# Identical particles/ Time dependent perturbation theory

• dreinh
In summary: However, the problem should be amended to state that the system is in the singlet state.In summary, the problem involves two identical spin-1/2 particles interacting with a Hamiltonian H0=ω0 S1.S2 where ω0>0. A time dependent perturbation is applied, H'=ω1 (S1z-S2z) θ(t) Exp[-t/τ], where ω1>0 and ω1<<ω0. The question asks for the probabilities of the system starting in the ground state to be excited into each of the three excited states to first order. The system is restricted to be in the singlet state, which is the only allowed anti-symmetric state for identical particles
dreinh

## Homework Statement

Two identical spin-1/2 particles interact with Hamiltonian H0=ω0 S1.S2 where ω0>0. A time dependent perturbation is applied, H'=ω1 (S1z-S2z) θ(t) Exp[-t/τ], where ω1>0 and ω1<<ω0. What are the probabilities that a system starting in the ground state will be excited into each of the three excited states to first order?

## Homework Equations

P[initial state -> final state] = Norm[ Integrate[ < final l H'[t'] l initial > Exp[ I (Ef - Ei) t'], {t',0,t}]]^2

## The Attempt at a Solution

Sorry if the statement is unclear, but I just wanted to make sure my thinking on this is correct. We have two identical fermions interacting through a spin-spin interaction, so the first thing to do in my head is to use the coupled basis states representing the particles. We only have spin interactions so spatial states aren't needed. In a non-identical particle situation, we would have four possible states:

l 1 1 > = l + +>
l 1 0 > = 1/Sqrt[2] (l + - > + l - + >)
l 0 0 > = 1/Sqrt[2] (l + - > - l - + >)
l 1 -1 > = l - - >

But since we have two identical spin-1/2 particles, only the anti-symmetric state is allowed, so our initial state must be the l 0 0 > state.

Now here is where I get confused. Our perturbation Hamiltonian is still only dependent on spin interactions, so our final state should still be one of the four kets listed above. If this is the case I don't see why all probabilities aren't zero (besides the l0 0> -> l0 0> probability) since we can't find fermions in a symmetric state. For all I know this is a trick question and I could be thinking correctly about it, but I feel like there should be some sort of computation that goes along with this.

Why do the particles have to have the same spatial wavefunction? (This would be what restricts the system to be in the singlet state.) Is this stated in the problem? An example of a typical spin-spin interaction which gives the Hamiltonian you supplied might be that between two electrons on neighboring lattice sites in a ferromagnetic crystal (c.f. Ising model). There is no reason why the system could not be in any of the four possible states.

## What are identical particles?

Identical particles are particles that cannot be distinguished from one another based on their physical properties, such as mass or charge. This means that they have the same quantum numbers and behave in the same way, making it impossible to track the individual particles.

## What is time dependent perturbation theory?

Time dependent perturbation theory is a mathematical technique used to study the behavior of quantum systems that are subject to a time-varying external influence. It allows for the calculation of the probability of a quantum system transitioning from one state to another over time.

## How is time dependent perturbation theory applied to identical particles?

In the case of identical particles, time dependent perturbation theory is used to study the effects of external perturbations on the behavior of the particles. This can help to understand how identical particles interact with each other and how their behavior may change over time.

## What are the limitations of time dependent perturbation theory?

Time dependent perturbation theory is only applicable to systems that are close to equilibrium and have small perturbations. It also assumes that the perturbation is time-dependent and that the system is linear. It may not accurately describe systems with strong interactions or large perturbations.

## What are some practical applications of time dependent perturbation theory?

Time dependent perturbation theory has many applications in physics, chemistry, and engineering. It is used to study the behavior of atoms, molecules, and other quantum systems. It also has practical applications in fields such as quantum computing, nuclear physics, and solid-state physics.

Replies
1
Views
550
Replies
1
Views
1K
Replies
2
Views
694
Replies
13
Views
2K
Replies
1
Views
936
Replies
2
Views
988
Replies
3
Views
1K
Replies
8
Views
2K
Replies
3
Views
879
Replies
16
Views
2K