Quantum Harmonic Oscillator

At sufficiently small energies, theharmonic oscillatoras governed by the laws ofquantum mechanics, known simply as thequantum harmonic oscillator, differs significantly from its description according to the laws ofclassical physics. Whereas the energy of the classical harmonic oscillator is allowed to take on any positive value, the quantum harmonic oscillator has discrete energy levels $E_n$given by

$E_n = \hbar \omega \left(n + \frac{1}{2}\right),$

where $\hbar$is thePlanck constant, $\omega$is the (classical)angular frequency, and $n$is a non-negative integer. Furthermore, whereas the classical harmonic oscillator is confined to a finite region of space, the quantum harmonic oscillator has a nonzero (but asymptotically vanishing) probability of being found anywhere.

As one of the few important quantum mechanical systems whose dynamics can be determined exactly, the quantum harmonic oscillator frequently serves as a basis for describing many real-world phenomena, such as molecular vibrations.

The energies $E$and their correspondingeigenfunctions $\psi(x)$of the one-dimensional quantum harmonic oscillator satisfy the time-independentSchrödinger equation:

$\left(\frac{\hat{p}^2}{2m} +\frac{1}{2} m \omega^2 \hat{x}^2\right) |\psi\rangle = E |\psi\rangle,$

where $\hat{x}$and $\hat{p}$的位置和momentum operators, respectively. Theladder operatormethod to obtain the eigenfunctions and energies (due toDirac) is to factor the equation by using the so-calledlowering operator

$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + i \frac{1}{\sqrt{2m\omega\hbar}} \hat{p}$

and itsadjoint,raising operator

$\hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \hat{x} - i \frac{1}{\sqrt{2m\omega\hbar}} \hat{p},$

which satisfy thecommutation relation

$\big[\hat{a}, \hat{a}^\dagger\big] = 1.$

The Schrödinger equation can then be factored as

$\hbar \omega \left( \hat{a}^\dagger \hat{a} + \frac{1}{2}\right) |\psi\rangle = E |\psi\rangle$

或

$\hbar \omega \left( \hat{a} \hat{a}^\dagger - \frac{1}{2}\right) |\psi\rangle = E |\psi\rangle.$

Using the commutation relation, one can show that if $| \psi \rangle$is an eigenfunction with eigenvalue $E,$then so must also $\hat{a}^\dagger | \psi \rangle$with corresponding eigenvalue $E + \hbar \omega$and $\hat{a} | \psi \rangle$with corresponding eigenvalue $E - \hbar \omega.$In essence, given an eigenfunction and corresponding energy, applying the raising operator obtains a new eigenfunction $\big($namely $\hat{a}^\dagger | \psi \rangle\big)$with corresponding energy increased by $\hbar \omega$while applying the lowering operator obtains a new eigenfunction $\big(\hat{a} | \psi \rangle\big)$with energy decreased by $\hbar \omega.$

The lowering cannot proceed indefinitely; there must exist some lowest energy level $E_0$corresponding to an eigenfunction $| \psi_0 \rangle$for which

$| \帽子{一}\ psi_0 \纠正= 0。$

Therefore, the spectrum of all eigenfunctions can be enumerated by a non-negative integer $n:$

$| \psi_n \rangle = \left(\hat{a}^\dagger\right)^n | \psi_0 \rangle$

with corresponding eigenvalues

$E_n = n \hbar \omega + E_0.$

From the time-independent Schrödinger equation, it is straightforward to show that the wavefunction corresponding to $| \psi_0 \rangle$(that is, the ground state wavefunction) is

$\psi_0(x) = \left( \frac{m\omega}{\pi\hbar} \right) \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$

with corresponding energy eigenvalue

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

so that

$E_n = \hbar \omega \left( n + \frac{1}{2} \right).$

In general, the closed-form expression for the wavefunctions $\psi_n(x)$(which can be found by solving the Schrödinger wave equation analytically) is

$\psi_n(x) = \left( \frac{m\omega}{\pi\hbar} \right) \frac{1}{\sqrt{2^n n!}} H_n\left(\sqrt{m\omega/\hbar} x\right) \exp\left(-\frac{m\omega x^2}{2\hbar}\right),$

where $H_n$is the $n^\text{th}$Hermite polynomial.A plot of the first few wavefunctions is shown below:

自从harmonic potential is non-vanishing, all of the eigenstates areboundand the energy spectrum is discrete, although, as is characteristic in quantum mechanics, the wavefunctions extend to all space (zero only at nodes). This should be contrasted with the classical harmonic oscillator, whose probability density is bounded by the amplitude of its oscillation and whose energies are continuous. Below is the probability density of the ground state of the quantum harmonic oscillator compared with the U-shaped density of the classical oscillator.

Using the ladder operators, many dynamical quantities can be calculated for the harmonic oscillator without direct integration. One can express the position and momentum operators as follows:

$\hat{x} = \sqrt{\frac{\hbar}{2m\omega}} \left( \hat{a}^\dagger + \hat{a} \right)$

and

$\hat{p} = i \sqrt{\frac{\hbar m\omega}{2}} \left( \hat{a}^\dagger - \hat{a} \right).$

Furthermore, one can show that the action of the ladder operators on the eigenstates works according to

$\hat{a}^\dagger | \psi_n \rangle = \sqrt{n + 1} | \psi_{n+1} \rangle$

and

$\hat{a} | \psi_n \rangle = \sqrt{n} | \psi_{n-1} \rangle.$

With these relations, as well as the fact that the eigenfunctions are orthogonal, that is

$\langle \psi_i | \psi_j \rangle = \delta_{ij},$

one can compute many quantities of interest.

It is easy to show that the expectation values of the position and momentum must vanish for any $n$since the eigenfunctions are orthogonal:

$\begin{aligned} \langle x \rangle &= \sqrt{\frac{\hbar}{2m\omega}} \big\langle \psi_n \big| \big( \hat{a}^\dagger + \hat{a} \big) \big| \psi_n \big\rangle \\ &= \sqrt{\frac{\hbar}{2m\omega}} \big( \sqrt{n+1} \langle \psi_n | \psi_{n+1} \rangle + \sqrt{n} \langle \psi_n | \psi_{n-1} \rangle \big) \\ &= 0. \end{aligned}$

Similarly,

$\begin{aligned} \langle p \rangle &= i \sqrt{\frac{\hbar m\omega}{2}} \big\langle \psi_n \big| \big( \hat{a}^\dagger - \hat{a} \big) \big| \psi_n \big\rangle \\ &= i \sqrt{\frac{\hbar m\omega}{2}} \big( \sqrt{n+1} \langle \psi_n | \psi_{n+1} \rangle - \sqrt{n} \langle \psi_n | \psi_{n-1} \rangle \big) \\ &= 0. \end{aligned}$

Show that the ground state is the only eigenstate of the harmonic oscillator that is a minimum-uncertainty state in position-momentum space $($i.e., equality holds for theHeisenberg uncertainty relationin $x$and $p).$

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First, we compute the expectation of the square of the position

$\begin{aligned} \big\langle x^2 \big\rangle &= \left(\frac{\hbar}{2m\omega}\right) \big\langle \psi_n \big| \big( \hat{a}^\dagger + \hat{a} \big)^2 \big| \psi_n \big\rangle \\ &= \left(\frac{\hbar}{2m\omega}\right) \big\langle \psi_n \big| \big( {\hat{a}^\dagger}^2 + \hat{a}^\dagger \hat{a} + \hat{a} \hat{a}^\dagger + \hat{a}^2 \big) \big| \psi_n \big\rangle \\ &= \left(\frac{\hbar}{2m\omega}\right) \big( 0 + \sqrt{n} | \psi_{n-1} \rangle + \sqrt{n+1} | \psi_{n+1} \rangle \big) \\ &= \left(\frac{\hbar}{2m\omega}\right) (2n + 1) \end{aligned}$

and square of the momentum

$\begin{aligned} \big\langle p^2 \big\rangle &= -\left(\frac{\hbar m\omega}{2}\right) \big\langle \psi_n \big| \big( \hat{a}^\dagger - \hat{a} \big)^2 \big| \psi_n \big\rangle \\ &= -\left(\frac{\hbar m\omega}{2}\right) \big\langle \psi_n \big| \big( {\hat{a}^\dagger}^2 - \hat{a}^\dagger \hat{a} - \hat{a} \hat{a}^\dagger + \hat{a}^2 \big) \big| \psi_n \big\rangle \\ &= -\left(\frac{\hbar m\omega}{2}\right) \big( 0 - \sqrt{n} | \psi_{n-1} \rangle - \sqrt{n+1} | \psi_{n+1} \rangle \big) \\ &= \left(\frac{\hbar m\omega}{2}\right) (2n + 1). \end{aligned}$

Combining with the previously calculated result $\langle x \rangle = \langle p \rangle = 0$yields

$\left( \Delta x \right)\left( \Delta p \right) = \sqrt{\big\langle x^2 \big\rangle - \big\langle x \big\rangle^2}\sqrt{\left\langle x^2 \right\rangle - \left\langle x \right\rangle^2} = \frac{\hbar}{2} (2n + 1),$

which achieves the mininmum uncertainty of $\hbar/2$only for the ground state $n = 0.$

Many practical potentials can be treated (or at least closely approximated) as harmonic potentials. The internuclear potential well of a molecule of diatomic gas, such as molecular oxygen or nitrogen, can be taken as a harmonic potential, in which case the vibrational energy levels $\epsilon_n$are given by

$\epsilon_n = \hbar \omega \left( n + \frac{1}{2} \right),$

where the vibrational angular frequency $\omega$can be computed from the force constant of the molecule and itsreduced mass.

Thepartition functionfor the system is

$q = \sum_{n=0}^\infty e^{\epsilon_n/kT} = \frac{e^{-\hbar \omega/2kT}}{1 - e^{-\hbar\omega/kT}},$

where $k$is theBoltzmann constantand $T$thetemperature.

From the partition function, one can compute many thermodynamics quantities of interest, such as the thermodynamic vibrational energy

$E = NkT^2 \frac{d \ln{q}}{dT} = Nk\left(\frac{\hbar\omega}{2k} + \frac{\hbar\omega/k}{e^{\hbar\omega/kT} - 1} \right)$

或vibrational heat capacity

$C_v = \left( \frac{\partial E}{\partial T} \right) = Nk \left( \frac{\hbar\omega}{kT} \right)^2 \frac{e^{\hbar\omega/kT}}{\left(e^{\hbar\omega/kT} - 1\right)^2}.$

For $kT \gg \hbar\omega,$one obtains the classical result $E = NkT$and $C_v = Nk.$

[1]格里菲斯D.J.必要的介绍ntum Mechanics. Second edition. Pearson, 2004.

[2] McQuarrie, D.A. Statistical Mechanics. University Science Books, 2000.

[3] Shankar, R. Principles of Quantum Mechanics. Second edition. Plenum, 1994.