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

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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 we have encountered in the processing of Harmonic Oscillator. So by analogy we could conclude

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