Derivation

Derivation - The derivation of the exercise problem in...

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The derivation of the exercise problem in Chapter 1.3.3: ( 29 ( 29 ( 29 1 2 2 1 2 ˆ | | ˆˆ | | | | ˆˆ | | | | ˆˆˆˆ | | | | | | | | ˆˆˆˆ | | | | | | | | | | ˆ ˆˆˆ | | | | | | | ˆ ˆˆˆ | | | | | | | ˆ ˆ | n m n n m n n n m n n m n m n n m n n m n n q m E E n q m n q m E E m q n n q m E E E m q n n q m E m q n n q m m q E n n q m E m q n n q m m q H n n q m E m q m m qH n n q m E m q m m qH = - = - - = - = - = - = - = 142 43 142 43 2 ˆˆ| | | m q m E m q m - (1) Now let us see what ˆ ˆˆ qHq is: ( 29 ( 29 2 2 ˆˆˆˆ ˆ ˆˆˆˆˆˆˆˆ 2 2 p qp q qHq q V q q qV q q m m = + = + ÷ (2) Because the potential ( 29 ˆ V q is only a function of ˆ q , so it could be taken Taylor expansions as following: ( 29 ( 29 ( 29 0 0 ˆˆ ! k k k V V q q k = = (3) You see from equation (3) that ( 29 ˆ V q is commutative with ˆ q , and thus the equation (2) could be written as: ( 29 2 2 ˆˆˆ ˆ ˆˆˆˆ 2 qp q qHq q V q m = + (4)
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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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Derivation - The derivation of the exercise problem in...

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