Creation/Anhilation Operator Commutation Relation

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Homework Help Overview

The discussion revolves around simplifying a commutator involving creation and annihilation operators, specifically the expression [a^{\dagger}a,a \sqrt{a^\dagger a}]. The subject area is quantum mechanics, focusing on operator algebra.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply normal ordering and has derived a related commutator but is uncertain about the current problem. Some participants suggest breaking down the commutator using a specific identity, while others question the correctness of signs in the results.

Discussion Status

The discussion is active, with participants exploring different approaches to the problem. Some guidance has been provided regarding the use of commutation identities, and there is acknowledgment of potential errors in sign, indicating a collaborative effort to clarify the reasoning.

Contextual Notes

Participants are navigating the complexities of operator algebra and are mindful of the implications of normal ordering and commutation relations. There is an ongoing examination of assumptions related to the operators involved.

teroenza
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Homework Statement


Simplify the following commutator involving the creation and annihilation operators.

[a^{\dagger}a,a \sqrt{a^\dagger a} ]

Homework Equations


I know that [a,a^\dagger] = 1.

The Attempt at a Solution


I think I should be trying to put the creation operators to the left (normal ordering). I have also worked out
[a^{\dagger}a,a]=a, but can't seem to figure out what to do in this case.
 
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There's no need to try too hard with normal ordering - notice that you have an operator and its square root in the commutation relation! Just break up the commutator using the identity \left[A, BC\right] = \left[A, B\right]C + B\left[A, C\right]
 
I see. Then the result is just:
[a^{\dagger}a,a \sqrt{a^\dagger a} ]=a\sqrt{a^\dagger a} + (a^\dagger a)^{3/2}-(a^\dagger a)^{3/2}=a\sqrt{a^\dagger a}
 
Hmm...I think you are off by a minus sign. [a^{\dagger}a,a] = - a
 
Yep, you're right. My original post above is off by a negative too.
 

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