Projection Operators: Pi, Pj, δij in Quantum Mechanics

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    Operators Projection
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Discussion Overview

The discussion revolves around the mathematical properties of projection operators in quantum mechanics, specifically focusing on the operators Pi and Pj, and their relationship through the Kronecker delta δij. Participants explore the implications of these relationships and seek clarification on the mathematical steps involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant cites Shankar's text stating that Pi is defined as |i>
  • Another participant suggests that once δij is obtained, one can replace |i⟩ with |j⟩, leading to the expression δij |i⟩⟨j| = δij |j⟩⟨j|.
  • A participant expresses the belief that they can substitute Pi for Pj in the final result without affecting the outcome, seeking confirmation of this idea.
  • Another participant agrees with the substitution and notes that there are multiple valid equalities in this context, referencing the potential for further clarification in Shankar's work.

Areas of Agreement / Disagreement

Participants express differing views on the implications of substituting Pi for Pj, with some agreeing that it is valid while others focus on the mathematical reasoning behind the original expressions. The discussion remains unresolved regarding the broader implications of these substitutions.

Contextual Notes

The discussion highlights potential ambiguities in the definitions and relationships of projection operators, as well as the assumptions underlying the use of the Kronecker delta in this context.

RohanJ
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TL;DR
Product of two projection operators
In Principles of Quantum mechanics by shankar it is written that
Pi is a projection operator and Pi=|i> <i|.
Then PiPj= |i> <i|j> <j|= (δij)Pj.
I don't understand how we got from the second result toh the third one mathematically.I know that the inner product of i and j can be written as δijbut how do we get to Pj in the final result from the second result?
 
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Once you get the ##\delta_{ij}##, you can change ##|i\rangle## to ##|j\rangle##:
$$
\delta_{ij} |i\rangle \langle j| = \delta_{ij} |j\rangle \langle j|
$$
 
DrClaude said:
Once you get the ##\delta_{ij}##, you can change ##|i\rangle## to ##|j\rangle##:
$$
\delta_{ij} |i\rangle \langle j| = \delta_{ij} |j\rangle \langle j|
$$
I was thinking that only. That means I can write Pi in place of Pj in the final result too and it won't make a difference.
Am I right?
 
RohanJ said:
I was thinking that only. That means I can write Pi in place of Pj in the final result too and it won't make a difference.
Am I right?
Right. This is not the only equality that is valid here. I don't have Shankar's book with me at the moment, but it can be that he uses that particular form later to make a point.
 

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