Question

3 Problem Three [10 points] (The Quantum Oscillator) We have seen in class that the Hamiltonian of a particle of a simple Harmonic oscillator potential in one dimension can be expressed in term of the creation and annihilation operators àt and à, respectively, as: or with In >, n = 0,1,..) are the nth eigenstates of the above Hamiltonian. Part A A.1. Show that the energy levels of a simple harmonic oscillator are E,' Aw (nti), n=0, 12, A.2. Calculate the expectation value of the operators in a state In >. A.3. Calculate the expectation value of i and f in a state In> Part B Now, consider two independent simple harmonic oscillators with the same fre- quency w. This is equivalent to one simple harmonic oscillator in two dimensions? described by the Hamiltonian B.1. What are the energies of the first three lowest-lying states? B.2. Make a table where you show the quantum numbers(n,ny) and the degeneracy for the first five lowest-lying states? Can you deduce the degeneracy of state Ing, ny>? l cimnle harmonic oscillator.

Answer 1

Given H_0^^= hw(a^+ a + 1/2) with ( Vertical-bar n > , n = 0, 1, 2,...) are the nth eigen state of above Hamiltonian. We know that, a Vertical-bar n > = squareroot n Vertical-bar n - 1 > a^+ Vertical-bar n > = squareroot n + 1 Vertical-bar n + 1 > & < n Vertical-bar m > = delta_nm orthogonality condition Part(A) A.1 we know that - H^^ Vertical-bar psi_n > = E_n Vertical-bar psi_n > Here Vertical-bar psi_n > = Vertical-bar n > H_0^^ Vertical-bar n > = (a^+ a + 1/2)hw Vertical-bar n > = (a^+ a Vertical-bar n > + 1/2 Vertical-bar n >)hw = (squareroot n a^+ Vertical-bar n - 1 > + 1/2 Vertical-bar n >)hw = (squareroot n squareroot n -1 + 1 Vertical-bar n > + 1/2 Vertical-bar n >)hw = (squareroot n middot squareroot n + 1/2)hw/n > = (n + 1/2)hw Vertical-bar n > H_0^^ Vertical-bar n > = E_n Vertical-bar n > E_n = (n + 1/2)hw n = 0, 1, 2, 3... \r\n (A.2) < x^^ > = < n Vertical-bar x^^ Vertical-bar n > = squareroot pi/2mw < n Vertical-bar (a + a^+) Vertical-bar n > =