{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

notes20

# notes20 - 5.73 Lecture#20 Density Matrices II 20 1 Read...

This preview shows pages 1–4. Sign up to view the full content.

20 - 1 5.73 Lecture #20 revised October 21, 2002 Density Matrices II Last time: ψ,  ⟩, ρ =  ⟩⟨  coherent superposition vs. statistical mixture populations along diagonal, coherences off diagonal A = Trace( ρ A ) = Trace( A ρ ) Today: Quantum Beats prepared state ρ detection as projection operator D What part of D samples a specific off-diagonal element of ρ ? Optimize magnitude of beats [partial traces] system consisting of 2 parts — e.g. coupled oscillators motion in state-space vs. motion in coordinate space. Read CTDL, pages 643-652. The material on pages 20–2, –3, –5, and –7 is an exact duplication of pages 19–5, –6, –7 and –8.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
20 - 2 5.73 Lecture #20 revised October 21, 2002 Example: Quantum Beats Preparation, evolution, detection magically prepare some coherent superposition state Ψ (t) Ψ t ( ) = N a n ψ n e iE n t h n ρ t ( ) = Ψ (t) Ψ (t) Several eigenstates of H. Evolve freely without any time-dependent intervention N = a n 2 n −1 / 2 normalization Thus D = ψ 1 ψ 1 = 1 0 L 0 0 0 M 0 0 ρ = N 2 a 1 2 a 1 a 2 * e E 1 E 2 ( ) t h L a 2 2 a 3 2 O ρ 12 = 1 Ψ Ψ 2 ρ 12 = N 2 a 1 e E 1 t h a 2 * e + iE 2 t h D D t 1 2 1 2 * i t 2 1 = Trace Trace a a a e stuff 0 0 0 0 N a 12 ρ ( ) = = N 2 2 ω L M M M M D picks out only 1st row of ρ . Case (1): Detection: only one of the eigenstates, ψ 1 , in the superposition is capable of giving fluorescence that our detector can “see”. a projection operator (designed to project out only | ψ 1 part of state vector or ρ 11 part of ρ . no time dependence!
20 - 3 5.73 Lecture #20 revised October 21, 2002 case (2): a particular linear combination of eigenstates is bright: the initial state 2 –1/2 ( ψ 1 + ψ 2 ) has D = 1. D = 1 2 ψ 1 + ψ 2 ( ) ψ 1 + ψ 2 ( ) = 1 2 ψ 1 ψ 1 + ψ 2 ψ 2 + ψ 1 ψ 2 + ψ 2 ψ 1 [ ] = 1 2 1 0 0 L 0 0 0 0 0 0 0 0 M 0 0 0 + 0 0 0 L 0 1 0 0 0 0 0 0 M 0 0 0 + 0 1 0 L 0 0 0 0 0 0 0 0 M 0 0 0 + 0 0 0 L 1 0 0 0 0 0 0 0 M 0 0 0 = 1 2 1 1 0 L 1 1 0 0 0 0 0 0 M 0 0 0 if the bright state had been 2 then = 1 1 -1 0 -1 1 0 0 0 -1/2 ψ ψ 1 2 0 0 0 0 0 ( ) , D

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}