HW14_Solutions - PHYS851 Quantum Mechanics I, Fall 2008...

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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 14: Solutions 1. [15 pts] Consider the two-state quantum system described by H = H + V , where H = planckover2pi1 2 ( | 1 )( 1 | | )( | ) , and V = planckover2pi1 2 ( | )( 1 | + | 1 )( | ) . In the Schrodinger picture we have | S ( t ) ) = c ( t ) | ) + c 1 ( t ) | 1 ) . Solve Schrodingers equation for the coefficients c ( t ) and c 1 ( t ) with the initial conditions c (0) = 1 and c 1 (0) = 0. Let the operator W be defined in the Schrodinger picture as W = | 1 )( 1 | | )( | . Compute ( W ) as a function of time. Answer: Starting from Schrodingers equation, i planckover2pi1 d dt | ) = H | ) , and applying ( | and ( 1 | from the left gives, respectively i planckover2pi1 d dt c = planckover2pi1 2 c + planckover2pi1 2 c 1 i planckover2pi1 d dt c 1 = planckover2pi1 2 c 1 + planckover2pi1 2 c Putting these equations in matrix form gives: d dt parenleftbigg c c 1 parenrightbigg = i 2 parenleftbigg parenrightbiggparenleftbigg c c 1 parenrightbigg From det vextendsingle vextendsingle vextendsingle vextendsingle 2 2 vextendsingle vextendsingle vextendsingle vextendsingle = 0 we find the characteristic equation 4 2 2 2 = 0 so that the eigenvalues are = 1 2 radicalbig 2 + 2 The solution is then of the form parenleftbigg c c 1 parenrightbigg = a parenleftbigg e +0 e +1 parenrightbigg e- i + t + b parenleftbigg e- e- 1 parenrightbigg e- i- t , 1 where a and b are free constants to be determined by initial conditions. Due to the presence of the free constants, it is not necessary to find normalized eigenvectors, so we can safely take e = 1. The eigenvalue equation is then: 2 + e 1 = 0 so that e 1 = + 2 Putting this together gives c = ae- i + t + be- i- t c 1 = a + 2 + e- i + t + b + 2 - e- i- t From the initial conditions c = 1 and c 1 = 0 we find a + b = 1 ( + 2 + ) a + ( + 2 - ) b = 0 the latter can be rewritten using - = + as ( a + b ) + 2 + ( a b ) = 0 which gives a b = 2 + with b = 1 a this becomes 2 a = 1 2 + or a = + / 2 2 + and therefore b = + + / 2 2 + so that finally we arrive at c = cos( +...
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This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.

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HW14_Solutions - PHYS851 Quantum Mechanics I, Fall 2008...

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