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5 Probability and Uncertainty Principle
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)|^2 δV (1)
Basic requirements for probability function:
- Positive semidefinite. δP ≥ 0 since ψψ∗^ ≥ 0 always.
- Normalized.
dP = 1. Particle must be somewhere! Wave func- tion is therefore normalized to 1: ∫ d^3 r|ψ(r, t)|^2 = 1 (2)
If
d^3 r|ψ|^2 is not 0 or ∞, can always normalize given eigensolution ψ of S.’s eqn. by multiplying with const., since ih∂ψ/∂t¯ = Hψ is linear.
- Conserved. Conservation of probability: if ψ normalized at t, must stay normalized at all times:
0 = d dt
ψ∗ψ(r, t)d^3 r =
∂ψ∗ ∂t
ψ + ψ∗^ ∂ψ ∂t
Is this consistent with S.-equation? The motion of ψ is determined by i¯h∂ψ∂t = Hψ, while the motion of ψ∗^ is determined by the complex conj. eqn. −ih¯∂ψ
∗ ∂t =^ Hψ
∗. Note H = H∗ (^) since potential V (r) in H = − ¯h 2 2 m∇
(^2) + V (r) is real. Therefore t-derivative of |ψ| (^2) is
∂t |ψ(r, t)|^2 = ψ∗^
∂ψ ∂t
∂ψ∗ ∂t =
i¯h 2 m
ψ∗∇^2 ψ −
ih¯ 2 m
ψ∇^2 ψ∗^ (4)
Now note we can write ψ∗∇^2 ψ − (∇^2 ψ∗)ψ = ∇ · (ψ∗∇ψ − (∇ψ∗)ψ) (5)
so we get ∂ ∂t |ψ|^2 + ∇ · [−
ih¯ 2 m [ψ∗∇ψ − ((∇ψ∗)ψ)] = 0, (6)
which just looks like a quantum-mechanical continuity equation: ∂ρ ∂t
where
ρ = |ψ|^2 = prob. density (8) j = − ih¯ 2 m
[ψ∗∇ψ − ((∇ψ∗)ψ)] = prob. current (9) or prob. flux density
- 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
just Gedanken experiments, but now doable!) by assuming there is a probability amplitude ψ 1 for particle to go through slit 1, ψ 2 to go through slit 2. With slit 2 closed, easy to see P 1 = |ψ 1 |^2 gives the smooth intensity profile observed, by extension P 2 = |ψ 2 |^2 when slit 1 is closed. But in order to explain interference pattern, deduce probability profile with both open is
P 12 = |ψ 1 + ψ 2 |^2 = |ψ 1 |^2 + |ψ 2 |^2 + ψ 1 ψ 2 ∗ + ψ 1 ∗ ψ 2. (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 ψ 1 + ψ 2 , but just, e.g. ψ 1. We must regard particle, if we do not measure it, as being a superpo- sition 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 Schr¨odinger.
example, determine A 0 by
A^20
∫ (^) a/ 2
−a/ 2
cos^2 πx/a dx = A^20 ·
a 2
so that A 0 =
2 /a. 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. ψ 0 , ψ 2 , etc. are even functions of x, whereas ψ 1 , ψ 3 , 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 = h¯^2 k^2 2 m
h¯^2 n^2 π^2 2 ma^2
Let’s use this example to discuss some basic concepts of quantum- mechanical 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,
〈x〉 =
xP (x)dx =
−∞
x|ψ(x)|^2 dx, (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 ψ 1 (x) (usually we call ψ 0 the ground state, but here the lowest standing wave
has n = 1, so we stick with this convenient notation), the expectation value of position is
〈x〉 = A^20
∫ (^) a/ 2
−a/ 2
x cos^2 (πx/a)dx = 0. (20)
You can see that the answer must be zero, because x is an odd function, cos^2 (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 〈x〉 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 〈x^2 〉 for the ground and 1st excited states ψ 1 , 2 –your answer should not be zero! This will give a measure of the amplitude.
5.3 Free particle
5.3.1 Fourier transform
- 1D Fourier series Suppose f (x) is periodic, f (x + a) = f (x), (21)
and f (x) is square integrable, ∫ (^) a
0
|f (x)|^2 = finite (22)
then f (x) can be expanded in the series
f (x) =
∑^ ∞
n=−∞
fn e ︸ ︷︷ ︸^2 πinx/a (23)
with n integer this has period a
f (x) =
n
e^2 πinx/a
∫ (^) a
0
dy a
f (y)e−^2 πiny/a^ (31)
∫ (^) a
0
f (y)dy
a
n
e^2 πin(x−y)/a
δ(x − y)
or (^) ∫ (^) a
0
f (y)δ(x − y)dy = f (x) (33)
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 ∫ δ(x)dx = 1 (34)
Can think of δ(x) as limit of sequence of functions
δm(x) =
∑^ m
n=−m
a
e−^2 πinx/a^ (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
- 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.
0
2
4
6
8
10
12
-3 -2 -1 (^1) x 2 3
Now define kn = 2πn/a, afn = g(kn) (36)
Using this notation, Eqs. (23) and (28) become
f (x) =
kn
a g(kn)eikx, (37)
g(kn) =
∫ (^) a/ 2
−a/ 2
dx f (x) e−iknx^ (38)
(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
where “coefficients” f (p) fixed by initial conditions up to overall normal- ization, fixed by ∫ (^) ∞
−∞
dx |ψ|^2 =
dpdp′ 2 πh ¯
f (p)f (p′)∗e−ip
(^2) t/ 2 m¯h eip
′^2 t/ 2 mh¯ (47)
×
−∞
dx ei(p−p ′)x/¯h ︸ ︷︷ ︸ 2 πhδ¯ (p − p′) =
−∞
dp |f (p)|^2 (48)
so ψ normalized ⇐⇒
−∞
dp |f (p)|^2 = 1. (49)
Check now that we can rewrite Eq. (40) as
ψ(x) = ei
mx2¯ht^2
dp √ 2 πh¯
f (p) e−^ 2 mit¯h (p−mx/t)^2
. (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:
ψ '
eimx^2 /2¯ht √ 2 π¯h
f (mx/t)
dp e−^ 2 mit¯h (p−mx/t)^2 (51)
Let’s shift p such that p ≡ mx/t +
2 m¯h/tz, change variables to z:
ψ '
eimx^2 /2¯ht √ 2 π¯h
f (mx/t)
2 mh¯ t
−∞
dz e−iz 2 (52)
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
-0.
0
1
-5 (^5) k 10 15
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 ∫ (^) ∞
−∞
dz e−iz
2
A+B+C
dz e−iz
2 (53) ∫
︸︷︷︸^ C
dz e−iz
2
1 − i √ 2
−∞
dy e−y
2
1 − i √ 2
π
(A & B don’t contribute as R → ∞)
Plug into Eq. (46), find |ψ(x, t)|^2 = m t
|f (mx/t)|^2. (54)
So prob. of finding particle at x is, for t → ∞,
dP = |ψ(x, t)|^2 dx = |f (mx/t)|^2 m dx t
Recall for free particle p = mx/t, so we may write prob. of particle having momentum p: dP = |f (p)|^2 dp (56)
πh¯ 2 x ø ∆p
Here eipx/h¯^ oscillates rapidly over ∆p =⇒
ψ(x) =
dp f (p) eipx/¯h^ ' 0. (59)
Then ψ(x) must look like “bump” of width ∼ h/¯ ∆p:
Was true for particular choice of f (p), what about in general? Could have f (p) varying rapidly within “spread” ∆p: At x = 0 ψ large since
dpf (p) 6 = 0. As x increases, faster oscillations eipx/¯h^ 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/¯h significantly faster than those in f (p), i.e. when ∆x ¿ h/δp¯ (¿ h/¯ ∆p!). For wiggly f (p), ∆x bigger than for “bump” f (p) =⇒ ∆x we found
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>∼h/¯ ∆p, or, as Heisenberg put it:
∆x∆p>∼h¯ (Heisenberg Uncertainty Principle) (60)
