Is Braket Operation Legal? Elaborate Here

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Discussion Overview

The discussion centers around the legality of a specific braket operation in quantum mechanics, particularly the expression involving operators and eigenvectors. Participants explore the validity of manipulating these expressions and the implications of operator properties.

Discussion Character

  • Debate/contested, Technical explanation

Main Points Raised

  • One participant questions the legality of the operation \langle A_1|BA|A_2\rangle=A_2\langle A_1|B|A_2\rangle and seeks clarification.
  • Another participant suggests that if |A_2 \rangle is an eigenvector of the operator \mathbf{A} with eigenvalue A_2, then the operations presented are correct.
  • A different participant attempts to manipulate the expression further, suggesting \langle A_1|BA|A_2\rangle=\langle A_1|B\rangle \langle A|A_2\rangle=A_2\langle A_1|B\rangle, seeking confirmation.
  • In contrast, another participant argues that there is a confusion between operators and bras/kets, stating that the expressions involving |A\rangle and

Areas of Agreement / Disagreement

Participants express differing views on the legality and correctness of the operations involving the braket notation, indicating that the discussion remains unresolved with multiple competing interpretations.

Contextual Notes

There is a potential ambiguity regarding the definitions of operators and their representations in bra-ket notation, which may affect the validity of the claims made.

intervoxel
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Is this operation legal?

[itex]\langle A_1|BA|A_2\rangle=A_2\langle A_1|B|A_2\rangle[/itex]

If so, please elaborate on that.

Thanks.
 
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If [itex]|A_2 \rangle[/itex] is an eigenvector of the operator [itex]\mathbf{A}[/itex] with eigenvalue [itex]A_2[/itex] then your operations are correct.
 
So

[tex] \langle A_1|BA|A_2\rangle=\langle A_1|B\rangle \langle A|A_2\rangle=<br /> \langle A_1|B\rangle A_2=A_2\langle A_1|B\rangle[/tex]

right?

Thank you.
 
No. You are mixing up operators with bras and kets. It looks like A and B are supposed to be operators. If so, the expressions |A>, <A|, |B>, or <B| make no sense. Operators are neither bras nor kets.
 

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