The quantum **harmonic oscillator** is worth our attention for several reasons. One of those reasons is that it provides an instructive example of the application of several of concepts of previous sections and chapters. The potential-energy function 𝑉(𝑥) of harmonic oscillator is a reasonable approximation for other potential energy functions in the vicinity of a potential minimum. This means that the harmonic oscillator, although idealized in this treatment, has a strong connection to several real-world configurations.

We may review the basics of the behavior of a system such as a mass sliding on a frictionless horizontal surface while attached to a spring. In the classical case, this type of system oscillates with constant total energy, continuously exchanging potential and kinetic energy as it moves from the equilibrium position to the "turning points" at which its direction of motion reverses. The potential energy of that object is zero at the equilibrium position and maximum at the turning points at which the spring is maximally compressed or extended. Conversely the kinetic energy is maximum as the object passes through equilibrium and zero when object's velocity passes through zero at the turning points. The object moves fastest at equilibrium and slowest at the turning points, which means that measurements of position taken at random times are more likely to yield results near the turning points, because the object spends more time there.

Several aspects of classical harmonic oscillator are relevant to the quantum case.

One of them is the quadratic form of potential energy, usually written as

(5.56) 𝑉(𝑥) = 1/2 𝑘𝑥^{2},

where 𝑥 represents the distance of the object from the equilibrium position and 𝑘 represents the "spring constant" (the force on the object per unit distance from the equilibrium position). The quadratic relationship between potential energy and position pertain to any restoring force that increases linearly with distance, that is, any force that obeys **Hook's Law**:

(5.57) 𝐹 = -𝑘𝑥,

where the minus sign indicates that the force is always in the direction toward the equilibrium point. We can see the relationship between Hooke's Law and quadratic potential energy by writing force as the negative gradient of the potential energy:

(5.58) 𝐹 = -∂𝑉/∂𝑥 = -∂(1/2 𝑘𝑥)/∂𝑥 = -2𝑘𝑥/2 = -𝑘𝑥,

Another useful result from classical harmonic oscillator is that the motion of the object is sinusoidal, with angular frequency 𝜔 give by

(5.59) 𝜔 = √(𝑘/𝑚),

where 𝑘 represents the spring constant and 𝑚 represents the mass of the object.

We can see a plot of the potential energy of a harmonic oscillator in **Fig. 5.25**. As in the potential wells we can find the energy levels and wavefunctions of the quantum harmonic oscillator by using separation of variables and solving the **TISE** (Eq. 3.40). So we have

(5.60) -(ℏ^{2}/2𝑚) 𝑑^{2}𝜓(𝑥)/𝑑𝑥^{2} + 1/2 𝑘𝑥^{2}𝜓(𝑥) = 𝐸𝜓(𝑥).

It's customary to use angular frequency 𝜔 rather than spring constant 𝑘 in quantum mechanics. Plugging 𝑘 = 𝑚𝜔^{2} into the **TISE** gives

𝑑^{2}𝜓(𝑥)/𝑑𝑥^{2} - 2𝑚/ℏ^{2}[1/2 𝑚𝜔^{2}𝑥^{2}𝜓(𝑥)] = - 2𝑚/ℏ^{2}𝐸𝜓(𝑥) 𝑑^{2}𝜓(𝑥)/𝑑𝑥^{2} - [(𝑚^{2}𝜔^{2}/ℏ^{2})𝑥^{2}𝜓(𝑥)] + (2𝑚/ℏ^{2})𝐸𝜓(𝑥) = 0

__(5.61) 𝑑 ^{2}𝜓(𝑥)/𝑑𝑥^{2} + [(2𝑚/ℏ^{2})𝐸 - (𝑚^{2}𝜔^{2}/ℏ^{2})𝑥^{2}]𝜓(𝑥) = 0.__

There are two different approaches to solve it, called the "analytic" approach and the "algebraic" approach. The former uses power series to solve Eq. 5.61 and the latter approach involves factoring the equation and an operator called a

(5.62) 𝜖 ≡ 𝐸/𝐸

where 𝜔 is the angular frequency of the ground-state wavefunction, and ℏ𝜔/2 turns out to be the ground-state energy of the quantum harmonic oscillator.

The dimensionless version of position is called 𝜉, defined by

(5.63) 𝜉 ≡ 𝑥/𝑥

where √(ℏ/𝑚𝜔) represent is the distance to the classical turning point of a harmonic oscillator for a particle in the ground state. Both 𝐸

Now we may start by solving Wqs. 5.62 and 5.63 for 𝐸 and 𝑥, respectively:

(5.64, 65) 𝐸 = 𝜖𝐸

Next, it's necessary to work on the second-order spatial derivative 𝑑

(5.66) 𝑑𝑥/𝑑𝜉 = √(ℏ/𝑚𝜔), 𝑑𝑥 = √(ℏ/𝑚𝜔) 𝑑𝜉, and so

(5.67) 𝑑𝑥

Plugging in the these expressions for 𝐸, 𝑥, and 𝑑𝑥

𝑑

Differential equation of this type called

If we consider the asymptotic behavior of the solution 𝜓(𝜉) (that is, the behavior at very large or very small values of 𝜉), we can separate out the behavior of the solution in one regime frm that in another, and the differential equation may be simpler to solve in those regimes. For large 𝜉, the Eq. 5.68 looks like this:

(5.69) 𝑑

where the 𝜖 term is negligible relative to the 𝜉

The solutions in this equation for large 𝜉 are

(5.70) 𝜓(𝜉 → ∓∞) = 𝐴𝑒

but for the harmonic oscillator, if 𝜉 → ∓∞ the, then the potential energy 𝑉(𝑥) increases infinitely and wavefunction 𝜓(𝜉) must go zero. So the coefficient 𝐴 must be zero and we can write

(5.71) 𝜓(𝜉) = 𝑓(𝜉)𝑒

where the 𝑓(𝜉) represents a function that determines the behavior of 𝜓(𝜉) at small values of 𝜉, and the coefficient 𝐵 has been absorbed into the function. Plugging Eq. 5.71 into Eq. 5.68:

(5.72) 𝑑

The first spatial dervative:

and taking another spatial derivative gives

Plugging this into Eq. 5.72 gives

𝑒

(5.73) 𝑒

Since the equation must be true for all values of 𝜉, the term in square brackets must equal zero:

This equation is amenable to solution by the power-series approach. First we may write the function 𝑓(𝜉) as a power series:

𝑓(𝜉) = 𝑎

It's customary to start the index at n = 0, so the ground-state wavefunction will be called 𝜓

𝑑𝑓(𝜉)/𝑑𝜉 = ∑

Inserting these into Eq. 5.74 gives

(5.75) ∑

Since at the first summation for which the 𝑛 = 0 and 𝑛 = 1 terms both contribute nothing to the sum, we can simplify the indices by letting 𝑛 → 𝑛+2. So Eq. 5.75 may be written as

∑

∑

which means that the coefficients of 𝜉

(𝑛 + 2)(𝑛 + 1)𝑎

This is a recursion relation that relates any coefficient 𝑎

(5.77) 𝑎

and for large values of 𝑛 this ratio converges to

(5.78) 𝑎

Because 2/𝑛 is exactly what the ratio of the even or odd terms in the power series for the function 𝑒

𝜓(𝜉) = 𝑓(𝜉)𝑒

which cannot be normalized if 𝜉 → ∓∞ without any limit. So we must find what condition can cause this series terminate. Accirding to Eq. 5.77 the coefficient 𝑎

[2𝑛 + (1 - 𝜖)]/(𝑛 + 2)(𝑛 + 1) = 0. so 2𝑛 + (1 - 𝜖) = 0 and

This means that 𝜖 is quantized, taking on discrete values that depend on the value of 𝑛. Denoting this quantization by using subscript 𝑛, so we have

𝐸

These are the allowed value for the energy of a quantum harmonic oscillator. So the ground-state energy is 𝐸

From this point foward the summation index for wavefunction will be labeled as 𝑚 like this:

(5.81) 𝑓(𝜉) = ∑

It's helpful to separate this into two series - one with the even powers and the other with the odd power:

(5.82) 𝑓(𝜉) = ∑

We know that the summation terminates whenever the energy parameter 𝜖

which means the series terminates when 𝑚 = 𝑛. When 𝑚 = 𝑛 = 0 (so all even term with 𝑚 > 𝑛 are zero), To ensure all the odd terms zero we must set 𝑎

Now consider the first excited (𝑛 = 1) case (so all odd term with 𝑚 > 𝑛 are zero). And to make sure that the even series doesn't blow up we must set 𝑎

For the second excited state 𝑛 = 2, the energy parameter 𝜖

𝑎

𝑎

For the third excited states 𝑛 = 3, 𝜖

𝑎

For 𝑚 = 3 and 𝑛 = 3

𝑎

For the fourth excited states 𝑛 = 4, 𝜖

𝑎

𝑎

Finally for 𝑚 = 4 and 𝑛 = 4

𝑎

For the fifth excited states 𝑛 = 5, 𝜖

𝑎

For 𝑚 = 3 and 𝑛 = 5

𝑎

Lastly for 𝑚 = 5 and 𝑛 = 5

𝑎

Now in order to relate to the Hermite polynomials mentioned earlier we may arrange the 𝑓

𝑓

𝑓

𝑓

𝑓

𝑓

𝑓

Referring to mathematical handbook we can find these expression:

𝐻

𝐻

So calling the constant factors involving 𝑎

where constants will be determined later by normalizing the wavefunction 𝜓

The quantum harmonic oscillator wavefunctions are comprised of the product of Hermite polynomials 𝐻

To accomplish that normalization, set the integrated probability density over all space to unity. For 𝜓

and Eq. 5.66 relates 𝑑𝑥 to 𝑑𝜉, so

(5.92) ∫

(5.93) √(ℏ/𝑚𝜔) ∫

(5.94) √(ℏ/𝑚𝜔) ∣𝐴

But mathematicans works on weber equation and Hermite polynomials have given us a very neat integral identity:

Inserting this expression into Eq. 5.94 yields

√(ℏ/𝑚𝜔) ∣𝐴

and taking the square root gives the normalization constant 𝐴

(5.96) ∣𝐴

With 𝐴

These wavefntions 𝜓

the probability densities 𝑃

We should also bear in mind that these wavefunctions are the eigenfunctions of the Hamiltonian operator achieved by separation of variables, so they represent stationary states for which the expectation values of observables such as position, momentum, and energy do not change over time. To determine the behavior of particles in other states, we must include the time function 𝑇(t), shich makes 𝜓

Knowing the allowed energy levels 𝐸

The algebraic approach involves a dimensionless version of Schrödinger equation using dimensionless versions of position and momentum operators 𝑋^ and 𝑃^. Starting by defining a momentum reference value 𝑝_{𝑟𝑒𝑓} using

𝑝_{𝑟𝑒𝑓}^{2}/2𝑚 = 𝐸_{𝑟𝑒𝑓} = 1/2 ℏ𝜔 𝑝_{𝑟𝑒𝑓} = √(𝑚ℏ𝜔),

we produce a dimensionless version of momentum called 𝓟 as

__(5.99, 100) 𝓟 = 𝑝/𝑝 _{𝑟𝑒𝑓} = 𝑝/√(𝑚ℏ𝜔)__ or

To produce a dimensionless version of the

(3.40) -(ℏ

which can be written in terms of the momentum operator 𝑃^ and position operator 𝑋^ for quantum harmonic oscillator as

[𝑃^

Using dimensionless operators 𝓟^ = 𝑃^/𝑝

[[𝓟^ (𝑝

To solve this equation we begin with two new operators, which are combinations of the dimensionless position and momentum operators. The first one is

(5.101) 𝑎^

(5.102) 𝑎^ = 1/√2 (𝜉^ + 𝑖𝓟^).

Their product is

𝑎^

where [𝜉^, 𝓟^] represents the commutator of the operators 𝜉^ and 𝓟^. Thhis makes the product 𝑎^

(5.105) 𝑎^

This can be simplified by writing the commutator in terms of 𝑋^ and 𝑃^.

𝑖[𝜉^, 𝓟^] = 𝑖[𝑋^/𝑥

Recall from Chapter 4 that the canonical communication relation Eq. 4.68 tells us that [𝑋^, 𝑃^] = 𝑖ℏ, which means

(5.106) 𝑖[𝜉^, 𝓟^] = 𝑖/ℏ [𝑖ℏ] = -1.

Plugging this into Eq. 5.105 gives

(5.107) 𝑎^

This makes the

[𝓟^

Plugging the definition of 𝑎^

(5.108) 2[{1/√2 (𝜉^ - 𝑖𝓟^)}{1/√2 (𝜉^ + 𝑖𝓟^)}][𝜓(𝜉)] = (𝜖 - 1)[𝜓(𝜉)] or

One way that this equation can be satisfied is for 𝜖 = 1 while 1/√2 (𝜉^ + 𝑖𝓟^)[𝜓(𝜉)] = 0.

So if 𝜖 = 1, the total energy is 𝐸 = 𝜖𝐸

in agreement which 𝐸

𝑃^ = -𝑖ℏ 𝑑/𝑑𝑥 = -𝑖ℏ 𝑑/[√(ℏ/𝑚𝜔) 𝑑𝜉] = -𝑖√(𝑚ℏ𝜔) 𝑑/𝑑𝜉 and 𝓟^ = 𝑃^/𝑃

(5.109) (𝜉^ + 𝑖𝓟^)[𝜓(𝜉)] = [𝜉 + 𝑖(-𝑖𝑑/𝑑𝜉)][𝜓(𝜉)] = 0 [𝜉 + 𝑑/𝑑𝜉][𝜓(𝜉)] = 0

The solution is this equation is 𝜓(𝜉) = 𝐴𝑒

Hence the algebraic approach gives the lowest-energy eigenfunction

exactly as found for 𝜓

The operator 𝑎^

𝑎^

For this reason 𝑎^

Also the operator 𝑎^ performs the complementary function, producinga wavefunction proportional to the wavefunction with the quantum number lowered by one. Hence 𝑎^ is called a "lowering operator", and for the lowering oprator, the constant of proportionality is √𝑛. Thus

This is why 𝑎^

^{※} attention: some rigorous derivation might be required