Undergrad Deriving the Commutator of Exchange Operator and Hamiltonian

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SUMMARY

The discussion focuses on deriving the commutator of the exchange operator ##\hat{P}_{12}## and the Hamiltonian ##H(1,2)## in quantum mechanics. It establishes that applying the Hamiltonian to the wave function ##\psi(1,2)## followed by the exchange operator does not yield the same result as applying the Hamiltonian in a swapped particle configuration. The equation $$P_{12}H(1,2)\psi(12) = H(2,1)\psi(2,1)$$ is derived using the properties of the exchange operator, specifically ##P_{12}^2 = I## and ##{P_{12}}^\dagger = P_{12}##.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically operators and wave functions.
  • Familiarity with the Hamiltonian operator in quantum systems.
  • Knowledge of the properties of exchange operators in quantum mechanics.
  • Basic grasp of commutation relations and their significance in quantum theory.
NEXT STEPS
  • Study the derivation of the commutation relations in quantum mechanics.
  • Explore the role of exchange symmetry in identical particles using the exchange operator.
  • Learn about the implications of the Hamiltonian in quantum statistical mechanics.
  • Investigate the mathematical properties of operators, including adjoint and identity operators.
USEFUL FOR

Quantum physicists, graduate students in physics, and researchers focusing on quantum mechanics and particle statistics will benefit from this discussion.

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}##.
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

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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