quantum harmonic oscillator - grandinetti.org · kinetic and potential energy operators harmonic...

Quantum Harmonic OscillatorChapter 13

P. J. Grandinetti

Chem. 4300

March 7, 2019

P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator March 7, 2019 1 / 25

Kinetic and Potential Energy OperatorsHarmonic oscillator well is model for small vibrations of atoms about bond as well as othersystems in physics and chemistry.

re

m2m1

Model bond as spring that acts as restoring force whenever two atoms are squeezed togetheror pulled away from equilibrium position.Kinetic and potential energy operators are

K =p2

2𝜇and V(x) = 1

2𝜅fx2

x = r − re and 𝜇 is reduced mass1𝜇= 1

m1+ 1

m2


Force constants for selected diatomic moleculesBond 𝜅f/(N/m) 𝜇∕10−28 kg ��/cm−1 Bond length/pmH2 570 8.367664 4401 74.1D2 527 16.72247 2990 74.1

H35Cl 478 16.26652 2886 127.5H79Br 408 16.52430 2630 141.4H127I 291 16.60347 2230 160.9

35Cl35Cl 319 290.3357 554 198.879Br79Br 240 655.2349 323 228.4127I127I 170 1053.649 213 266.716O16O 1142 132.8009 1556 120.714N14N 2243 116.2633 2331 109.412C16O 1857 113.8500 2143 112.814N16O 1550 123.9830 1876 115.1

23Na23Na 17 190.8770 158 307.823Na35Cl 117 230.3282 378 236.139K35Cl 84 306.0237 278 266.7


Schrödinger equation for harmonic oscillatorInsert Kinetic and Potential Energy operators

H𝜓(x) =(K + V

)𝜓(x) = −ℏ

2

2𝜇d2𝜓(x)

dx2 + 12𝜅fx2𝜓(x) = E𝜓(x)

Defining k2 =2𝜇Eℏ2 and 𝛼4 =

𝜇𝜅fℏ2 and substituting above, the Schrödinger equation becomes

d2𝜓(x)dx2 + (k2 − 𝛼4x2)𝜓(x) = 0

This ODE doesn’t have simple solutions like particle in infinite well.Harmonic oscillator potential becomes infinitely high as x goes to ∞Wave function is continuous and single valued over x = −∞ to ∞.P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator March 7, 2019 4 / 25

Schrödinger equation for harmonic oscillatorNotice at large values of |x| one can approximate ODE:

d2𝜓(x)dx2 + (k2 − 𝛼4x2)𝜓(x) = 0 at large |x| becomes d2𝜓(x)

dx2 − 𝛼4x2𝜓(x) ≈ 0

With solutions

𝜓(x) ∼ Ae±𝛼2x2∕2 but only accept 𝜓(x) ∼ Ae−𝛼

2x2∕2 as physical

Don’t forget, these are NOT solutions for all x — only large |x|However, in light of this asymptotic solution we further define

𝜉 = 𝛼x and 𝜓(x)dx = 𝜒(𝜉)d𝜉

to transform Schrödinger equation into

d2𝜒(𝜉)d𝜉2 +

(k2

𝛼2 − 𝜉2)𝜒(𝜉) = 0


Schrödinger equation for harmonic oscillatorWe expect solution to this ODE to have asymptotic limits

lim|𝜉|→∞𝜒(𝜉) = Ae−𝜉

2∕2

We propose a general solution𝜒(𝜉) = AH(𝜉)e−𝜉2∕2

Substituting into ODE gives

d2H(𝜉)d𝜉2 − 2𝜉

dH(𝜉)d𝜉

+ 2nH(𝜉) = 0 where n = k2

𝛼2 − 1

9 out of 10 math majors recognize this as Hermite’s differential equationIts solutions are the Hermite polynomials.Only solutions with n = 0, 1, 2,… are physically acceptable for harmonic oscillator.


First seven Hermite polynomials and approximate roots.

n Hn(y) Roots0 1

1 2y 0

2 4y2 − 2 ±0.707107

3 8y3 − 12y 0 ±1.224745

4 16y4 − 48y2 + 12 ±0.5246476 ±1.650680

5 32y5 + 160y3 + 120y 0 ±0.958572 ±2.020183

6 64y6 − 480y4 + 720y2 − 120 ±0.436077 ±1.335850 ±2.350605

7 128y7 − 1344y5 + 3360y3 − 1680y 0 ±0.816288 ±1.673552 ±2.65196


Harmonic Oscillator Wave FunctionNormalized solutions to Schrödinger equation for harmonic oscillator are

𝜒n(𝜉) = AnHn(𝜉)e−𝜉2∕2, where An ≡ 1√

2nn!𝜋1∕2

Condition that n only be integers leads to harmonic oscillator energy levels

En = ℏ𝜔0(n + 1∕2), n = 0, 1, 2,… where 𝜔0 =√𝜅f∕𝜇

Energy levels are equally spaced at intervals of ℏ𝜔0.

In spectroscopy vibrational frequencies are given in terms of the spectroscopic wavenumber,

�� =𝜔0

2𝜋c0


Harmonic Oscillator Energy Levels

-2 -1 10 2

Ener

gy

Ground state with n = 0 has zero point energy of 12ℏ𝜔0.


Maximum displacement of classical harmonic oscillator

-1.0 -0.5 0.0 0.5 1.0

2

0

4

6

8

10

Recall: Maximum displacement ofclassical harmonic oscillator in termsof energy

xmax = 1𝜔0

√2Em

Combined with En = ℏ𝜔0(n + 1∕2) weobtain corresponding xmax : classicalturning point for each oscillator state

𝜉maxn = 𝛼xmax

n =√

2n + 1


Harmonic Oscillator Wave Functions

-4 -2 0 2 4-1.0

-0.5

0.5

0.0

1.0

-4 -2 0 2 4

-1.0

-0.5

0.5

0.0

1.0-1.0

-0.5

0.5

0.0

1.0-1.0

-0.5

0.5

0.0

1.0-1.0

-0.5

0.5

0.0

1.0

-1.0

-0.5

0.5

0.0

1.0-1.0

-0.5

0.5

0.0

1.0-1.0

-0.5

0.5

0.0

1.0-1.0

-0.5

0.5

0.0

1.0-1.0

-0.5

0.5

0.0

1.0

nth oscillator state has n nodes.

▶ approximate roots of the Hermite polynomials inearlier slide.

Note similarities and differences with 1D infinite well.

▶ nth oscillator state corresponds to (n + 1)th infinitewell state.

Tiny horizontal arrows represent classical oscillatordisplacement range for same energy.

▶ Gray represents classically excluded region,𝜉 > 𝜉max

n▶ Finite potential leads to wave function penetration

into classically excluded region, i.e., tunneling.


Classical Oscillator Turning PointsAs n increases probability density function approaches that of classical harmonic oscillatordisplacement probability (gray line) shown with the n = 112 oscillator

-15 -10 -5 0 5 10 15

0.2

0.0

0.4

0.6

0.8

1.0classical limit


Web App - 1D QM simulation of single bound particleLink here: 1D Quantum Wells

Web app instructions:Solves Schrödinger equation and shows solutions.Default is infinite square well (zero everywhere inside, infinite at edges).Top is graph of potential and horizontal lines show energy levels.Below is probability distribution of particle’s position, oscillating back and forth in a combinationof two states.Below particle’s position is graph of momentum.Bottom set of phasors show magnitude and phase of lower-energy states.To view state, move mouse over energy level on potential graph.To select a single state, click on it.Select single state by picking one phasor at bottom and double-clicking.Click on phasor and drag value to modify magnitude and phase to create combination of states.Select different potentials from Setup menu at top right.P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator March 7, 2019 13 / 25

http://www.falstad.com/qm1d/

Harmonic Oscillator Wave Functions in terms of x

𝜒n(𝜉) = AnHn(𝜉)e−𝜉2∕2, where An ≡ 1√

2nn!𝜋1∕2

Rewriting in terms of x gives

𝜓n(x) = NnHn(𝛼x)e−(𝛼x)2∕2 where Nn ≡ √𝛼An =

√𝛼

2nn!𝜋1∕2

Lowest energy wave function has form of Gaussian function

𝜓0(x) =√

𝛼𝜋1∕2

e−(𝛼x)2∕2

and probability distribution that is Gaussian

𝜓∗0 (x)𝜓0(x) =

𝛼√𝜋

e−(𝛼x)2

with standard deviation of Δx = 1∕(𝛼√

2).P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator March 7, 2019 14 / 25

(Avoiding) Integrals involving Hermite polynomialsHermite polynomials with even n are even functions while those with odd n are odd functions.Keep this in mind when evaluating integrals.

ExampleCalculate ⟨x⟩ for harmonic oscillator wave function.

Starting with integral

⟨x⟩ = ∫∞

−∞𝜓∗

n (x)x𝜓n(x)dx = ∫∞

−∞x𝜓∗

n (x)𝜓n(x)dx,

Note that 𝜓n are real, so 𝜓∗n (x) = 𝜓n(x).

Since 𝜓n(x) is either even or odd depending on whether n, then product,𝜓∗

n (x)𝜓n(x) = 𝜓n(x)𝜓n(x) is always even.Therefore x𝜓2

n (x) is always odd and we obtain ⟨x⟩ = 0.P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator March 7, 2019 15 / 25

(Avoiding) Integrals involving Hermite polynomials

ExampleCalculate ⟨p⟩ for harmonic oscillator wave function.

Calculate ⟨p⟩ = ∫∞

−∞𝜓∗

n (x)p𝜓n(x)dx = −iℏ∫∞

−∞𝜓n(x)

𝜕𝜓n(x)𝜕x

dx.

The derivative of an odd function is even

The derivative of an even function is odd

Integrand is product of even and odd functions

Thus, integrand is odd function

and therefore integral is zero, that is, ⟨p⟩ = 0.


(Avoiding) Integrals involving Hermite polynomials

Another useful result for avoiding integrals involving the Hermite polynomials is

AmAn ∫∞

−∞Hm(𝜉)Hn(𝜉)e−𝜉

2d𝜉 = 𝛿m,n

Hermite polynomials also obey two useful recursion relations

Hn+1(𝜉) − 2𝜉Hn(𝜉) + 2nHn−1(𝜉) = 0

anddHn(𝜉)

d𝜉= 2n𝜉Hn−1(𝜉)


ExampleUsing the recursion relations above show that ⟨x⟩ = 0.

Since 𝜉 = 𝛼x this is identical to ⟨𝜉⟩ = 0.Start with ⟨𝜉⟩ = ∫

∞

−∞(Hne−𝜉

2∕2)𝜉(Hne−𝜉2∕2)d𝜉 = ∫

∞

−∞Hn𝜉Hne−𝜉

2d𝜉

Using recursion relation we find𝜉Hn = 1

2Hn+1 + nHn−1,

Substituting back into our integral we obtain

⟨𝜉⟩ = ∫∞

−∞Hn

(12

Hn+1 + nHn−1

)e−𝜉

2d𝜉

= 12 ∫

∞

−∞HnHn+1e−𝜉

2d𝜉 + n∫

∞

−∞HnHn−1e−𝜉

2d𝜉

Since n ≠ n ± 1, both integrals are zero and thus ⟨𝜉⟩ = 0.P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator March 7, 2019 18 / 25

Creation and Annihilation Operators

Nearly every potential well can be approximated as harmonic oscillator

Describes situations from molecular vibration to nuclear structure.

Quantum field theory starting point–basis of quantum theory of light.

Consider harmonic oscillator Hamiltonian written in form

= −ℏ2

2𝜇d2

dx2 + 12𝜇𝜔2

0x2

We now define two non-hermitian operators

a+ =√𝜇𝜔02ℏ

(x − i

p𝜇𝜔0

)and a− =

√𝜇𝜔02ℏ

(x + i

p𝜇𝜔0

)


Creation and Annihilation OperatorsCalculate product a+a−

a+a− =𝜇𝜔02ℏ

(x2 +

ixp − ipx𝜇𝜔0

+p2

𝜇2𝜔2

)= 1ℏ𝜔0

(p2

2𝜇+𝜇𝜔2

0x2

2

)⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟

+ i2ℏ

(xp − px)⏟⏞⏟⏞⏟[x,p]=iℏ

Recognizing the commutator in the last term we obtain

a+a− = ℏ𝜔0

+ i2ℏ

(iℏ) = ℏ𝜔0

− 12

and obtain = ℏ𝜔0

(a+a− + 1

2

)Both and a+a− have same 𝜓n(x) as eigenstates. We further define the number operator as

N = a+a− where N𝜓n(x) = n𝜓n(x)

and = ℏ𝜔0

(N + 1

2

)P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator March 7, 2019 20 / 25

Creation and Annihilation OperatorsLook at effect of applying a+ on eigenstates of .

If 𝜓 ′ = a+𝜓n then 𝜓 ′ = (a+𝜓n) = ℏ𝜔0

(a+a− + 1

2

)⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

(a+𝜓n)

and𝜓 ′ = ℏ𝜔0

(a+a−a+ + 1

2a+

)𝜓n = a+ℏ𝜔0

(a−a+ + 1

2

)⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟

ℏ𝜔0

+1

𝜓n = a+( + ℏ𝜔0

)𝜓n

Then we can write

𝜓 ′ = (a+𝜓n) = a+( + ℏ𝜔0

)𝜓n = a+

(E + ℏ𝜔0

)𝜓n =

(E + ℏ𝜔0

) (a+𝜓n

)We just learned that energy of a+𝜓n is ℏ𝜔0 higher than E, the energy of 𝜓n.Effect of a+ on 𝜓n is to change it into 𝜓n+1 with En+1 = En + ℏ𝜔0


Creation (Raising) and Annihilation (Lowering) operators

Similarly, one can show that a− applied to 𝜓n turns it into 𝜓n−1.

a+ and a− are called Creation and Annihilation operators, respectively.

a+ and a− are also called Raising and Lowering operators, respectively.

Without proof, coefficients that maintain normalization of the wave functions whenapplying a± are

a+𝜓n =√(n + 1)𝜓n+1 and a−𝜓n =

√n𝜓n−1


Fun trick with Creation and Annihilation operatorsAs we can’t go any lower than n = 0 we must have a−𝜓0 = 0We can use this to determine 𝜓0Since

a−𝜓0 =√𝜇𝜔02ℏ

(x + i

p𝜇𝜔0

)𝜓0 = 0

Expanding and rearranging givesd𝜓0dx

= −𝜇𝜔0ℏ

x𝜓0

Integrating

∫d𝜓0𝜓0

= −𝜇𝜔0ℏ ∫ xdx

gives𝜓0 = A0e(𝜇𝜔0∕ℏ)x2∕2

Recalling (𝜇𝜔0∕ℏ) =√𝜇𝜅f∕ℏ2 = 𝛼2

gives𝜓0 = A0e−𝛼

2x2∕2 = A0e−𝜉2∕2


Fun trick with Creation and Annihilation operators

Normalizing 𝜓0 with integral

∫∞

−∞𝜓∗

0 (x)𝜓0(x)dx = ∫∞

−∞|A0|2e−𝛼

2x2dx = 1

gives |A0|2 = 𝛼∕√𝜋

so we have𝜓0 = 𝛼1∕2

𝜋1∕4e−𝛼

2x2∕2

From 𝜓0 we can use a+ to generate all higher energy eigenstates.


ExampleUse a+ to generate 𝜓1(x) from 𝜓0(x).

𝜓1 = a+𝜓0 =

[𝛼√2

(x − i

p𝜇𝜔

)]𝛼1∕2

𝜋1∕4e−𝛼

2x2∕2 = 𝛼3∕2√2𝜋1∕4

[1 + ℏ𝛼2

𝜇𝜔

]xe−𝛼

2x2∕2

Check thatℏ𝛼2

𝜇𝜔0= ℏ𝜇𝜔0

(𝜇𝜅fℏ2

)1∕2=

(𝜇𝜅f)1∕2

𝜇𝜔0=

(𝜇𝜅f)1∕2

𝜇

(𝜇𝜅f

)1∕2= 1

Thus we obtain𝜓1(x) =

𝛼3∕2√2𝜋1∕4

2xe−𝛼2x2∕2 = 𝛼1∕2√

2𝜋1∕4⏟⏞⏟⏞⏟

A1

(2𝛼x)⏟⏟⏟H1(𝛼x)

e−𝛼2x2∕2

Although tedious you can find all Hermite polynomials this way.


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