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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 .


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] .


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 .


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.

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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

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40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
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40cm=40.0×10^-2m =400.0×10^-3m =400mm.
this msg is out of mistake. sorry friends​.
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