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The quantum harmonic oscillator

One problem with this classical formulation is that it is not general. We cannot use it, for example, to describe vibrations of diatomic molecules, where quantum effects are important. A first step toward a quantum formulation is to use the classical expression k = m ω 2 to limit mention of a “spring” constant between the atoms. In this way the potential energy function can be written in a more general form,

U ( x ) = 1 2 m ω 2 x 2 .

Combining this expression with the time-independent Schrӧdinger equation gives

2 m d 2 ψ ( x ) d x 2 + 1 2 m ω 2 x 2 ψ ( x ) = E ψ ( x ) .

To solve [link] —that is, to find the allowed energies E and their corresponding wave functions ψ ( x ) —we require the wave functions to be symmetric about x = 0 (the bottom of the potential well) and to be normalizable. These conditions ensure that the probability density | ψ ( x ) | 2 must be finite when integrated over the entire range of x from to + . How to solve [link] is the subject of a more advanced course in quantum mechanics; here, we simply cite the results. The allowed energies are

E n = ( n + 1 2 ) ω = 2 n + 1 2 ω , n = 0 , 1 , 2 , 3 , . . .

The wave functions that correspond to these energies (the stationary states or states of definite energy) are

ψ n ( x ) = N n e β 2 x 2 / 2 H n ( β x ) , n = 0 , 1 , 2 , 3 , . . .

where β = m ω / , N n is the normalization constant, and H n ( y ) is a polynomial of degree n called a Hermite polynomial . The first four Hermite polynomials are

H 0 ( y ) = 1 H 1 ( y ) = 2 y H 2 ( y ) = 4 y 2 2 H 3 ( y ) = 8 y 3 12 y .

A few sample wave functions are given in [link] . As the value of the principal number increases, the solutions alternate between even functions and odd functions about x = 0 .

The harmonic potential V of x and the wave functions for the n=0 through n=4 quantum states of the potential are shown. Each wave function is displaced vertically by its energy, measured in units of h nu sub zero. The vertical energy scale runs from 0 to 5. The potential V of x is an upward opening parabola, centered and equal to zero at x = 0. The region below the curve, outside the potential, is shaded. The energy levels are indicated by horizontal dashed lines and are regularly spaced at energy of 0.5, 1.2, 2.5, 3.5 and 4.5 h nu sub 0. The n=0 state is even. It is symmetric, positive and peaked at x=0. The n = 1 state is odd. It is negative for x less than zero, positive for x greater than zero, zero at the origin. It has one negative minimum and one positive minimum. The n=2 state is even. It is symmetric, with a negative minimum at x=0 and two positive maxima, one at positive x and the other at negative x. The n = 3 state is odd. It is zero at the origin. It has, from left to right, a negative minimum and positive maximum on the left of the origin, then a positive maximum and negative minimum to the right of the origin. The n=4 state is even. It has a maximum at the origin, a negative minimum on either side, and a positive maximum outside of the minima. All of the states are clearly nonzero in the shaded region and go asymptotically to zero as x goes to plus and minus infinity. The minima and maxima are all inside the potential, in the unshaded region. Vertical dashed lines show the values of x where the potential is equal to the energy of the state, that is, where the horizontal dashed lines cross the V of x curve.
The first five wave functions of the quantum harmonic oscillator. The classical limits of the oscillator’s motion are indicated by vertical lines, corresponding to the classical turning points at x = ± A of a classical particle with the same energy as the energy of a quantum oscillator in the state indicated in the figure.

Classical region of harmonic oscillations

Find the amplitude A of oscillations for a classical oscillator with energy equal to the energy of a quantum oscillator in the quantum state n .

Strategy

To determine the amplitude A , we set the classical energy E = k x 2 / 2 = m ω 2 A 2 / 2 equal to E n given by [link] .

Solution

We obtain

E n = m ω 2 A n 2 / 2 A n = 2 m ω 2 E n = 2 m ω 2 2 n + 1 2 ω = ( 2 n + 1 ) m ω .

Significance

As the quantum number n increases, the energy of the oscillator and therefore the amplitude of oscillation increases (for a fixed natural angular frequency. For large n , the amplitude is approximately proportional to the square root of the quantum number.

Got questions? Get instant answers now!

Several interesting features appear in this solution. Unlike a classical oscillator, the measured energies of a quantum oscillator can have only energy values given by [link] . Moreover, unlike the case for a quantum particle in a box, the allowable energy levels are evenly spaced,

Δ E = E n + 1 E n = 2 ( n + 1 ) + 1 2 ω 2 n + 1 2 ω = ω = h f .

When a particle bound to such a system makes a transition from a higher-energy state to a lower-energy state, the smallest-energy quantum carried by the emitted photon is necessarily hf . Similarly, when the particle makes a transition from a lower-energy state to a higher-energy state, the smallest-energy quantum that can be absorbed by the particle is hf . A quantum oscillator can absorb or emit energy only in multiples of this smallest-energy quantum. This is consistent with Planck’s hypothesis for the energy exchanges between radiation and the cavity walls in the blackbody radiation problem.

Questions & Answers

For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
Evan Reply
as a free falling object increases speed what is happening to the acceleration
Success Reply
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
Rafi Reply
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
Mohammed Reply
give any fix value to wave length
Rafi
40 cm into change mm
Arhaan Reply
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
sisay Reply
why we have physics
Anil Reply
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
is this a physics forum
Physics Reply
explain l-s coupling
Depk Reply
how can we say dirac equation is also called a relativistic equation in one word
preeti Reply
what is the electronic configration of Al
usman Reply
what's the signeficance of dirac equetion.?
Sibghat Reply
what is the effect of heat on refractive index
Nepal Reply
As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
Wahid
if we heated the ice then the refractive index be change from natural water
Nepal
can someone explain normalization condition
Priyojit Reply
please tell
Swati
yes
Chemist
1 millimeter is How many metres
Darling Reply
1millimeter =0.001metre
Gitanjali

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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