Classical Limit
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Transcript of Classical Limit
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8/6/2019 Classical Limit
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The Classical Limit of Quantum Mechanics
If the theory of quantum mechanics is correct, then when quantum mechanics is applied to big things, it
must give the results of classical mechanics.
Because quantum mechanics emphasizes potential energy V(x), and classical mechanics emphasizes forceF, lets remind ourselves how theyre related. The definition of potential energy (in one dimension) is
V(x) V(x0) =
x
x0
F(x) dx, (1)
where F(x) is the force function i.e., the force that would be exerted on the particle if the particle were at
position x. (Im not saying that the particle is at x, Im saying thats what the force would be if the particle
were at x.) Taking the derivative of both sides with respect to x (and using the fundamental theorem of
calculus on the right)dV(x)
dx= F(x). (2)
In the classical limit, the quantal spread in x will be small. We ask how the expected position (thecenter of the wavepacket) moves:
x =
+
x(x, t)(x, t) dx, (3)
sodx
dt=
+
x(x, t)
t(x, t) dx +
+
x(x, t)(x, t)
tdx. (4)
But the Schrodinger equation tells us how wavefunction (x, t) changes with time:
(x, t)
t=
i
h
h2
2m
2(x, t)
x2+ V(x)(x, t)
(5)
and(x, t)
t= +
i
h
h2
2m
2(x, t)
x2+ V(x)(x, t)
. (6)
(From here on Im going to write (x, t) as and V(x) as V.) Thus
dx
dt=
i
h
+
x
h2
2m
2
x2+ V
dx
+
x
h2
2m
2
x2+ V
dx
(7)
= i
h
h2
2m
+
x2
x2 dx
+
x2
x2dx
+
+
xV dx
+
xV dx
(8)
= ih
2m
+
x2
x2 dx
+
x2
x2dx
(9)
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Can we say anything about integrals such as the second integral in square brackets above? Surprisingly,
the answer is yes. If we define
f(x) = x and g(x) =
x(10)
then
+
x2
x2dx =
+
f(x)g(x) dx (11)
which suggests integration by parts:
+
f(x)g(x) dx = [f(x)g(x)]+
+
f(x)g(x) dx. (12)
Now remember what the wavefunction is normalized, so it has to fall to zero at both infinity and negative
infinity. Typically the slope /x also falls to zero at both infinity and negative infinity, and does so very
rapidly much more rapidly than linearly. (There are exceptions to these typical behaviors, such as the
Coulomb scattering wave functions, and in these atypical cases this argument has to be rethought.) The
upshot is that in typical situations
[f(x)g(x)]+
= 0 (13)
so +
x2
x2dx =
+
(x)
x
xdx. (14)
Well use this trick several times. . . Ill just call it the integration-by-parts trick.
Applying this trick to both integrals of equation (9) gives
dx
dt= i
h
2m
+
(x)
x
xdx
+
(x)
x
xdx
= ih
2m
+
x
x
x
dx + +
x
dx +
x
x
x
dx +
x
dx= i
h
2m
+
+
xdx
+
xdx
= h
mm
+
xdx
. (15)
Notice that dx/dt is real, as it must be. (This isnt proof that weve made no algebra errors, but if our
expression for dx/dt had been complex, that would have been proof that we had made algebra errors.)
All this is fine and good, but it takes us only part way to our goal. This is clearly not a classical
equation. . . it contains h right there! Since the classical F = ma involves the second derivative of position
with respect to time, we take one more derivative of x.
d2x
dt2=
h
mm
+
t
xdx +
+
x
tdx
. (16)
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The second-order derivative on the right looks particularly grotesque, so we use the integration-by-parts
trick to get rid of it:
d2x
dt2=
h
mm
+
t
xdx
+
x
tdx
(17)
=2h
mm
+
x
tdx
. (18)
Now use Schrodingers equation:
d2x
dt2=
2h
mm
+
x
i
h
h2
2m
2
x2+ V
dx
= 2
me
+
x
h2
2m
2
x2+ V
dx
. (19)
Look at that. . . two of the hs have cancelled out! Were not home yet because theres still an h within the
square brackets, but were certainly making progress. We have that
d2
xdt2 = 2me h
2
2m+
x2
x2
dx ++
xV dx
, (20)
but lets apply the integration-by-parts trick to the first integral:+
x
2
x2dx =
+
2
x2
xdx. (21)
Think about this for a minute: if the integral on the left is z, this equation says that z = z, whence z is
pure imaginary or e{z} = 0. Thus
d2x
dt2=
2
me
+
xV dx
, (22)
an expression devoid of hs! Apply the integration-by-parts trick to this integral:+
xV dx =
+
(V )
xdx
+
xV dx =
+
V
xdx
+
V
x dx
+
xV dx +
+
V
xdx =
+
V
x dx
2e
+
xV dx
=
+
V
x dx (23)
Plugging this result back into equation (22) gives
d2x
dt2 =
1
m+
V
x dx. (24)
But remember that the force function is F(x) = V/x, so
d2x
dt2=
1
m
+
F(x)(x, t)(x, t) dx =1
mF(x). (25)
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There it is
F(x) = md2x
dt2. (26)
This result is called Ehrenfests theorem.
Note that the result is notF(x) = m
d2x
dt2. (27)
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