Introduction to Quantum Mechanics Unit 2. Time Independent Schroedinger Equation A. Stationary States 1. Time independent Schroedinger Equation (i) Separation of variables on Schroedinger equation:

(ii) The first equation is readily soluble.

(iii) Therefore, the general solution of the Schroedinger equation must be in the form

2. Properties of the Stationary States ψ(x)

(x)E(x)Hsimply or (x),E(x)V(x)m2

- R.H.S.

(t)E(t)t

i L.H.S.

:equations twobecomesdinger oSchr NowE.constant a toequalonly can they so x,offunction a is H.S. R. andonly t offunction a is .S.H.L

(x)V(x)m2

-(x)1(t)

t(t)i

(x)V(x)m2

-(t)(t)t

(x)i

(t)(x)V(t)(x)m2

-(t)(x)t

i)t,x(V)t,x(m2

-)t,x(t

i

:equationdinger oSchr theinto thisSubstitute(t)(x)t)(x, Let

2

2

2

22

ψ=ψψ=ψ+ψ⇒

ϕ=ϕ∂∂

⇒

⎥⎦

⎤⎢⎣

⎡ψ+ψ

ψ=ϕ

∂∂

ϕ⇒

⎥⎦

⎤⎢⎣

⎡ψ+ψϕ=ϕ

∂∂

ψ⇒

ϕψ+ϕψ=ϕψ∂∂

⇒Ψ+Ψ=Ψ∂∂

ϕψ=Ψ

h

h

&&

hh

hh

hh

hh

&&

h

hhh

E wheree(t) Therefore,

Ei Eeei (t)E(t)t

i

e(t)Let

ti

t t

t

=ω=ϕ

−=Ω⇒=Ω⇒ϕ=ϕ∂∂

=ϕ

ω−

ΩΩ

Ω

(x)E(x)V(x)dxd

m2

:equationr Schrodinget independen timeby the determined are (x) and E and E where

(x)e t)(x,

2

22

ti

ψ=ψ+ψ−

ψ=ω

ψ=Ψ ω−

h

h

(i) Suppose we find all solutions of the time independent Schroedinger equation and

label them as n, i.e. Hψn(x)=Enψn(x). In here, for simplicity, we assume n is discrete (i.e. these solutions are countable one by one). In reality, n can be continuous and E will be a continuous variable in that case.

(ii) Note that since Hψn(x)=Enψn(x), ψn(x) is an eigenfunction of the Hamiltonian. (iii) Since H is Hermitian, all the solution of the time independent Schroedinger

equation form an orthonomal basis, and let us call this basis E. So that this time, for the same linear space, there are three bases we know well. They are X, P, and E. Unlike linear momentum, the basis are always formed by eigenstates (in x-representation):

The basis E depend on the potential and are solution of the equation:

(iv) Solving above equation to determine ψn(x) is the same as diagonalizing the Hamiltonian H.

(v) Any linear combination of ψn with the time factor is a solution of the original time

dependent Schroedinger equation, i.e.

is a solution of the original time dependent Schroedinger equation, even though

it is NOT an eigenstate of the Hamiltonian.

(vi) Eigenstates of H (or ψn(x)) will evolve with time as

Note that ONLY ψn(x) vary with time in this simple harmonic form. They are also called stationary state, because its probability density is independent of time:

xpie

21

h

π

(x)E(x)V(x)m2

- nnnn

2

ψ=ψ+ψh

[ ][ ]

>Ψ>=Ψ∂∂

∴

ψ++ψ+ψ=

ψω++ψω+ψω=

ψω++ψω+ψω>=Ψ∂∂

ψ++ψ+ψ>=Ψ∴

ψ++ψ+ψ>=Ψ

ω−ω−ω−

ω−ω−ω−

ω−ω−ω−

ω−ω−ω−

ω−ω−ω−

|H|t

i

eEaeEaeEa

)eaeaiea

)ea-i()ea(-i)ea(-ii|t

i

eEaeEaeEa|H

aaea| If

tiNNN

ti222

ti111

tiNNN

ti222

ti111

tiNNN

ti222

ti111

tiNNN

ti222

ti111

Nti

Nti

22ti

11

N21

N21

N21

N21

N21

h

L

hLhh

Lhh

L

L

Nti

Nti

22ti

11N21 aaea| ψ++ψ+ψ>=Ψ ω−ω−ω− L

NN2211 aaa ψ++ψ+ψ L

tinn

ne)x()t,x( ω−ψ=ψ

)x()x(e)x(e)x()t,x()t,x()t,x(P n*

nti

nti*

nn*

nnn ψψ=ψψ=ψψ= ω−ω

Any other arbitrary state function will depend on time in a more complex way:

and the probability density will NOT be a constant of time. (vii) Above show why the diagonalization of the Hamiltonian is so important. We

know exactly how these stationary states vary with time. Now, given any arbitrary wave function at t=0 (i.e. Ψ(x,0)), we will know how it will evolve with time if we can express it as a linear combination of ψn at t=0:

(viii) How to determine a1, a2, …, aN? Do not forget that a Hermitian H will ensure an

orthonormal basis E, in other words, <ψn|ψm>=δnm.

B. Particle in a box 1. General solution (i) Consider potential

This is the case when a particle is trapped in the box 0≤x≤a.

Nti

Nti

22ti

11N21 aaea| ψ++ψ+ψ>=Ψ ω−ω−ω− L

tiN

titiN

N

Nexexex

txtxtxtxxxxx

ωωω ψψψ

ψψψψψψ

−−− +++=

+++=Ψ=+++=Ψ

)(a)(a)(a

),(a),(a),(a),( then0,at t )0,(a)0,(a)0,(a)0,( If

N2211

N2211

N2211

21 L

L

L

dxxx

or

xxei

xxxxxxxxxxx

j

j

ijijijj

iN

)0,()0,(a

case, continuousin

)0,(|)0,(a ..

aa)0,(|)0,(a)0,(|a|)0,()0,(|)0,( )0,(|a )0,(|a)0,(|a)0,(|a)0,(|

*j

j

jiii

iN2211

Ψ∫=

>Ψ=<

=∑>=<∑>=<∑>=Ψ<⇒>∑=>++>+>>=Ψ

ψ

ψ

δψψψψψψψψψ L

⎩⎨⎧

∞≤≤

=otherwise

ax0 0 )x(V

x

x=a x=0

∞ ∞

(ii) By insight, we should know that ψn(x)=0 for x≤0 and x≥a. ψn(x) is continuous at

x=0 and x=a, but ψn’(x) is NOT continuous at these two points because V is infinity at these two points.

(iii) By insight from standing waves, we should be able to write down the solution

directly:

(iv) Together with the time dependent part, the energy eigenvectors are:

(v) By insight, we know the integration (why?)

This immediate gives

(vi) At this point we know all the eigenvectors in basis E for a particle in a box. Since H is a Hermitian operator, these vectors are orthonormal.

(vii) All eigenvectors in basis E satisfy boundary condition ψn(a)=ψn(0)=0. Any state

vector |Ψ> can be expressed as linear combination of ψn and this will ensure |Ψ> also satisfy the boundary condition Ψ(a)= Ψ(0)=0.

2

2222n

2

n

nn

n

ma2n

m2kE

negative?)or 0not (why 1,2,3,...)(n a

n k 0)x(

aor x 0x 0ax0 kx sin A

)x(

hh π==

=π

=⇒=ψ

⎩⎨⎧

≥≤≤≤

=ψ

0dx xk cos xk sin and

2adx xk cos dx xk sin

nnbox

n2

boxn2

box

=

==

∫∫∫

a2 A 1

2aA2 =⇒=⋅

2

22n

n

-i

2E

aor x 0x 0

ax0 kx sin Ae ),(

n

man

txt

n

h

h

πω

ψω

==

⎩⎨⎧

≥≤≤≤

=

(viii) For any state function |Ψ>, we can easily express it as a linear combination of ψn at any time (say, t=0) because of the orthonormal property of basis E:

(ix) And now we can calculate how the wave function evolves over time:

(x) The following are constant of time, or “stationary”, or time independent: (a) Expectation of energy <E>

(b) Pobability density of the energy eigenstates | ψn (x, t)|2. For this reason, the Hamiltonian eigenstates are also known as stationary states.

(c) Energy eigenvalues En. (d) Operator (e.g. H and p) and their corresponding eigenvalues (Schroedinger

picture). The following are NOT constant of time: (a) Eigenstates | ψn> of the Hamiltonian or energy (don’t forget the e-iωt factor

in it). (b) Any state vector |Ψ> (c) <x> (it oscillates!) (d) <p> and so on.

dxxxa

dxxx

or

xxeixxxxxx

xxxxx

j

j

ijijijj

iN

)0,(k sin 2 )0,()0,(a

case, continuousin

)0,(|)0,(a ..aa)(|)0,(a)0,(|a|)0,()0,(|)0,(

)0,(|a )0,(|a)0,(|a)0,(|a)0,(|

n*

j

j

jiii

iN2211

Ψ∫=Ψ∫=

>Ψ=<

=∑>=<∑>=<∑>=Ψ<⇒>∑=>++>+>>=Ψ

ψ

ψ

δψψψψψψψψψ L

[ ]( ) ti

ti

ti

tiN

titi

N

i

i

N

exdxxxa

exa

dxxxa

ex

exexex

txtxtxtx

ω

ω

ω

ωωω

ψ

ψψψ

ψψψ

−

−

−

−−−

Ψ∫∑=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛Ψ∫∑=

∑=

+++=

+++=Ψ

iii

iii

iii

N2211

N2211

k sin')0,'('k sin2

k sin2')0,'('k sin 2

)0,(a

)0,(a)0,(a)0,(a

),(a),(a),(a),(

1

21 L

L

(xi) Example.

A particle of mass m is confined to a one dimensional region 0 ≤ x ≤ a with the potential

At t=0 its normalized wave function is

The Schroedinger eigenfunctions and eigenvalues for the above potential are given as

(a) What is he wave function at a later time t = t0?

⎪⎩

⎪⎨

⎧

>∞<<

<∞=

0 x a x 00

0 x V(x)

⎟⎠⎞

⎜⎝⎛ π

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ π

+==Ψaxsin

axcos 1

5a8 )0t,x(

⎟⎠⎞

⎜⎝⎛ π

⎟⎟⎠

⎞⎜⎜⎝

⎛ π−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ π−⎟⎠⎞

⎜⎝⎛ π

+

⎟⎟⎠

⎞⎜⎜⎝

⎛ π−⎟⎠⎞

⎜⎝⎛ π

+⎟⎟⎠

⎞⎜⎜⎝

⎛ π−⎟⎠⎞

⎜⎝⎛ π

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ π−Ψ+⎟⎟

⎠

⎞⎜⎜⎝

⎛ π−Ψ=

Ψ+Ψ=Ψ∴

Ψ+Ψ=⎟⎠⎞

⎜⎝⎛ π

+⎟⎠⎞

⎜⎝⎛ π

=

⎟⎠⎞

⎜⎝⎛ π

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ π

+==Ψ

−−

axsin

ma2ti

expma2

t3iexp

axcos 1

5a8or

mat2i

expa

x2sin 5a2

ma2ti

expaxsin

5a8

mat2i

exp5

1ma2

tiexp

52

e5

1e5

2 )t,x(

51

52

ax2sin

5a2

axsin

5a8

axsin

axcos 1

5a8 )0t,x(

20

2

20

2

20

2

20

2

20

2

220

2

1

t)/E(i2

t)/E(i1

21

21

hh

hh

hh

hh

2

222

nn ma2nE ;

axnsin

a2 hπ

=⎟⎠⎞

⎜⎝⎛ π

=Ψ

(b) What is the average energy of the syatem at t=0 and at t=t0?

(c) What is the probability that the particle is found in the left half of the box

(i.e., in the region 0 ≤ x ≤ a/2) at t=t0?

.t tand 0ofr t valuesame thehas andmotion ofconstant a is This

ma54

ma24

51

ma254

E51E

54

dxe5

Ee5

E2e5

1e5

2

dxe5

1e5

2He5

1e5

2E

e5

1e5

2 )t,x(

0

2

22

2

22

2

22

21

t)/E(i2

2t)/E(i1

1t)/E(i*2

t)/E(i*1

t)/E(i2

t)/E(i1

t)/E(i*2

t)/E(i*1

t)/E(i2

t)/E(i1

2121

2121

21

==

π=

π+

π=

+=

⎥⎦

⎤⎢⎣

⎡Ψ+Ψ⎥

⎦

⎤⎢⎣

⎡Ψ+Ψ=

⎥⎦

⎤⎢⎣

⎡Ψ+Ψ⎥

⎦

⎤⎢⎣

⎡Ψ+Ψ>=<

Ψ+Ψ=Ψ∴

−−

−−

−−

∫

∫

h

hh

hhhh

hhhh

hh

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

∫⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎭⎬⎫

⎩⎨⎧+

⎭⎬⎫

⎩⎨⎧=

∫⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

∫ ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−+∫ ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

∫ ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛+

∫⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −⎟⎠⎞

⎜⎝⎛=

∫ Ψ=≤≤

2/

0

2/

020

2

2/

020

2

2/

020

2

2/

0

2/

0

20

22/

0

2/

0

22

2/

0

2

20

2

20

2

2/

0

20

ax3sin

3axsin

23cos

54

21

ax3cos

axcos

23cos

54

45a2

45a8

ax

ax2cos

ax

ax2cos

23cos

54

ax4cos

21

21

5a2

ax2cos

21

21

5a8

23cos

ax2sin

5a2

axsin

5a82

ax2sin

5a2

axsin

5a8

2expax2sin

5a2

2exp

axsin

5a8

)t(x, )2/0(

aa

a

a

aa

a

a

a

a

aama

ta

dxma

ta

aa

dxma

ta

dxdx

dxma

t

dx

dxma

tima

ti

dxaxP

ππ

ππ

π

πππ

πππππ

ππ

πππ

ππ

ππππ

h

h

h

h

hh

)(E )(21 )(

22

2

22

xxkxxdxd

mψψψ =+−

h

C. Harmonic oscillator – Hermite polynomials

(i) The harmonic potential is V(x)=kx2/2. Substitute this into the Schroedinger

equation: (ii) Importance of Harminic oscillators: Locally all equilibrium potential always look

like this: Any potential like this can be approximated locally with an open upward

parabola. In other words, a simple harmonic oscillator potential is the first order approximation of equilibrium potential. That is the reason why we can approximate between atoms in a crystal with springs:

Equilibrium position

x

V(x)

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎭⎬⎫

⎩⎨⎧ +⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

20

2

20

2

23

cos1516

21

323

cos54

21

mat

aama

ta

h

h

ππ

πππ

(iii) Insight: From our previous knowledge, we should know that the energy

eigenvalues are:

where the angular frequency ω is defined in the classical way:

Schroedinger equation becomes

(iv) Simplifying the Schroedinger equation:

(v) Solution at large x. At large x (or ξ), above Schroedinger equation becomes

ω+= h)21(n E n

mk =ω

)x(E )x(xm21 )x(

dxd

m222

2

22

ψ=ψω+ψ−h

:becomesequation er Schroeding E,2 K and ,xm Let

)x(E2 )x(xm )x(

xmd

d-

)x(Em

2m

)x(xm )x(dxm

d-

)x(E2m )x(xm )x(dxd- )x(E )x(xm

21 )x(

dxd

m2

2

2

2

22

2

2

22

2

22

2

222

2

22

ω=

ω=ξ

ψω

=ψ⎟⎟⎠

⎞⎜⎜⎝

⎛ ω+ψ

⎟⎟⎠

⎞⎜⎜⎝

⎛ ω⇒

ψω

=ψω

+ψω

⇒

ψ=ψω

+ψ⇒ψ=ψω+ψ−

hh

hh

h

h

hh

h

hh

h

)x(K )x( )x(dd- 2

2

2

ψ=ψξ+ψξ

(vi) Complete solution The complete solution of the Schroedinger equation has to be in the form

Where H(x) is some polynomials which can be solved by series method (see p.

52-54 of textbook). We will not try to do the solution here. Instead, we will focus on some important properties of H(x) here.

(vii) Since series solution is involved here, we know that H(x) is a polynomial. By

convergence of ψ(x), we know H(x) should be finite in the highest order of x. We can use this to label H(x). For example, the highest order of x in Hn(x) is xn. This requirement will “force” E (or K) to taking discrete values En only. As expected,

(viii) Hn(x) is known as the Hermite polynomials and all the coefficients are real. i.e.

H*(x)=H(x). (viii) Insight: Since the Hamiltonian is Hermitian, All ψn(x) must be complete and

orthonormal. Attention: Hermite polynomials itself are not orthogonal, but ψn’s are. (ix) Integration remarks: Integration involving ψn(x) and power of x can be readily calculated using the

following relationship (for generality we start our limit of integration from 0 instead of -∞):

±∞→→⇒

±∞→→⇒=+

−

±

ξψψ

ξψψξψξ

ξ

ξ

as e )( )( of eConvergenc

as e )( 0 )( )(-

2

222

2

2

2

xx

xxxdd

e )H(A )( 2

2ξ

ξξψ−

=

x)()21n(x)(Ex)(xm

21x)(

dxd

2- i.e.

)21n(E

nnnn22

n2

22

n

ωψψψωψ

ω

hh

h

+==+

+=

m

nm

xm-

mn2

nmm*

n dx x)em(x)Hm(HAor (x)dx (x)2

δωωδψψω

=∫=∫∞

∞−

∞

∞−

h

hh

(ix) The normalization constant can be calculated as: Unlike particle in a box, different eigenfunction has different normalization

constant. (x) Hermite polynomials can be generated by the following triangles:

απ

αππθ

αα

ααα

ααα

πααα

α

αα

ααα

αααααα

α

=⇒==∫∫=∫∫=

=⎥⎦⎤

⎢⎣⎡−=

=∫=∫=

−=∫

−=⎥⎦

⎤⎢⎣⎡=∫=∴

−+=⇒−+=

∫=

∞∞∞

∞

∞∞

∞

∞−

∞∞

∞−

∞

01

2

0

R -

0

y -

0

x-

0

20

0

u -

2u -

0

x-

01

10

2-n x-2-n

0

x-1-n x-nn

x-2-n x-1-n x-n x-2-n x-n x-1-n

x-n

0n

I I2RdRedyedxe I

21e

21

)x(u du e21dxxeI

:I and I know toneed weend, At the

I2

)1n(dxex2

)1n( )e(x2-1dxex I

ex2

)1n()ed(x2-1e xex)1n(ex-2)ed(x that Note

dxex I Define

222

2

222

222222

2

d

n!21mA n

41

n ⎟⎠⎞

⎜⎝⎛=

hπω

0 1 2 3 4 5

0 1

1 0 2

2 -2 0 4

3 0 -12 0 8

4 12 0 -48 0 16

5 0 120 0 -160 0 32

(xi) If we denote (n,m) as the coefficient of ξm in Hermite polynomials Hn. Above

construction comes from the recurrence relation: In specific, denote y=(n,m), z=(n+1,m+1), and x=(n-1,m+1) with the following

relative position in the triangle:

m)1,-2n(n-1)-m2(n,m)1,(n )(2nH- )(H2)(H

1)-m1,-2n(nm)m(n, )(2nH d

dH

1-nn1n

1-nn

=+⇒=

=⇒=

+ ξξξξ

ξξ

x

z

y n

n-1

n+1

m m+1

Above recurring relations imply: We just keep on using this equation to construct the element two rows directly

below. The last two elements in a row cannot be done by this way (because there is nothing above!), just add a zero (the second last element) and 2 times the last element of the previous row.

(xii) Some properties of Hermit polynomials (hence ψn): a. If n is even:

Hn is in all even power of x, and hence Hn and ψn are even in x (ψn (x)= ψn (-x)). If n is odd:

Hn is in all odd power of x, and hence Hn and ψn are odd in x (ψn (x)= -ψn (-x)). b. Starting from a positive highest order term anxn, The sign within a Hermite

polynomial alternates. c. Coefficient of xn in Hn = 2 × Coefficient of xn-1 in Hn-1. d. All coefficients are even numbers. The only exception is the “1” in H0=1. e. The complete wave function of a simple harmonic oscillator is:

ψn form a complete orthonomal set. Any function of x satisfying boundary conditions f(∞)=f(-∞)=0 can be expressed as a linear combination of ψn.

f. For a wavefunction in polynomial form:

xn-m1)2n(nz

1)x2n(n1)z(n1)z(m 22nx z1)2(n1)z(m

:(1) into (2) Substitute(2)--- 2nx -2yz

1)m1,-2n(n-m)2(n,1)m1,(n m)1,-2n(n-1)-m2(n,m)1,(n (1)--- 1)y 2(n1)z(m

m)1)(n,2(n1)m1,1)(n(m 1)-m1,-2n(nm)m(n,

+=⇒

+++=+⇒⎟⎠⎞

⎜⎝⎛ +

+=+

=⇒+=++⇒=+

+=+⇒+=+++⇒=

( ) ( )

( ) ω+=ψ

ω=ξξ⎟

⎠⎞

⎜⎝⎛

πω

=ψ ξ−

h

hh

)21n(E is x of eigenvalue Energy

xm with eH!n2

1m x

nn

2/nn

41

n

2

We just need to express it as linear combination of ψ1,ψ2,…. up toψn. Eigenfunction of order larger than n will not contribute to the combination because they will give power of x with order larger than n.

(xiii) For example:

( ) ( ) ( )

( ) ( ) ( )

( ) 2/241

2/241

2/41

2/41

2/241

2/41

2/41

210

2

222

222

2m

12mm 1m

248

1m222

1m2

1 1m

22

1f(x)construct us

ξ

ξξξ

ξξξ

ξξπω

ξπωξ

πω

πω

ξπωξ

πω

πω

ψψψ

−

−−−

−−−

+⎟⎠⎞

⎜⎝⎛=

−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

++=

e

eee

eee

Let

h

hhh

hhh

[ ] 2/01

1n1-n

nn

2

ea xa xaxa )x(f ξ−− ++++= L

( )

( )

( )

22

2

21

22132

22

de)2(22

dm

)48(8

1em

dxe2me)24(!22

1mdx)x(f)x(H

21

21

de21

de2

dm

2em

dxe2me)2(!12

1mdx)x(f)x(H

1 1

de1

de2

dm

2em

dxe2memdx)x(f)x(H

: work thinghow is Here

2

2

22

2

2

2

22

2

2

2

22

24

2421

2/241

2/2

2

41

*2

2

2

221

2/241

2/

1

41

*1

2

221

2/241

2/41

*0

==

π⎥⎦⎤

⎢⎣⎡ −

⋅⋅

⋅π

=

ξξ−ξπ

=

ξω

ξ−ξ⎟⎠⎞

⎜⎝⎛

πω

=

ξ+ξ⎟⎠⎞

⎜⎝⎛

πω

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−ξ

⋅⎟⎠⎞

⎜⎝⎛

πω

=

=ππ

=

ξξπ

=

ξξπ

=

ξω

ξ⎟⎠⎞

⎜⎝⎛

πω

=

ξ+ξ⎟⎠⎞

⎜⎝⎛

πω

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ξ

⋅⎟⎠⎞

⎜⎝⎛

πω

=

=ππ

=

ξπ

=

ξξπ

=

ξω

ξ⎟⎠⎞

⎜⎝⎛

πω

=

ξ+ξ⎟⎠⎞

⎜⎝⎛

πω

⎟⎠⎞

⎜⎝⎛

πω

=

ξ−∞

∞−

ξ−∞

∞−

ξ−ξ−∞

∞−

∞

∞−

ξ−∞

∞−

ξ−∞

∞−

ξ−∞

∞−

ξ−ξ−∞

∞−

∞

∞−

ξ−∞

∞−

ξ−∞

∞−

ξ−∞

∞−

ξ−ξ−∞

∞−

∞

∞−

∫

∫

∫∫

∫

∫

∫

∫∫

∫

∫

∫

∫∫

h

h

hh

h

h

hh

h

h

hh

D. Simple Harmonic Oscillator - Ladder operator. (ii) (iii) When this happen, we can always write the operator in expression of ladder operators. (iv) For simplicity, let us consider operator Z, with eigenvectors |ψ0>, |ψ1>,|ψ2>,…..

and corresponding eigenvalues B, A+B, A+2B, …… In its own (Z) representation, operator Z is simply the matrix

(iv) Ladder operators are formed by step up and step down operators. Definition of step up operaptor a+ (in Z-representation): Where f(i) is some functional form of index i. Under this definition, for finite cases, we can assume the “boundary values” f(n)=f(0)=0 (since they are not used). The self adjoint of a+ is known as the step down operator:

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

++

+

=

BBA

BA2

BnA

ZO 0

0

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=+

00000)1(f0000

0)2(f000000000)1n(f0

aOO

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−

=

0)1(f00000)2(f000000000)1n(f00000

a O

O

From these we have: (v) Meaning of step up and step down operators:

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=

+

+

)1(00000)2(0000000000)1(000000

,000000)1(00000)2(0000000000)1(

0)1(00000)2(000000000)1(00000

00000)1(0000

0)2(000000000)1(0

a

ff

nfaa

Similarly

ff

nf

ff

nf

ff

nf

a

O

O

O

OOO

0|

|)1(

0

1

0

0)1(00000)2(000000000)1(00000

|

0|

|)(

0

1

0

00000)1(0000

0)2(000000000)1(0

|

imin

1-ii

imax

1ii

>=

>−=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−

>=

>=

>=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=>

+

+

++

ψ

ψψ

ψ

ψψ

a

if

ff

nfa

a

iff

f

nf

a

M

M

O

O

M

MOO

(vi) a+ and a are not Hermitian. They are just a mathematical tool! (vii) a+a and aa+ are diagonal matrices. From this we know ladder operators have to satisfy the general condition: [Z, a+a]=[Z, aa+]=0, where a and are ladder operators correspond to operator Z. (viii) a+ and a do not commute:

0).f(0)f(n) that (Note

)1()0(00000)2()1(0000000000)1()2(00000)()1(

a],[

)1(00000)2(0000000000)1(000000

;

000000)1(00000)2(0000000000)1(

a

==

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−−

−−−−−

=∴

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=

+

++

ffff

nfnfnfnf

a

ff

nfaa

ff

nf

a

O

O

O

(ix) [Z,a+] or [Z,a] provides a recurrence relationship:

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

−

+

00000)1(0000

000000)2(00000)1(0

00000)1(0000

0)2(0000000000)1(0

00000000000000000000

Za

2

1

1

2

3

fZ

nfZnfZ

ff

nf

ZZ

Z

Z

x

x

n

O

OO

>−>=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−

−−−−

=∴

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−−

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛ −

=

+++

−−

−

+

−

−

+

11

12

22

1

1

2

1

1

2

3

|)()(|]a,[or

00000)1()(0000

000000)2()(00000)1()(0

]a,[

00000)1(0000

000000)2(00000)1(0

00000000000000000000

00000)1(0000

0)2(0000000000)1(0

Za

iiii

xx

xn

x

x

n

ifZZZ

fZZ

nfZZnfZZ

Z

fZ

nfZnfZ

ZZ

Z

Z

ff

nf

ψψ

O

O

OO

(x) So far the formulation is quite general. We do not even put any requirement on the explicit form of f(i). To make use of the above recurrence relationship, we need to know [Z, a+] explicitly. Note that [Z, a+] itself looks like a step up operator. (xi) For example, let us consider the simplest situations under which the ladder operators are most effectively used:

We require Z to have eigenvalues in the form A+B, 2A+B, 3A+B, …., nA+B and so on, where A and B are some constants. We can use n to label the corresponding eigenvector of the operator (n=1,2,…n in this case).

Examples of these operators are the z-component of angular momentum, number operator in second quantization, and of course, the Hamiltonian of the simple harmonic oscillator. In the case of simple harmonic Hamiltonian, A=nhω and B= hω/2. (xi) Now Z|ψj>=jA+B|ψj>, we can calculate [Z,a+] explicitly:

) |Aa( |f(j)

|)(f(j)|f(j)B]1)A[(j

|)(a|f(j) Z

|a|Za

|]a[Za |]aZ,[

1

11

1

>=>=

>+−>++=

>+−>=

>−>=

>−=>

++

++

++

++

+++

jj

jj

jj

jj

jj

A

BjA

BjA

Z

Z

ψψ

ψψ

ψψ

ψψ

ψψ

Put this result into our recurrence relationship:

AZZifZZA

ifZZZA

iiiiij

iiiijj

+=⇒>−>=∴

>−>=>=>

++++

+++

++

1111

111

| )()(|f(j)

|)()(|]a,[ and |f(j) |]aZ,[

ψψ

ψψψψ

(xi) Above looks obvious, because we are working in Z-representation itself! In a real problem, Z is posted in another representation and we have no idea what are its eienvalues and eigenvectors. In situation like this, we will follow the above procedure in the given representation: (a) In the given representation, construct step up or step down operators by investigating operators satisfying the commutation relation [Z, a+a] = 0. Show that it is a proper ladder operators by showing

>−=>>=> −++

1ii1ii |)1( | and |)( | ψψψψ ifaifa

It is useful to know f(i). (b) Once we know a+ and a, we can construct [Z, a+] (or [Z, a]). If it is one of the simple case of eigenvalues with equal separation, [Z,a] will be proportional to a+ and the “proportional constant” will be equal to the equal increment A in eigenvcalues (as we have shown above). (c) We can then plug [Z, a+] into the recurrence relationship

>−>= +++

11 |)()(|]a,[ iiii ifZZZ ψψ

to get a relationship between Zi+1 and Zi. (d) If Z is infinite dimension and have no upper AND lower bounds, we need to know some eigenvalues of Z to begin with. If Z has upper or lower bounds, we can determine the limiting eigenvalues (at least in most simple cases) by writing Z in terms of a and a+, and make use of the equation a+|ψmax>=0 and a|ψmin>=0. (xii) Now a practical case for simple harmonic oscillator. From our previous knowledge we know that it is a simple case with consecutive eigenvalues differ by ħω. Step 1 (the most difficult) is to construct a+ and a in x-representation. In x-representation, the Hamiltonian is

If we can “factorize” H in the form (Ax+iBp)(Ax-iBp) where A and B are some real constants, then one factor is the complex conjugate of the other and it just looks like a+a (or aa+ which we do not know at this point) and H must commute with it (since they are equal!). So we now work out the algebra:

21

2p

21

222

222

2

22

xmm

xmdxd

mH ωω +=+−=

h

( )( )

( )( )

operator). (a)down step toapplied becan argument Similar operator. down" step" a is a hence and E eigenvalue with H ofr eigenvectoan also is |a s,other wordIn

|aE |Ha

)(*tricky! |21aaa

|21aaa

|a21aaa|a

21aa|Ha

:operator down) step(not up step areally is a sure make to|Ha calculate uslet Nowplan. nalplan.origi original

the toaccordancein H with commute still a)a(or aa hence andconstant ajust isit

but ,21 ermconstant t theof because original thefrom offslightly isplan The

21aa

21aaH :Also

],[i2

i-2

i2

p2m1i x

2p,

2m1i x

2],[a know wenotation, With this

p2m1i x

2 a)or (a and

p2m1i x

2 )a(or asuspect can point we At this

21p

2m1i x

2p

2m1i x

2

21

2p H

21AB and

2m1B and

21Aset weif So

],[

iBpAxiBpAx

nn

nn

n

n

n

nnn

n

22

2

2

2222

2

2

2222

2222

2222

++

+

+

++

++

++++++

++

++

++

+

+

+

+>

>+=

>+=

>⎟⎠⎞

⎜⎝⎛ −+=

>⎟⎠⎞

⎜⎝⎛ +=

>⎟⎠⎞

⎜⎝⎛ +>=⎟

⎠⎞

⎜⎝⎛ +>=

>

+=−=

−==⎥⎦⎤

⎢⎣⎡=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

−⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛+=+=

===

++=

−+=

−++=−+

ωψ

ψω

ψω

ψωω

ψω

ψωψωψ

ψ

ω

ωω

ωωωω

ωω

ω

ω

ωωωω

ωω

h

h

h

hh

h

hh

h

hh

h

h

h

pxpxxp

mma

m

m

mmxmm

m

iABpBxApxiABpBxA

iABxpiABpxpBxA

Since we now know that a+ and a are step up and step down operators, above relationshop provide us a recurrence relationship to calculate En+1 from En:

( ) ωψωψ hh +=⇒>+>= +++

n1nnnn E E |aE|Ha as expected. (xii) However, there are two things we have to address before the problem is solved: (a) We need to know E0 ! (b) We know that a+|ψn> is ( )|ψn+1>, but what is in that factor ( )? (i.e. what is f(i)?) To answer the first question, we need to work on the ground state with the fact that a|ψ0>=0.

ωωωωωω

ω

ψωψωψ

ωω

hhLLhhhh

h

hh

hh

21n E

23.

21E EE

.21E Therefore,

|210|

21aa|H

21aa

21aaH

1n1n1n

0

000

+==+=⇒+=

=

>+>=⎟⎠⎞

⎜⎝⎛ +>=∴

+=−=

++

+

++

To answer the second question, let a+|ψi >=√(fi) |ψi+1 >:

>|>=|

>|+>=|

+=⇒+=∴

>|+>=|⎥⎦⎤

⎢⎣⎡ ++>=|⎟

⎠⎞

⎜⎝⎛ +>=|∴

+=⇒−=

>=| =< >| <∴

−

++

+

++

+++

1ii

1ii

ii

iiii

ii1i1i

a 1,-it with coefficien in the i thereplacingby Similarly,

)1(a

)1(f )1(f

)1(21)

21(

21Haa

21Haa

21aaHbut

1a| f

ψωψ

ψωψ

ωω

ψωψωωψωψ

ωω

ψψψψ

h

h

hh

hhhh

hh

i

ior

ii

ii

ai

E. Delta function potential (i) We know that any wavefunction has to be continuous. The second derivative in the Schroedinger equation cannot be calculated at points where the wavefunction is not continuous. (ii) For the first derivative of the wavefunction, integrate the Schroedinger equation:

.0)()( that so continuous is V(x) if

)continuous is )( (i.e. 0)(2

)()()()(2

)()()()(2

2

2

2

22

=Ψ∫

Ψ=⎥⎦⎤

⎢⎣⎡ Ψ−⇒

Ψ∫=Ψ∫+⎥⎦⎤

⎢⎣⎡ Ψ−⇒Ψ=Ψ+Ψ−

+−

+

−

+−

+−

+

−

dxxxV

xdxdx

dxd

m

dxxEdxxxVxdxd

mxExxVx

dxd

m

xx

x

x

xx

xx

x

x

εε

ε

ε

εε

εε

ε

ε

h

hh

(iii) The first derivative of the wavefunction does not need to be continuous at points where V(x) is infinite. At points where V(x) is infinite,

0)()( ≠Ψ∫+− dxxxVx

xεε

Example is particle in a box:

1

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=+

000000000

02000003000000

a

form,matrix in So,

ωω

ω

h

h

h

O

∞

x

x=a x=0

∞ ∞

Wavefunction is not continuous at these two points.

(iv) Consider a delta function potential at x=0: (v) We should now know that the first derivative of the wavefunction will not be continuous at x=0. Instead, at x=0:

)0(2)(')('

)0()(2

)()()(2

)()()()(2

)()()()(2

2

2

2

2

2

22

Ψ−=−Ψ−+Ψ⇒

Ψ=⎥⎦⎤

⎢⎣⎡ Ψ−⇒

Ψ∫=⎥⎦⎤

⎢⎣⎡ Ψ−⇒

Ψ∫=Ψ∫−⎥⎦⎤

⎢⎣⎡ Ψ−⇒Ψ=Ψ−Ψ−

+

−

+−

+

−

+−

+−

+

−

h

h

h

hh

αεε

α

αδ

αδαδ

ε

ε

εε

ε

ε

εε

εε

ε

ε

m

xdxd

m

dxxxxdxd

m

dxxEdxxxxdxd

mxExxx

dxd

mx

x

(vi) We can now solve the Schroedinger equation with the above boundary conditions. First consider the bound state situation:

⎩⎨⎧

>=Ψ<=Ψ

=Ψ

Ψ+=Ψ⇒

−=Ψ=Ψ

<

Ψ−=Ψ⇒Ψ=Ψ−

=≠

−

−

0 xD)(0 xC)(

)(

:integrable be to)(For )(

2 re whe )()(

0)(E state Bound 1. Case

)(2)()()(2

:0V(x) 0,point xany At

II

I

222

2

2

22

2

2

22

x

x

xx

exex

x

xBeAex

mExxdxd

xmExdxdxEx

dxd

m

κ

κ

κκ

κκh

h

h

x

V(x)

V(x)=-αδ(x)

2

22

II

I

II

I

20

22

220

2

0

202

II

I

III

2)()()0(2)(')('

0 x)('0 x)(')('

:conditionboundary theof because any value ecannot tak Now0 x)(0 x)()(

C 1C

121-2C

1dxC2

1dxCdxC 1dx)(

:ionnormalizatby determined is C0 xC)(0 xC)(

)(

DC )0()0( 0at x continuous is

h

hhακ

κακκκκαεε

κκκκ

κκκ

κκ

κ

κ

κ

κ

κ

κ

κ

κκ

κ

κ

m

mmex

exx

exexx

e

e

eex

exex

x

x

x

x

x

x

x

xx

x

x

−=⇒

−=−−⇒Ψ−=−Ψ−+Ψ

⎩⎨⎧

>−=Ψ<=Ψ

=Ψ

⎩⎨⎧

>=Ψ<=Ψ

=Ψ∴

=⇒=⇒

=⎥⎦⎤

⎢⎣⎡⇒

=∫⇒

=∫+∫⇒=Ψ∫

⎩⎨⎧

>=Ψ<=Ψ

=Ψ∴

=⇒Ψ=Ψ⇒=Ψ

−

−

∞−

−∞

−∞∞−

∞∞−

−

(vii) There is only one bound state for delta function potential, no matter what is the value of α. In other word, there is only one eigenfunction with E<0 in the Hilbert space that can satisfy the boundary conditions required by the delta function potential.

x

Ψ(x)

V(x)=-αδ(x)

(viii)

negative. isenergy theexpected, As state. bound theofenergy theis This2

2

2

)()(2

)()(2

2

2

4

222

22

22

2

22

h

h

h

h

hh

α

α

κ

κ

mE

mm

E

mE

xExm

xExdxd

m

−=⇒

−=⇒

−=⇒

Ψ=Ψ−⇒Ψ=Ψ−

(ix) Now consider the continuous case (E>0).

( ) ( ) 2

22

22

22

ikx-ikxII

ikx-ikxI

III

ikx-ikxII

ikx-ikxI

ikx-ikx

222

2

2

22

2

2

22

k where i21B-i21AG-F

k21B-

k21AG-F

ik21B-

ik2-1AG-F

)(2Bik)-(Aik-Gik)-Fik()0(2)(')('

0 xikik)('0 xikik)('

)('

:0at x )(' ofity discontinu heconsider t NowGFBA )0()0(

:0at x Continuity0 x)(0 x)(

)(

)(

2 re whe )()(

0)(E states Scattering 2. Case

)(2)()()(2

:0V(x) 0,point xany At

h

hh

hh

hh

h

h

h

αβββ

αα

αα

ααεε

κ

m

mimi

mm

BAmmeGeFxeBeAx

x

x

GeFexBeAex

x

BeAex

mExkxdxd

xmExdxdxEx

dxd

m

=−+=⇒

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛ +=⇒

⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛=⇒

+−=⇒Ψ−=−Ψ−+Ψ

⎩⎨⎧

>−=Ψ<−=Ψ

=Ψ

=Ψ+=+⇒Ψ=Ψ

=⎩⎨⎧

>+=Ψ<+=Ψ

=Ψ⇒

+=Ψ⇒

=Ψ−=Ψ

>

Ψ−=Ψ⇒Ψ=Ψ−

=≠

(x) There are two boundary conditions for the coefficients to follow, and then there is one normalization condition. There are a total of four undetermined coefficients, so we have the freedom to assign one arbitrarily. (xi) There are four components involve in the above solution. Physical meaning of these components: (xii) In here, we can assume the wave is incoming to the potential from the left and “arbitrarily” choose G=0. Under this assumption:

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) )1

i(or i1

ii1B2

i2i1B2i21B-i21ABA

)1

i1(or i1

2i12i21AA i21FF i21:Ffor Solving

k e wher i21B-i21AF

FBA

2

2

2

2

AA

A

AAF

AF

m

βββ

βββ

ββββββ

β

ββββ

αβββ

++

−=−⇒

=−⇒−+=+++

−=⇒

=−⇒++−=+−

=−+=

=+

h

(xiii) The reflection and transmission coefficients are defined as:

( )

2

22

2

2

2

2

2

2

2

22

4222

2

2

22

2

2

2E1

1

k1

11

1i1

1AFT

2E1

1k1

1

k1

111

11i1

iABR

hh

hhh

ααββ

αααββ

ββ

β

mm

mmm

+=

⎟⎠⎞

⎜⎝⎛+

=+

=−

==

+=

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=+

=+

=−

==

Aeikx

Be-ikx Ge-ikx

Feikx

Notes: (a) R+T=1 (b) R and T depends on E. The higher the energy, the smaller the R and the

larger the T. Does it make sense? (xiv) We can define Transfer matrix M in relating the coefficients at right (F, G) to that at left (A, B):

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

ββββi1i

ii1MMMM

M

:potential delta of case In theBA

MMMM

BA

MGF

2221

1211

2221

1211

Properties of M for symmetric potential: (a) det(M)=1 (b) M11=M22* (c) M12= -M12* (i.e. pure impurity) = - M21 = M21* (i.e. M12 and M21 are

conjugate of each other). (xv) It is more appropriate to define another scattering matrix S as:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛GA

SSSS

GA

SFB

2221

1211

This relates the outgoing waves (B and F after scattering) from the potential to the incoming waves (A and G).

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−

−−=⎟⎟⎠

⎞⎜⎜⎝

⎛=

ββ

β

βββ

i1i

i11

i11

i1i

SSSS

S

:potential delta of case In the

2221

1211

Properties of S for symmetric potential: (a) S+S = 1 (b) S11=S22

(c) S12= S12 F. Finite square well (i)

x

-V0

a -a

V(x)

I II III

⎩⎨⎧

><<

=a |x| 0axa- V-

V(x) 0

Note that the potential is symmetric in this problem. This introduces special properties to the energy eigen functions. (ii) For symmetric potential V(x)=V(-x):

[ ] [ ]

odd. is )( andeven is )( (iii) other.each oft independenlinearly are )( and )( (ii)

E. eigenvalue same with theionseigenfunct are )( and )( (i) clear that isIt

)(- )(21)( and )( )(

21)(

constructcan weHowever, x.of functions oddor even benot may twoThese E. eigenvalue same for the solutionst independenlinearly twoare )( and )( Suppose

E. eigenvaluefor ion eigenfunctt independenlinear one than more are There :case Degenerate 2. Casepotential.

symmetric a of econsequencdirect a is This function. )1( oddor )1(even either is )( 1 1 )()( )()( x ofsign theChanging

)()(Let other.each ofdependent linearly bemust )( and )(

E. eigenvaluefor ion eigenfunctt independenlinear oneonly is There :case degenerate-Non 1. CaseE. eigenvalue same theofequation er Schroeding original theofsolution are )( and )(Both

)()(V(x))(2

)()(V(-x))(2

)()(V(x))(2

2121

21

22

2

22

2

22

2

22

xxxx

xx

xxxxxx

xx

xxx

xxxx

xx

xx

xExxdxd

m

xExxdxd

mxExx

dxd

m

−+

−+

−+

−+

ΨΨΨΨ

ΨΨ

ΨΨ=ΨΨ+Ψ=Ψ

ΨΨ

−=+=Ψ±=⇒=⇒−Ψ=−Ψ∴

−Ψ=Ψ⇒Ψ=−Ψ

−ΨΨ

−ΨΨ∴

−Ψ=−Ψ+−Ψ−⇒

−Ψ=−Ψ+−Ψ−⇒Ψ=Ψ+Ψ−

εεεεε

εε

h

hh

In summary: If the potential is symmetry such that V(x)=V(-x), then the solution of the time independent Schroedinger equation must be even or odd for non-degenerate case, and can be even or odd for degenerate case. (iii) Case 1 – bound state E<0

( )

( )

(odd).x sin orD (even)x cos C beonly can )( hence ),()( that know weSincex sin D x cos C )(

ore D e C )(

assolution theteeither wrican We

VE2 where )()(

)(VE2)()()(V)(2

:-VV(x) a,aFor A]B hence ),()( that know weSince [

a xB)(a x)(

)(

:integrable be to)(For )(

2 re whe )()(

)(2)()()(2

:0V(x) a,|x|For

II

II

x i-xiII

022

2

2

022

2

02

220

III

I

222

2

2

22

2

2

22

ll

ll

hll

h

h

h

h

h

ll

xxxx

x

mxxdxd

xmxdxdxExx

dxd

m

xxx

exAex

x

xBeAex

mExxdxd

xmExdxdxEx

dxd

m

x

x

xx

ΨΨ±=−Ψ+=Ψ

+=Ψ

+=Ψ−=Ψ⇒

Ψ+−=Ψ⇒Ψ=Ψ−Ψ−

=<<−±=Ψ±=−Ψ

⎩⎨⎧

−>=Ψ−<=Ψ

=Ψ

Ψ+=Ψ⇒

−=Ψ=Ψ⇒

Ψ−=Ψ⇒Ψ=Ψ−

=>

+

−

−

κ

κ

κκ

κκ

(iv) Even solution

( ) z) inside"" is E that (Note a V2z and z a VE2 zaLet

atan )1/()2(-(2)--- a sin Ce A- aat x )x(' of Continuity

-(1)--- a cos CeA aat x )x( of Continuity-a.or x aeither xat conditionsboundary

heconsider t toneedjust wesolution, above in the incluced are proporties symmetric theBecauseeA )x(

x cos C )x(eA )x(

02002

a

a

xIII

II

xI

hhl

ll

ll

l

l

mm==+⇒=

=⇒−=⇒=Ψ

=⇒=Ψ

==

=Ψ

=Ψ=Ψ

−

−

−

κκ κ

κ

κ

κ

1z

z z tan ztan

azzz

a1 atan

zza1

az

az

E2

2022

0

220

220

2

−⎟⎠⎞

⎜⎝⎛=⇒=−⇒=

−=⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=−=∴

ll

h

κ

κ m

This can be solved graphically:

(v) Odd solution

( )

1zz

- zcot zcot azzz

a1 a cot

zza1

az

az

E2

z) inside"" is E that (Note a V2z and z a VE2 zaLet

acot - )1/()2(-(2)--- a cos Ce A- aat x )x(' of Continuity

-(1)--- asin CeA aat x )x( of Continuity

eA )x(

xsin C )x(eA - )x(

2022

0

220

220

2

02002

a

a

xIII

II

xI

−⎟⎠⎞

⎜⎝⎛=⇒−=−⇒−=

−=⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=−=∴

==+⇒=

=⇒=⇒=Ψ

=⇒=Ψ

=Ψ

=Ψ=Ψ

−

−

−

ll

h

hhl

ll

ll

l

l

κ

κ

κκ κ

κ

κ

κ

m

mm

This can be solved graphically:

y

π/2 π 3π/2 2π 5π/2z

y = tan z

1zz0 −⎟

⎠⎞

⎜⎝⎛=y

z = z0

(iv) From these graphs, we can see: (a) As z0 (i.e. V0) is increased, the red curve will be pulled taller and wider at the same time. (b) The number of bound states depends on the value of z0 (i.e. V0). The deeper the well (i.e. larger V0), the more bound states are there. (c) For very shallow well (i.e. small z0 when z0 < π/2), there will be only one bound state. However, no matter how shallow is the will, there will be at least one bound state. As z0 →0 (i.e. V0→0), there is still one bound state with E → 0. (d) As z0 → ∞ (i.e. V0→ ∞), the well become infinite.

n)(even 2

n z

m zcot

)z (large zz

zcot 1zz

zcot

:solution Odd

n) (odd 2

n z

)z (large zz

z tan 1zz

ztan

:solutionEven

00

20

00

20

ππ

π

≈⇒

≈⇒

−=⇒−⎟⎠⎞

⎜⎝⎛−=

≈⇒

=⇒−⎟⎠⎞

⎜⎝⎛=

z

y

π/2 π 3π/2 2π 5π/2

y = -cot z

1zz0 −⎟

⎠⎞

⎜⎝⎛=y

z = z0

( ) ( )

02

222

2

22

0202

V2m(2a)nE

(2a)nVE2 a VE2z

−≈⇒

≈+⇒+=

h

hh

π

πmm

This is the same as our previous results on infinite square well of width 2a: (a) n takes odd integers for even solution, even integers for odd solution. (b) The energy value is pressed down by an amount of V0 as expected. (vi) Case 2 – continuous state E>0

( )

(10)--- DC

)k

(1)k

1(

)k

-1()k

1(

21

GF

(9) and (8)

(9)--- )k

G(121)

k1C(

21G

)Di(k)-Ci(k2Gik (7)ik(6)

(8)--- )k

-1D(21)

k1C(

21F

)-Di(k)Ci(k2Fik (7)ik(6)

(7)--- Di Ci Gik Fik )a(')a('

(6)--- D C G F )a()a(:aat x conditions continuityconsider Now

-(5)--- BA

)k(1)k(1

)k(1)k(1

21

DC

(4) and (3)

(4)--- )kB(121)kA(1

21D

k)Bi(k)-Ai(2Di (2)(1)

(3)--- )kB(121)kA(1

21C

k)-Bi()Ai(k2Ci (2)(1)

(2)--- Bik Aik Di Ci )a(')-a('

(1)--- B AD C )a()-a(

E2k and VE2 where

a x G F)(axa- D C)(a x B A)(

)(

:)(for Solution

k)a(k)a(

k)a(k)a(

k)a(k)a(

aaka

k)a(k)a(

aaka

aakakaIIIII

aakakaIIIII

a)-k()ak(

a)k(k)a(

a)-k()ak(

kakaa

a)k(k)a(

kakaa

kakaaaIII

kakaaaIII

202

kxkxIII

II

kxkxI

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛⇒

++−=⇒

++=⇒−×

++=⇒

++=⇒+×

−=−⇒Ψ=Ψ

+=+⇒Ψ=Ψ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−

−+=⎟⎟

⎠

⎞⎜⎜⎝

⎛⇒

++−=⇒

++=⇒−×

−++=⇒

++=⇒+×

−=−⇒−Ψ=Ψ

+=+⇒−Ψ=Ψ

=+=

⎪⎩

⎪⎨

⎧

>+=Ψ<<+=Ψ−<+=Ψ

=Ψ

Ψ

−−+

+−−

−−+

−−

+−−

−

−−

−−

+−

+−

+−

−

+−

−−

−−

−−

−

−

−

ll

ll

ll

ll

ll

ll

ll

ll

ll

ll

ll

l

ll

l

ll

ll

ll

ll

ll

ll

ll

ll

ll

ll

ll

ll

ll

llll

ll

llll

ll

hhl

ii

ii

ii

iii

ii

iii

iiii

iiii

ii

ii

ii

iii

ii

iii

iiii

iiii

ii

xixi

ii

ee

ee

ee

eee

ee

eee

eeee

eeee

ee

ee

ee

eeei

ee

eeei

eeee

eeee

mm

eexeexeex

x

x

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ −

−⎟⎠⎞

⎜⎝⎛ +

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

++−+++

=

=+=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+++−−−+++−

++−+−+++=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−

−+⋅

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⇒

−

−−+−

−+−−

−−+−

−+−−

+−

+−

−−+

+−−

ka2

ka2

k)a(2k)a(2a2a2

a2a2k)a(2k)a(2

k)a(2k)a(2a2a2

a2a2k)a(2k)a(2

a)-k()ak(

a)k(k)a(

k)a(k)a(

k)a(k)a(

a2sin2

ia2cosa2sin2i

a2sin2ia2sin

2ia2cos

)(2)-(2-)-(2)(2

41 M

k -k and k

kLet

)k

)(1k

(1)k)(1k

1()k)(1k

(1)k)(1k

1(

)k)(1k

-1()k)(1k

1()k)(1k

-1()k)(1k

1(

41

)k(1)k(1

)k(1)k(1

21

)k

(1)k

1(

)k

-1()k

1(

21M

BA

MGF

)10(&)5(

i

i

iiii

iiii

iiii

iiii

ii

ii

ii

ii

e

e

eeeeeeee

eeee

eeee

ee

ee

ee

ee

lll

lll

l

ll

l

ll

l

l

l

l

l

ll

l

l

l

l

l

l

l

ll

llll

ll

llll

llll

llll

llll

ll

ll

ll

ll

ηε

εη

ηηεεεεηη

εη

Note this satisfy the following properties of M: (a) det(M)=1 (b) M11=M22* (c) M12= -M12* (i.e. pure impurity) = - M21 = M21* (i.e. M12 and M21 are

conjugate of each other). R and T can be determined from M by putting G=0:

22

211211

12111211

2

22

212

22

212221

2221

1211

2221

1211

MMMM

AF

ABMM

AFBMAMF

MM

ABR

MM

AB 0BMAM

0BMAMBMAMF

BA

MMMM

0F

−=⇒

+=⇒+=

==⇒

−=⇒=+

⎩⎨⎧

=++=

⇒⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛∴

( )

1 a2sina2cos4a2sina2cos4

a2sina2cos4a24)sin-(4

a2sina2cos4a2sin4

a2sina2cos4a2sin

a2sina2cos44 RT

a2sin11

a2sin)4(a2cos44

a2sina2cos44

a2sin4

a2cos

1 a2sin

4a2cos

a2sina2cos T

4 k

-k and kk

that Note

a2sin4

a2cos

a2sin4

a2sin4

a2cos

a2sin2

ia2cos

a2sin2i

a2sin2ia2sin

2ia2cos

MMMMT

a2sina2cos4a2sin

a2sin2

ia2cos

a2sin2i

MM

R

:details in the plug NowMMMM

AFT

222

222

222

22

222

22

222

22

222

22222

2222

222

22

222

22

22

2

22

22

22

2

ka2

ka22

22

211211

222

22

2

ka2

2

22

21

2

22

211211

2

=++

=

++

=

++

=

++

+=+

+=

++=

+=

+=

+

+=∴

=−⇒=+=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=

⎟⎠⎞

⎜⎝⎛ −

+⎟⎠⎞

⎜⎝⎛ +=−=

+=

⎟⎠⎞

⎜⎝⎛ −

==

−==⇒∴

−

ll

ll

ll

l

ll

l

ll

l

ll

lll

llllll

ll

l

ll

l

ll

lll

ll

llll

ll

l

ll

l

ηηη

ηη

εη

εη

εε

ηηη

εηεη

η

εη

η

εεη

ηε

η

ε

i

i

i

ee

e

(2)--- BMAMG (1)--- BMAMF

BA

MMMM

GF

:follow as derived becan matrix Scattering

2221

1211

2221

1211

+=+=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ −

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛∴

=+=

+=

+−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛++=+=

+=⇒

−

a2sin2i-1

1a2sin2i-

a2sin

2ia2cos

MM

M1

M1

MM-

S

:details in the plug Now

GA

MM

M1

M1

MM-

GA

SSSS

FB

1) (det(M) G MM

AM

1

GMMA

Mdet(M)

GMMA

MMMMM

GM

1AMM

-MAMBMAMF

:(1) into thisSubstitute

GM

1AMM-B (2)

:G andA of in terms F and B Write

ka2

22

12

22

2222

21

22

12

22

2222

21

2221

1211

22

12

22

22

12

22

22

12

22

21122211

2222

2112111211

2222

21

l

l

llε

ε

η

ie

(vii) Now look at

4a. of differencepath a with ceinterferen veconstructi a toscorrespond This

.4nna22

2let well,e within th wave theofnumber wavey theeffectivel is

integers) (n na2 whenmaximum a with a, with oscillates T

a2sin11 T 22

a=⇒=∴

=

==

+=

λπλπ

λπ

π

ε

ll

l

l

T is also energy dependence. When energy is small, there is a higher chance for the particle to get trapped by the well and less chance to be transmitted.

x

-V0

a -a

V(x)

I II III

Path difference = 4a