Operator Method

X

n=0

anΨn(x, t) . (2.52)

Since the a_n are not functions of time, we can evaluate them at t = 0 by multiplying Equation (2.52) by Ψ^∗_mand integrating to obtain

Z ∞

−∞

Ψ^∗_mΨ(x, 0) dx =

∞

X

n=0

an

Z ∞

−∞

Ψ^∗_mΨndx = am, (2.53)

where we have used the orthogonality of the eigenfunctions for the final result.

With the coefficients thus defined, Equation (2.52) then gives the evolution of Ψ(x, t) given Ψ(x, 0).

Problem 2.21 Mixed states. A harmonic oscillator is initially in the state Ψ(x, 0) = Ae^−α2x2/2αx(2αx + i) .

(a) Find the wave function as a function of the time.

(b) Find the average energy of the oscillator.

2.4 Operator Method

There is an entirely independent method for solving the harmonic oscillator problem without any apparent solution of a differential equation. Using this method, all of the pertinent physical quantities can be calculated without ever knowing the particular form of the wave function. The technique of using raising and lowering operators will be useful here and again later when we study angular momentum.

2.4.1 Raising and Lowering Operators

We begin by introducing the operator notation with ξ = q ˆ

q ≡ ξ = q (2.54)

ˆ p ≡ −id

dξ = −id

dq. (2.55)

Then ˆp²= −d²/dq² and Equation (2.32) may be written

(ˆp²+ ˆq²)ψ = 2λψ . (2.56)

Now ˆp and ˆq do not commute, since [ˆp, ˆq] = −i d

dqq + iq d

dq = −i , (2.57)

so that the usual factors of the operator ˆp²+ ˆq² become

(ˆq + iˆp)(ˆq − iˆp) = ˆq²+ ˆp²+ 1 (2.58) (ˆq − iˆp)(ˆq + iˆp) = ˆq²+ ˆp²− 1 . (2.59) If these two expressions are added, then we find

ˆ

p²+ ˆq²=¹₂[(ˆq + iˆp)(ˆq − iˆp) + (ˆq − iˆp)(ˆq + iˆp)] . If we now define two new operators based on the two factors as

ˆ a ≡ 1

√2(ˆq + iˆp) = 1

√2

 q + d

dq



(2.60) ˆ

a^†≡ 1

√2(ˆq − iˆp) = 1

√2

 q − d

dq



, (2.61)

then

ˆ

p²+ ˆq²= ˆaˆa^†+ ˆa^†ˆa , and Equation (2.56) may be written

(ˆaˆa^†+ ˆa^†ˆa)ψ = 2λψ . (2.62) The operators satisfy the commutation relation

[ˆa, ˆa^†] = ˆaˆa^†− ˆa^†ˆa = 1 . (2.63) Using Equation (2.63), we may write the Schr¨odinger equation, Equation (2.62), in either of the two forms:

ˆ

aˆa^†ψ = λ +¹₂
ψ (2.64)

ˆ

a^†ˆaψ = λ −¹₂
ψ . (2.65) The task now is to construct an infinite set of eigenfunctions from any solution of Equation (2.62), Equation (2.64), or Equation (2.65). To do this, we first operate on Equation (2.64) with ˆa^† on the left so that

ˆ

a^†ˆaˆa^†ψ = λ +¹₂
ˆa^†ψ . (2.66) Then using the commutation relation ˆa^†ˆa = ˆaˆa^†− 1, we have

(ˆaˆa^†− 1)ˆa^†ψ = λ +¹₂
ˆa^†ψ

The Schr¨odinger Equation in One Dimension 67 and moving the ˆa^†ψ term on the left over to the right, we obtain

ˆ

aˆa^†(ˆa^†ψ) = λ +³₂
ˆa^†ψ . (2.67) If we now add Equation (2.66) and Equation (2.67), we get the result

(ˆaˆa^†+ ˆa^†a)(ˆˆ a^†ψ) = 2(λ + 1)(ˆa^†ψ) . (2.68) We thus have the surprising result that if ψ is an eigenfunction of Equation (2.62) with eigenvalue λ, then ˆa^†ψ is also an eigenfunction but with eigenvalue λ + 1.

We may do this again, operating on Equation (2.67), to obtain ˆ

a^†(ˆaˆa^†)(ˆa^†ψ) = (ˆa^†ˆa)(ˆa^†)²ψ = (ˆaˆa^†− 1)(ˆa^†)²ψ = λ +³₂
(ˆa^†)²ψ . (2.69) From this it follows that

ˆ

aˆa^†(ˆa^†)²ψ = λ +⁵₂
(ˆa^†)²ψ , (2.70) and adding these two equations, the result is

(ˆaˆa^†+ ˆa^†ˆa)(ˆa^†)²ψ = 2(λ + 2)(ˆa^†)²ψ . (2.71) We may continue this process indefinitely, so the operator ˆa^† has the ef-fect of raising any eigenfunction ψ_n with eigenvalue λ_n to the next higher eigenfunction ψ_n+1with eigenvalue λ_n+1= λ_n+ 1.

In similar fashion, the operator ˆa is a lowering operator. We see this by operating on the left by ˆa on Equation (2.65), obtaining (with the use of the commutator again)

ˆ

aˆa^†ˆaψ = (1 + ˆa^†a)ˆˆ aψ = λ −¹₂
ˆaψ , (2.72) or

ˆ

a^†ˆa(ˆaψ) = λ −³₂
(ˆaψ) . Adding, this leads to

(ˆaˆa^†+ ˆa^†ˆa)(ˆaψ) = 2(λ − 1)(ˆaψ) . (2.73) Hence, ˆa is a lowering operator just as ˆa^† is a raising operator.

2.4.2 Eigenfunctions and Eigenvalues

For the lowering operator, there is a limit, since we may not lower the eigen-value forever or it will go negative. To demonstrate this, we calculate the expectation value of the energy (or of λ), such that

hλi = Z ∞

−∞

ψ^∗λψ dq = ¹₂ Z ∞

−∞

ψ^∗(ˆp²+ ˆq²)ψ dq

= −¹₂ Z ∞

−∞

ψ^∗d²ψ dq² dq + ¹₂

Z ∞

−∞

ψ^∗q²ψ dq .

Integrating the first term on the right by parts, hλi = −¹₂

 ψ^∗dψ

dq

∞

−∞

+¹₂ Z ∞

−∞

dψ^∗ dq

dψ dq dq +¹₂

Z ∞

−∞

|ψ|²q²dq .

The first term vanishes since ψ must vanish as |q| → ∞, and the remaining terms give

hλi = ¹₂ Z ∞

−∞

   

dψ dq    

2

+ |ψ|²q²

!

dq ≥ 0 ,

since the integrand is positive definite. There exists then a lowest possible value of λn = λ0 and a corresponding wave function such that

ˆ

aψ0= 0 , (2.74)

since any further lowering would lead to λ < 0. We can establish this lowest eigenvalue from Equation (2.65), where

ˆ

a^†ˆaψ₀= 0 = λ₀−¹₂
ψ0, so that

λ0= ¹₂ =E0

~ω , or

E0= ¹₂~ω . (2.75)

Since all of the other eigenvalues increase by 1 each time we use the raising operator, the general expression for the eigenvalues is

λn= n +¹₂, n = 0, 1, 2, . . . , (2.76) which is equivalent to

E_n = ~ω(n +¹2) , n = 0, 1, 2, . . . . (2.77) The lowest energy state of a system with discrete energy states is called a ground state and we refer to both the ground state eigenvalue and the ground state eigenfunction. The ground state eigenfunction for the harmonic oscillator may be obtained from the form of the lowering operator where

ˆ

aψ0= 1

√ 2

 q + d

dq



ψ0= 0 ,

or dψ₀

dq + qψ0= 0 . (2.78)

This is a first order differential equation whose solution is

ψ0= C0e^−q2/2, (2.79)

The Schr¨odinger Equation in One Dimension 69 where the normalization constant is given by

hψ0|ψ0i = 1 = |C0|²

While the successive application of the raising operator will produce all of the higher-order wave functions, they are not necessarily normalized. For the normalization constants, we define the normalization constants in terms of the raising operator such that

ψn+1=Cn+1

C_n ˆa^†ψn. (2.80)

Starting at the bottom,

ψ₁ = C1 The steps then follow as

hψn|ψni =

In going from Equation (2.81) to Equation (2.82), we used the fact that the adjoint of ˆa^† is ˆa, which may be verified by integrating by parts. In going from Equation (2.82) to Equation (2.83), we used Equation (2.64). In going

from Equation (2.83) to Equation (2.84), we used λ_n = 2n + 1. The result is that

Cn =

 1

n!√ π

1/2

. (2.87)

The relationship to the Hermite polynomials is illustrated by the relations ψn= Nne^−q2/2Hn(q) = Cn(ˆa^†)ⁿe^−q2/2= Cn

√ 2ⁿ

 q − d

dq

n

e^−q2/2, (2.88) where the Hermite polynomials are given by

Hn(q) = e^q2/2

 q − d

dq

n

e^−q2/2. (2.89)

By comparing Equation (2.87) and Equation (2.88), we find N_n= C_n

√2ⁿ =

 1

2ⁿn!√ π

^1/2

. (2.90)

2.4.3 Expectation Values

Returning to Equation (2.80), we may now write that equation as ˆ

a^†ψ_n = Cn

C_n+1ψ_n+1=√

n + 1ψ_n+1. (2.91)

Lowering the index by one and then operating by ˆa on the left leads to ˆ

aˆa^†ψn−1=√

nˆaψn= nψn−1,

where the latter equality is from Equation (2.64). This may be written as ˆ

aψ_n =√

nψ_n−1. (2.92)

With these expressions, it is convenient to write expressions for ˆp and ˆq in terms of the raising and lowering operators, such that

ˆ q = 1

√2(ˆa + ˆa^†) (2.93)

ˆ p = i

√2(ˆa^†− ˆa) . (2.94)

We also note that using Equations (2.91) and (2.92) leads to hai = hψn|ˆaψ_ni =√

nhψ_n|ψn−1i = 0 hˆa^†i = hψn|ˆa^†ψni =√

n + 1hψn|ψn+1i = 0 hˆaˆa^†i = hψ_n|ˆaˆa^†ψ_ni

=√

n + 1hψn|ˆaψn+1i = (n + 1)hψn|ψni = n + 1 hˆa^†ˆai = hψn|ˆa^†ˆaψni

=√

nhψn|ˆa^†ψn−1i = nhψn|ψni = n .

The Schr¨odinger Equation in One Dimension 71 From these we may readily establish that:

hqi = 1

√2(hˆai + hˆa^†i) = 0 hpi = i

√2(hˆa^†i − hˆai) = 0

hq²i = ¹₂(hˆaˆai + hˆaˆa^†i + hˆa^†ˆai + hˆa^†ˆa^†i)

= ¹₂(2n + 1) = n +¹₂

hp²i = −¹₂(hˆa^†ˆa^†i − hˆaˆa^†i − hˆa^†ˆai + hˆaˆai)

= n +¹₂. Therefore,

hxi = hqi α = 0 hp_xi = ~αhpi = 0 hx²i = hq²i

α² = 1

α²(n +¹₂) hp²_xi = ~²α²hp²i = ~²α²(n +¹₂)

hT i = hp²_xi

2m = ¹₂~ω(n +¹₂) hV i = khx²i

2 =¹₂~ω(n +¹2)

hEi = ~ωhλi = ¹₂~ω(hp²i + hq²i) = ~ω(n + ¹₂) .

Problem 2.22 Expectation values from the raising and lowering operators.

(a) Use the raising and lowering operators to evaluate hψn|x|ψmi. (For a given n, find for which values of m there is a nonzero result, and find the result for those special cases.)

(b) Use the raising and lowering operators to evaluate hψn|x²|ψmi.

(c) Use the raising and lowering operators to prove hψ_n|x³|ψ_mi = 1

2√ 2α³[p

(n + 3)(n + 2)(n + 1)δ_n,m−3 +3(n + 1)^3/2δn,m−1+ 3n^3/2δn,m+1

+p

n(n − 1)(n − 2)δn,m+3]. (2.95) (d) Use the raising and lowering operators to evaluate hψn|x⁴|ψmi. [The re-sult should be similar in form to Equation (2.95), but with more terms.]

Problem 2.23 Uncertainty product. Find the minimum uncertainty product (∆x)(∆px) for the n^th state of the harmonic oscillator.

Problem 2.24 Adjoint relationships. Using the definition of the adjoint of an operator, and integrating by parts,

(a) Prove that ˆa^† is the adjoint of ˆa.

(b) Prove that ˆa is the adjoint of ˆa^†. 2.4.3.1 Concluding Remarks

Since we have seen that the ground state of the harmonic oscillator has a gaussian wave function, we can now develop a physical interpretation for the wave packet in Section 2.1.2 and Figure 2.1. If we imagine that we are moving at velocity v0 = p0/m past a stationary harmonic oscillator system (or that the system were moving by us, it’s all just relative motion) and that suddenly at time t = 0, the spring breaks so that the particle is suddenly a free particle, the wave function will evolve in time as in the figure.

If instead, we were standing still with respect to the system when the spring broke, the packet would simply spread in time (p₀= 0), retaining its gaussian envelope that gets broader and lower over time so that the area is conserved.

In such a case, there would neither be a phase velocity nor a group velocity.

3

The Schr¨ odinger Equation in Three

In document Quantum Mechanics Foundations and Applications (Page 79-87)