Harmonic Oscillator

Anything which moves back and forth is an oscillator. A harmonic oscillator is one which moves back and forth in a smooth rhythm, with a single frequency. A pendulum, a mass on a spring, or an LC circuit are all examples of a classical harmonic oscillator.

Classical Harmonic Oscillator

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Quantum Harmonic Oscillator

We must find the Hamiltonian $\hat{H}$ , and this corresponds to finding $T$ and $V$ . For $V$ , we must discover a general form of the potential energy, as although the potential for a mass on a spring is $V(x) = \frac{1}{2}kx^2$ , potential for a pendulum is $V(h) = mgh$ , however $mgh$ is only an approximation, and this leads us to a general form of $V(x)$ .

Consider that $V(x) \approx V(0) + V'(a)(x - a) + \frac{V''(a)}{2!}(x-a)^2+...$ .

\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2 \hat{x}^2

- \frac{\hbar^2}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} + \frac{1}{2} m \omega^2 x^2 \psi(x) = E \psi(x)

E_n = \hbar \omega(n + \frac{1}{2}) \quad n=0,1,2,3...

H \ket{n} = E_n \ket{n}

Raising and Lowering Operators

In independent basis, the lowering operator is:

a = \sqrt{\frac{m \omega}{2 \hbar}} \Biggr(\hat{x} + i \frac{\hat{p}}{m \omega} \Biggr)

In independent basis, the raising operator is:

a^\dagger = \sqrt{\frac{m \omega}{2 \hbar}} \Biggr(\hat{x} - i \frac{\hat{p}}{m \omega} \Biggr)

a^\dagger a = \frac{m \omega}{2 \hbar} \Biggr( \hat{x}^2 + \frac{\hat{p}^2}{m^2 \omega^2} \Biggr) - \frac{1}{2}

H = \hbar \omega (a^\dagger a + \frac{1}{2})

a a^\dagger = \frac{m \omega}{2 \hbar} \Biggr( \hat{x}^2 + \frac{\hat{p}^2}{m^2 \omega^2} \Biggr) + \frac{1}{2}

[a, a^\dagger] = a a^\dagger - a^\dagger a = 1

[H,a] = - \hbar \omega a

[H,a^\dagger] = \hbar \omega a^\dagger

E_0 = \frac{1}{2} \hbar \omega

The number operator $N$ is:

N = a^\dagger a

N \ket{n} = n \ket{n}

\frac{d}{dx} \psi_0(x) = - \frac{m \omega}{\hbar} x \, \psi_0(x)

\psi_0(x) = \Biggr( \frac{m \omega}{\pi \hbar} \Biggr)^{1/4} e^{-m\omega x^2 / 2 \hbar}

a\ket{n} = \sqrt{n} \ket{n - 1}

a^\dagger \ket{n} = \sqrt{n + 1} \ket{n + 1}

\ket{1} = \frac{1}{\sqrt{1}} a^\dagger \ket{0}

\ket{2} = \frac{1}{\sqrt{2}} a^\dagger \ket{1} = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1}} (a^\dagger)^2 \ket{0}

We generalize to find $\ket{n}$ :

\ket{n} = \frac{1}{\sqrt{n!}} (a^\dagger)^n \ket{0}

In the spatial, as opposed to the energy basis we have

\psi_n(x) = \frac{1}{\sqrt{n!}} \Biggr[ \sqrt{ \frac{m \omega}{2 \hbar} }\Biggr( x - \frac{\hbar}{m \omega} \frac{d}{dx} \Biggr) \Biggr]^n \psi_0(x)

We use Hermite polynomials to help us:

Practice

Exercises

7 problems

1 of 7

A quantum harmonic oscillator has ω = 2.0×10¹⁴ rad/s. Observe the parabolic potential and Gaussian ground state. Find the zero-point energy E₀ = ℏω/2 in eV.

The Gaussian ground state ψ₀ (green) oscillates inside the parabolic potential V(x) = ½mω²x². The yellow dashed line shows the zero-point energy E₀. Enter E₀ = ℏω/2 in eV.

E₀ = eV

2 of 7

Using the energy ladder diagram for the same oscillator (ω = 2.0×10¹⁴ rad/s), find E₃ in eV (n=3 state).

The ladder shows equally-spaced levels Eₙ = (n + ½)ℏω separated by ℏω = 0.1317 eV. The green level (n=3) is highlighted. An animated photon marks a downward transition. Enter E₃ in eV.

E₃ = eV

3 of 7

Find the energy spacing ΔE = Eₙ₊₁ - Eₙ for ω = 2.0×10¹⁴ rad/s in eV.

Unlock Exercise 3