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Unformatted text preview: QUANTU M MECHANICS , < 0, in which case E lies
is condition is also sufﬁcient
dimensions (see Prob. 10, Fig. 5a is sufﬁciently broad
n corresponding to a larger
re 6a, b, and 0 show a series
in Fig. 5c, d, and e for suc
re) values of E; both signs
6b show the eigenfunctions
73/ levels of a particle bound
by an extension of the fore
any higher discrete energy
than that corresponding to (b) ../ ial and larger (less negative)
from (a) to (b) to (c) and is
ilope can both be continuous THE SCHRODINGER WAVE EQUATION 37 Thus, for a potential energy that approaches a ﬁnite constant value
as a: —) i 00, there may be a ﬁnite number of discrete energy levels, or in
some cases an inﬁnite number [if V(x) falls off slowly enough for large
[ml], depending on V(r) and the mass of the particle. However, if
V(z) ——> + so as a: —> i on, an argument like that given above shows that
there will always be an inﬁnite number of discrete energy levels; apart
from arbitrary multiplying constants there will be just one eigenfunction
u(:z:) for each of these. CONTINUOUS ENERGY EIGENVALUES It is possible to ﬁnd eigenfunctions that obey the boundary and continuity
conditions for all energy eigenvalues that exceed the smaller of the two
numbers V(+ 00) and V(— 00 ). If, for example, the potential energy has
the form illustrated in Fig. 5a, then solutions of the wave equation can be
found for all positive values of E. This is because the solutions for large
III are of the form A sin alzl + B cos alzl a = + (22?), (8.6)
and there is no reason Why both terms should not be kept. Thus it is
always possible to adjust the phase of each of the wave functions for large
[ml (which is equivalent to adjusting the ratios A /B for the solutions for
large positive and negative 2:) so that they join together smoothly when
continued in to a: = 0. K The classical terms periodic (or multiply periodic) and aperiodic are
sometimes used to designate the particle motions associated With discrete
and continuous energy eigenvalues, respectively. DISCRETE AND CONTINUOUS EIGENVALUES IN THREE DIMENSIONS We shall assume Without further discussion that all the foregoing results
can be taken over in a natural way for the three—dimensional wave equa
tion (8.2). We can expect that, if V(r) ——> + on as r —> oo in all directions,
there will be an inﬁnite set of discrete energy levels extending to + 00.
If V(r) is bounded as r —> oo in some direction, there may be a ﬁnite or an
inﬁnite number of discrete levels, depending on the form of V. In this
case, the discrete energy levels cannot exceed the smallest value that
V(oo) has in any direction. For values of E larger than this smallest
V( 00 ), the energy eigenvalues cover a continuous range extending to + co . SHONE—DIMENSIONAL SQUARE WELL POTENTIAL As a simple explicit example of the calculation of discrete energy levels
of a particle in quantum mechanics, we consider the onedimensional QUANTU M MECHANCS (15) Fly. 7 Onedimensional square well potential with (a) perfectly
rigid walls and (b) ﬁnite potential step. motion of a particle that is restrained by reﬂecting walls that terminate a
region of c'onstant potential energy. Two simple types of potential
energy are considered. Figure 7a shows a situation in which V(:c) = 0
for —a < a: < a, and V(:c) = +00 for Isl > a, corresponding to per
fectly rigid, impenetrable walls at the points a: = is. In Fig. 7b the
increase in potential energy at the walls is abrupt but ﬁnite, so that
V(2:) = V0 for [ml > a; because of its appearance, this is often called a
square well potential. The motion of a classical particle with total energy
E less than V0 is the same for both these potentials; but, as we shall see,
the quantummechanical behavior is different. In general, an abrupt
ﬁnite increase in potential energy at the boundaries of a region forces a
particle toward the interior of the region. Such a potential may be
thought of as a limiting case of a potential of the type shown in Fig. So, for
which the force —dV/dx is always directed in toward x = 0. The force
associated with a square well potential is zero except at the boundaries,
so that the particle is acted on by no force except a sudden impulse
directed toward the origin as it passes the points a: = ia. PERFECTLY RIGID WALLS It was shown in Sec. 8 that the wave function must vanish at the points
a: = ia, when the potential energy has the form shown in Fig. 7a.
From Eq. (8.5) the wave equation for Izl < a is simply (9.1) 2m *
ME) (9.2) which has the general solution “($)=Asina:c+Bcos:z:c a=+( QUANTU M MECHANICS (b) it}: (a) perfectly ﬂeeting walls that terminate a
v0 simple types of potential
. situation in which V(a:) = 0
:l > a, correSponding to per—
nts 2: = id. In Fig. 7b the
is abrupt but ﬁnite, so that
earance, this is often called a
iical particle with total energy
otentials; but, as we shall see,
rent. In general, an abrupt
oundaries of a region forces a
1. Such a potential may be
f the type shown in Fig. 5a, for
l in toward a: = 0. The force
zero except at the boundaries,
‘rce except a sudden impulse
points a: = id. .ion must vanish at the points
the form shown in Fig. 7a.
1 a is simply (9.1) = + (9.2) THE SCHRGDINGER WAVE EQUATION 39 Application of the boundary conditions at a: = ia gives Asinaa+BcOSaa=0
—Asinaa+BcOSaa=0 from which we obtain Asinaa=0 Bc05aa=0 Now we do not want both A and B to be zero, since this would give
the physically uninteresting solution 11. = 0 everywhere. Also, we cannot
make both sin ac and cos aa zero for a given value of a or E. There are
then two possible classes of solutions: For the ﬁrst class A = 0 and cos em = 0
and for the second class
B = 0 and sin aa = 0 Thus aa = mr/ 2, where n is an odd integer for the ﬁrst class and an even integer for the second class. The two classes of solutions and their
energy eigenvalues are then u(a:) = B cos 13—255 n odd
u(x) = A sin 7% n even
2 2 2
E = «8;; in both cases It is evident that n = 0 gives the physically uninteresting result u = 0
and that solutions for negative values of n are not linearly independent
of those for positive n. The constants A and B can easily be chosen in
each case so that the eigenfunctions 14(2) are normalized. There is thus an inﬁnite sequence of discrete energy levels that cor
respond to all positive integer values of the quantum number n. There
is just one eigenfunction for each level, and the number of nodes of the
nth eigenfunction that are within the potential well is n — 1. These
results are in agreement with the discussion of Sec. 8. It is interesting to
note that the order of magnitude of the lowest or groundstate energy
level is in agreement with the uncertainty relation (3.1). The position
uncertainty of order a implies a momentum uncertainty at least of order
ﬁ/a, which in turn implies a minimum kinetic energy of order 712/ ma”. FINITE POTENTIAL STEP When the potential energy has the form shown in Fig. 7b, it is necessary
to supplement the general solution (9.2), which is still valid for lzl < a QUANTUM MECHANICS since Eq. (9.1) is unaltered there, by a solution for M > a. The wave
equation in this region is 2
—[email protected]+Vuu=Eu 2m d1:2 which has the general solution for E < V0 (bound states) 2m(Vg — E) F u(:c) = Cr” + D65” {3 = + I: h, (9.3) The boundary conditions at 2: = i no discussed in Sec. 8 require that we
set D = 0 if Eq. (9.3) is to represent the solution for :c > a, and C = 0
if the solution is for a: < ——a. We now impose on the solutions (9.2) and (9.3) the requirements
that u and du/dz be continuous at :v = j; a. A sin aa + B cos aa = Ce‘“ aA cos aa — aB sin aa = —ﬁCe—5“
—A sin aa + B cos aa = De‘“
aA cos aa + «:3 sin aa = BDeﬁ“ from which we obtain 2A sin aa = (C — D)e‘5“ 2aA cos aa = —B(C — D)e—5“
"‘ 2B cos aa = (C + D)e—5“ 2aB sin aa =’ 5(C + D)e"" (9.5) Unless A = O and C = D, Eqs. (9.4) have as their consequence Similarly, unless B = 0 and C = —D, Eqs. (9.5) give l
l
l
l 1‘
1!5: a cot aa = —B (95)
l
l
l
l
l a tan aa = 13 (9.7) Now it is impossible for Eqs. (9.6) and (9.7) to be valid at once, since / on elimination of )3 this would require that tan” out = — 1, which in turn
would make a imaginary and 13 negative, contrary to Eq. (9.3). Also, we do not want A, B, C, and D all to vanish. Thus the solutions may again be divided into two classes: For the ﬁrst class ....V We ....,.9 A=0 C=D and a tan em = {3 and for the second class B=0 C=—D acot out = —{3 QUANTU M MECHANICS ition for [x] > a. The wave (bound states) n(Vo — E)]§ ,. (9.3) ssed in Sec. 8 require that we
)lution for :c > a, and C’ = 0 ) and (9.3) the requirements os aa = —ﬂ(C’ — D)e‘ﬁ“ (9.4)
in aa = 6(0' + D)e‘ﬂ“ (9.5)
as their consequence (9.6)
(9.5) give (9.7) :97) to be valid at once, since
;an2 aa = —1, which in turn
.trary to Eq. (9.3). Also, we
[‘hus the solutions may again
ss ztanaa=ﬂ acot aa = —ﬂ THE SCHRODINGER WAVE EQUATION 41 EN ERGY L EVELS The energy levels are found from a numerical or graphical solution of
Eqs. (9.6) and (9.7) with the deﬁnitions for a and 6 given in Eqs. (9.2)
and (9.3). A simple graphical method for effecting this solution is
described here, since it shows quite clearly the way in which the number
of discrete levels depends on V0 and a. We put E = aa, 1; = 6a, whence
Eq. (9.7) becomes E tan E = n, with £2 + 112 = 2me1? ....—_———i” Since E and g are restricted to ositive es, the ener levels ma be
ound in this case tom the intersections in the first quadrant of the curve of E tan E plotted against E,_Vji_t_h_the circle of knowwdmw
F, "—— '1 he construction is drawn in Fig". 8 for three values of V002; for each of
the two smaller of these values there is one solution of Eq. (9.7) and for
the largest there are two. Figure 9 is a similar construction for the solution of Eq. (9.6) in
‘which the energy levels are obtained from the intersections of the same
circles with the curve of —E cot E in the ﬁrst quadrant. The smallest
value of Voa,2 gives no solution, and the two larger values each give one.
Thus the three increasing values of Voa2 give altogether one, two, and
three energy levels, respectively. Fig. 8 Graphical solution of Eq. (9.7) for three values of Vac“; the vertical dashed
lines are the ﬁrst two asymptotes of 'q = Etan E. QUANTUM MECHANICS Fly. 9 Graphical solution of Eq. (9.6) for three values of Voa’; the vertical
dashed line is the ﬁrst asymptote of 1, = —s cot s. It is clear from Figs. 8 and 9 that, for a given particle mass, the
energy levels depend on the parameters of the potential energy through
the combination Voa”. For Voa2 between zero and #271,2/ 8m, there is just
one energy level of the ﬁrst class; for Van between rzhz/ 8m and four times
this value, there is one energy level of each class, or two altogether. As
Voa2 increases, energy levels appear successively, ﬁrst of one class and then
of the other. It is not difﬁcult to see from Eq. (9.2) that, when ordered according to increasing eigenvalues, the nth eigenfunction has n — 1
nodes. PARITY It follows from the foregoing discussion that the eigenfunctions of the ﬁrst
class are even with respect to change in sign of :c [u( :c) = u(:c)], whereas
the eigenfunctions of the second class are odd [u(—:c) = —u(:c)]. This
division of the eigenfunctions into even and odd types is not accidental
and will now be shown to be a direct consequence of the fact that the
potential energy function V(z) is symmetric about a: = 0. If we change
the sign of a: in the wave equation (8.5) — %d2;(f) + V(:c)u(x) = Em) (9.3) QUANTU M MECHANICS values of Van“; the vertical 5. for a given particle mass, the
>f the potential energy through
. zero and «aha/8m, there is just
between «ﬁzz/8m and four times
:h class, or two altogether. As
sively, ﬁrst of one class and then
In Eq. (9.2) that, when ordered
: nth eigenfunction has n — 1 .at the eigenfunctions of the ﬁrst
gn of a: [u(—x) = u(a:)], whereas
e odd [u(—x) = —u(x)]. This
and odd types is not accidental
)nsequence of the fact that the
trio about x = 0. If we change (9.8) THE SCHRéDINGER WAVE EQUATION 43 and if V(—x) = V(x), we obtain it: d’u( — 2:) _ 2m dxz + V($)u(—x) = Eu(_x) Then 24(2) and u(—x) are solutions of the same wave equation with
the same eigenvalue E. Suppose at ﬁrst that there is only one linearly
independent eigenfunction that corresponds to this energy level; then
u(:z:) and u(—x) can differ only by a multiplicative constant: u(—:z:) = eu(:z:) ' (9.9) Changing the sign of :1: in Eq. (9.9) gives u(:z:) = eu(—x). From these
two equations it follows at once that ez=1 or e=i1 Thus all such eigenfunctions of a symmetric potential are either even or
odd with respect to changes of sign of 2:. Such wave functions are said
to have even or odd parity. If an eigenvalue has more than one linearly independent eigenfunc
tion, the foregoing argument breaks down, and these eigenfunctions need
not have a deﬁnite parity; they need not be even or odd. However, we
can easily see that linear combinations of such eigenfunctions can be
found such that each has even or odd parity. Suppose that an eigenfunc—
tion u(x) does not have a deﬁnite parity. It can always be written u(x) = u.(x) + we) where u.(x) = &[u(x) + u(—:c)] is even, uo(x) = 11;[u(:c) — u(—x)] is odd,
and u. and u, are linearly independent. Then if the wave equation (9.8)
is symmetric, we can write it as h? dzu. h: dzuo _
_ in dzz + (V — E)u. — ﬁn dxz + (V — E)uo _ 0 (9.10)
0’1 Changing the Slign of a: in Eq. (9.10), we obtain
hz dine h’ d’ua ..
_ in dx2 + (V — E)“, 5,7, dz.  (V  E)u., — 0 (9.11) Addition and subtraction of Eqs. (9.10) and (9.11) show that u. and u,
are separately solutions of the wave equation with .the same eigenvalue E. A SIMPLIFIED SOLUTION Knowledge that the solutions possess a deﬁnite parity sometimes simpliﬁes
the determination of the energy levels, since then we need only find the
solution for positive 2:. Even solutions have zero slope and odd solutions
have zero value at x = 0. If, for example, we wish to ﬁnd the even solu 44 QUANTUM MECHANICS tions, Eqs. (9.2) and (9.3) can be replaced at once by u(x)=BcOSaz 0<x<a
u(2:) = 06—5” 2: > 0. Instead of making both u and du/dx continuous at a: = a, it is enough to
make the ratio (1 /u) (du/ dz) continuous at a: = (1, since the normalizing
constants B and C are eliminated thereby. This gives Eq. (9.7) at once.
Similarly, the odd solutions are u(2:)=Asina2: 0<x<a
u(x)=Cc—’ x>a Then continuity of (1 / u) (du/ dz) at :1: = a immediately gives Eq. (9.6). PROBLEMS 1. Use the arguments of Sec. 6 to set up a differential equation for \b that involves
a second time derivative of ‘p, in the case of a free particle. Discuss any solutions that
this equation has that are not shared by the freeparticle Schrodinger equation. 2. Show that the freeparticle onedimensional Schrodinger wave equation (6.8) is
invariant with respect to galilean transformations. Do this by showing that, when
the transformation x’ = x —— at, t' =t is applied, the transformed wave function
¢’(:z:’,t’) = f (1:,t)‘p(:z:,t) satisﬁes Eq. (6.8) with respect to the primed variables, where f
involves only 2, t, h, m, and 9. Find the form of f, and show that the traveling wave
solution ¢(:c,t) = Ae“""“’" transforms as expected. 3. How must a wave packet \6 fall off for large r in order that the volume integral
of P and the surface integral of S. in Eq. (7.4) converge? 4. Show directly that (1),) is real for a wave packet.
5. Show that for a three—dimensional wave packet is» = its.) + <pzx>> 6. Calculate the energy levels and plot the eigenfunctions for the three bound states
in the potential of Fig. 7b when Voag = (BM/m. Compare with the ﬁrst three states
for the potential of Fig. 7a. 7. Discuss the relation between the energy levels for the potential of Fig. 7b‘ and
those for the potential V(:z:) = + an, :2: < 0; 17(2) = 0, 0 < z < a; V(.1:) = V0, 1: > a.
8. Show that if the potential energy V(r) is changed everywhere by a constant,
the timeindependent wave functions are unchanged. What is the effect on the
energy eigenvalues? 2.
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 Fall '08
 AsokRay
 mechanics

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