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.. 226 25 One-dimensi
Trang 1Next 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) = C1 cos(k · x) + C2 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)− ; C2= 0) and
odd parity, i.e., u( −x) = −u(x) , (⇒ B(1)
− =−B(2)
− ; C1 = 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 > 0 and κ > 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, where n is integer, and afterwards they diverge to
+∞ from −∞, at k ± · 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 dimensions1
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
Trang 2226 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
V0is just deep enough to generate a bound state, whereas for a “di-neutron”
it is just not deep enough.)
Nodal theorem: The ground state, ψ0, has no “nodes” (i.e., no zeroes) at all (between the interval limits, i.e., here for−∞ < x < ∞) In contrast, an eigenstate ψ n , for n = 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) , has even 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, , exists iff the quantum well is sufficiently deep and broad, i.e., for the present case iff
2m |V0|a2
2 > n · π
2 .
25.2 Reflection and Transmission at Steps
in the Potential Energy; Unitarity
For simplicity we assume firstly that
V (x) ≡ 0 for x < 0 and ≡ 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 Otherwise we have total reflection; this case can be treated separately.)
We thus have, with ω := E/:
ψ(x, t) = A ·ei(k − x −ωt) + r · ei(−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) = −V0for
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
Trang 3k − 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 , see below) The coefficients r and t follow in fact from the two continuity conditions for ψ(x) and dψ(x) dx The reflectivity R(E) and the transmittivity T (E) them-selves are functions of r(E) and t(E), i.e.,
R = |r|2, T = k+
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 =2k2−
2m ≡
2k2+
T can be directly calculated from R using the so-called unitarity relation2
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).
Trang 4228 25 One-dimensional Problems in Quantum Mechanics
25.3 Probability Current
All these statements follow explicitly (with ˆv = m −1(ˆp − e ˆ A)) from a gauge
invariant relation for the probability current density:
j w(r, t) := Re {ψ ∗(r, t)ˆvψ(r, t)} =
2im (ψ
∗ ∇ψ − ψ∇ψ ∗)
− e
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 conservaequa-tion theorem of the total probability:
∞
d3 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 nmay 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 V0(> 0) will be considered Assume that
V (x) = 0 for x < 0 and x > a, but V (x) = V0 for 0 ≤ x ≤ a; furthermore assume that the energy E is smaller than the barrier height, i.e., 0 < E < V0 (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
1 +14(κ k +k κ)2sinh2(κa)) . (25.14)
Trang 5k(E) =
2m
2E
1/2 , and κ(E) =
2m
2(V0− E)
1/2 For barriers with κa
tially small, T 1, but finite, as can be seen from
sinh x = (1/2)( e x+ e−x)= e4 x /2 for x
T (E)=4 k 16
κ+κ k
The factor in front of the exponential in (25.15) is of the order ofO(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) > E only in the interval a < x < b) Then one obtains (as long as the result is 1):
T (E)=4O(4) · exp
*
−2
b
a
dx
2m
2 (V (x) − E)
+
(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 factorO(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 prefactorO(4) is obtained from the
relation 4 = 22 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 ≡ 0 for x < 0 and x > a, but V (x) ≡ −|V0| 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:
1 + 1
k0 + k2
sin2(k0· a)
Trang 6230 25 One-dimensional Problems in Quantum Mechanics
where
k0 (E) =
2m
2(E + |V0|) Complete transmission (T = 1) is only obtained for the special energy values
k0· a = nπ (where n is an integer); otherwise T < 1, i.e., there is a finite
reflectivity
Trang 7This is one of the most important problems of quantum mechanics The Hamilton operator is
ˆ
H = pˆ2 2m+
mω02xˆ2
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) = V0+1
2V
(x
0)(x − x0)2+
We thus have
mω02≡ 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 x0= 0 and V0= 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 = ˆHψ(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 · mω2x2
2 E
· u(x) =,mω0
2
− 2m
2 E
-· u(x) (26.2)
It is now advantageous to use reduced variables without physical
dimen-sion, i.e.,
ε := E/( ω0/2), ξ := x/
/(mω0) and
˜
u(ξ) := u(x)
mω .
Trang 8232 26 The Harmonic Oscillator I
Equation (26.2) is thus simplified to
d2u(ξ)˜
and the normalization condition
∞
∞ |u(x)|2dx= 1! becomes
∞
−∞ |˜u(ξ)|2dξ = 1 ! Using the ansatz
˜
u(ξ) =: v(ξ) · e −ξ2/2
the (asymptotically dominating) exponential behavior ∼ e −ξ2/2 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
Although Schr¨odinger’s differential equation (26.3) is now satisfied, one
must additionally demand that the recursion relation terminates at a finite
ν(= ν0 ) , i.e., ε = 2ν0 + 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
i.e., for [a0= 0, a1= 0], the divergency would be asymptotically as∼ e +ξ2,
whereas for odd parity, i.e., for [a1= 0, a0 = 0], v(ξ) ∼ ξ · e +ξ2) This would violate the condition that
u(ξ)
= v(ξ) · e − ξ2
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/
ω0/2
Trang 9
is identical to one of the eigenvalues 2n + 1 with n = 0, 1, 2, ,
˜
u(ξ)
= v(ξ) · e −ξ2/2
is square-integrable.
In this way the following energy levels for the harmonic oscillator are obtained:
for n = 0, 1, 2, , with the eigenfunctions
˜
u n (ξ) ∼ H n (ξ) e −ξ2/2
The H n (ξ) are the so-called Hermite polynomials; e.g.,
H0 (ξ) = 1 , H1 (ξ) = ξ , H2 (ξ) = 1 − 2ξ2 and H3 (ξ) = ξ ·
1−2
3ξ 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 completeness1, 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 account2
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 −x2/x2
, 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; and n = 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?
Trang 1027 The Hydrogen Atom according to
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 = ˆp
2
The explicit expression for the potential energy,
V (r) = − Z e2
4πε0r ,
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 II1)
This product ansatz can be made essentially because the operators ˆL2
and ˆL z commute with each other and (as conserved quantities) also with ˆH,
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 ˆL2 and ˆL z, i.e.,
ˆ
L2Y lm (θ, ϕ) =2l · (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