Finding probability of changing states

  • Context: Undergrad 
  • Thread starter Thread starter TheCanadian
  • Start date Start date
  • Tags Tags
    Probability States
Click For Summary
SUMMARY

The discussion focuses on the derivation of the probability of changing states during adiabatic passage, specifically analyzing the expression $$ \langle \hat{U}^\dagger(t_1,t_0)\hat{U}(t_1,t_0)\rangle - \langle \hat{U}^\dagger(t_1,t_0)\rangle\langle \hat{U}(t_1,t_0)\rangle $$. The first term represents the unitarity condition, which equals 1, while the last term indicates the probability of remaining in the initial state. The difference between these two terms quantifies the total probability of transitioning to a different state.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly adiabatic passage.
  • Familiarity with unitary operators and their properties.
  • Knowledge of expectation values in quantum mechanics.
  • Basic mathematical skills to interpret quantum expressions and derivations.
NEXT STEPS
  • Study the concept of adiabatic theorem in quantum mechanics.
  • Learn about the properties of unitary operators in quantum systems.
  • Explore expectation values and their significance in quantum state transitions.
  • Investigate applications of adiabatic passage in quantum computing and quantum state manipulation.
USEFUL FOR

Quantum physicists, students of quantum mechanics, and researchers interested in quantum state transitions and adiabatic processes.

TheCanadian
Messages
361
Reaction score
13
I am following the derivation shown in this link on adiabatic passage.

I have posted one part below:

Screen Shot 2016-07-15 at 11.53.04 AM.png


I am simply wondering how this expression was derived and how it indicates the probability of being in a state that is different from the initial state? How exactly is this represented by:

$$ \langle \hat{U}^\dagger(t_1,t_0)\hat{U}(t_1,t_0)\rangle - \langle \hat{U}^\dagger(t_1,t_0)\rangle\langle \hat{U}(t_1,t_0)\rangle $$
 
Physics news on Phys.org
The last term in your last expression is the probability that you will remain in the initial state. The first term is 1 by unitarity. The difference of these terms is the total probability of not being in the initial state.
 
  • Like
Likes   Reactions: Jilang and TheCanadian

Similar threads

Graduate Asymptotic states in the Heisenberg and Schrödinger pictures
  • · Replies 8 ·
Replies
8
Views
2K
Graduate Heisenberg picture and Path integrals (Zee QFT)
  • · Replies 15 ·
Replies
15
Views
3K
Undergrad Realizing a Quantum Mechanical Watch-Stop
  • · Replies 4 ·
Replies
4
Views
2K
Graduate Changing Hamiltonian with some eigenvalues constant
  • · Replies 5 ·
Replies
5
Views
2K
Graduate Matrix representation of a unitary operator, change of basis
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Instability of hydrogen ground state if the time-reversal operator is unitary
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad How to solve 2nd order TDSE for a Gaussian-kicked harmonic oscillator?
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Feynman's Lectures volume III (Ch:8) -- Resolution of vector states
  • · Replies 4 ·
Replies
4
Views
1K
Undergrad Physical interpretation of this coherent state
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Operations regarding tensor-product states
  • · Replies 2 ·
Replies
2
Views
2K