Quantum Mechanics 1 Lecture 5: The Harmonic Oscillator

Quantum Mechanics 1

Lecture 5: The Harmonic Oscillator

Massimiliano Grazzini

University of Zurich

Outline

Analytic solution

Algebraic solution

Coherent states

- Hermite polynomials

- Correspondence with classical theory

- Zero point energy

Harmonic Oscillator

The Hamiltonian of the classical harmonic oscillator with mass and

frequency is

H = p²

2m + 12 mω²x² V(x)

x

which can be rewritten as

Corresponds to a “force”

with _{ω = k/m} ^{F = − kx}

The classical equation of motion is

and its generic solution is

x(t) = C cos(ωt + c)

while

is

p(t) = mx(t) =· − mωC sin(ωt + c) T + V = 1

2 mω²C² (cos²(ωt + c) + sin²(ωt + c)) = 1

2 mω²C²

and the energy is

··x(t) = − ω²x(t)

Harmonic Oscillator

The Hamiltonian of the classical harmonic oscillator with mass and

frequency is

H = p²

2m + 12 mω²x² V(x)

x

Since the potential is not bounded from above, there is only a discrete spectrum with

The time-independent Schrödinger equation reads

− ℏ²

2m ψ(x)′′+ 1

2 mω²x²ψ(x) = Eψ(x)

There are two methods to solve this equation, and we will discuss them in turn

Corresponds to a “force”

with _{ω = k/m} ^{F = − kx}

Analytic solution

We start from the time-independent Schrödinger equation

− ℏ²

2m ψ′′(x) + 1

2 mω²x²ψ(x) = Eψ(x)

and we define the dimensionless variable

characteristic length of the problem

x₀ = ℏ mω

The equation becomes

− ℏ²

2m mω

ℏ ψ′′(ξ) + 1

2 mω² ℏ

mω ξ²ψ(ξ) = Eψ(ξ) ψ′′(ξ) − ξ²ψ(ξ) + 2E

ℏω ψ(ξ) = 0

In the limit

the term proportional to can be neglected and we get

This equation is approximately solved by

ψ(ξ) = Ae^−ξ2/2 + Be^ξ2/2

We have to drop this term in order to obtain _{ψ ∈ L}²_[−∞,∞]

Indeed and

as is large

d

dξ exp{±ξ²/2} = ± ξexp{±ξ²/2}

d²

dξ² exp{±ξ²/2} = ξ² exp{±ξ²/2} ± exp{±ξ²/2} ∼ ξ² exp{±ξ²/2} ξ

We can then look for a solution in the form

Plugging this ansatz in the differential equation we find

ψ′(ξ) = exp{−ξ²/2}(H′(ξ) − ξH(ξ))

ψ′′(ξ) = exp{−ξ²/2}((−ξ)(H′(ξ) − ξH(ξ)) + (H′′(ξ) − ξH′(ξ) − H(ξ))) H′′(ξ) − 2ξH′(ξ) + (λ − 1) H(ξ) = 0

This is the Hermite equation and can be solved by using the Frobenius method (recall MMP1 !)

H(ξ) = ξ^s∑^∞

k=0

c_kξ^k = c₀ξ^s + c₁ξ^s+1 + c₂ξ^s+2 + . . . H′(ξ) = c₀sξ^s−1 + c₁(s + 1)ξ^s + c₂(s + 2)ξ^s+1 + . . .

H′′(ξ) = c₀s(s − 1)ξ^s−2 + c₁(s + 1)sξ^s−1 + c₂(s + 2)(s + 1)ξ^s . . . . c₀ ≠ 0 s ≥ 0 λ = 2E

ℏω

Plugging these expressions in the differential equation the coefficients of each power of have to vanish

c₀s(s − 1)ξ^s−2 + c₁(s + 1)sξ^s−1 + c₂(s + 2)(s + 1)ξ^s . . . .

−2c₀sξ^s − 2c₁(s + 1)ξ^s+1 − 2c₂(s + 2)ξ^s+2 + . . . . . +(λ − 1)(c₀ξ^s + c₁ξ^s+1 + c₂ξ^s+2 + . . . ) = 0

s(s − 1)c₀ = 0 s(s + 1)c₁ = 0

(s + 1)(s + 2)c₂ − (2s + 1 − λ)c₀ = 0

……….

(s + k + 2)(s + k + 1)c_k+2 − (2s + 2k + 1 − λ)c_k = 0

Since

from the first equation we must have either

or

The second equation tells us that either

or

or both

The third equation gives us in terms of and so on

Since the Hamiltonian

is invariant under the parity transformation

V(−x) = V(x)

P : x → − x

we can choose eigenfunctions of the Hamiltonian with definite parity

− 2ξH′ (ξ) H′ ′ (ξ)

+(λ − 1)H (ξ) = 0

In order for the function to be well behaved at infinity the series has to terminate beyond a given

2/2}

k

This requires the condition

to be integer

E = ℏω (n + 1 2 )

The functions

are the Hermite polynomials encountered in MMP1

, ,

…..

They form a basis of with the

weight

2[−∞, ∞]

exp{−ξ²/2}

n = 0,1,2....

H(ξ) = ξ^s∑^∞

k=0

c_kξ^k

Since we have seen that or we can

consistently set all odd coefficients to zero

: even

s = 0 H(x) = c₀ + c₂x² + . . . . .

: odd

s = 1 H(x) = x(c₀ + c₂x² + . . . )

At large the relation between and is and it is at least as bad as for the expansion of

thus spoiling the boundary condition we have set

c_k+2/c_k ∼ 2/k

This means

or

ℏω − 1 = 2n

(1)

Correspondence with classical theory

The classical motion satisfies

We can compare the quantum mechanical results with what happens for a classical oscillator

x = q₀ sinωt

and the energy is

2 mω²q₀²

We can define a classical position probability

T

Where is the amount of time spent within and is the oscillation period

We have

Comparing with the definition of

we find

2πq₀ 1 − (x/q₀)²

To compare the classical and quantum results we have to compute from

For example:

E₁₀ = (21/2)ℏω

q₀ = 3ℏω

mω² = 3x₀ q = 21ℏω = 21x

or

2πq₀ 1 − (x/q₀)² = dt T

Correspondence with classical theory

W_class W_class

−q₀/x_{0 q0/x0} x/x₀

|ψ₁|²

|ψ₁₀|²

x/x₀

q₀/x₀

−q₀/x₀

The classical motion is confined within the range while the quantum probability distribution, although exponentially suppressed, extends to the entire real axis

[−q₀/x₀, q₀/x₀]

In the large limit the quantum probability distribution approaches the shape of the classical distribution, despite the rapid oscillations

The classical and quantum probability distributions look quite different

Algebraic solution

We want to represent the Hamiltonian as a square of an operator: we thus define using the dimensionless variable the non-hermitian

operators

a = 1

2 (ξ + ∂_ξ) = 1

2 ( x

x₀ + x₀∂_x

) = ωmx + ip

2mωℏ a^† = ωmx − ip

2mωℏ

Inverting these relations we find

x = ℏ

2ωm (a + a^†) p = − i ℏωm

2 (a − a^†)

By using

we find

[a, a^†] = 12mωℏ (ωm(−i)[x, p] + imω[p, x]) = 1

Replacing the expressions for and in the Hamiltonian we find

H = − ℏωm

2 (a² + (a^†)² − aa^† − a^†a) 1

2m + ℏ

2mω (a² + (a^†)² + aa^† + a^†a) 1

2 mω²

p² x²

Algebraic solution

We want to represent the Hamiltonian as a square of an operator: we thus define using the dimensionless variable the non-hermitian

operators

a = 1

2 (ξ + ∂_ξ) = 1

2 ( x

x₀ + x₀∂_x

) = ωmx + ip

2mωℏ a^† = ωmx − ip

2mωℏ

Inverting these relations we find

x = ℏ

2ωm (a + a^†) p = − i ℏωm

2 (a − a^†)

By using

we find

[a, a^†] = 12mωℏ (ωm(−i)[x, p] + imω[p, x]) = 1

Replacing the expressions for and in the Hamiltonian we find

H = 1

2ℏω (aa^† + a^†a) = ℏω (a^†a + 12)

using _[a,a^†_{] = 1}

The problem is thus reduced to the one of finding the eigenvalues of the hermitian operator

N = a^† a

Let us denote with the eigenfunction of with eigenvalue

Nψ_ν = νψ_ν

We have

The lowest possible eigenvalue is thus

If

from the previous equation we have

( x

x₀ + x₀∂_x

) ψ₀ = 0

Therefore the function fulfils the first order differential

equation

The normalised solution of this equation is

2 ( x x₀)

2

} (2)

The remaining eigenfunctions can be obtained by exploiting the commutation relations

[N, a^†] = a^†aa^† − a^†a^†a = a^† [N, a] = a^†aa − aa^†a = − a

which stem from

We can use these relations to evaluate

and

We have

So

is an eigenvector of with eigenvalue

We have

So

is an eigenvector of with eigenvalue

For these reasons the operators and are called raising and lowering operators

We also have

So if and

are normalised to unity we have

Analogously we can show that

Starting from we can thus construct and infinite set of states

ψ_n = 1

n a^†ψ_n−1 = 1

n! (a^†)ⁿψ₀ ^N

Is there anything else ?

Are there eigenstates of with non-integer eigenvalues ?

Let us suppose that there exist an eigenvector with eigenvalue with

integer and

n 0 < α < 1

and

N(aⁿψ_ν) = α(aⁿψ_ν) N(aⁿ⁺¹ψ_ν) = (α − 1)(aⁿ⁺¹ψ_ν)

But (recall Eq. ) the norm of is non-vanishing: we would then have an eigenvector of with finite norm and negative eigenvalue, contradicting the positivity of the eigenvalues shown before

(2) aⁿ⁺¹ψ_ν

N

Zero-point energy

The infinite sequence of levels has the equal spacing

postulated by Planck, which is in agreement with what one would find with the old quantum theory (recall Lecture 1)

E_n = ℏω(n + 1/2)

The finite value of the ground state energy is called zero-point

energy is instead a genuine characteristic of quantum mechanics and is related to the uncertainty principle, as it happens for the lowest energy level for the square well with perfectly rigid walls

E₀ = ℏω/2

Classically the lowest level of the harmonic oscillator is

To see this we compute the product

For the average values we have

⟨p⟩ = ⟨ψ_n|p|ψ_n⟩ ∼ ⟨ψ_n|a − a^†|ψ_n⟩ = 0

For the position and momentum uncertainties we get

(Δx)² = ⟨x²⟩ = ℏ

2mω ⟨ψ_n|a² + aa^† + a^†a + (a^†)²|ψ_n⟩ (Δp)² = ⟨p²⟩ = − ℏωm

2 ⟨ψ_n|a² − aa^† − a^†a + (a^†)²|ψ_n⟩

But we have seen that

and

We find

and

x₀² (n + 1/2)

recall that _{x0 = ℏ/mω}

Then

and

And for the uncertainty product we obtain

This expression is minimum for the ground state

2 ( x x₀ )

2

}

The wave function is not concentrated at but it has a spacial extension

of order and a corresponding uncertainty

We can also derive a condition on the zero-point energy by using the uncertainty relation

ΔxΔp ≥ ℏ/2

Since

this relation implies

But

2m + 12 mω²⟨x²⟩ ≥ ⟨p²⟩

2m + 12 mω² ℏ²

4 1

⟨p²⟩

Differentiating with respect to

and setting to zero we obtain

1

2m − mω²ℏ²

8 1

⟨p²⟩²_min = 0

That is

2

The energy then has the lower bound That is

4m + 12 mω² ℏ²

4 4

mℏω = ℏω 2

The zero-point energy is the smallest energy eigenvalue

which is consistent with the uncertainty principle

Coherent states

The general solution of the time-dependent Schrödinger equation is (

) The position expectation value vanishes for

the stationary states of the harmonic oscillator

These states, therefore, individually seem to have

nothing in common with the classical oscillatory motion

ψ(x, t) = ∑^∞

k=0

c_nψ_n(x)e^−iEnt/ℏ = e^−iωt/2∑^∞

k=0

c_nψ_n(x)e^−iωnt

Thus, apart from the phase factor the function is periodic with the classical period

−iωt/2 ψ(x, t) T = 2π/ω

This suggests that it might be possibile to find a solution of the Schrödinger

equation in the form of a wave packet whose center of gravity oscillates with the period of the classical motion

To construct it we look for eigenstates

of the annihilation operator

a|β⟩ = β|β⟩ β ∈ ℂ

(recall: is not hermitian !)

V(x)

x

Expanding over the stationary states we obtain

a|β⟩ = ∑^∞

n=1

a|ψ_n⟩⟨ψ_n|β⟩ = ∑^∞

n=1

n |ψ_n−1⟩⟨ψ_n|β⟩ ⁼ _∑^∞

n=0

n + 1|ψ_n⟩⟨ψ_n+1|β⟩

send _{n → n + 1}

a|β⟩ = β∑^∞

n=0

|ψ_n⟩⟨ψ_n|β⟩

since

is eigenstate of

Since the above expressions for must be

equal the coefficients of

must also be equal

It follows that

n ⟨ψ_n−1|β⟩ ₌ ^β n

β

n − 1 ⟨ψ_n−2|β⟩ . . . = βⁿ

n! ⟨ψ₀|β⟩

We can thus write

|β⟩ = ∑^∞

n=0

|ψ_n⟩⟨ψ_n|β⟩ = ∑^∞

n=0

|ψ_n⟩ βⁿ

n! ⟨ψ₀|β⟩

Recalling that

we find

n! (a^†)ⁿψ₀

|β⟩ = ∑^∞ (βa^†)ⁿ

n! |ψ₀⟩⟨ψ₀|β⟩ = e^βa†|ψ₀⟩⟨ψ₀|β⟩

The normalisation of this state can be fixed as follows

⟨β|β⟩ = |⟨ψ₀|β⟩|² ∑^∞

n,m=0

⟨ψ_m|ψ_n⟩βⁿ(β*)^m

n!m! = |⟨ψ₀|β⟩|²∑^∞

n=0

|β|²ⁿ n!

=

δ_nm

= |⟨ψ₀|β⟩|²e^|β|2

⟨ψ₀|β⟩ = e^−|β|2/2

and

The states that we have constructed were studied by Schrödinger already in 1926 and are called coherent states

To understand their properties, we compute the expectation values of position and momentum over these state

⟨β|x|β⟩ = ℏ

2mω⟨β|a + a^†|β⟩ = ℏ

2mω (β + β*)

Because _|β⟩ is eigenvector of with eigenvalue _{a β}

⟨β|p|β⟩ = − i ℏmω

2 ⟨β|a − a^†|β⟩ = − i ℏmω

2 (β − β*)

For the uncertainties we have to evaluate:

(Δp)² = ⟨β|(p − ⟨p⟩)²|β⟩

Relevant in optics for the representation of light waves

Proceeding as above we find

(Δx)² = ⟨β|(x − ⟨x⟩)²|β⟩ = ℏ

2mω (Δp)² = ⟨β|(p − ⟨p⟩)²|β⟩ = ℏmω 2 (Δx)²(Δp)² = ℏ²/4

The coherent states are thus states characterised by a minimal uncertainty and fully identified by the complex number

The most interesting properties of coherent states, which justify their name, emerge when we consider their time evolution

A state

prepared at becomes at time

n=0

|ψ_n⟩ βⁿ

n! t = 0 t

ψ_β(x, t) ∼ |β; t⟩ = e^−|β|2/2∑^∞

n=0

|ψ_n⟩ βⁿ

n! e−iω(n+1/2)t = e^{−|β|2/2−iωt/2}∑^∞

n=0

|ψ_n⟩ βⁿ

n! e^−iωnt

= e^{−|β|2/2−iωt/2}∑^∞

n=0

|ψ_n⟩(βe^−iωt)ⁿ

n! = e^−iωt/2|β(t)⟩

with

We thus see that the time evolution of a coherent state can be expressed by the time dependence of the parameter that characterises it

that is a complex number that rotates along the circumference of modulus in the

complex plane

β(t) = βe^−iωt

This implies that a coherent state remains coherent with time evolution, but it is characterised by a different

In particular, this means that its uncertainty remains constant, and equal to the minimum allowed by the uncertainty principle

We can now compute the average value of the position at time

⟨x(t)⟩ = ℏ

2mω (β(t) + β*(t))

and setting

we get

= 2x₀|β|cos(ωt − δ)

The expectation value of the position evolves as the classical oscillation !