# Permutation operator and Hamiltonian

• I
In summary, the permutation operator ##\hat{P}_{21}## commutes with the Hamiltonian when considering identical particles, as shown by the relation $$[\hat{P}_{21}, \hat{H}] = 0 \tag{1}$$ This implies that for a general eigenvector ##{\lvert} {\psi} {\rangle}##, the term ##(\hat{P}_{21} \hat{H}){\lvert} {\psi}{\rangle}## is equal to 0. This can be seen by expanding the operator ##\hat{P}_{21}## to act on each term in the state determined by ##(\hat{H}{\lvert}{\psi}{\rangle
The permutation operator commutes with the Hamiltonian when considering identical particles, which implies:

$$[\hat{P}_{21}, \hat{H}] = 0 \tag{1}$$

Now given a general eigenvector ##{\lvert} {\psi} {\rangle}##, where

$$\hat{P}_{21} (\hat{H}{\lvert}{\psi}){\rangle} = (\hat{P}_{21} \hat{H}) {\lvert} {\psi}{\rangle} + \hat{H}(\hat{P}_{21}{\lvert}{\psi}{\rangle})$$

Using (1):
$$(\hat{P}_{21} \hat{H}){\lvert} {\psi}{\rangle} = 0$$

But how exactly does this last relation follow? Why does acting the permutation operator on the Hamiltonian result in 0? In this case, if the Hamiltonian is symmetric with respect to permutation, how does this term going to 0 indicate that?

Now given a general eigenvector ##{\lvert} {\psi} {\rangle}##, where

$$\hat{P}_{21} (\hat{H}{\lvert}{\psi}){\rangle} = (\hat{P}_{21} \hat{H}) {\lvert} {\psi}{\rangle} + \hat{H}(\hat{P}_{21}{\lvert}{\psi}{\rangle})$$
I don't understand how you get that equation.

DrClaude said:
I don't understand how you get that equation.

Yikes. Looking back at it, I don't quite understand my step either. I was under the assumption the operator ##\hat{P}_{21}## was acting on the total state determined by ##(\hat{H}{\lvert}{\psi}{\rangle})## and thus expanded it out to act on each term in that state, but switching the order of operation here seems very wrong.

## 1. What is the permutation operator in quantum mechanics?

The permutation operator in quantum mechanics is a mathematical operator that represents the exchange of two identical particles in a system. It is denoted by the symbol P and is used to describe the symmetries of the system.

## 2. How does the permutation operator affect the wavefunction of a system?

The permutation operator acts on the wavefunction of a system to swap the positions of identical particles. This exchange of particles may result in a change in the overall phase of the wavefunction, but it does not change the probability distribution of the system.

## 3. What is the significance of the permutation operator in quantum chemistry?

In quantum chemistry, the permutation operator is used to describe the symmetries of molecular systems. It is a crucial tool in the study of molecular bonding and electronic structure, as it helps to determine the symmetry properties of molecular orbitals.

## 4. How is the permutation operator related to the Hamiltonian in quantum mechanics?

The permutation operator and the Hamiltonian are closely related in quantum mechanics. The Hamiltonian operator includes the kinetic and potential energies of a system, while the permutation operator accounts for the symmetries of the system. Together, they describe the total energy of a quantum system.

## 5. Can the permutation operator be used to solve the Schrödinger equation?

Yes, the permutation operator is an important tool in solving the Schrödinger equation for systems with identical particles. It helps to simplify the equations and allows for the application of symmetry principles to find solutions.

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