Heisenberg's Uncertainty Principle using Linear Algebra

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SUMMARY

The discussion centers on the application of Heisenberg's Uncertainty Principle within the context of linear algebra, specifically referencing the equation AB - BA = I for infinite matrices A and B. The user, Bill, seeks clarification on the relationship between the expressions xTx and xTABx - xTBAx, as well as an elegant proof for the premise AB - BA = I. The conversation highlights the conditions for matrices A and B, where A is symmetric (A = AT) and B is skew-symmetric (B = -BT), and notes that the identity matrix I may be replaced by iC, where C is a real constant.

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  • Understanding of linear algebra concepts, particularly matrix operations.
  • Familiarity with infinite matrices and their properties.
  • Knowledge of symmetric and skew-symmetric matrices.
  • Basic grasp of the Heisenberg Uncertainty Principle in quantum mechanics.
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  • Research the properties of infinite matrices in linear algebra.
  • Study proofs related to the Heisenberg Uncertainty Principle using linear algebra.
  • Learn about symmetric and skew-symmetric matrices and their applications.
  • Explore advanced topics in linear algebra, such as operator theory and commutation relations.
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Students and researchers in mathematics, particularly those studying linear algebra and its applications in quantum mechanics, as well as anyone interested in the theoretical foundations of the Heisenberg Uncertainty Principle.

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

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

My question is how does
x^{ T }x=x^{ T }ABx-x^{ T }BAx?
 
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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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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