Basic Theoretical Physics: A Concise Overview P23 pdf

25.1 Bound Systems in a Box (Quantum Well); Parity 225
Next we assume E<0 (i.e., for bound states). Then we have the following
solutions of the Schr¨odinger equation
u(x)=B
(1)
−
exp(−κ|x|)+B
(1)
+
exp(+κ|x|) , −x ≥ a, (25.3)
u(x)=C
1
cos(k ·x)+C
2
sin(k · x) , |x| <a, (25.4)
u(x)=B
(2)
−
exp(−κx)+B
(2)
+
exp(+κx) , +x ≥ a. (25.5)
a) Firstly, we recognise that the coefficents B
(1)
+
and B
(2)
+
must vanish, since
otherwise it would not be possible to satisfy the condition


∞
−∞
dx|u(x)|
2
!
=1.
b) The remaining coefficients are determined (apart from a common factor,
where only the magnitude is fixed by the normalization condition) from
the continuity conditions for u and u

at the potential steps, x = ±a.
The calculation is thus much easier for symmetrical potentials, V (x)=
V (−x), since then all solutions can be divided into two different classes:
even parity, i.e.,u(−x)=u(x) , (⇒ B
(1)
−
= B
(2)
−
; C
2
=0) and
odd parity, i.e.,u(−x)=−u(x) , (⇒ B
(1)
−
= −B
(2)
−
; C
1

=0).
One then only needs one continuity condition, i.e. the one for u

/u at
x =+a. From this condition (see below) one also obtains the discrete energy
values E = E
n
, for which continuity is possible (for k>0andκ>0):
κ(E)
k(E)
=tan(k(E) ·a) , for even parity, (25.6)
−
k(E)
κ(E)
=tan(k(E) ·a) , for odd parity . (25.7)
These equations can be solved graphically (this is a typical exercise), by
plotting all branches of tan(k·a) as a function of k·a (these branches intercept
the x-axis at k ·a = n ·π,wheren is integer, and afterwards they diverge to
+∞ from −∞,atk
±
·a =(2n ± 1) ·π/2 ∓0
+
). Then one can determine the
intersections of this multi-branched curve with the line obtained by plotting
the l.h.s. of (25.6) or (25.7) as a function of k(E) ·a.
In this way one obtains the following general statements which are true
for a whole class of similar problems:
Existence I :
There is always at least one bound state. (This statement is true for
similar problems in one and two dimensions, but not in three dimensions

1
.
1
In d=3 dimensions one can show, see below, that for so-called s-states, i.e., if the
state does not depend on the angular coordinates ϑ and ϕ, the wave-function
226 25 One-dimensional Problems in Quantum Mechanics
For example, for the analogous three-dimensional “potential box model” for
the mutual binding of a neutron and proton in the deuteron nucleus the depth
V
0
is just deep enough to generate a bound state, whereas for a“di-neutron”
it is just not deep enough.)
Nodal theorem: The ground state, ψ
0
, has no “nodes” (i.e., no zeroes) at
all (between the interval limits, i.e., here for −∞ <x<∞). In contrast, an
eigenstate ψ
n
,forn =1, 2, , if existent, has exactly n nodes.
If parity is a “good” quantum number, i.e., for symmetric potentials,
V (x) ≡ V (−x), the following principle is additionally true:
Alternating parity: The ground state, ψ
(n=0)
,haseven parity, the first
excited state odd parity, the 2nd excited state again even parity, etc
Existence II :
Quantitatively one finds that the nth bound state, n =1, 2, ,existsiff
the quantum well is sufficiently deep and broad, i.e., for the present case iff

2m|V

0
|a
2

2
>n·
π
2
.
25.2 Reflection and Transmission at Steps
in the Potential Energy; Unitarity
For simplicity we assume firstly that
V (x) ≡ 0forx<0and ≡ ΔV (x)(> 0) for all x ≥ 0 ,
with a barrier in a finite range including x = 0. Consider the reflection and
transmission of a monochromatic wave traveling from the left. We assume
below that E is sufficiently high (e.g., E>V(∞) in Fig. 25.2, see below.
Otherwisewehavetotal reflection; this case can be treated separately.)
We thus have, with ω := E/:
ψ(x, t)=A ·

e
i(k
−
x−ωt)
+ r · e
i(−k
−
x−ωt)

for x<0 , and

= A · t · e
i(k
+
x−ωt)
for x>a; (25.8)
is quasi one-dimensional in the following sense: The auxiliary quantity w(r):=
r ·ψ(r) satisfies the same “quasi one-dimensional” Schr¨odinger equation as noted
above, see the three equations beginning with (25.3). As a consequence, one only
needs to put x → r and u(x) → w(r), and can thus transfer the above “one-
dimensional” results to three dimensions. But now one has to take into account
that negative r values are not allowed and that w(0)
!
= 0 (remember: w(r)=
r ·ψ(r)). The one-dimensional solutions with even parity, i.e. u(x) ∝ cos kx,are
thus unphysical for a “three-dimensional quantum box”, i.e. for V (r)=−V
0
for
r ≤ a, V ≡ 0 otherwise. In contrast, the solutions of odd parity, i.e., w(r)=
r · ψ(r) ∝ sin kr, transfer to d=3. – This is a useful tip for similar problems in
written examinations.
25.2 Reflection and Transmission at a Barrier; Unitarity 227
k
−
and k
+
are the wave numbers on the l.h.s. and r.h.s. of the barrier (see
below).
The amplitude A is usually replaced by 1, which does not lead to any
restriction. The complex quantities r and t are the coefficients of reflection
and transmission (not yet the reflectivity R and transmittivity T ,seebelow).

The coefficients r and t follow in fact from the two continuity conditions
for ψ(x)and
dψ(x)
dx
.Thereflectivity R(E)andthetransmittivity T (E)them-
selves are functions of r(E)andt(E), i.e.,
R = |r|
2
,T=
k
+
k
−
|t|
2
. (25.9)
The fraction
k
+
k
−
in the formula for T is the ratio of the velocities on the
r.h.s. and l.h.s. of the barrier in the potential energy; i.e.,
E =

2
k
2
−
2m

≡

2
k
2
+
2m
+ ΔV. (25.10)
T can be directly calculated from R using the so-called unitarity relation
2
R + T ≡ 1. (25.11)
Fig. 25.2. Scattering of a plane wave by a barrier (schematically). A wave ∝ e
ik·x
traveling from −∞ with a certain energy meets a rectangular barrier, where it is
partially reflected and transmitted (indicated by the straight lines with arrows, i.e.,
the corresponding complex amplitudes r and t are also associated with cosine-like
behavior). The conditions determining r and t are that the wave function and its
derivative are continuous. The velocities on each side are different. The energy is
also allowed to be higher than that of the barrier
2
The name unitarity relation follows by addition of an incoming wave from the
right, i.e., the incoming wave is now a two-component vector with indices l
and r (representing left and right, respectively). This is also the case with the
outgoing wave. The incoming and outgoing waves are related, as can be shown,
by a unitary matrix (the so-called S-matrix ), which generalizes (25.11).
228 25 One-dimensional Problems in Quantum Mechanics
25.3 Probability Current
All these statements follow explicitly (with
ˆ
v = m

−1
(
ˆ
p − e
ˆ
A)) from a gauge
invariant relation for the probability current density:
j
w
(r,t):=Re {ψ
∗
(r,t)
ˆ
vψ(r,t)} =

2im
(ψ
∗
∇ψ − ψ∇ψ
∗
)
−
e
m
A|ψ|
2
. (25.12)
Together with the scalar probability density

w

(r,t)=|ψ(r,t)|
2
,
the current density j
w
satisfies (as one can show) the continuity equation
∂
w
(r,t)
∂t
+divj
w
(r,t) ≡ 0 . (25.13)
As shown in Part II in the context of electrodynamics, this continuity equa-
tion is equivalent to the conservation theorem of the total probability:

∞
d
3
r
w
(r,t) ≡ 1, ∀t.
Ultimately it is this fundamental conservation theorem, which stands behind
the unitarity relation (25.11).
For a series of different steps in potential energy the complex coefficients
r
n
and t
n
may be calculated sequentially. This gives rise to a so-called “trans-

fer matrix”.
25.4 Tunneling
In this section the probability of tunneling through a symmetrical rectangu-
lar barrier of width a and height V
0
(> 0) will be considered. Assume that
V (x)=0forx<0andx>a, but V (x)=V
0
for 0 ≤ x ≤ a;furthermore
assume that the energy E is smaller than the barrier height, i.e., 0 <E<V
0
(but see below!). Classically, in such a situation a particle will be elastically
reflected at the barrier. In contrast, quantum mechanically, with the methods
outlined above, it is straightforward to show that one obtains a finite tunnel-
ing probability, given by
T (E)=
1
1+
1
4
(
κ
k
+
k
κ
)
2
sinh
2

(κa))
. (25.14)
25.4 Tunneling 229
Here
k(E)=

2m

2
E

1/2
, and κ(E)=

2m

2
(V
0
− E)

1/2
.
For barriers with κa  1 the tunneling transmittivity is therefore exponen-
tially small, T  1, but finite, as can be seen from
sinh x =(1/2)(e
x
+e
−x
)=e

x
/2
for x  1:
T (E)=
16

k
κ
+
κ
k

2
· e
−2κa
. (25.15)
The factor in front of the exponential in (25.15) is of the order of O(4), if
k and κ are comparable. Therefore the following result for a non-rectangular
tunneling barrier is plausible (here we assume that V (x) >Eonly in the
interval a<x<b). Then one obtains (as long as the result is  1):
T (E)=O(4) ·exp

−2

b
a
dx

2m


2
(V (x) − E)

. (25.16)
(The exponent in this expression is essentially proportional to the product
of the width and the square-root of the height of the barrier. This yields
a rough but systematic approximation for tunnelling through a barrier.)
Furthermore, the factor 2 in the exponent and the factor O(4) in (25.16)
have an obvious meaning. They result from the correspondence T ∝|ψ|
2
;
(i), the factor 2 in the exponent in front of the integral immediately follows
from the exponent 2 in |ψ|
2
, and, (ii), the prefactor O(4) is obtained from the
relation 4 = 2
2
by the fact that the wave-function must decay exponentially
on both sides of the barrier. Furthermore, the reciprocal length appearing
in the exponent of (25.16) is, as expected, proportional to , i.e., this decay
length vanishes in the classical limit  → 0.
Quantum mechanical reflection at a depression in the potential energy,
i.e., at a barrier with negative sign (e.g., a “quantum well” as in figure 25.1) is
also of interest: We now assume an incoming plane wave with E>0(instead
of the problem of bound states, E<0); in any case (as already mentioned)
for V (x) we have the same situation as in Fig. 19.1, i.e. V ≡ 0forx<0
and x>a, but V (x) ≡−|V
0
| for 0 ≤ x ≤ a. For such a “quantum well” (as
already stated) there is at least one bound state for E<0. On the other

hand, classically an electron with E>0 does not “see” the quantum well.
Quantum mechanically, however, even in this case there is a finite reflection
probability. In fact the transmittivity T for this case is analogous to (25.14),
again without proof:
T (E)=
1
1+
1
4

k
0
k
+
k
k
0

2
sin
2
(k
0
· a)
, (25.17)
230 25 One-dimensional Problems in Quantum Mechanics
where
k
0
(E)=


2m

2
(E + |V
0
|) .
Complete transmission (T = 1) is only obtained for the special energy values
k
0
· a = nπ (where n is an integer); otherwise T<1, i.e., there is a finite
reflectivity.
26 The Harmonic Oscillator I
This is one of the most important problems of quantum mechanics. The
Hamilton operator is
ˆ
H =
ˆp
2
2m
+
mω
2
0
ˆx
2
2
. (26.1)
The harmonic oscillator is important inter alia because a potential energy
V (x) in the vicinity of a local minimum can almost always be approximated

by a parabola (which is the characteristic potential energy of the harmonic
oscillator),
V (x)=V
0
+
1
2
V

(x
0
)(x −x
0
)
2
+
We thus have
mω
2
0
≡ V

(x
0
) ,
and it is assumed that anharmonicities, i.e., the correction terms of higher
order, which are denoted by the dots, + , can be neglected. In contrast,
the assumptions x
0
=0andV

0
= 0 impose no restrictions on generality.
In the position representation (wave mechanics), we thus have to solve
the Schr¨odinger equation
−

i
∂ψ(x, t)
∂t
=
ˆ
Hψ(x, t) .
Using the ansatz for stationary states
ψ(x, t)
!
= u(x) ·e
−iEt/
,
we obtain the following time-independent Schr¨odinger equation:
u

(x)
!
=

2m

2
·
mω

2
0
x
2
2
−
2m

2
E

· u(x)=


mω
0


2
−
2m

2
E

·u(x) (26.2)
It is now advantageous to use reduced variables without physical dimen-
sion, i.e.,
ε := E/(ω
0

/2),ξ := x/

/(mω
0
)and
˜u(ξ):=u(x)


mω
0
.
232 26 The Harmonic Oscillator I
Equation (26.2) is thus simplified to
d
2
˜u(ξ)
dξ
2
=(ξ
2
− ε) · ˜u(ξ) , (26.3)
and the normalization condition

∞
∞
|u(x)|
2
dx
!
= 1 becomes


∞
−∞
|˜u(ξ)|
2
dξ
!
=1.
Using the ansatz
˜u(ξ)=:v(ξ) · e
−ξ
2
/2
the (asymptotically dominating) exponential behavior ∼ e
−ξ
2
/2
can now be
separated out; the differential equation for v(ξ) (the so-called Hermitian dif-
ferential equation) is solved by means of a power series,
v(ξ)=
∞

ν=0,1,
a
ν
ξ
ν
.
Finally after straightforward calculations, for the coefficients a

ν
the following
recursion relation is obtained:
a
ν+2
a
ν
=
2ν +1− ε
(ν +1)(ν +2)
. (26.4)
Although Schr¨odinger’s differential equation (26.3) is now satisfied, one
must additionally demand that the recursion relation terminates at a finite
ν(= ν
0
) , i.e., ε =2ν
0
+1.
If it were not terminated, i.e., if for all non-negative integers ν the identity
ε =2ν + 1 were violated, then v(ξ) would diverge for |ξ|1(forevenparity,
i.e., for [a
0
=0,a
1
= 0], the divergency would be asymptotically as ∼ e
+ξ
2
,
whereas for odd parity, i.e., for [a
1

=0,a
0
=0],v(ξ) ∼ ξ · e
+ξ
2
). This would
violate the condition that
u(ξ)

= v(ξ) · e
−
ξ
2
2

must be square-integrable. In contrast, if there is termination at a finite ν,
then the question of the convergence of the ratio a
ν+2
/a
ν
becomes obsolete.
In conclusion, iff the power series for v(ξ) terminates at a finite n, i.e.
iff the reduced energy
ε = E/


ω
0
/2


26 The Harmonic Oscillator I 233
is identical to one of the eigenvalues 2n +1with n =0, 1, 2, ,
˜u(ξ)

= v(ξ) · e
−ξ
2
/2

is square-integrable.
In this way the following energy levels for the harmonic oscillator are
obtained:
E
n
= ω
0
·(n +1/2) , (26.5)
for n =0, 1, 2, , with the eigenfunctions
˜u
n
(ξ) ∼ H
n
(ξ)e
−ξ
2
/2
. (26.6)
The H
n
(ξ) are the so-called Hermite polynomials; e.g.,

H
0
(ξ)=1,H
1
(ξ)=ξ, H
2
(ξ)=1− 2ξ
2
and H
3
(ξ)=ξ ·

1 −
2
3
ξ
2

(only the first and second should be kept in mind). The missing normalization
factors in (26.6) are also unimportant.
Eigenfunctions of a Hermitian operator, corresponding to different eigen-
values of that operator, are necessarily orthogonal, as can easily be shown.
We thus have
˜u
i
|˜u
j
 = u
i
|u

j
 = δ
i,j
,
as expected.
The completeness
1
, i.e., the basis property of the function system {u
n
},
is also given, if all polynomial degrees n =0, 1, 2, are taken into account
2
.
In fact, the probability that an oscillating particle of a given energy E
n
is
found outside the “inner” region which is classically allowed, is very small,
i.e., ∝ e
−x
2
/x
2
0
, but finite even for large n.
As we shall see in a later section (→ 28.1), the harmonic oscillator can
also be treated in a purely abstract (i.e., algebraic) way.
1
We remember that the completeness property is not satisfied for the bound
functions of an arbitary quantum well.
2

Here the following examination questions suggest themselves: (i) What do the
eigenfunctions u
n
(x) of a harmonic oscillator look like (qualitatively!) for the
following three cases: n =0;n =1;andn = 256? (ii) What would be obtained,
classically and quantum mechanically, for the probability of a harmonically os-
cillating particle of energy E to be found in a small interval Δx?
27 The Hydrogen Atom according to
Schr¨odinger’s Wave Mechanics
27.1 Product Ansatz; the Radial Function
The electron in a hydrogen atom is treated analogously to the previous cases.
The Hamilton operator is written
ˆ
H =
ˆ
p
2
2m
+ V (r) . (27.1)
The explicit expression for the potential energy,
V (r)=−
Z e
2
4πε
0
r
,
is not immediately important: one only requires that V (r) should be rota-
tionally invariant.
Next, the usual ansatz for stationary states is made:

ψ(r,t)=u(r) · e
−iEt/
.
Then, in spherical coordinates a product ansatz (separation of variables)fol-
lows:
u(r)=R(r) ·Y
lm
(θ, ϕ) ,
with the spherical harmonics Y
lm
(θ, ϕ) (which have already been introduced
in Part II
1
).
This product ansatz can be made essentially because the operators
ˆ
L
2
and
ˆ
L
z
commute with each other and (as conserved quantities) also with
ˆ
H,
such that all three operators can be diagonalised simultaneously.
As for the Y
lm
, here we only need the property that they are eigenfunctions
of the (orbital) angular momentum operators

ˆ
L
2
and
ˆ
L
z
, i.e.,
ˆ
L
2
Y
lm
(θ, ϕ)=
2
l · (l +1)Y
lm
(θ, ϕ) , (27.2)
ˆ
L
z
Y
lm
(θ, ϕ)=mY
lm
(θ, ϕ) . (27.3)
1
It is important to remind ourselves of other instances where the same or similar
ideas or equations are used.