement

Participants express differing views on the symmetry of creation and destruction operators and their implications for physical states. There is no consensus on the physical interpretation of their non-commutation, and multiple competing perspectives remain present in the discussion.

Contextual Notes

Some statements rely on specific interpretations of quantum mechanics and the mathematical framework of operators, which may not be universally accepted. The discussion includes assumptions about the nature of states and the properties of operators that are not fully resolved.

carllacan
Messages
272
Reaction score
3
Hi.

I was wondering why creation and destruction operators a+ and a- do not conmute.

Of course, I can show that they don't conmute by computing the conmutator [a+, a-] = -1. But I want to know the "physical" meaning of this.

Isn't destruction/creation a symmetric transformation? We "go up the ladder" with a+ and we "go down the ladder" with a-. Shouldn't they therefore cancel each other, i.e. a+a-=a-a+ = I?

Mathematically [itex]a_- a_+ \vert n \rangle = a_-\vert n+1 \rangle = \vert n \rangle[/itex]

Thank you for your time.
 
Last edited:
Physics news on Phys.org
I would say that since a+ and a- are not Hermitean, they thus represent no physical measurable observables. The number operator is a physical observable, though.
 
Last edited:
  • Like
Likes   Reactions: 1 person
carllacan said:
Isn't destruction/creation a symmetric transformation?

No. For example, suppose you are starting in the ground state/vacuum state, with no particles. Creation, then destruction takes you to a 1-particle state, then back to the vacuum state. Destruction, then creation is physically impossible: the destruction operator applied to the vacuum state gives zero, which doesn't represent any physical state.

More generally, if you are in an n-particle state, there is no guarantee that creation + destruction will take you back to the *same* n-particle state as destruction + creation, because there are many possible n-particle states.
 
  • Like
Likes   Reactions: 1 person
carllacan said:
Mathematically [itex]a_- a_+ \vert n \rangle = a_-\vert n+1 \rangle = \vert n \rangle[/itex]

Try this line of thinking on ##a_+a_-\vert{0}\rangle##. What happens?

##a_+\vert n \rangle = \alpha \vert n+1 \rangle## not ##\vert n+1 \rangle## and you can't ignore the value of ##\alpha##.
 
  • Like
Likes   Reactions: 1 person
Ok, thank you guys, I get it now.
 
This might be of interest if you can understand some of it.

http://math.ucr.edu/home/baez/qg-fall2003/

Quote from there:

"Andre Joyal invented his theory of "espèces des structures" - translated as "species" or "structure types" - in order to understand more deeply how people use generating functions to count structures on finite sets. It turns out that just as a natural number is a watered-down or "decategorified" version of a finite set, a generating function is a decategorified version of a structure type.

Recently, James Dolan and John Baez realized that structure types and more general "stuff types" can also be used to more deeply understand the role of annihilation and creation operators, Feynman diagrams and the like in quantum theory. It turns out that some of the mysteries of quantum mechanics are really just decategorified versions of simple facts about structures on finite sets. For example, the fact that position and momentum don't commute has a purely combinatorial interpretation! "

Here's the punchline, which you can meditate on as a combinatorics problem, whether or not you're interested in category theory or structure types:

"Ultimately, it boils down to the fact that there's one more way to put a ball in a box and then take one out than to take one out and then put one in."
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K