Let Ïn be normalized stationary states of the

Physics 4301: Homework 5. Problem 1 This exercise is intended to ease your possible “pain” and illustrate how algebraically straightforward it is to do calculations using the ladder operators b a and b a† of the 1D harmonic oscillator problem. Let ψn be normalized stationary states of the oscillator (n = 0, 1, 2, . . .). We established in class the following basic relations: √ √ b aψn = n ψn−1 , b a† ψn = n + 1 ψn+1 . (1) b can now be represented as combinations of ladder operators to ﬁnd the Other operators A result of their actions on ψn : ∑ b n= Aψ bm ψm . (2) m

The right-hand side of (2) is just an expansion of the result, which is some wave function, over the set of our eigenstates, bm being the expansion coeﬃcients. You will ﬁnd there would be only very few non-zero bm in our examples. In the following, use the representation of the position x and momentum pb operators via ladder operators as discussed in class. (a) Find (in a form similar to Eqs. (1) and (2)) xψn

and pbψn .

(3)

(b) Use now only results of (3) to ﬁnd x2 ψn ,

pb 2 ψn

and xb pψn .

(4)

(c) And, come on!, why don’t you calculate - with the least expense x3 ψn ? (d) Use the derived in (4) to check if indeed b n = (n + 1/2) ~ω ψn , Hψ

(5)

b is, of course, the Hamiltonian for our oscillator. Can you also conﬁrm that the where H expectation values of the kinetic and potential energies in any stationary state are equal to each other? Problem 2 Let ψn (n = 0, 1, 2, . . .) be standard normalized stationary states of the 1D harmonic oscillator characterized by spatial displacement x, mass m and frequency ω. You may want to use the algebraic approach in your calculations. 1

(a) At time t = 0, the oscillator is prepared in (normalized) state i 1 Ψ(t = 0) = √ ψ2 + √ ψ4 . 2 2

(i2 = −1)

(6)

Find the time dependence of the expectation value of the kinetic energy operator ⟨ Tb ⟩(t),

Tb = pb2 /2m,

(7)

in this state Ψ(t) at times t > 0. (b) What would be the result for time dependence (7) if the initial state was i 1 Ψ(t = 0) = √ ψ2 + √ ψ6 , 2 2 instead of Eq. (6)?

2

(8)

The right-hand side of (2) is just an expansion of the result, which is some wave function, over the set of our eigenstates, bm being the expansion coeﬃcients. You will ﬁnd there would be only very few non-zero bm in our examples. In the following, use the representation of the position x and momentum pb operators via ladder operators as discussed in class. (a) Find (in a form similar to Eqs. (1) and (2)) xψn

and pbψn .

(3)

(b) Use now only results of (3) to ﬁnd x2 ψn ,

pb 2 ψn

and xb pψn .

(4)

(c) And, come on!, why don’t you calculate - with the least expense x3 ψn ? (d) Use the derived in (4) to check if indeed b n = (n + 1/2) ~ω ψn , Hψ

(5)

b is, of course, the Hamiltonian for our oscillator. Can you also conﬁrm that the where H expectation values of the kinetic and potential energies in any stationary state are equal to each other? Problem 2 Let ψn (n = 0, 1, 2, . . .) be standard normalized stationary states of the 1D harmonic oscillator characterized by spatial displacement x, mass m and frequency ω. You may want to use the algebraic approach in your calculations. 1

(a) At time t = 0, the oscillator is prepared in (normalized) state i 1 Ψ(t = 0) = √ ψ2 + √ ψ4 . 2 2

(i2 = −1)

(6)

Find the time dependence of the expectation value of the kinetic energy operator ⟨ Tb ⟩(t),

Tb = pb2 /2m,

(7)

in this state Ψ(t) at times t > 0. (b) What would be the result for time dependence (7) if the initial state was i 1 Ψ(t = 0) = √ ψ2 + √ ψ6 , 2 2 instead of Eq. (6)?

2

(8)