introduction to quantum mechanics unit 2. time independent...
Post on 11-Aug-2020
8 Views
Preview:
TRANSCRIPT
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
top related