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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.

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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

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
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bill
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bill
-24m+3+3mÁ^2
Susan
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Amira
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Amira
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Aphelele
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Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
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Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
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Grace Reply
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Grace
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Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
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state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
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Method
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Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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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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