Halved Harmonic Oscillator

Consider a particle of mass m moving in one dimension (see Figure 1.1) in a potential

Figure 1.1

a) Using the normalized trial function

find the value of λ which minimizes the ground state energy and the resulting estimate of the ground state energy. How is this value related to the true ground state energy?

b) What is the exact ground state wave function and energy for this system (neglect the normalization of the wave function)? Do not solve the Schrodinger equation directly. Rather, state the answer and justify it.

Hint: You may need the integral

SOLUTION

a) Using the Rayleigh–Ritz variation principle, calculate ⟨E(λ)⟩, the expectation value of the ground state energy as a function of λ:

(1)

First calculate the denominator of (1):

(2)

So our trial function is already normalized. Continuing with the numerator of (1), we have

(3)

where we set A ≡ h²/2m and B = mω²/2. Evaluate the integral I ≡ ⟨E(λ)⟩/4λ³:

(4)

Now,

(6)

So, we have

(6)

Finally,

(7)

To minimize this function, find λ₀ corresponding to

(8)

(9)

(10)

(11)

Therefore,

(12)

We should have the inequality

(13)

where E₀ is the true ground state energy.

b) To find the exact ground state of the system, notice that odd wave functions of a symmetric oscillator problem (from -∞ to ∞) will also be solutions for these boundary conditions since they tend to zero at x = 0.

Therefore, the ground state wave function of this halved oscillator will correspond to the first excited state wave function of the symmetrical oscillator. The wave function can easily be obtained if you take the ground state |0⟩ and act on it by the creation operator a^†:

The ground state energy of our halved oscillator will in turn correspond to the first excited state energy  of the symmetrical oscillator:

Comparing this result with that of (a), we see that the inequality (13) holds and that our trial function is a fairly good approximation, since it gives the ground state energy to within 15% accuracy.