quantum physics - ustc

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27

Quantum PhysicsYiJing Yan and Shuiming Hu

Textbook and Main References:

[Le] Ira N. Levine, Quantum Chemistry, 7th ed. (2014)

[Gr] David J. Griffith, Introduction to Quantum Mechanics, 3rd ed. (2018)

[Sa] J. J. Sakurai, Modern Quantum Mechanics, 2nd ed. (2017)


28 Quantum Physics (003155.e1) / USTC / Fall 2020


Chapter 3

Elementary Systems

In this chapter, we will study three elementary systems: (i) Angular mo-

mentum along the z–direction, concerned with the angular momentum op-

erator, Lz, and its eigenequation. To facilitate the mathematic complexity

in later chapters, I will also discuss some basic knowledge and derivations

on angular momentum in the spherical coordinates; (ii) Piecewise con-

stant potential system. By solving the Schrodinger equation, we will see

some interesting tunneling phenomena; (iii) Harmonic oscillator, with its

Schrodinger equation being solved via the elegant commutator algebraic

approach. The involving the creation and annihilation operators are basic

ingredients in quantum field theory. The algebraic technique will also be

exploited in the general angular momentum theory in Chapter §4.

3.1 Angular momentum

3.1.1 Angular momentum along the z–direction

• The form of Lz, the angular momentum along the z–direction, for the

circular motion of a particle in the xy–plane:

Lz = xpy − ypx =~i

(x∂

∂y− y ∂

∂x

)(3.1.1)

In the circular coordinate frame, x = r cosφ and y = r sinφ, we have

Lz =~i

∂

∂φ(3.1.2)

29



——

Proof of Eq. (3.1.2): The involving Jacobian transformation is evaluated as follow.[∂∂r∂∂φ

]≡ A

[∂∂x∂∂y

]=

[∂x∂r

∂y∂r

∂x∂φ

∂y∂φ

][∂∂x∂∂y

]=

[cosφ sinφ

−r sinφ r cosφ

][ ∂∂x∂∂y

](3.1.3a)[

∂∂x∂∂y

]= A−1

[∂∂r∂∂φ

]= r−1

[r cosφ − sinφ

r sinφ cosφ

][ ∂∂r∂∂φ

](3.1.3b)

Evaluate then

x∂

∂y− y

∂

∂x=

[−yx

]T [ ∂∂x∂∂y

]=

[− sinφ

cosφ

]T [r cosφ − sinφ

r sinφ cosφ

][ ∂∂r∂∂φ

](3.1.4)

We obtain

x∂

∂y− y

∂

∂x=

[01

]T [ ∂∂r∂∂φ

]=

∂

∂φ(3.1.4′)

Substituting it into Eq. (3.1.1) results in Eq. (3.1.2). Q.E.D.

——

• Angular momentum along z-direction is quantized

This is concerned with the eigenequation, Lz|Φ〉 = λ|Φ〉. That is

~i

∂

∂φΦ(φ) = λΦ(φ) ⇒ Φ(φ) = Ceiλφ/~ (3.1.5)

Now consider the single–valued constraint,

Φ(φ+ 2π) = Φ(φ) ⇒ ei2πλ/~ = 1 (3.1.6)

The resultant quantization of angular momentum along z-direction reads

λ = m~; m = 0,±1,±2, · · · (3.1.7)

It is easy to obtain the normalization constant in Eq. (3.1.5), so that

Φ(φ) =1√2πeiλφ/~

Note also that the above quantization is often reported as

Lz|Φ〉 = m~|Φ〉; m = 0,±1,±2, · · · (3.1.8)


Elementary Systems 31

3.1.2 Angular momentum in three dimension space

L = r× p =

∣∣∣∣∣∣i j k

x y z

px py pz

∣∣∣∣∣∣ (3.1.9)

In Chapter §4, we will discuss in details the angular momentum theory.

Involved are three important commutator relations, as listed in Eq. (3.1.10)

below, that can be readily obtained via the canonical [α, pα′ ] = i~δαα′ ,

where α, α′ = x, y, z. To facilitate the later discussion on such at H-

like atoms, I will also present the angular momentum and related quanti-

ties, Eqs. (3.1.11)–(3.1.14), in spherical coordinates frame, followed by their

derivations.

• Exercises: Show that (Hint: Using canonical relations)

[Lx, Ly] = i~Lz; (also the two cyclic counterparts) (3.1.10a)

[L2, Lz] = [L2, Lx] = [L2, Ly] = 0 (3.1.10b)

• Angular momentum in spherical coordinates:

Lx ≡~i

Λx, Ly ≡~i

Λy, Lz ≡~i

Λz =~i

∂

∂φ(3.1.11)

where [cf. Levine’s Eq. (5.65)–(5.67)]

Λx = y∂

∂z− z ∂

∂y= − sinφ

∂

∂θ− cot θ cosφ

∂

∂φ(3.1.12a)

Λy = z∂

∂x− x ∂

∂z= cosφ

∂

∂θ− cot θ sinφ

∂

∂φ(3.1.12b)

• Angular momentum square operator [cf. Levine’s Eq. (5.68)]

L2 = −~2Λ2, where Λ2 =∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2 θ

∂2

∂φ2(3.1.13)

• Laplacian operator:

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2

In spherical coordinates [cf. Levine’s Eq. (6.6)],

∇2 =1

r

∂2

∂r2r +

1

r2Λ2 (3.1.14)

To derive Eqs. (3.1.12)–(3.1.14), we engage the Jacobian transformation

similar to Eq. (3.1.3). The details are as follows.



————

Derivations of Eqs. (3.1.12)–(3.1.14) (Self-study if you have time)

• The Jacobian matrix for (x, y, z) = r(sin θ cosφ, sin θ sinφ, cos θ) [cf. Eq. (3.1.3)]:

A ≡

∂x∂r

∂y∂r

∂z∂r

∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

=

sin θ cosφ sin θ sinφ cos θ

r cos θ cosφ r cos θ sinφ −r sin θ−r sin θ sinφ r sin θ cosφ 0

Evaluate then

A−1 =1

detA

A11 A21 A31

A12 A22 A33

A13 A23 A33

(3.1.15)

where

A11 =

∣∣∣∣r cos θ sinφ −r sin θ

r sin θ cosφ 0

∣∣∣∣ , A21 = −∣∣∣∣ sin θ sinφ cos θ

r sin θ cosφ 0

∣∣∣∣ , A31 =

∣∣∣∣ sin θ sinφ cos θ

r cos θ sinφ −r sin θ

∣∣∣∣and so on. We obtain

A11 = r2 sin2 θ cosφ, A21 = r sin θ cos θ cosφ, A31 = −r sinφ,

A12 = r2 sin2 θ sinφ, A22 = r sin θ cos θ sinφ, A32 = r cosφ,

A13 = r2 sin θ cos θ, A23 = −r sin2 θ, A33 = 0

and

detA = cos θA13 − r sin θA23 = r2 sin θ

The resultant Eq. (3.1.15) reads

A−1 =

sin θ cosφ 1r

cos θ cosφ − 1r sin θ

sinφ

sin θ sinφ 1r

cos θ sinφ 1r sin θ

cosφ

cos θ − 1r

sin θ 0

(3.1.15′)

This determines [cf. Eq. (3.1.4) and Levin’s Eqs. (5.62)–(5.64)]∂∂x∂∂y∂∂z

= A−1

∂∂r∂∂θ∂∂φ

=

sin θ cosφ cos θ cosφ − sinφ

sin θ sinφ cos θ sinφ cosφcos θ − sin θ 0

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

(3.1.16)

• Derivations of Eq. (3.1.12): Consider for example,

y∂

∂z− z

∂

∂y=

0

−zy

T∂∂x∂∂y∂∂z

= −r

0

cos θ

− sin θ sinφ

T∂∂x∂∂y∂∂z

Applying Eq. (3.1.16) results in

y∂

∂z− z

∂

∂y= −r

0

cos θ− sin θ sinφ

T sin θ cosφ cos θ cosφ − sinφ


∂∂r

1r∂∂θ

1r sin θ

∂∂φ

= −r

0sinφ

cos θ cosφ

T

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

We obtain Eq. (3.1.12a); i.e.,

Λx = y∂

∂z− z

∂

∂y= −

(sinφ

∂

∂θ+ cot θ cosφ

∂

∂φ

)



• Derivations on Eq. (3.1.13): By using Eq. (3.1.12), we have

Λ2x = sinφ

∂

∂θ

(sinφ

∂

∂θ+ cot θ cosφ

∂

∂φ

)+ cot θ cosφ

∂

∂φ

(sinφ

∂

∂θ+ cot θ cosφ

∂

∂φ

)Λ2y = cosφ

∂

∂θ

(cosφ

∂


∂

∂φ

)− cot θ sinφ

∂

∂φ

(cosφ

∂


∂

∂φ

)Consequently,

Λ2x + Λ2

y =∂2

∂θ2+ cot θ

(cosφ

∂

∂φsinφ− sinφ

∂

∂φcosφ

) ∂∂θ

+ cot2 θ(

cosφ∂

∂φcosφ+ sinφ

∂

∂φsinφ

) ∂

∂φ(3.1.17)

Note that∂

∂φsinφ = cosφ+ sinφ

∂

∂φand

∂

∂φcosφ = − sinφ+ cosφ

∂

∂φ(3.1.18)

Hence,

cosφ∂

∂φsinφ− sinφ

∂

∂φcosφ = 1

cosφ∂

∂φcosφ+ sinφ

∂

∂φsinφ =

∂

∂φ

(3.1.19)

We obtain Eq. (3.1.17) the expression,

Λ2x + Λ2

y =∂2

∂θ2+ cot θ

∂

∂θ+ cot2 θ

∂2

∂φ2(3.1.20)

This together with Λ2z = ∂2

∂φ2 result in Eq. (3.1.13); i.e.,

Λ2 = Λ2x + Λ2

y + Λ2z =

∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2 θ

∂2

∂φ2

• Derivations on Eq. (3.1.14): Let us start with

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2=

∂∂x∂∂y∂∂z

T

∂∂x∂∂y∂∂z

(3.1.21)

The transpose to Eq. (3.1.16) goes through its individual elements as follows.

∂

∂x= sin θ cosφ

∂

∂r+

1

rcos θ cosφ

∂

∂θ−

1

r sin θsinφ

∂

∂φ

=( ∂∂r

sin θ cosφ+1

r

∂

∂θcos θ cosφ−

1

r sin θ

∂

∂φsinφ

)+

1

rsin θ cosφ+

1

r sin θcosφ (3.1.22a)

∂

∂y= sin θ sinφ

∂

∂r+

1

rcos θ sinφ

∂

∂θ+

1

r sin θcosφ

∂

∂φ

=( ∂∂r

sin θ sinφ+1

r

∂

∂θcos θ sinφ+

1

r sin θ

∂

∂φcosφ

)+

1

rsin θ sinφ+

1

r sin θsinφ (3.1.22b)

∂

∂z= cos θ

∂

∂r−

1

rsin θ

∂

∂θ=( ∂∂r

cos θ −1

r

∂

∂θsin θ

)+

1

rcos θ (3.1.22c)



We obtain∂∂x∂∂y∂∂z

T

=

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

T

UT +1

r

sin θ cosφ+ 1sin θ

cosφ

sin θ sinφ+ 1sin θ

sinφ

cos θ

T (3.1.23)

where

U ≡

sin θ cosφ cos θ cosφ − sinφ


(UTU = 1) (3.1.24)

with which Eq. (3.1.16) reads ∂∂x∂∂y∂∂z

= U

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

(3.1.25)

By substituting Eqs. (3.1.23)–(3.1.25) into Eq. (3.1.21), we have

∇2 =

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

T

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

+1

r

sin θ cosφ+ 1sin θ

cosφ

sin θ sinφ+ 1sin θ

sinφ

cos θ

T U

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

=

∂2

∂r2+

1

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂φ2+

1

r

2cot θ

0

T

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

=

∂2

∂r2+

1

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂φ2+

2

r

∂

∂r+

cot θ

r2∂

∂θ(3.1.26)

Together with Eq. (3.1.13), we obtain [Levine’s Eq. (6.6)]

∇2 =∂2

∂r2+

2

r

∂

∂r+

1

r2Λ2 (3.1.27)

Moreover,1

r

∂2

∂r2r =

1

r

(r∂2

∂r2+ 2

∂

∂r

)=

∂2

∂r2+

2

r

∂

∂r(3.1.28)

Substituting this into Eq. (3.1.27) results in Eq. (3.1.14).

Q.E.D. ————



3.2 Piecewise potential systems

First of all, tunneling is a universal quantum phenomenon describes the fact

that particle can appear in classical forbidden region. This section deals

with the class of piecewise potential cases (Levine §2.4 & §2.5; Sakurai §B.2

& §B.3), whose Schrodinger equations are analytically solvable. The solu-

tions are then analysized in different scenarios, including both the bound–

and unbound–state cases.

The general procedure involves the following four steps: (i) Consider

Hψ(x) = Eψ(x), as a mathematical second–order differential equation,

with E > Vmin = 0. This gives rise to a mathematical ψ(x), in each region

of the piecewise V (x) such as Eq. (3.2.1); (ii) Apply the square–integrable

requirement to eliminate unphysical term(s), if any, in individual region [cf.

Eq. (3.2.2) with Eq. (3.2.4)]; (iii) Exploit the boundary–values conditions,

i.e., the continuous and/or single–valued constraints, to determine those

involving parameters, up to the normalization that is the task of step (iv).

3.2.1 The bound–state solutions with square-well potential

V (x) =

{0 for |x| < a

V0 for |x| > a (V0 > 0)(3.2.1)

The well depth and width are given by V0 and 2a, respectively. The bound–

state solutions are concerned with E < V0.

To proceed, we set

κ =√

2m(V0 − E)/~ and k =√

2mE/~ (3.2.2)

Denote further

ξ ≡ ka = (2mE/~2)12 a

χ ≡ (κ2 + k2)12 a = (2mV0/~2)

12 a > ξ

(3.2.3)

Note also the symmetry in the potential function, V (−x) = V (x) of E-

q. (3.2.1). The resultant bound–state wavefunctions must be either even

or odd functions, ψeven/odd(−x) = ±ψeven/odd(x), respectively, which needs

only be considered in the x > 0 region.

• Now, the aforementioned steps (i) and (ii), where the Schrodinger equa-

tion is integrated, result in [cf. Sakurai’s Eq. (B.2.6)]

ψeven/odd(x > 0) =

{cos(kx) or sin(kx) for 0 < x < a

Ae−κx or Be−κx for x > a(3.2.4)



• Next, the aforementioned step (iii) leads to the boundary–values,

ψeven(x = a) : Ae−κa = cos(ka) and κAe−κa = k sin(ka)

ψodd(x = a) : Be−κa = sin(ka) and κBe−κa = −k cos(ka)(3.2.5)

We obtain

Even parity: κ cos ξ = k sin ξ; Odd parity: κ sin ξ = −k cos ξ

which read in terms of the dimensionless parameters in Eq. (3.2.3) as

Even parity : ξ = χ| cos ξ|Odd parity : ξ = χ| sin ξ|

(3.2.6)

These determine the quantized energies [cf. Eq. (3.2.3)],

E =~2ξ2

2ma2

Hence,

k = ξ/a and κ = (χ2 − ξ2)12 /a

The coefficients in Eq. (3.2.4) via the first identities of Eq. (3.2.5) are

A = eκa cos(ka) and B = eκa sin(ka)

• Step (iv): The normalization constant Neven/odd is given via

1

2N2

even =

∫ a

0

cos2(kx)dx+ |A|2∫ ∞a

e−2κxdx

1

2N2

odd =

∫ a

0

sin2(kx)dx+ |B|2∫ ∞a

e−2κxdx

——Exercises:

(a) Reproduce the particle in the box limit, where V0 →∞.

(b) Solve Eq. (3.2.6) graphically and identify the pair of parameters, V0 and a, that cansupport only two bounded states.

(c) Identity of the zero–point energy.

(d) Plot the ground state and the first–excited state wave functions and mark down thetunneling regions.

——



3.2.2 Transmission–reflection scattering problems

The transmission–reflection is a scattering process. It involves an incident

wave from x→ −∞, under the influence of potential V (x) that takes effect

within a finite x–region. A partial of incident wave is transmitted forward

to x → ∞ and the remaining part is reflected backward to x → −∞. The

incident ψin and transmitted ψtran are forward waves, whereas the reflected

ψrefl is backward wave. Their amplitude squares are proportional to the

fluxes. The transmission and reflection coefficients, T and R, are

T = |t|2 = |ψtran|2/|ψin|2 and R = |r|2 = |ψrefl|2/|ψin|2 (3.2.8)

satisfying the flux conservation law of

T +R = 1 (3.2.9)

It is noticed that in transmission–reflection problems, wave functions are

asymmetric and unbounded in general. As exemplified below, we will

learn how to handle the quantum scattering problems that satisfy the flux

conservation law, Eq. (3.2.9). We will also see the quantum interference

between transmitted and reflected waves. The total transmission could

also occur at resonance scenario.

A. Transmission–reflection with a square–well potential

Consider the scattering against the square–well potential V (x), E-

q. (3.2.1) but shifted to the asymmetric region of x ∈ [0, L], since the in-

volving waves are asymmetric and unbounded. The incident wave energy

is above the well depth; i.e., E > V0. To proceed, we denote

k′ ≡√

2m(E − V0)/~ and k ≡√

2mE/~ (3.2.10)

• The integration of the Schrodinger equation gives rise to the scattering

wave function the form of

ψ(x) =

eik

′x + re−ik′x, x ∈ (−∞, 0)

A cos(kx) +B sin(kx), x ∈ (0, L)

teik′x, x ∈ (L,∞)

(3.2.11)

Remarks: In x ∈ (−∞, 0), the before potential well region, there are the incident wave

ψin(x) = eik′x and the reflected wave, ψrefl(x) = re−ik

′x. The former comes from

x→ −∞ and moves in the forward direction. The reflected one is a backward wave, in

response to the potential variation. In the region of x ∈ (L,∞), the potential does not

change, and there is only the forward transmitted wave, ψtran(x) = teik′x. Q.E.D.

———



• The complex parameters t and r involved in Eq. (3.2.11) follow the

transmission–reflection setup, as defined in Eq. (3.2.8) that satisfies the

flux conservation law, Eq. (3.2.9); i.e., |t|2 + |r|2 = 1. Moreover, from the

boundary conditions on the wave function of Eq. (3.2.11), we will obtain

|r|2

|t|2=

(k2 − k′2)2

(2kk′)2sin2(kL) (3.2.12)

This describes the quantum interference between the transmitted and re-

flected waves. Transmission resonances, where T = |t|2 = 1, occur at

k ≡√

2mE/~2 = nπ/L; n = 1, 2, · · · (3.2.13)

which are formally identical to those of Eq. (2.3.2) for the standing waves

of a particle in box. Further remarks will be given later after Eq. (3.2.20).———Derivations on Eq. (3.2.12): To complete the scattering wave function, Eq. (3.2.10),we apply the continuity requirements, which read

1 + r = A, A cos(kL) +B sin(kL) = teik′L

ik′(1− r) = kB, −k[A sin(kL)−B cos(kL)

]= ik′teik

′L(3.2.14)

Introduce for bookkeeping

t ≡ teik′L and α ≡ kL (3.2.15)

We obtain from the left column of Eq. (3.2.14) the relations,

A =t

k(k cosα− ik′ sinα) and B =

t

k(k sinα+ ik′ cosα) (3.2.16)

Combining the right column of Eq. (3.2.14), we complete the solutions further with

t =2ikk′

(k2 + k′2) sinα+ 2ikk′ cosαand r = −

(k2 − k′2) sinα

(k2 + k′2) sinα+ 2ikk′ cosα(3.2.17)

It is noticed that the relative phase between r and t is π/2, whenever r 6= 0. Moreover,the transmission and reflection coefficients are obtained to be T = |t|2 and R = |r|2,

T ≡ |t|2 =(2kk′)2

(2kk′)2 + (k2 − k′2)2 sin2 α(3.2.18)

and

R ≡ |r|2 =(k2 − k′2)2 sin2 α

(2kk′)2 + (k2 − k′2)2 sin2 α(3.2.19)

Apparently, T +R = 1 and

R

T=|r|2

|t|2=

(k2 − k′2)2

(2kk′)2sin2 α (3.2.20)

This is Eq. (3.2.12), since α = kL as defined in Eq. (3.2.15). Q.E.D.

——



Comments on Eq. (3.2.13): This is the transmission resonances scenario, occurring

at α = kL = nπ; n = 1, 2, · · · . Interestingly, this expression is formally identical toEq. (2.3.2) for the standing waves of a particle in box. Note that the transmission reso-

nance, where T = 1, resembles the classical case. The quantum wave–particle scenario

is rather t ≡ teik′L = 1, according to Eq. (3.2.17), and also ψ(x = 0) = ψ(x = L) = 1,

as inferred from Eq. (3.2.11).

———

B. Transmission–reflection with a square–barrier potential

You can find the resultant transmission coefficients in Sakurai’s book,

page 526. Derive them by yourself, together with your study teammate(s).



C. Transmission–reflection with a stepwise potential

V (x) =

{0, x < 0

V0, x > 0(3.2.26)

Case (i): E > |V0|

ψ(x) =

{eikx + re−ikx, x < 0; k =

√2mE/~

Aeik′x, x > 0; k′ =

√2m(E − V0)/~

(3.2.27)

The boundary conditions lead to

r =k − k′

k + k′and A =

2k

k + k′(3.2.28)

It is noticed that the flux is proportional to the wavevector amplitude. The

resultant flection and transmission coefficients, R and T , are

R = r2 =(k − k′)2

(k + k′)2and T =

k′

kA2 =

4kk′

(k + k′)2(3.2.29)

respectively, satisfying T +R = 1.

Case (ii): 0 < E < V0

ψ(x) =

{eikx + re−ikx, x < 0; k =

√2mE/~

Ae−κx, x > 0; κ =√

2m(V0 − E)/~(3.2.30)

There is no transmission but pure tunnelling in the region of x > 0. We

have [cf. Eq. (3.2.28)]

r =k − iκk + iκ

and A =2k

k + iκ(3.2.31)

Here, R = |r|2 = 1; thus T = 1−R = 0, as anticipated.

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3.2.3 The delta function potential systems

Consider a potential well with a delta–function (Griffiths, 3rd ed., §2.5.2):

V (x) = −αδ(x) (α > 0) (3.2.32)

It is noticed that as V (x = 0)→∞, the wave function derivative, ψ′(x) ≡dψ(x)/dx, would be discontinued at x = 0.

The value of ψ′(0+) − ψ′(0−) for the δ(x)–potential is determined as

follows. Consider the Schrodinger equation the limiting expression,

limε→0

∫ ε

−ε

[− ~2

2mψ′′(x) + V (x)ψ(x)

]dx = E lim

ε→0

∫ ε

−εψ(x)dx = 0 (3.2.33)

with noting that∫ ε

−εψ′′(x)dx = ψ′(ε)− ψ′(−ε)∫ ε

−εV (x)ψ(x)dx = −α

∫ ε

−εδ(x)ψ(x)dx = −αψ(0)

(3.2.34)

Substitute them into Eq. (3.2.33) resulting in the boundary condition,

ψ′(0+)− ψ′(0−) = −2mα

~2ψ(0) (3.2.35)

for the wave function in the δ(x)–potential system.

A. The normalized bound state (E < 0) has the form of

ψ(x) =√κ e−κ|x|; κ =

√−2mE/~ (3.2.36)

with the boundary values satisfying

ψ(0) =√κ and ψ′(0+)− ψ′(0−) = −2κ

√κ

Apply then Eq. (3.2.35), resulting in the final expressions of

ψ(x) =

√mα

~e−mα|x|/~

2

; E = −mα2

2~2(3.2.37)

Evidently, there is only one bound state in any given δ(x)–well potential.

B. Scattering (E > 0) setup for δ(x)–potential:

ψ(x) =

{eikx + re−ikx; x < 0

teikx; x > 0k =√

2mE/~ (3.2.38)

with the boundary values satisfying

ψ(x = 0) = 1 + r = t

ψ′(x = 0+)− ψ′(x = 0−) = ik(t− 1 + r) = −2mα

~2ψ(0)

(3.2.39)

The last identity arises from Eq. (3.2.35). We obtain

r =iβ

1− iβand t =

1

1− iβ, where β ≡ mα

k~2=

(mα2

2E~2

) 12

(3.2.40)

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3.3 Solutions to harmonic oscillator: Algebraic approach

The Hamiltonian of a one–dimensional harmonic oscillator of mass m and

frequency ω reads

H =p2

2m+

1

2mω2x2 =

~ω2

(P 2 + Q2) (3.3.1)

The second identity is expressed in terms of the dimensionless coordinate

and momentum, defined as

Q ≡√mω

~x and P ≡ 1√

m~ωp (3.3.1′)

which satisfy [Q, P ] = i. Introduce now

a =1√2

(Q+ iP ) =

(mω

2~

) 12(x+

ip

mω

)a† =

1√2

(Q− iP ) =

(mω

2~

) 12(x− ip

mω

) (3.3.2)

They satisfy

[a, a†] = 1 ⇒ [a†a, a] = −a and [a†a, a†] = a† (3.3.3)

• Number operator versus annihilation and creation operators

It is easy to show that

N ≡ a†a =1

2(P 2 + Q2 − 1) (3.3.4)

This leads to the harmonic oscillator Hamiltonian, Eq. (3.3.1), the form of

H = ~ω(N +

1

2

)(3.3.5)

Hence, eigenfunctions of H must also be eigenfunctions of N , and vice versa.

Physically, N ≡ a†a represents the number operator, for its satisfying

N |n〉 = n|n〉; n = 0, 1, 2, · · · (3.3.6)

We will show this result, together with

a|n〉 =√n |n− 1〉 and a†|n〉 =

√n+ 1 |n+ 1〉 (3.3.7)

These represent the physical meanings of a† and a† to be the annihilation

and creation operators, respectively.———

Derivations of Eqs. (3.3.6) and (3.3.7): Let us start with the eigenequation for a†a,which is Hermitian and nonnegative:

a†a|n〉 = λn|n〉 (λn ≥ 0) (3.3.8)



Together with Eq. (3.3.3), we obtain

a†a(a|n〉) = a(a†a− 1)|n〉 = (λn − 1)(a|n〉)

a†a(a†|n〉) = a†(a†a+ 1)|n〉 = (λn + 1)(a†|n〉)(3.3.9)

These conclude that

a|n〉 ∝ |n− 1〉 and a†|n〉 ∝ |n+ 1〉, whereas λn±1 = λn ± 1 (3.3.10)

On the other hand, applying the annihilation operator on the ground state should phys-

ically results in

a|0〉 = 0 ⇒ a†a|0〉 = 0 ⇒ λ0 = 0 (3.3.11)

Together with the last expression in Eq. (3.3.11), we obtain

λn = λn−1 + 1 = n (3.3.12)

Consequently,

(〈n|a†)(a|n〉) = n and (〈n|a)(a†|n〉) = n+ 1 (3.3.13)

These give rise to Eq. (3.3.6) and Eq. (3.3.7), respectively. Q.E.D.

———

• Steady–state wave functions, {ψn(q) ≡ 〈q|n〉; n = 0, 1, · · · }Adopted below is the dimensionless coordinate representation, in which

Q.= q and P

.= −i ∂

∂q, where q =

√mω

~x (3.3.14)

The annihilation and creation operators [Eq. (3.3.2)] are then

a.=

1√2

(q +

∂

∂q

)and a†

.=

1√2

(q − ∂

∂p

)(3.3.15)

1) Consider first the ground state via a|0〉 = 0 [Eq. (3.3.11)]. That is(q +

∂

∂q

)ψ0(q) = 0 (3.3.16)

This gives rise to the normalized ground–state wave function,

ψ0(q) = π−14 e−q

2/2 (3.3.17)

2) Next, from the second identity of Eq. (3.3.7), we obtain

ψn(q) =1√2n

(q − ∂

∂q

)ψn−1(q) =

1

(2nn!)12

(q − ∂

∂q

)nψ0(q) (3.3.18)

3) Note that ψ0(q) of Eq. (3.3.17) is Gaussian. The resultant Eq. (3.3.18)

acquires the Hermite polynomial–Gaussian expression of [cf. Eqs. (3.3.24)–

(3.3.29)]

ψn(q) =1

(2nn!)12

Hn(q)ψ0(q) =1

(2nn!√π)

12

Hn(q)e−q2/2 (3.3.19)



———

• Hermite polynomials (a specific type of orthogonal polynomials in mathematics):The Hermite polynomials, {Hn(q);n = 0, 1, · · · }, satisfy

H′′n(q) = 2qH′n(q)− 2nHn(q) (3.3.20)

and [in line with Eq. (3.3.19)]∫ ∞−∞

dq Hm(q)Hn(q)e−q2

= 2nn!√π δmn (3.3.21)

The above properties will be used in obtaining Eq. (3.3.19).

The first two Hermite polynomials are

H0(q) = 1 and H1(q) = 2q (3.3.22)

The higher–order Hermite polynomials can be evaluated via the recurrence relation,

Hn+1(q) = 2qHn(q)− 2nHn−1(q) (3.3.23)

The Hermite polynomials are of the parity, Hn(−q) = (−)nHn(q). Moreover,

Hn(q) = (−1)neq2 dn

dqne−q

2and e−z

2+2zq =

∞∑n=0

zn

n!Hn(q)

• Derivations of Eq. (3.3.19):Consider the Schrodinger equation, in terms of the eigenequation for the occupation

number operator [cf. Eq. (3.3.8)]:

N |ψ〉 = λ|ψ〉 (3.3.24)

In the dimensionless coordinate representation of Eq. (3.3.15), we have

N ≡ a†a .=

1

2

(q2 −

∂2

∂q2− 1

)(3.3.25)

This leads to Eq. (3.3.24) the q-representation expression of

1

2

(q2 −

∂2

∂q2− 1

)ψ(q) = λψ(q) (3.3.26)

To proceed, we first recognize that e−q2/2 is a solution. This in fact is the ground eigen-

state, since e−q2/2 contains no node. The general solutions are of the same asymptotic

behavior. Let f(q) be the undetermined polynomial in

ψ(q) = f(q)e−q2/2 (3.3.27)

that satisfies Eq. (3.3.26). Evaluate then

∂2

∂q2

[f(q)e−q

2/2]

= f ′′(q)e−q2/2 + 2f ′(q)

(e−q

2/2)′

+ f(q)(e−q

2/2)′′

=[f ′′(q)− 2qf ′(q) + (q2 − 1)f(q)

]e−q

2/2

Rearrange it as(q2 −

∂2

∂q2− 1

)[f(q)e−q

2/2]

=[2qf ′(q)− f ′′(q)

]e−q

2/2

Substituting it into Eq. (3.3.26) with Eq. (3.3.27) results in

2qf ′(q)− f ′′(q) = 2λf(q) (3.3.28)

Compare it with Eq. (3.3.20), resulting in

f(q) ∝ Hn(q) and λn = n (3.3.29)

This determines the wave function, Eq. (3.3.27), as ψn(q) ∝ Hn(q)e−q2/2. Applying

then the normalization, Eq. (3.3.21), we obtain Eq. (3.3.19). (Q.E.D.)

———



• Importance of annihilation/creation operators:

General speaking, the algebraic approach, based on annihilation and

creation operators that satisfy [a, a†] = 1, is more fundamental, due to

the underlying rich physical picture, used widely in bosonic quantum field

theory. The resultant Eqs. (3.3.3), (3.3.6) and (3.3.7) are also importan-

t, and used frequently in various tasks/evaluations for harmonic systems.

These include the expectation values, the uncertainty relations, the optical

selection rule and also the dynamics of a(t) in the Heisenberg picture.

———

Exercise 3.3.1: Consider harmonic oscillator, with a given |n〉.Show that (a) 〈x〉 = 〈p〉 = 0

(b) The kinetic energy and potential energy are of

〈K〉 = 〈V 〉 =1

2~ω(n+

1

2

)(3.3.30)

(c) The uncertainty relation reads

∆x∆p = ~(n+

1

2

)(3.3.31)

Exercise 3.3.2: (a) Show that for harmonic oscillator,

a(t) = a(0)e−iωt (3.3.32)

(b) Write down also the result on a†(t)

(c) Derive x(t) and p(t) and compare them with the classical results.

Exercise 3.3.3: Prove H ′n(q) = 2nHn−1(q), for the Hermite polynomials.

Hint: Use Eqs. (3.3.18) and (3.3.19). Note also that a|n〉 =√n|n− 1〉.

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