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Schrödinger We already know how to find the momentum eigenvalues of a system. How about the energy and the evolution of a system? Schrödinger Representation: qi , t ˆ i H qi , t t The evolution of a state of a system is given by the application of the Hamiltonian operator to the state wavefunction. In the Schrodinger representation, the evoultion is given by the wavefunction r , t , and not by the Hˆ the Hˆ is the operator corresponding to the Energy of a system... if the question is What is the En erg y of a stat e? the operator re presenting the obs ervable i s Hˆ For a time-independent Hˆ , although qi , t evolves in time, the Energy remains a co nsant What is the Hamiltonian? Hˆ Tˆ Kinetic operator 1 2 pˆ 2 T mv = 2 2m Vˆ potential and considering p pˆ i qi 2 Tˆ 2m qi2 2 2 2 2 2 2 2 for a 3D system, Tˆ 2m x 2 y 2 z 2 2m V Vˆ qi or Vˆ x, y , z this is the charactertistic potential defining the boundary conditions in each system, examples: V qi 0 free particle V qi ) 1 2 m qi qo Harmonic Oscillator 2 Separation of variables i qi , t ˆ H qi , t t If V(qi) is t-independent, then the Hamiltonian is also t-independent, and we can proposed a separation of variables separation of variables: f x, t g x x t f x, t x t f x, t =g x h t f x, t h t t x It is not possible to separate instead, for f x, t x t x t f x , t into 2 functions of f x, t g x h t independent variables Time and coordinate separation Starting with i qi , t ˆ H qi , t , and proposing qi , t qi t t i qi t t Hˆ qi t qi t ˆ ˆ i t qi T V qi qi t t t 2 2 qi t ˆ t i qi V qi qi t 2 t 2m qi 2 2 qi t ˆ q q i qi t t V i i t 2m qi2 2 2 qi t i qi t t Vˆ qi qi 2 qi t t qi t 2m qi t qi Cont. 2 i t 1 2 qi 1 ˆ V qi qi constan t W 2 t t qi 2m qi qi depends only on t depend s only on qi 2 1 2 qi 1 ˆ V qi qi W 2 qi 2m qi qi i t W t t t iW t t t e iW 2 2 qi 2m qi2 Vˆ qi qi W qi 2 2 ˆ V qi qi W qi 2 2m qi t Hˆ qi W qi qi , t qi e iW t where W is the t -independent en energy of the system! The phase function qi , t qi e iW t When Hˆ does not depend on time V V t , then the system is in a stationary state of energy W. Its time dependence is simply a phase iW t W W W function e cos t i sin t , with an oscillatory frequency Probability density: qi , t 2 qi t qi e 2 * qi e iW t qi e iW iW t 2 t * qi qi qi 2 Even if the wavefunction evolves in time, the probability density remains constant! Expectation value Consider a state which is a superposition of eigenstates of the t indep. Hˆ Hˆ k qi Wk k qi and qi , t ck e iWk t k qi k qi , t Hˆ qi , t H cm e iWm t m qi Hˆ m ck e iWk t k cm* e iWm t m ck e iWk t m qi Hˆ k qi k cm* ck e i Wm Wk t m qi Hˆ k qi m,k 0 for k m 1 for k m cm* ck e m,k i Wm Wk t ck Wk The expectation value of the Hamiltonian does not depend on time. A superposition of eigenstsates has constant E (E≠t). Wk m qi k qi 0 for k m 1 fo r k m 2 k k qi This is TRUE for any observable whose [A,H]=0 Heisenberg Representation In the Heisenberg (interaction) representation, the evolution of the system is given by the evolution of an operator d A dt A H , A t i But… how do we get to this equation? Consider the time derivative of an expectation value: d A dt A t We can evaluate the partial derivatives for each state and for the operator A A A A t t t We recognize that 2 of these terms correspond to the evolution of states i H t and i H t cont d A dt A i A H t i HA i HA AH i HA AH i H , A i H , A A t A t A t A t Schrödinger equation of a free particle Consider again a FREE PARTICLE. We know how to evaluate the momentum, by solving the eigenvalue-eigenvector eq. i P p P x …what about the Energy of a free particle? We need to construct the Hamiltonian operator to learn about the energy of the system Ĥ E Hˆ Tˆ Vˆ and for a free particle Vˆ 0 ˆ 2 2 2 P Hˆ Tˆ 2m 2m x 2 There are 2 ways to solve this problem solve the 2nd order differential equation and find 2 2 E 2 2m x HARD WAY!!!! How useful are []? ˆ 2 Pˆ 2 ˆ 3 Pˆ 3 P P ˆ ˆ consider the P, H Pˆ Pˆ 0 2m 2m 2m 2m Pˆ , Hˆ 0, there is a complete set which are eigenfunctions of both operators! Since we already know the eigenfunction of P, we can try to see if it is also an eigenfunction of H 1 ikx for P e what about Hˆ P ? 2π 2 2 Hˆ P 2m x 2 1 2π eikx 2 2 2 eikx 2 k 2 ikx ik e 2 2m 2m 2p x 2m 2p 2 1 2π eikx p2 EASY WAY!! P 2m The momentum eigenfunctions are also eigenfunctions of the Hamiltonian, with eigenvalues = classical kinetic energy Even though the quantum and classical energies coincide, the particle behavior is very different, i.e. the Quantum particle can never be at rest, because DpDx≽h/4p Good solutions of the Schrödinger eq. Solving the Schrödinger eq (most times) solve a differential eq., but not all mathematical solutions will be good physical solutions for the probability density. well - behaved functions : is finite (must not go to at any point) Single valued continuous 1st derivative, also continuous