notes13 - 5.73 Lecture #13 1. 13 - 1 End Matrix Solution of...

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13 - 1 5.73 Lecture #13 updated 9/27/02 9:22 AM End Matrix Solution of H-O, a + a Operators 1. starting from H = p 2 2 m + 1 2 k x 2 and [ x , p ] = i h 2. showed p nm = m i h x nm E m E n () x nm = i h k p nm E m E n x nn =0 ,p nn and x nm p nm if E n = E m 3. x nm 2 =− 1 km p nm 2 E m E n i h ωω = km 1/2 the only non-zero x and p elements are between states whose E’s differ by ± h ω 4. combs of connected states, block diag. of H , x , p , x 2 , p 2 E n (i) = h ω n i 6. Recursion Relationship 5. lowest index must exist because lowest E must exist. Call this index 0 x 01 2 = h 2 (km) p 01 2 = h 2 + from phase choice x 01 =+ i p 01 Today x nn + 1 2 in terms of x nn-1 2 general matrix elements x nn + 1 2 nn + 1 2 7. general x and p elements 8. only blocks correspond to ε i = 1 2 h ω Dimensionless x, p, H and a (annihilation) and a (creation) operators
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13 - 2 5.73 Lecture #13 updated 9/27/02 9:22 AM phase ambiguity: we can specify absolute phase of x or p BUT NOT BOTH because that would affect value of [ x , p ] matrix elements of x are REAL p are IMAGINARY BY CONVENTION: x 01 =+ i km () −1 / 2 p 01 and plug this into x 01 p 01 * p 01 x 01 * = i h try x 01 2 = h 2 km 1/2 p 01 2 = h 2 km + get If we had chosen we would have obtained x which is impossible! 01 xk m p km 01 12 01 2 2 =− i / / h check for self-consistency of seemingly arbitrary phase choices at every opportunity: * Hermiticity * 2 0 Recursion Relation for x ii + 1 2 6. start again with gerand equation derived in #3 above using the phase choice that worked above x n + 1n * = i km p n + 1n * going up Hermiticity c.c. of both sides x nn +1 = i(km) −1 / 2 p nn going down x i km p nn + + 1 1 /
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13 - 3 5.73 Lecture #13 updated 9/27/02 9:22 AM x nn ± 1 i km () 1/2 p nn ± 1 now the arbitrary part of the phase ambiguity in the relationship between x and p is eliminated Apply this to the general term in [ x,p ] algebra NONLECTURE : from terms in [ x , p ] = i h x nn + 1 p n + 1n = x nn + 1 p nn + 1 * = x nn + 1 km i x nn + 1 * = x nn + 1 2 + i km p nn + 1 x n + 1n =− km i x nn + 1 x nn + 1 * = x nn + 1 2 + i km x nn 1 p n 1n = x nn 1 p nn 1 * = x nn 1 + km i x nn 1 * = x nn 1 2 i km p nn 1 x n 1n =− − km i x nn 1 x nn 1 * = x nn 1 2 i km i h = 2 i km x nn + 1 2 x nn 1 2 [] x nn + 1 2 = h (km) 2 + x nn 1 2 but x 01 2 = x 10 2 = h 2 km x nn + 1 2 = n + 1 h 2 km p nn + 1 2 = n + 1
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notes13 - 5.73 Lecture #13 1. 13 - 1 End Matrix Solution of...

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