By Kim S. Ponce, MS Physics, MSU-IIT Problem Given that the Hamiltonian is . Show that and the mean value of the momentum () is zero. Solution Substituting our Hamiltonian operator, Using the commutator of and , this can be written in the form: But , so Therefore, […]

### Finding the commutator of the Hamiltonian operator, H and the position operator, x and finding the mean value of the momentum operator, p

Friday, May 24th, 2019Posted in Quantum Physics **|** No Comments »

### Position and Momentum operators derived from annihilation and creation operators

Tuesday, May 21st, 2019By: Levine T. Poralan, MS Physics, MSU-IIT Given the annihilation and creation operators below, Show that (a) (b) (c) and are Hermitian but not and Solution: (a) Adding the annihilation and creation operators this results to and this becomes . (b) Now, finding the difference of the two operators we can obtain multiplying the whole equation by […]

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### Commutators of the Position and Momentum Operators

Tuesday, May 21st, 2019By: Levine T. Poralan, MS Physics, MSU-IIT Show that (a) (b) Solution: (a) Given that we let an arbitrary function Now solving for the commutator, The first and second terms cancel and we are left with Therefore, (b) Introducing dimensionless observable and , where Now, The first and second terms cancel and we are left […]

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### Properties of Quantum Oscillators 1

Friday, April 3rd, 2009by SIMON JUDE BURGOS In this post we investigate the properties of a quantum oscillator by using an algebraic tool in quantum mechanics called ‘ladder operators’. Using the ladder operator it becomes easy to find the following properties for a quantum oscillator in a given energy level: the average position and momentum and the square […]

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