7.5 The quantum harmonic oscillator  (Page 2/4)

 Page 2 / 4

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 $k=m\text{ }{\omega }^{\text{ }2}$ 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,

$U\left(x\right)=\frac{1}{2}m{\omega }^{\text{ }2}{x}^{\text{ }2}.$

Combining this expression with the time-independent Schrӧdinger equation gives

$-\frac{\hslash }{2m}\phantom{\rule{0.2em}{0ex}}\frac{{d}^{\text{ }2}\psi \left(x\right)}{d\text{ }{x}^{2}}+\frac{1}{2}m{\omega }^{\text{ }2}{x}^{\text{ }2}\psi \left(x\right)=E\psi \left(x\right).$

To solve [link] —that is, to find the allowed energies E and their corresponding wave functions $\psi \left(x\right)$ —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 $|\psi \left(x\right){|}^{\text{ }2}$ must be finite when integrated over the entire range of x from $\text{−}\infty$ to $+\infty$ . 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

${E}_{n}=\left(n+\frac{1}{2}\right)\hslash \omega =\frac{2n+1}{2}\hslash \omega ,\phantom{\rule{0.5em}{0ex}}n=0,1,2,3,...$

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

${\psi }_{n}\left(x\right)={N}_{n}{e}^{\text{ }-{\beta }^{\text{ }2}{x}^{\text{ }2}\text{/}2}{H}_{n}\left(\beta x\right),\phantom{\rule{0.5em}{0ex}}n=0,1,2,3,...$

where $\beta =\sqrt{m\text{ }\text{ }\omega \text{/}\hslash }$ , ${N}_{n}$ is the normalization constant, and ${H}_{n}\left(y\right)$ is a polynomial of degree n called a Hermite polynomial . The first four Hermite polynomials are

$\begin{array}{c}{H}_{0}\left(y\right)=1\hfill \\ {H}_{1}\left(y\right)=2y\hfill \\ {H}_{2}\left(y\right)=4{y}^{2}-2\hfill \\ {H}_{3}\left(y\right)=8{y}^{3}-12y.\hfill \end{array}$

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 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 $E=k{x}^{2}\text{/}2=m\text{ }{\omega }^{2}{A}^{2}\text{/}2$ equal to ${E}_{n}$ given by [link] .

Solution

We obtain

${E}_{n}=m\text{ }{\omega }^{2}{A}_{n}^{\text{ }2}\text{/}2\phantom{\rule{0.5em}{0ex}}⇒\phantom{\rule{0.5em}{0ex}}{A}_{n}=\sqrt{\frac{2}{m\text{ }{\omega }^{\text{ }2}}{E}_{n}}=\sqrt{\frac{2}{m\text{ }{\omega }^{\text{ }2}}\phantom{\rule{0.2em}{0ex}}\frac{2n+1}{2}\hslash \omega }=\sqrt{\left(2n+1\right)\frac{\hslash }{m\text{ }\omega }}.$

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.

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,

$\text{Δ}E={E}_{n+1}-{E}_{n}=\frac{2\left(n+1\right)+1}{2}\hslash \omega -\frac{2n+1}{2}\hslash \omega =\hslash \omega =h\text{ }f.$

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.

what is cathodic protection
its just a technique used for the protection of a metal from corrosion by making it cathode of an electrochemical cell.
akif
what is interferometer
Show that n1Sino1=n2Sino2
what's propagation
is it in context of waves?
Edgar
It is the manner of motion of the energy whether mechanical(requiring elastic medium)or electromagnetic(non interference with medium)
Edgar
determine displacement cat any time t for a body of mass 2kg under a time varrying force ft=bt³+csinkt
A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
how does colour appear in thin films
hii
Sonu
hao
Naorem
hello
Naorem
hiiiiii
ram
🎓📖
Deepika
yaaa ☺
Deepika
ok
Naorem
hii
PALAK
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
derivation of lateral shieft
hi
Imran
total binding energy of ionic crystal at equilibrium is
How does, ray of light coming form focus, behaves in concave mirror after refraction?
Refraction does not occur in concave mirror. If refraction occurs then I don't know about this.
Sushant
What is motion
Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
and what's time? is time everywhere same
Chemist
No
Sushant
how can u say that
Chemist
do u know about black hole
Chemist
Not so more
Sushant
DHEERAJ
These substance create harmful radiation like alpha particle radiation, beta particle radiation, gamma particle radiation
Sushant
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
DHEERAJ
explain cavendish experiment to determine the value of gravitational concept.
Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
Triio
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?   By      By By