I Deriving the Commutator of Exchange Operator and Hamiltonian

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The discussion focuses on deriving the commutator of the exchange operator and the Hamiltonian. It emphasizes that applying the Hamiltonian to the wave function followed by the exchange operator does not yield the expected result of swapping particle positions. The equation presented illustrates the relationship between the exchange operator and the Hamiltonian, highlighting the importance of applying the operators in the correct order. The use of properties like ##P_{12}^2 = I## and ##{P_{12}}^\dagger = P_{12}## is crucial for the derivation. Understanding these operations is essential for accurate quantum mechanical calculations involving indistinguishable particles.
Samama Fahim
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In the boxed equation, how would you get the right hand side from the left hand side? We know that ##H(1,2) = H(2,1)##, but we first have to apply ##H(1,2)## to ##\psi(1,2)##, and then we would apply ##\hat{P}_{12}##; the result would not be ##H(2,1) \psi(2,1)##. ##\hat{P}_{12}## is the exchange operator and ##H(1,2)## is the hamiltonian.

Source: https://books.google.com.pk/books?i...inction as to which particle is which&f=false
 
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$$P_{12}H(1,2)\psi(12) = P_{12}H(1,2){P_{12}}^\dagger P_{12}\psi(1,2) = H(2,1)\psi(2,1)$$
using ##P_{12}^2 = I## and ##{P_{12}}^\dagger = P_{12}##.
 

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