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Textbook 1
Introduction to Quantum Mechanics
by David J. Griffiths
Part I Theory
Chap.1 The Wave Function
Chap.2 The Time-Independent Schrodinger Equation
Chap.3 Formalism
Chap.4 Quantum Mechanics in Three Dimensions
Chap.5 Identical Particles
Part II Applications
Chap.6 Time-Independent Perturbation Theory
Chap.7 The Variational Principle
Chap.8 The WKB Approximation
Chap.9 Time-Dependent Perturbaion Theory
Chap.10 The Adiabatic Approximation
Chap.11 Scattering
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Bibliography
in English
D. J. Griffiths, Introduction to Quantum Mechanics.
S. M. McMurry, Quantum mechanics.
P. Roman, Advanced Quantum Theory-an outline of the fundamental ideas.
J. J. Sakurai, Modern quantum mechanics.
R. Shankar, Principles of quantum mechanics.
R. P. Feynman, The Feynman Lectures on Physics, Vol. 3.
Cohen-Tannoudji, Quantum Mechanics, Vol. 1, 2.
L. I. Schiff, Quantum mechanics.
L. D. Landau and E. M. Lifshitz, Quantum Mechanics.
R. Robinett’s Quantum mechanics: classical results, modern systems and
visualized examples.
J. Townsend, A modern approach to quantum mechanics.
D. Park, Introduction to quantum theory.
P. A. M. Dirac, The Principles of Quantum Mechanics.
L. E. Ballentine，Quantum Mechanics: A Modern Development.
T. Haar, Problems in Quantum Mechanics.
……
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in Chinese
周世勋：《量子力学教程》，高等教育出版社，1979.
曾谨言：《量子力学》，科学出版社，1984.
曾谨言，《量子力学》(I,II),《量子力学导论》，2003.
钱伯初，《量子力学基本原理和计算方法》，甘肃人民出版社，1984.
苏汝铿: 《量子力学》,复旦大学出版社，1997.
张永德：《量子力学》，科学出版社，2002.
关
洪：《量子力学基础》，高等教育出版社，1999.
曾谨言、钱伯初：《量子力学专题分析（上,下）》，1990,1999.
钱伯初、曾谨言：《量子力学习题精选与剖析》，科学出版社，1988.
哈兴林：《高等量子力学》，高等教育出版社，1999.
倪光炯、陈苏卿：《高等量子力学》，复旦大学出版社，2000.
刘连寿：《理论物理基础教程》，高等教育出版社，2003.
汪德新：《量子力学》，科学出版社，2008.
……
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Outline
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Imagine a particle of mass m, constrained to move
along the x-axis, subject to some specified force F(x,t):
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Classical Mechanics (or Newton’s mechanics):
The position of the particle at any given time x(t) can
be determined by Classical Mechanics.
How does it do by CM or NM?
Then, the velocity [ v(t) = dx(t)/dt ], the momentum
[ p(t)=mv(t) ], the kinetic energy ( T=mv2/2 ), or any
other dynamical variable of interest can be figured out !
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However, Quantum Mechanics approaches this
same problem quite differently!
Quantum Mechanics:
In this case what we are looking for is the wave
function, (x,t), of the particle, and we get it by solving
the Schrödinger Equation:
where i is the square root of 1, and ħ is Planck’s
constant:
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The Schrödinger Equation plays a role logically
analogous to Newton’s second law:
Given suitable initial conditions [ typically, (x,t =
0) ], the Schrödinger equation determines (x,t) for all
future time, just as, in classical mechanics, Newton’s
law determines x(t) for all future time.
NOTE: The Schrödinger Equation, not derived but
guessed at intuitively, would then be a postulate of new
theory, and its validity would have to be checked by
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experiment !
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Fundamental principles of Quantum Mechanics：
Assumption 1: Wave function (r, t) describes the
state of a particle !
Assumption 2: Schrödinger Equation
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Now, Question is:
What exactly is this “wave function” ?
or what does it do for you once you are got it ?
or how can such “wave function” be said to
describe the state of a particle ?
14
The wave function (r,t) itself is complex, not any
physical means, but |(r,t)|2=* ( where * is the
complex conjugate of  ) is real and nonnegative — as
a probability, of course, must be.
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The answer is provided by Born’s statistical
interpretation of the wave function, which says that
|(x,t)|2 gives the probability of finding the particle at
point x, at time t — or more precisely :
16
For example:
One would be quite likely to find that the particle in the
vicinity of point A, and relatively unlikely to find it near
point B.
17
The statistical interpretation introduces a kind of
indeterminacy into quantum mechanics.
Even if you know everything the theory has to tell you
about the particle, you cannot predict with certainty the
outcome of a simple experiment to measure its position.
Quantum Mechanics has to offer is statistical
information about the possible results !
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This indeterminacy has been profoundly disturbing
to physicists and philosophers alike.
Is it peculiarity of nature, a deficiency in the theory,
a fault in the measuring apparatus, or what ?
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“Measurement” of Microscopic Particle (or System)
Suppose we do measure the position of the particle,
and we find it to be at the point C.
Question: Where was the particle just before we
made the measurement ?
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Answer 1. The realist position: The particle was at C.
This certainly seems like a sensible response, and it
is the one Einstein advocated.
However, that if this is true then quantum mechanics
is an incomplete theory, since the particle really was at C,
and yet quantum mechanics was unable to tell us so.
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Answer 2. The orthodox position: The particle was
not really anywhere.
It was the act of measurement that forced the particle
to “take a stand” . This view is associated with Bohr
and his followers --- Copenhagen interpretation.
However, that if it is correct there is something very
peculiar about the act of measurement — something that
over half a century of debate has done precious little to
illuminate.
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Answer 3. The agnostic position: Refuse to answer.
This is not quite as silly as it sounds --- after all, what
sense can there be in making assertions about the
status of a particle before a measurement, when the
only way of knowing whether you were right is
precisely to conduct a measurement, in which case
what you get is no longer “before the measurement”?
It is metaphysics to worry about something that cannot,
by its nature, be tested.
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For decades this was the “fall back” position of most
physicists: They’d try to sell you answer 2, but if you
were persistent they’d switch to 3 and terminate the
conversation.
Until fairly recently, all three position (realist, orthodox,
and agnostic) had their partisans.
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But in 1964 John Bell astonished the physics
community by showing that it makes an observable
difference if the particle had a precise (though
unknown) position prior to the measurement.
Bell’s discovery effectively eliminated agnosticism as
a viable option , and made it an experimental question
whether 1 or 2 is the correct choice.
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For now, suffice it to say that the experiments have
confirmed decisively the orthodox inter- pretation:
A particle simply does not have a precise position
prior to measurement, any more than the ripples on a
pond do.
It is the measurement process that insists on one
particular number, and thereby in a sence creates the
specific result, limited only by the statistical weighting
imposed by the wave function.
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What if we made a second measurement, immediately
after the first?
Would we get C again, or does the act of measurement
cough up some completely new number each time?
On this question everyone is in agreement: A repeated
measurement (on the same particle) must return the
same value.
Indeed, it would be tough to prove that the particle was
really found at C in the first instance if this could not be
confirmed by immediate repetition of the measurement.
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How does the orthodox interpretation account for
the fact that the second measurement is bound to give
the value C ?
Evidently the first measurement radically alters
the wave function, so that it is now sharply peaked
about C (see Figure 1.3).
28
We say that the wave function collapses upon
measurement, to a spike at the point C (  soon spreads
out again, in accordance with the Schrödinger equation,
so the second measurement must be made quickly).
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A measurement on the given state wave function (x,t)
before the measurement
make a measurement
30
There are, then, two entirely distinct kinds of
physical processes:
(i) “ordinary” ones, in which the wave function
evolves in a leisurely fashion under the Schrödinger
equation.
(ii) “measurements”, in which  suddenly and
discontinuously collapses.
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Fundamental principles of Quantum Mechanics：
One of the most fundamental principles of quantum
mechanics is the principle of linear superposition of state
or, for short, the superposition principle.
Assumption 3:
A quantum mechanical system, which can take on the
discrete states n ( nN ), is also able to occupy the
state
and the probability density of taking on state n is then
given by
32
Because of the statistical interpretation of wave
function, probability plays a central role in quantum
mechanics.
So we digress now for a brief discussion of the
theory of probability. It is mainly a question of
introducing some notation and terminology.
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(1) Discrete Variable
Consider a room containing 14 people, whose ages are:
1 person aged 14,
1 person aged 15,
3 person aged 16,
2 person aged 22,
2 person aged 24,
5 person aged 25.
If we let N(j) represent the number of people of age j, then
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The total number of people in the room is
In this instance, N=14, and a histogram of the data is
showed by Figure 1.4.
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Question 1. If you selected one individual at random
from this group, what is the probability that this
persion’s age would be 15 ?
Answer: the probability that the persion’s age would
be 15 is P(15)=1/14 in above example.
In general, if let P(j) be the probability of value j,
then, the definition of probability is
In particular, the sum of all the probability is 1, the
normalization conditaion:
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Question 2. What is the most probability age ?
Answer: age 25.
In general, the most probability j is the j for
which P( j ) is a maximum.
Question 3: What is the median age ?
Answer: age 23.
In general, the median is the value of j such that
the probability of getting a larger result is the same
as the probability of getting a smaller result.
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Question 4: What is the average (or mean, or expection)
age ?
Answer: [14×1+15×1+16×3+22×2+24×2
+25×5]/14=294/14=21.
In general, the average (or mean, or expection)
value of j is given by
Question 5: What is the average of square of the ages ?
Answer: One could get 142 =196 with probability 1/14,
152 =225 with probability 1/14, 162 =256 with
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probability 3/14, and so on.
Then, the average of square of the ages can be calculated
by
In general, the average value of any fuction f(j) of
j is given by
for example, f (j) = jn, then
where n is called the n-th moment of j.
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Question 6: What is the deviation of age j from the
average ?
Answer: One could calculated each indicidual deviates
from the average by using of following formula
In general, the deviation means the amount of
“spread” in a distribution with respect to the average.
Note that the average of j is zero:
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Therefore, one should define the average of square j by:
This quantity 2 is known as the variance of the
distribution.
 itself is called the standard deviation, which is the
customary measure of the spread about  j .
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(2) Continuous Variable
In general, a “random number” or “stochastic variable”
is an object X defined by
(a) a set of possible values (called “range”, “set of
states”, “sample space”, or “phase space”);
(b) a probability distribution over this set.
The probability distribution (or probability density),
in the case of a continuous one-dimensional range, is
given by a function (x) that is nonnegative
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And normalized in the sense
The probability that X has a value between x and x+dx
is (x)dx, i.e.
The probability that x lies between a and b (a finite
interval) is given by the integral of (x):
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and the rules we deduced for discrete distributions
translate in the obvious way:
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Homework:
Problem 1.1, Problem 1.3.
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We return now to the statistical interpretation of the
wave function (Eq.[l.3]), which says that |(x,t)|2 is the
probability density for finding the particle at point x,
at time t. It follows (Eq.[1.16]) that the integral of ||2
must be l (the particle’s got to be somewhere):
Without this, the statistical interpretation would be
nonsense.
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The wave function is supposed to be determined
by the Schrödinger equation — We can’t impose an
extraneous condition on  without checking that the
two are consistent.
Schrödinger equation [1.1] reveals that if (x,t)
is a solution, so too is A(x,t), where A is any
(complex) constant.
What we must do, then, is pick this undetermined
multiplicative factor so as to ensure that Eq.[1.20]
is satisfied. This process is called normalizing the
wave function.
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For some solutions to the Schrödinger equation,
the integral is infinite; in that case no multiplicative
factor is going to make it 1.
The same goes for the trivial solution  = 0.
Such non-normalizable solutions cannot represent
particles, and must be rejected.
Conclusion :
Physically realizable states correspond to the
“square-integrable”, solutions to Schrödinger’s
equation !
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Above normalization only fixes the modulus of A；
the phase remains undetermined.
However, as we shall see, the latter (phase) carries
no physical significance anyway.
Suppose we have normalized the wave function at
time t = 0.
Question: How do we know that it will stay normalized,
as time goes on and  evolves ?
You can’t keep renormalizing the wave function, for
then A becomes a function of t , and you no longer
have a solution to the Schrödinger equation.
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Fortunately, the Schrödinger equation has the property
that:
It automatically preserves the normalization of the
wave function .
Without this crucial feature the Schrödinger equation
would be incompatible with the statistical interpretation,
and the whole theory would crumble !
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So we’d better pause for a careful proof of this point:
Note that the integral is a function only of t, so we use a
total derivative (d/dt) in the first term, but the integrand
is a function of x as well as t, so it’s a partial derivative
(/t) in the second one.
By the product rule,
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Now the Schrödinger equation says that
and hence also (taking the complex conjugate of Eq.
[1.23])
so
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The integral (Eq.[1.21]) can now be evaluated explicitly:
But (x,t) must go to zero as x goes to  , otherwise the
wave function would not be normalizable.
It follows that
and hence that the integral on the left is constant
(independent of time); if  is normalized at t = 0, its
stays normalized for all future time. QED
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Homework:
Problem 1.4, Problem 1.5.
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For a particle in state ，the expectation value of x is
What exactly does this mean ?
It emphatically does not mean that if you measure
the position of one particle over and over again, x||2dx
is the average of the results you’ll get.
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However, the first measurement will collapse the
wave function to a spike at the value actually obtained,
and the subsequent measurements (if they’re performed
quickly) will simply repeat that same result.
Rather, x is the average of measurements
performed on particles all in the state .
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Which means that either you must find some way of
returning the particle to its original state after each
measurement , or else you prepare a whole ensemble of
particles, each in the same state ，and measure the
positions of all of them: x is the average of these
results.
In short, the expectation value is the average of
repeated measurements on an ensemble of identically
prepared systems, not the average of repeated
measurements on one and the same system.
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As time goes on, x will change (because of the time
dependence of ), we might be interested in knowing
how fast x moves.
Referring to Eqs.[1.25] and [l.28], and x is independent
of t, we see that
This expression can be simplified using integration by
parts:
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Performing another integration by parts on the second
term of Eq.[1.30], we conclude that
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What are we to make of this result ?
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Note that we’re talking about the “velocity” of the
expectation value of x, which is not the same thing as the
velocity of the particle.
Nothing we have seen so far would enable us to
calculate the velocity of a particle — it’s not even clear
what velocity means in quantum mechanics.
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If the particle doesn’t have a determinate position
(prior to measurement), neither does it have a well defined velocity. All we could reasonably ask for is the
probability of getting a particular value.
We’ll see in Chapter 3 how to construct the probability
density for velocity, given .
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For our present purposes it will suffice to postulate that
the expectation value of the velocity is equal to the time
derivative of the expectation value of position:
Eq.[1.31] tells us, then, how to calculate v directly
from ．
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Actually, it is customary to work with momentum
( p=mv ), rather than velocity:
Let we write the expressions for x and p in a more
suggestive way:
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We say that : in quantum mechanics,
the operator x “represents” position, and
the operator (ħ/i)(/x) “represents”, momentum.
To calculate expectation values, we “sandwich” the
appropriate operator between * and ，and integrate.
From above discussion, it can be found that the position
and momentum are represented by an operator,
respectively.
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Assumption 4: In quantum mechanics, all physical
quantities are represented by a linear and hermitian
operator !
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What about other dynamical variables ?
The fact is, all such quantities can be written in
terms of position and momentum.
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To calculate the expectation value of such a quantity
we simply replace every px by operator (ħ/i)(/x), insert
the resulting operator between * and ，and integrate:
Note: there is no ∧ on the symbol of dynamical variables
when it represents a operator in this textbook !
For example,
Eq.[1.36] is a recipe for computing the expectation
value of any dynamical quantity for a particle in state
；it subsumes Eqs.[1.34] and [1.35] as special cases. 68
NOTE:
Although all physical quantities are represented by a
linear and hermitian operator in quantum mechanics.
An operator itself has no any physical means.
Its mean is only reflected in the action on wave
function ! e. g. : Fu = v.
It is important to realize that the product of two
operators in general does not commute, i. e. AB≠BA.
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We have tried in this section to make Eq.[1.36] seem
plausible, given Born’s statistical interpretation.
but the truth is that the equation [1.36] represents
such a radically new way of doing business (as
compared with classical mechanics) that it’s a good
idea to get some practice using it before we come back
(in Chapter 3) and put it on a firmer theoretical
foundation.
In the meantime, if you prefer to think of it as an
axiom, that’s fine !
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Homework:
Problem 1.6, Problem 1.7.
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Imagine that you’re holding one end of a very long
rope, and you generate a wave by shaking it up and
down rhythmically (Figure 1.6).
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If someone asked you, “Precisely where is that wave?”
You’d probably think he was a little bit nutty:
The wave isn’t precisely anywhere —it’s spread out
over 50 feet or so.
If he asked you what its wavelength is ?
you could give him a reasonable answer: It looks like
about 6 feet.
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By contrast, if you gave the rope a sudden jerk
(Figure 1.7), you’d get a relatively narrow bump
traveling down the line.
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In this case,
the first question (Where precisely is the wave? ) is a
sensible one,
the second (What is its wavelength? ) seems nutty—it
isn’t even vaguely periodic.
how can you assign a wavelength to it?
Of course, you can draw intermediate cases, in which
the wave is fairly well localized and the wavelength is
fairly well defined, but there is an inescapable trade-off
here:
The more precise a wave’s position is, the less precise
is its wavelength, and vice versa .
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A theorem in Fourier analysis makes all this rigorous in
Chapt. 3, but for the moment we are only concerned with
the qualitative argument.
This applies to any wave phenomenon, and hence in
particular to the quantum mechanical wave function.
Now the wavelength of  is related to the momentum
of the particle by the De Broglie formula:
Thus, a spread in wavelength corresponds to a spread
in momentum.
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In fact, similarly, a spread in frequency corresponds to
a spread in energy.
E  h  
which is another of the De Broglie formulae.
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In general, the more precisely determined a
particle’s position is, the less precisely its momentum
is determined, quantitatively,
where x is the standard deviation in x, and p is the
standard deviation in p. This is Heisenberg’s famous
uncertainty principle.
We’ll prove it in Chapter 3, we mention it here so you
can test it out on the examples in Chapter 2.
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How to understand the uncertainty principle means ?
Like position measurements, momentum
measurements yield precise answers — the “spread”
here refers to the fact that momentum measurements
on identical systems do not yield consistent results.
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You can prepare a system such that repeated position
measurements will be very close together (by making
 a localized “spike”), but you will pay a price:
Momentum measurements on this state will be
widely scattered.
Or you can prepare a system with a reproducible
momentum (by making  a long sinusoidal wave), but
in that case position measurements will be widely
scattered.
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And, of course, if you’re in a really bad mood (心情) you
can prepare a system in which neither position nor
momentum is well defined:
Eq.[1.40] is an inequality, and there’s no limit on
how big x and p can be — just make  some long
wiggly line with lots of bumps and potholes and no
periodic structure.
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Homework:
Problem 1.9, Problem 1.14,
Problem 1.16, Problem 1.17.
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Summary :
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