Dyade Dirac Notation: Why Last Equation?

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Discussion Overview

The discussion revolves around the use of dyade notation in Dirac notation, particularly focusing on the representation and manipulation of kets and bras in quantum mechanics and tensor analysis. Participants explore the implications of different notations and their mathematical consistency.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant presents equations involving dyades and questions the notation used in the last equation, specifically why it is not expressed as \(\langle \vec{C} |\vec{A}\rangle\langle \vec{B}|\).
  • Another participant asserts that the notation should maintain consistency with the 'bra-keting' of \(\vec{C}\) on the left-hand side (LHS).
  • Some participants express confusion regarding the distinction between kets and bras in quantum mechanics versus their representation in tensor analysis, noting that they do not see a difference between \(|\vec{A}\rangle\) and \(\langle \vec{A}|\) in certain contexts.
  • A later reply acknowledges that \(\langle C|A\rangle\) is a number and questions why this is problematic, suggesting that the arrow notation may not be necessary if \(\vec{A}\) is considered a vector in a different vector space.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate notation and its implications, indicating that there is no consensus on the matter. Confusion persists regarding the treatment of kets and bras across different mathematical frameworks.

Contextual Notes

Participants highlight potential limitations in understanding the notation due to the context of quantum mechanics versus tensor analysis, as well as the implications of using vector notation.

LagrangeEuler
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[tex]\{\vec{A},\vec{B}\}\cdot \vec{C}=\vec{A}(\vec{B}\cdot \vec{C})[/tex]
[tex]\vec{C} \cdot \{\vec{A},\vec{B}\}=(\vec{C}\cdot \vec{A}) \vec{B}[/tex]

I want to write dyade in Dirac notation.

[tex](|\vec{A}\rangle\langle\vec{B}|)|\vec{C}\rangle= |\vec{A}\rangle\langle\vec{B}|\vec{C}\rangle[/tex]
[tex]\langle\vec{C}|(|\vec{A}\rangle\langle\vec{B}|)=< \vec{C} |\vec{A}\rangle|\vec{B}\rangle[/tex]

Why not

[tex]\langle \vec{C} |\vec{A}\rangle\langle \vec{B}|[/tex]

in last equation?
 
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It should be, it should have the same 'bra-keting' as the C in the LHS.
 
I'm confused. In QM [tex]|\psi \rangle[/tex] and [tex]\langle \psi |[/tex] are vectors from some vector space and his dual respectively. But in some tensor analyses I don't see difference between [tex]|\vec{A}\rangle[/tex] and [tex]\langle \vec{A}|[/tex].
In this case precisely. I have some number [tex]\langle|\rangle[/tex] which multiply vector.
 
LagrangeEuler said:
I'm confused. In QM [tex]|\psi \rangle[/tex] and [tex]\langle \psi |[/tex] are vectors from some vector space and his dual respectively. But in some tensor analyses I don't see difference between [tex]|\vec{A}\rangle[/tex] and [tex]\langle \vec{A}|[/tex].
In this case precisely. I have some number [tex]\langle|\rangle[/tex] which multiply vector.
I don't follow what you're saying here. Yes, ##\langle C|A\rangle## is a number. Why is that a problem?

There's no need to use the arrow notation for vectors here (unless what you have in mind is that ##\vec A## is a vector in some other vector space ##\mathbb R^3##), since kets are always vectors.
 

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