Harmonic Oscillator in Field

A particle of mass m and charge e moves in one dimension. It is attached to a spring of constant k and is initially in the ground state (n = 0) of the harmonic oscillator. An electric field E is switched on during the interval 0 < t < τ, where τ is a constant.

a) What is the probability of the particle ending up in the n = 1 state?

b) What is the probability of the particle ending up in the n = 2 state?

SOLUTION

Now we label the eigenstates with the index n for the harmonic oscillator state of energy E_n = hω(n + 1/2) and write the equation satisfied by the time-dependent amplitudes a_n(t),

(1)

(2)

We need to evaluate the matrix element ⟨n|x|m⟩ of x between the states |n⟩ and |m⟩ of the harmonic oscillator. It is only nonzero if m = n ±1. In terms of raising and lowering operators,

(3)

(4)

(5)

(6)

a) If the initial state is n = 0 at t = 0, then the amplitude of the n = 1 state for (t < τ) is given by

(7)

(8)

The last equation is the probability of ending in the state n = 1 if the initial state is n = 0. This expression is valid as long as it is less than 1 or if 2e|E|X₀ < hω.

b) The n = 2 state cannot be reached by a single transition from n = 0 since the matrix element ⟨2|x|0⟩ = 0. However, n = 2 can be reached by a two-step process. It can be reached from n = 1, and n = 1 is excited from n = 0. The matrix element is so we have that

(9)

(10)

Note that P₂ = P₁²/2!. Similarly, one can show that P_n = P₁ⁿ/n! However, the total probability, when summed over all transitions, cannot exceed 1. Therefore, we define a normalized probability

(11)