Probability and Uncertainty Principle
5
5.1
Probability Interpretation of ψ
Wave-particle “duality”: particles are supposed to be found where we
expect from interference pattern that ψ is large. e.g.: Davisson-Germer
experiment: Max Born: probability that particle described by ψ(r, t)
found at time t in volume elt. δV at r is
δP = |ψ(r, t)|2δV
(1)
Basic requirements for probability function:
1. Positive semidefinite. δP ≥ 0 since ψψ ∗ ≥ 0 always.
Z
2. Normalized. dP = 1. Particle must be somewhere! Wave function is therefore normalized to 1:
Z
d3r|ψ(r, t)|2 = 1
(2)
Z
If
d3r|ψ|2 is not 0 or ∞, can always normalize given eigensolution
ψ of S.’s eqn. by multiplying with const., since ih̄∂ψ/∂t = Hψ is
linear.
1
3. Conserved. Conservation of probability: if ψ normalized at t, must
stay normalized at all times:
Z
Z ³ ∗
´
d
∂ψ
∗
3
∗ ∂ψ
0=
ψ ψ(r, t)d r =
ψ+ψ
(3)
dt
∂t
∂t
Is this consistent with S.-equation? The motion of ψ is determined
by ih̄ ∂ψ
the motion of ψ ∗ is determined by the complex
∂t = Hψ, while
∗
∗
∗
conj. eqn. −ih̄ ∂ψ
∂t = Hψ . Note H = H since potential V (r) in
h̄2
H = − 2m
∇2 + V (r) is real. Therefore t-derivative of |ψ|2 is
∂
∂ψ ∗
2
∗ ∂ψ
|ψ(r, t)| = ψ
+ψ
∂t
∂t
∂t
ih̄ ∗ 2
ih̄
=
ψ ∇ ψ−
ψ∇2ψ ∗
2m
2m
(4)
Now note we can write
ψ ∗∇2ψ − (∇2ψ ∗)ψ = ∇ · (ψ ∗∇ψ − (∇ψ ∗)ψ)
(5)
∂
ih̄
|ψ|2 + ∇ · [− [ψ ∗∇ψ − ((∇ψ ∗)ψ)] = 0,
∂t
2m
(6)
so we get
which just looks like a quantum-mechanical continuity equation:
∂ρ
+ ∇ · j = 0,
∂t
(7)
where
ρ = |ψ|2 = prob. density
j=−
ih̄ ∗
[ψ ∇ψ − ((∇ψ ∗)ψ)] = prob. current
2m
or prob. flux density
2
(8)
(9)
Integrate over volume V enclosed by surface S, use Gauss’s law:
Z
Z
∇ · jd3r =
j · n̂ dS
(10)
V
| S {z }
flux of probability
out of S
Z
d
= −
|ψ|2d3r
dt V
|
{z
}
time rate of change of
the prob. of finding
particle within S
(11)
(12)
Let V , S → ∞, and
that ψ falls off at ∞ fast enough =⇒
R assume
d
2
j · n̂ will too=⇒ dt ∞ |ψ| → 0. Probability conserved!
4. Collapse of ψ. Suppose at time t particle position is measured and
found to be in particular volume of space δV . Now look for particle
at time t + dt, dt arb. small. Can always make dt small enough so
the particle is still in δV since it moves with finite velocity v < c.
Therefore at time t + dt wave fctn. must be
½
0
r not in δV
ψ(r, t + dt) =
(13)
const.
r in δV
Before t probability spread over entire universe. Time dt later restricted to δV . Measurement collapsed wave fctn.!
3
5. Meaning of ψ. Classically, particle moving along trajectory r(t).
Can it be that ψ(r, t) just reflects our ignorance (i.e., for t before we
measure, we just haven’t found the particle yet–ψ far away from r(t)
not physically meaningful.)? NO!
Send one particle at a time through this setup, find (expt. has been
done!) interference fringes. But: block one side and fringes go away!
So a given single particle in quantum mechanics really has probabil-
ity of being either at position of first slit or second slit. With two
slits, interferes with itself. We can explain experiments (originally
4
just Gedanken experiments, but now doable!) by assuming there is
a probability amplitude ψ1 for particle to go through slit 1, ψ2 to
go through slit 2. With slit 2 closed, easy to see P1 = |ψ1|2 gives
the smooth intensity profile observed, by extension P2 = |ψ2|2 when
slit 1 is closed. But in order to explain interference pattern, deduce
probability profile with both open is
P12 = |ψ1 + ψ2|2 = |ψ1|2 + |ψ2|2 + ψ1ψ2∗ + ψ1∗ψ2.
(14)
Last terms provide interference effects.
This is silly—why not just check to see which slit particle goes through?
If we do (see figure) interference pattern disappears due to collapse
of wave function, i.e. ψ is no longer ψ1 + ψ2, but just, e.g. ψ1. We
must regard particle, if we do not measure it, as being a superposition of a particle which went through slit 1 and one which goes
through slit 2.
5.2
Particle in a box
Let’s talk about the use of ψ as a probability function with more concrete
example. We talked about the allowed normal modes of a cavity already,
but let’s revisit the problem from the point of view of Herr Schrödinger.
5
One way to write the Hamiltonian of a cavity or “box” in 1D is to simply
say there is an infinite potential if the particle is outside the box:

 ∞ x < −a/2
(15)
V (x) = 0
− a/2 ≤ x ≤ a/2

∞ x > a/2
Outside the box, there is zero probablity amplitude to find the particle,
ψ = 0. Inside the box, however, there is no potential, so Schrödinger’s
equation has the form for a free particle, (h̄2∇2/2m)ψ = Eψ. We know
that the solutions are waves, but we also recall that they must satisfy the
correct boundary conditions, in this case that ψ go to zero at the ends
of the box (what we called standing wave boundary conditions). The
first (lowest energy) standing wave has no nodes, and has a maximum at
x = 0, so this must be cos πx/a. The next has a node at zero, so it must
be sin 2πx/a. Then we get cos 3πx/a, and sin 4πx/a, and so on.
So the solutions may be written therefore
½
An cos kx k = nπ/a, n = 1, 3, 5, ...
ψn =
(16)
An sin kx k = nπ/a, n = 2, 4, 6...
(Don’t be confused if you remember dimly that we used sin instead cos
before (we used a coordinate system before where 0 was the left side of
the cavity, now 0 is the middle! See, e.g. Griffiths 2.28 and make sure
you understand the difference.). Check to make sure that for n = 1, 2, 3...
you get theR usual standing waves. We should use our normalization
condition dx|ψn|2 = 1 condition to determine each of the An. For
6
example, determine A0 by
Z a/2
a
A20
cos2 πx/a dx = A20 · ,
2
−a/2
p
so that A0 = 2/a.
(17)
One final remark about the eigenfunctions. Note that each successive
function has a different symmetry with respect to reflection about the x
axis, or parity. ψ0, ψ2, etc. are even functions of x, whereas ψ1, ψ3,
etc. are odd functions of x. There are no functions which are neither
even nor odd. This is a special case of a more general theorem which
says if H displays a certain symmetry (in this case invariance under x →
−x), the eigenfunctions are eigenfunctions of a differential operator which
implements this symmetry. In other words, the ψn(x) are eigenfunctions
of parity, too. We’ll revisit this concept later.
Inserting any of the wavefunctions into the S-eqn., the energy levels are
found to be just
En
h̄2k 2 h̄2n2π 2
=
=
2m
2ma2
(18)
Let’s use this example to discuss some basic concepts of quantummechanical measurement. Once we know ψ(x), we claim to know the
probability |ψ(x)|2 of finding the particle at x if we make a measurement.
But what if we make many measurements–what will be the average value,
or expectation value of such a series of measurements? Fortunately since
we know the probability function, we know the answer immediately,
Z
Z ∞
hxi = xP (x)dx =
x|ψ(x)|2dx,
(19)
−∞
i.e. each possible value of the thing measured, here x, is weighted by
the probability of finding it. If the particle is in the ground state ψ1(x)
(usually we call ψ0 the ground state, but here the lowest standing wave
7
has n = 1, so we stick with this convenient notation), the expectation
value of position is
Z a/2
hxi = A20
x cos2(πx/a)dx = 0.
(20)
−a/2
You can see that the answer must be zero, because x is an odd function,
cos2(x) is an even function, so the product is odd and it’s being integrated
over a symmetric interval (draw a picture!). You should show that hxi
is zero for any of the eigenfunctions, for the same reason. This should
correspond to your classical intuition that the average position over an
entire period of an SHO is just zero (eq. position). Exercise: calculate
hx2i for the ground and 1st excited states ψ1,2–your answer should not be
zero! This will give a measure of the amplitude.
5.3
5.3.1
Free particle
Fourier transform
• 1D Fourier series Suppose f (x) is periodic,
f (x + a) = f (x),
and f (x) is square integrable,
Z a
|f (x)|2 = finite
(21)
(22)
0
then f (x) can be expanded in the series
f (x) =
∞
X
e|2πinx/a
{z }
fn
n=−∞
with n integer this
has period a
8
(23)
Orthogonality relations:
Z
Z
a
dx e
2πinx/a −2πimx/a
e
=
0
½0
=
a
dx e2πi(n−m)x/a
0 n 6= m
a n=m
(25)
= aδm,n
(26)
where δm,n is the Kronecker δ fctn,
½
0 n 6= m
δm,n =
1 n=m
to get fn in Eq. (23), multiply by e−2πimx/a, integrate
Z
1 a
fn =
dx f (x) e−2πinx/a,
a 0
(24)
(27)
Ra
0
dx. Find
(28)
the inverse of Eq. (23).
Relation we’ll use again:
Z
a
Z
a
2
dx|f (x)| =
0
dx
0
= a
X
X
fnfm∗ e−2πi(n−m)x/a
(29)
m,n
|fn|2.
(30)
n
• Dirac δ-fctn. Fourier transform may be expressed succinctly using
δ-fctn. To see how it enters, substitute Eq. (28) back into Eq. (23):
9
X
Z
a
dy
f (y)e−2πiny/a
0 a
n
!
Ã
Z a
X
1
=
f (y)dy
e2πin(x−y)/a
a n
0
|
{z
}
δ(x − y)
f (x) =
or
Z
e
2πinx/a
(31)
(32)
a
f (y)δ(x − y)dy = f (x)
(33)
0
True for any well behaved fctn. f (x). Note mathematically δ(x) itself
not function but distribution. For case f (x) = 1, get normalization
condition
Z
δ(x)dx = 1
(34)
Can think of δ(x) as limit of sequence of functions
m
X
1 −2πinx/a
δm(x) =
e
a
n=−m
(35)
which look like (m=0,3,6–Maple) In sequence width of main peak
given by a/m, height by 2m/a. Note for large m oscillations outside
of central peak die out. So true δ-fctn. infinitely sharp, but with area
1.
• 1D Fourier integral transform.
Want to do similar things with function f (x) which isn’t periodic but
which vanishes suff. rapidly at ∞. Crudely, replace f (x) with another
function which agrees with it over a large interval (−a/2, a/2), make
periodic.
10
12
10
8
6
4
2
-3
-2
0
-1
1
2
3
x
-2
Now define
kn = 2πn/a,
afn = g(kn)
Using this notation, Eqs. (23) and (28) become
X1
f (x) =
g(kn)eikx,
a
kn
Z a/2
g(kn) =
dx f (x) e−iknx
(36)
(37)
(38)
−a/2
(since new f (x) periodic, any interval of length a is ok.
Now let a get very large =⇒ in sum over kn, The difference between two
11
neighboring k’s gets small, ∆kn = 2π
a , so replace sum by integral as usual:
Z ∞
dk
f (x) =
g(k) eikx,
(39)
2π
Z−∞
∞
g(k) =
dx f (x) e−ikx
(40)
−∞
Again substituting Eq. for f into one for g, get relation
µ Z ∞
¶
Z ∞
1
f (x) =
dy f (y)
dkeik(x−y)
2π −∞
−∞
{z
}
|
≡ δ(x − y)
i.e. integral representation for Dirac delta-fctn:
Z ∞
1
δ(x) =
dk eikx
2π −∞
5.3.2
(41)
(42)
(43)
Momentum measurement
Let’s consider a free particle first, so we know classically its motion will
be uniform. Time of flight measurement ok: suppose at t = 0 we measure
& find it to be in a small volume around r = 0. At later time t remeasure,
found at r, so velocity is v = r/t, momentum mr/t. Note there was some
uncertainty in our determination of initial position, but if we make t large
this becomes unimportant. In 1D S.-eqn. reads
h̄2 ∂ 2ψ
∂ψ
(44)
=−
ih̄
∂t
2m ∂x2
Plane wave (free-particle) solution is
ψ ∼ ei(px/h̄−p
2 t/2mh̄)
,
(45)
and as in the wave packet example, gen. soln. is lin. comb. of such waves:
Z
dp
2
√
ψ(x) =
f (p) ei(px/h̄−p t/2mh̄)
(46)
2πh̄
12
where “coefficients” f (p) fixed by initial conditions up to overall normalization, fixed by
Z ∞
Z
dpdp0
2
02
2
f (p)f (p0)∗e−ip t/2mh̄eip t/2mh̄
(47)
dx |ψ| =
2πh̄
−∞
Z ∞
0
×
dx ei(p−p )x/h̄
| −∞ {z
}
0
2πh̄δ(p − p )
Z ∞
=
dp |f (p)|2
(48)
−∞
so
Z
∞
ψ normalized ⇐⇒
dp |f (p)|2 = 1.
(49)
−∞
Check now that we can rewrite Eq. (40) as
Z
2
mx
it
dp
2
√
ψ(x) = ei 2h̄t
f (p) e− 2mh̄ (p−mx/t) .
2πh̄
(50)
Rapid oscillations of exponential kill f (p) unless p = mx/t, so we can
extract value of f (p) at this point to get estimate for integral:
Z
2
it
eimx /2h̄t
2
ψ' √
f (mx/t) dp e− 2mh̄ (p−mx/t)
(51)
2πh̄
p
Let’s shift p such that p ≡ mx/t + 2mh̄/tz, change variables to z:
r
Z
imx2 /2h̄t
e
2mh̄ ∞
2
ψ' √
f (mx/t)
dz e−iz
(52)
t
2πh̄
−∞
Integral very nasty, handle however with cute trick:
Path of integration along the z axis, where integrand horrible oscillatory
fctn. Cauchy’s theorem: may distort contour anywhere in complex plane if
integrand has no singularities, and falls off more rapidly than 1/|z| at large
13
1
0.5
-5
0
5
10
15
k
-0.5
-1
Wiggly curve is Re part of exponential, plotted vs. p, with stationary point at p = mx/t. Curve
in middle is possible f (p).
◦
|z|. Convenient to distort contour as shown in Figure, rotate by 45√
such
that path C parametrized by real variable y, i.e. put z ≡ (1 − i)y/ 2, so
Z ∞
Z
2
2
dz e−iz
(53)
dz e−iz =
A+B+C
Z−∞
Z ∞
1
−
i
1 − i√
2
2
dz e−iz = √
dy e−y = √
π
2 −∞
2
C
|{z}
(A & B don’t contribute as R → ∞)
Plug into Eq. (46), find
m
|f (mx/t)|2.
(54)
t
So prob. of finding particle at x is, for t → ∞,
m dx
dP = |ψ(x, t)|2dx = |f (mx/t)|2
.
(55)
t
Recall for free particle p = mx/t, so we may write prob. of particle having
momentum p:
dP = |f (p)|2dp
(56)
|ψ(x, t)|2 =
14
Integration contour for
R
2
eiz dz integration.
and from Eqs. (41-42) this prob. is properly normalized
5.3.3
R
dp|f (p)|2 = 1.
Heisenberg uncertainty principle
At t = 0,
Z
ψ(x) =
dp
√
f (p) eipx,
2πh̄
(57)
and as usual assume f (p) peaks at some p0, width ∆p.
• Case I.
πh̄
À ∆p
2x
Then Re eipx/h̄ varies slowly over the range of p where f (p) appreciable
=⇒ eipx/h̄ ' eip0x/h̄ in this range, so
Z
Z
(58)
ψ(x) = dp f (p) eipx/h̄ ' eip0x/h̄ dp f (p)
and |ψ(x)|2 nearly ind. of x.
15
• Case II.
πh̄
¿ ∆p
2x
Here eipx/h̄ oscillates rapidly over ∆p =⇒
Z
ψ(x) = dp f (p) eipx/h̄ ' 0.
(59)
Then ψ(x) must look like “bump” of width ∼ h̄/∆p:
Was true for particular choice of f (p), what about in general? Could
have f (p) varying rapidlyRwithin “spread” ∆p:
At x = 0 ψ large since dpf (p) 6= 0. As x increases, faster oscillations
eipx/h̄ don’t kill ψ as above, since they can match up with fast oscillations
of f (p) within envelope. ψ(x) only gets small when oscillations of eipx/h̄
significantly faster than those in f (p), i.e. when ∆x À h̄/δp (À h̄/∆p!).
For wiggly f (p), ∆x bigger than for “bump” f (p) =⇒ ∆x we found
16
Dashed line is “envelope” of f (p), of width ∆p, solid line f (p) itself. Scale of fast oscillations within
envelope is δp
before is rough lower bound for the position uncertainty, ∆x>
∼h̄/∆p, or,
as Heisenberg put it:
∆x∆p>
∼h̄ (Heisenberg Uncertainty Principle)
17
(60)