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By Levi A.F.J.

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This gives † mω h ∂ h ∂ ( bˆ bˆ ) ψ =  --------  x + -------- -----  x – -------- -----  ψ  2h   mω ∂ x  mω ∂x  † mω 2 h h ∂ h ∂ h ------∂ - ( bˆ bˆ ) ψ =  --------  x + -------- + -------- x ----- – -------- x ----- – -----------ψ 2h mω mω ∂ x mω ∂x m2 ω 2 ∂x 2 and 2 † mω h ∂ h ∂ ( bˆ bˆ ) ψ =  --------  x – -------- -----   x + -------- -----  ψ  2h   mω ∂ x  mω ∂x  4 2 † mω 2 h h ∂ h ∂ h ∂  ( bˆ bˆ ) ψ =  --------  x – -------- – --------x ----- + -------- x ----- – ------------------- ψ 2h mω mω ∂x mω ∂x m 2 ω 2 ∂x 2 so that the commutation relation simplifies to just 2 2 † † mω 2h 2h ∂ 2h ∂ ( bˆ bˆ – bˆ bˆ ) ψ =  --------   -------- – -------- x ----- + -------- x -----  ψ = ψ 2h mω mω ∂x mω ∂ x or, in even more compact form, † † † [ bˆ , bˆ ] = bˆ bˆ – bˆ bˆ = 1 If we do not want to use differential operators and a dummy wavefunction then we could write ih pˆ ihpˆ ihpˆ ihpˆ mω † † † mω [ bˆ , bˆ ] = bˆ bˆ – bˆ bˆ =  --------  xˆ + ----------x  xˆ – ----------x  –  --------   xˆ – ----------x   xˆ + ----------x          2h   mω  mω mω 2h mω 2 2 2 † ihpˆ x xˆ ihxˆ pˆ x h pˆ x   mω  2 ihpˆ x xˆ ihxˆ pˆ x h pˆ x  mω – + + ----------------------– -----------------x ˆ – ------------- + ----------[ bˆ , bˆ ] =  --------  xˆ 2 + ------------ 2h   mω mω mω mω m 2 ω 2 m 2 ω 2  2h   2 † imω hpˆ x xˆ hxˆ pˆ x – ----------- = i ( pˆ x xˆ – xˆ pˆ x ) = i [ pˆ x,xˆ ] = i ( – i h ) = 1 [ bˆ , bˆ ] =  ----------   ----------h mω mω since [pˆ x ,xˆ ] = – i h .

Ryzhik, Table of integrals, series, and products, Academic Press, San Diego, 1980, p. 196 (ISBN 0 12 294760 6)) ∫e ax ∫e ax e ( a cos ( bx ) + b sin ( bx ) ) cos ( bx ) dx = -------------------------------------------------------------2 2 a +b ax e ( a sin ( b x ) – b cos ( b x ) -) sin ( bx ) dx = -------------------------------------------------------------2 2 a +b ψ ( p x, t ) ax ip L ⁄ 2 2 – ip L ⁄ 2 ip L ⁄ 2 2 –i p L ⁄ 2 2 1 Lπ ( e x + e x ) π( e x – e x ) + = ------------- -----------------------------------------------+ 4L --------------------------------------------------2 2 2 2 2 2 2πh L 16 π – p x L π – px L ip L ⁄ 2 –i p L ⁄ 2 –i p L ⁄ 2 ip L ⁄ 2  i ( ω4 – ω1 )t Lπ ( e x + e x ) 4 Lπ ( e x – e x )  ------------------------------------------------------------------------------------------------- 2Re  e π 2 – p 2x L 2 16 π 2 – p 2x L 2   ψ (p x, t ) 2 p xL p xL  = A cos 2  --------  + Bsin 2  -------+ C sin ( p x L ) sin ( ( ω 4 – ω 1 ) t ) 2 2  32L π –8L π 2Lπ - , B = ----------------------------------------2 , C = ----------------------------------------------------------------A = ------------------------------.

I 2 (d) Show that the electromagnetic energy density is U = G . 4 SOLUTIONS Solution 1 (a) Let the potential in the region – L ⁄ 2 < x < L ⁄ 2 be zero and infinity elsewhere. The ψ n ( x, t ) = eigenfunctions are 2 ⁄ L sin ( k n ( x + L ⁄ 2 ) )e – iω nt k n = nπ ⁄ L , , where ω n = hk ⁄ 2m , and n = 1 , 2, 3, … . Hence, 2 n ψ 1 (x , t ) = 2 ⁄ L sin ( ( π ⁄ L ) ( x + L ⁄ 2 ) ) e –iω 1 t = 2 ⁄ L cos ( ( πx ⁄ L ) e –i ω1 t –i ω4 t ) –i ω4 t ψ 4 ( x , t ) = 2 ⁄ L sin ( ( 4 π ⁄ L ) ( x + L ⁄ 2 ) ) e = 2 ⁄ L sin ( 4πx ⁄ L ) e and the probability distribution of the particle in the superposition state ψ ( x, t ) = is ψ ( x, t ) 1 = -- ( ψ 1 ( x ) 2 + ψ 4 ( x ) 2 + 2 ψ 1 ( x ) ψ 4 ( x ) cos ( ( ω 4 – ω 1 )t ) ) 2 2 ψ ( x, t ) 1 ⁄ 2 ( ψ 1 ( x, t ) + ψ 4 ( x , t ) ) 1 xπ 4 πx πx 4πx h 15π   = --- cos 2 ------  + sin2  ---------  + 2 cos  ------  sin  ---------  cos  ------- -----------t L  L  L  L   2m L 2   L 2 2 h 15 π where we note that ω 4 – ω 1 = ------- ----------.

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