Heisenberg's Uncertainty Principle using Linear Algebra

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Discussion Overview

The discussion revolves around the application of linear algebra to the Heisenberg Uncertainty Principle, specifically exploring the relationship between certain matrix operations and the principle itself. Participants are examining the mathematical expressions and seeking proofs related to the premise that the commutation relation AB - BA equals the identity matrix.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how the expression x^{ T }x = x^{ T }ABx - x^{ T }BAx is derived in the context of the uncertainty principle.
  • Another participant acknowledges the relation AB - BA = I but does not elaborate on its implications.
  • A participant expresses a desire to prove the premise AB - BA = I and inquires about an elegant method for doing so.
  • There is a mention that the premise might include a multiplicative constant, suggesting that the identity matrix could be replaced by iC, where C is a real constant.
  • Expressions of gratitude for assistance are noted, indicating a collaborative atmosphere.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the proof of the premise AB - BA = I, and there are competing views regarding the inclusion of a multiplicative constant in the expression.

Contextual Notes

The discussion includes unresolved mathematical steps and assumptions regarding the properties of the matrices involved, particularly concerning their symmetry and the implications of the commutation relation.

rpthomps
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I am working through linear algebra from MITs MOOC online courses. One of the question refers to the uncertainty principle. It states:
AB-BA=I can happen for infinite matrices with A

[tex]A=A^{ T }\\ and\\ B=-B^{ T }\\ Then\\ x^{ T }x=x^{ T }ABx-x^{ T }BAx\le 2\parallel Ax\parallel \parallel Bx\parallel[/tex]

My question is how does
[tex]x^{ T }x=x^{ T }ABx-x^{ T }BAx[/tex]?
 
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Like you said:
rpthomps said:
AB-BA=I

Thanks
Bill
 
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bhobba said:
Like you said:Thanks
Bill

Okay. I should have got that one. :) Okay, now I want to prove the premise AB-BA=I, is there an elegant way of doing that?
 
rpthomps said:
Okay. I should have got that one. :) Okay, now I want to prove the premise AB-BA=I, is there an elegant way of doing that?

That is the premise of the theorem except for a multiplicative constant - the I is replaced by iC - C a real constant.

Thanks
Bill
 
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Thanks again I really appreciate your attention.
 

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