Quantum Mechanics
Lecture 2
1
Recall from the previous lecture: the Schrödinger equation for a particle
of mass
m
in a 1D potential
V(x) is
∂ψ ( x, t )
= 2 ∂ 2ψ ( x, t )
+ V ( x)ψ ( x, t )
i=
=−
2
2m ∂x
∂t
(2.1)
and its bound state wave function is square integrable and normalisable:
∫
ψ dV = 1
2
(2.2)
all space
We shall now derive the time independent Schrödinger equation
and then apply it to particular cases.
2
Time independent Schrödinger equation (TiSE).
When the Hamiltonian has no explicit time dependence, the Schrödinger
eqn can be separated by the substitution
ψ ( x, t ) = u ( x)e −iEt / =
hence
∂
E
ψ ( x, t ) = −i u ( x)e − iEt / =
∂t
=
and after cancellation of the exponential factor we have
Hˆ u ( x) = Eu ( x)
where the function
u(x)
(2.3)
is called the spatial wave function.
This is the time independent Schrödinger eqn. Mathematically speaking
it is an eigenvalue equation: the solutions
u(x)
are the eigenfunctions
and the corresponding values of E are the eigenvalues.
A system that is described by the TiSE is said to be in a stationary state.
3
Continuity of the wave function and of its derivative.
We shall for the time being consider only stationary states, and we shall
say “wave function” meaning the spatial wfn
The 1D TiSE for a particle of mass
m
u(x).
in the potential
V(x)
is
= 2 d 2u ( x)
−
+ V ( x)u ( x) = Eu ( x )
2
2m dx
or
2m
u ′′ ( x ) = 2 ⎡⎣V ( x ) − E ⎤⎦ u ( x )
=
where
(2.4)
u′′ ( x ) ≡ d 2u ( x ) dx 2
Let us integrate this eqn from
x0
to
x:
4
x
2m
u ′ ( x ) = u ′ ( x0 ) + 2 ∫ ⎡⎣V ( t ) − E ⎤⎦ u ( t ) dt
= x0
Now we postulate that the wfn is continuous everywhere.
This is plausible since the probability of finding the electron at two
points separated by an infinitesimal distance should not change
discontinuously.
Then, if the P.E. is continuous in the interval [x0,x], we have
2m
⎡V x + θ Δx ) − E ⎤⎦ u ( x0 + θ Δx ) Δx
2 ⎣ ( 0
=
where θ ∈ [ 0,1] and Δx = x − x0
we get
and taking the limit x → x0 + i.e. Δx → 0
u ′ ( x ) = u ′ ( x0 ) +
lim ⎡⎣u ′ ( x ) − u ′ ( x0 ) ⎤⎦ = 0
x → x0+
i.e. the derivative of the wfn is also continuous.
5
The requirement of continuity of the P.E. fn V(x) is actually too strong:
continuity of the derivative of the wfn can be shown also for P.E. fs
with finite discontinuity – Exercise!
Only at points of infinite discontinuity of the P.E. fn does
the derivative of the wfn exhibit a discontinuity.
In the following examples we shall apply these general
results to particular cases.
6
Particle in infinite 1D square well
Recall the TiSE (2.2):
u ′′ ( x ) +
2m
⎡ E − V ( x ) ⎤⎦ u ( x ) = 0
2 ⎣
=
or
u ′′ ( x ) + ⎡⎣ε − v ( x ) ⎤⎦ u ( x ) = 0
where
ε = 2mE = 2 , v ( x ) = 2mV ( x ) = 2
For an infinite square well we have
⎧0 if
v ( x) = ⎨
⎩∞ if
0≤ x≤a
x ≤ 0 or x ≥ a
7
In region (I) and (III):
V(x)
uI , III ( x) = 0
and in region (II) :
(II)
(I)
(III)
uII′′ ( x ) + k 2uII ( x ) = 0
V=0
k 2 = ε = 2mE = 2 > 0
with general solution
O
Continuity at
Continuity at
a
x
uII ( x) = c1 cos kx + c2 sin kx
x=0:
uI ( 0 ) = uII ( 0 ) , i.e.
x=a:
uII ( a ) = uIII ( a ) , i.e. 0 = c2 sin ka
0 = c1
8
c2=0 ? No: if c1=0 and c2=0, then u(x)=0 identically, i.e. there is
no physical state.
Therefore we must put
sin ka = 0
hence
and
kn a = nπ , n = 1, 2,3,"
2 2
=2 2
2 π =
En =
kn = n
, n = 1, 2,3,"
2
2m
2ma
Example: electron in infinite sq well of atomic size
mec2=0.5 MeV=0.5x106 eV, ħc=200 eV nm
hydrogen atom in its ground state (Bohr radius): aB=0.05 nm
hence
E1=320 eV
This is fairly large on an atomic scale
but not unreasonable, considering the artificial P.E. fn.
9
E3
u3 ( x ) = sin ( 3π x a )
E2
u2 ( x ) = sin ( 2π x a )
E1
u1 ( x ) = sin (π x a )
10
Conclusions:
Continuity of the wave function, imposed on the general solution of the
TiSE, has led to a set of discrete energy levels En.
The corresponding wave functions are square integrable and normalisable:
∞
a
1 2
∫−∞ u ( x ) dx = c ∫0 sin ( kn x ) dx = 2 a c2 = 1
2
n
2
2
2
hence, with suitable choice of the phase of
c2:
c2 = a 2
We should note the difference with the behaviour of a
classical particle in an infinite square well:
the energy of a classical particle is continuous!
11
Particle in finite 1D square well
Consider the finite 1D square well potential
⎧0
⎪
V ( x ) = ⎨V0
⎪
⎩0
x ≤ −a
if
if
−a ≤ x ≤ a
if
x≥a
We are looking for solutions of the TiSE with
(I )
( II )
( III )
E<0 (bound states).
Because of the discontinuities of V(x) we must write down the
TiSE for the three regions separately:
uI′′ ( x ) − κ 2uI ( x ) = 0
uII′′ ( x ) + k 2uII ( x ) = 0
where
′′ ( x ) − κ 2uIII ( x ) = 0
uIII
κ = −2mE = ; k = ( 2m =
2
2
2
2
)[ E − V ]
0
12
Finite square well potential of width
2a
and depth
V0 <0
V(x)
-a
a
O
x
V0
13
We make use of the symmetry of the P.E. fn: V(-x)=V(x)
(i) If
x ≤ a then we are in region II, and if we apply the transformation
x → − x then we remain in region II. Therefore
uII′′ ( − x ) + k 2uII ( − x ) = 0
Now define
then
i.e.
u2( ± ) ( x ) =
1
( uII ( x ) ± uII ( − x ) )
2
u2( ± ) ( − x ) = ±u2( ± ) ( x )
u2( + ) ( x )
is an even fn and
u2( − ) ( x )
is an odd fn of x
and they satisfy the DEq.
u2( ± )′′ ( x ) + k 2u2( ± ) ( x ) = 0
14
with solutions
u2( + ) = C ( + ) cos x; u2( − ) ( x ) = C ( − ) sin x
(ii)
assume x>a, i.e. we are in region (III) and the wfn satisfies
u III ′′ ( x ) − κ 2u III ( x ) = 0
with the general solution
uIII ( x ) = Aeκ x + Be−κ x
But to be square integrable, the wfn must tend to zero for x→ ∞
hence
A=0, i.e.
uIII ( x ) = Be−κ x ,
( x > a)
Similarly we find in region (I), (x < -a)
uI ( x ) = Deκ x ,
( x < −a )
15
Now we impose symmetry:
(+)
uIII
( x ) = uI( + ) ( − x )
hence
B( ) = D(
+
+)
and if we impose antisymmetry:
( −)
u III
( x ) = −uI( −) ( − x )
then
B( ) = − D(
−
Summary:
uI( ± ) ( x ) = ± B ( ± ) eκ x ,
(±)
uIII
( x ) = B ( ± ) e −κ x ,
−)
( x < −a ) ;
( x > a);
The symmetric (antisymmetric) solutions are also called solutions
of positive (negative) parity.
16
To complete the calculation we must impose continuity of the wave functions
at
x=-a and x=a
Let us do this first for the positive parity solutions.
( −a )
+
+
uI( )′ (− a ) = u2( )′ ( − a )
uI( ) (− a ) = u2(
+)
+
and
and putting
B ( + ) e −κ a = C ( + ) cos ka
κ B ( + ) e−κ a = kC ( + ) sin ka
ξ = ka; η = κ a
and dividing the equations, we get
η = ξ tan ξ
Recall the definition of
and since
V0<0:
k
and
(2.5)
κ:
κ 2 = −2mE = 2 ; k 2 = ( 2m = 2 ) [ E − V0 ]
k 2 + κ 2 = ( 2m = 2 ) V0
17
ξ 2 +η 2 = R2
where
(2.6)
R 2 = ( 2ma 2 = 2 ) V0
The curves (2.5) and (2.6) are shown in the following figure:
The energy eigenvalues can
be found from the intersections
of the curves.
For example: for R=4 we find
two roots, one near η = 1.8
and a second one near 3.7
and with
η 2 = κ 2 a = −2ma 2 E = 2
E = −η 2 = 2 2ma 2
= − V0 (η 2 R 2 )
18
The greatest value of η corresponds to the lowest value of the energy,
i.e. to the ground state energy.
We see from the figure that there are two energy levels for
levels for
R=4
and three
R=8.
Taking the approximate values of η which we have read off the figure
for
R=4, we find
E1( ) = −0.86 V0
+
E2( ) = −0.20 V0
+
If more accurate values need to be found, then one has to resort to
numerical methods to solve the equations (2.5) and (2.6).
19
Negative parity solutions.
The continuity conditions applied to the negative parity solutions give
the following result (Exercise!):
η = −ξ cot ξ
(2.7)
This set of curves together with (2.6)
is shown in the figure.
For R=4 there is only one
negative parity level, corresponding
to η = 3.2 (approximately).
The energy eigenvalue is
E1( − ) = −0.64 V0
20
and we note that
E1( + ) < E1( − ) < E2( + )
Conclusions:
• In a finite square well there is at least one energy level.
• if the P.E. fn is symmetric, then the ground state has positive parity,
the first excited state has negative parity, etc.
An example of a square well with three energy levels is shown in the
next figure where I have also plotted the wave functions of the three
states corresponding to these levels.
21
Energy levels and wave functions
of an electron in a finite square well:
V0 = −40 eV
2a 2.5 nm
The constants were taken
with approximate values:
me = 0.5 MeV c 2
=c = 200 eV nm
22
Square Barrier Potential
(I)
(III)
(II)
u0
ut
ur
0
x
a
for x ≤ 0
(I )
⎧0
⎪
V ( x ) = ⎨V0 > 0 for 0 ≤ x ≤ a ( II )
⎪0
for x ≥ 0
( III )
⎩
u0 is the incident wave
ur is the reflected wave
ut is the transmitted wave
Two cases are of interest:
(i) E < V0
(ii) E > V0
23
(i) Case 1:
E < V0
TiSE:
uI′′ ( x ) + k 2uI ( x ) = 0, k 2 = 2mE = 2
uII′′ ( x ) − κ 2uII ( x ) = 0, κ 2 = 2m (V0 − E ) = 2
′′ ( x ) + k 2u III ( x ) = 0
uIII
General solution:
u I ( x ) = Aeikx + Be − ikx
uII ( x ) = Ceκ x + De −κ x
u III ( x ) = Feikx + Ge − ikx
No wave incident from the right:
Continuity at
x=0
and
G=0
x = a:
uI ( 0 ) = uII ( 0 )
uII ( a ) = uIII ( a )
uI′ ( 0 ) = uII′ ( 0 )
′ (a)
uII′ ( a ) = uIII
24
Ceκ a + De −κ a = Feika
A+ B = C + D
ik ( A − B ) = κ ( C + D )
κ ( Ceκ a − De −κ a ) = ikFeika
After a fairly lengthy calculation we get the following result:
V02 sinh 2 κ a
B
R=
=
4 E (V0 − E ) + V02 sinh 2 κ a
A
2
4 E (V0 − E )
F
=
T=
4 E (V0 − E ) + V02 sinh 2 κ a
A
2
R is the reflection coefficient
T is the transmission coefficient
T+R = 1
25
Discussion:
Contrary to classical expectation, the particle is not completely
reflected by the barrier:
partly it is reflected and partly it is transmitted.
But part of an electron is never observed, so our result
must have another meaning.
This lies in the probability interpretation proposed
by Max Born and universally accepted:
R is the probability of the particle being reflected
T
is the probability of the particle being transmitted
The result R+T = 1 means that the total probability of the particle
being reflected or transmitted is certainty.
26
Derivation:
Recall the equation which express continuity of the wfn
and its derivative:
Continuity at
Continuity at
x=0
A+ B = C + D
(3.1)
ik ( A − B ) = κ ( C − D )
(3.2)
x=a:
Ceκ a + De −κ a = Feika
(3.3)
κ ( Ceκ a − De −κ a ) = ikFeika
(3.4)
i.e. we have 4 equations for 5 coefficients (amplitudes).
Our strategy will be to express
B
and
F
in terms of
B is the amplitude of the reflected wave,
F is the amplitude of the transmitted wave.
C and D are of no interest; we shall eliminate them
A:
27
1
⎡⎣ A + B + iλ ( A − B ) ⎤⎦
2
1
( 3.2 ) κ − ( 3.1) : D = ⎡⎣ A + B − iλ ( A − B ) ⎤⎦
2
( 3.5)
1 (ik −κ )a
C = Fe
(1 + iλ )
2
1
D = Fe( ik +κ )a (1 − iλ )
2
( 3.7 )
( 3.2 ) κ + ( 3.1) : C =
( 3.4 )
κ + ( 3.3) :
( 3.4 )
κ − ( 3.3) :
( 3.6 )
( 3.8)
where λ=k/κ. Thus
Fe( ik −κ )a (1 + iλ ) = ⎡⎣ A + B + iλ ( A − B ) ⎤⎦ = A (1 + iλ ) + B (1 − iλ )
Fe( ik +κ )a (1 − iλ ) = ⎡⎣ A + B − iλ ( A − B ) ⎤⎦ = A (1 − iλ ) + B (1 + iλ )
28
Eliminate
Feika, hence
2
2
A (1 + λ 2 )( eκ a − e −κ a ) = B ⎡(1 + iλ ) e −κ a − (1 − iλ ) eκ a ⎤
⎣
⎦
or
1 + λ 2 )( eκ a − e −κ a )
(
B
=
A (1 + iλ )2 e −κ a − (1 − iλ )2 eκ a
1 + λ ) sinh κ a
(
=
2iλ cosh κ a − (1 − λ ) sinh κ a
2
2
and taking the mod-square we get
1 + λ ) sinh 2 κ a
(
B
=
2
A
4λ 2 cosh 2 κ a + (1 − λ 2 ) sinh 2 κ a
2
2 2
29
and using the identity
cosh 2 x − sinh 2 x = 1
we get
1 + λ ) sinh 2 κ a
(
B
=
2
A
4λ 2 + (1 + λ 2 ) sinh 2 κ a
2 2
2
Now recall:
k 2 = 2mE = 2 , κ 2 = 2m (V0 − E ) = 2
hence
λ2 =
V0
E
, and 1 + λ 2 =
V0 − E
V0 − E
and thus finally
V02 sinh 2 κ a
B
R=
=
A
4 E (V0 − E ) + V02 sinh 2 κ a
2
and similarly we get the formula for
T=|F/A|2
30
(ii) Case 2:
For
E>V0
E > V0
we can similarly find the formulae for
R
and
T (Exercise):
V02 sin 2 κ a
B
2
2
R=
m
E
V
;
κ
2
=
=
−
=
(
)
0
A
4 E ( E − V0 ) + V02 sin 2 κ a
2
But we do not need to rederive the formulae: just note that they
are related by the transformation κ2 → -κ2
(iii) Case 3:
E = V0
Exercise: derive the formulae for this case (a) directly from the TiSE
and (b) from the results of Cases 1 and 2 by taking the appropriate limit.
In the following figures I am showing two examples of curves of
for different choices of parameters.
R and T
31
32