Answer - PHY4221 Quantum Mechanics I(Fall 2004 Lecture...

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Unformatted text preview: PHY4221 Quantum Mechanics I (Fall 2004) Lecture Exercise (Answer) Question 2: Again, for simplicity I will suppose that 1 m ϖ = = = h . Now the Hamiltonian is: ( 29 ( 29 ( 29 2 2 2 2 2 2 2 2 2 1 2 1 2 2 1 2 1 1 2 2 2 1 2 2 2 2 1 2 1 2 2 2 H a a a a a a x a a a a x p x x p x λ λ λ λ λ λ λ λ + + + + + = + + + ÷ = + +-- ÷ =- + + ÷ =- + + ÷ - + = + (1) Latter you will see that the writing Hamiltonian in this form is not an arbitrary choice. We could redefine another two raising/lowering operators as following: 1/4 1/4 1/4 1/4 1 1 2 1 2 1 2 1 2 2 1 1 2 1 2 1 2 1 2 2 b x i p b x i p λ λ λ λ λ λ λ λ + +- =- ÷ ÷- + +- = + ÷ ÷- + (2) Also, this redefinition is not an arbitrary choice, as I will explain below. You could check that the two operators have the completely same properties of , a a + , that is: 2 , 1, , , , 1 1 4 2 b b b b b b b b b b H b b λ + + + + + + = = = - =- + ÷ (3) Equation (3) gives us the Hamiltonian of this problem in the form which is completely same as...
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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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Answer - PHY4221 Quantum Mechanics I(Fall 2004 Lecture...

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