text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
the lower bound can be. However, M cannot be ar-
bitrary larger because of the constraint (i). We are therefore facing a packing
problem w
here the goal is to “pack” as many Euclidean balls of radius propor-
tional to σ log(M )/n in Θ under the constraint that their centers remain close
ii)). If Θ = IRd, this the goal ... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
(cid:2)
1
2
−
γ
d
M (M 1)
2
−
≤
IP
X
d
2
−
> γd .
(cid:0)
(cid:1)
(cid:3)
(cid:0)
(cid:1)
Hoeffding’s inequality then yields
1)
M (M
2
−
IP
X
d
2
−
> γd
exp
≤
(cid:0)
(cid:1)
−
(cid:16)
2
2γ d + log
M (M 1)
2
−
(cid:0)
<
1
(cid:1)(cid:17)
6
6
6
5.5. Application to the Gaussian sequence model
114
as soon as
M (M
1) < 2 ... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
⌊
θ1, . . . , θM be such that
ed/32 and such that ρ(ωj, ωk)
βσ
θj = ωj √
n
for some β > 0 to be chosen later. We can check the conditions of Theorem 5.11:
≥
,
(i)
θ
|
j −
2
θk|2 =
(ii)
θj −
|
θk|
2
2 =
β2σ2
n
β2σ2
n
ρ(ωj, ωk)
ρ(ωj, ωk)
4
β2σ2d
≥ 16n
β2σ2d
n
≤
≤
32β2σ2
n
log(M ) =
2ασ2
n
log(M ) ,
for β = α . Applying n... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
Note that the vectors θ1, . . . , θM employed in the previous subsection are
not guaranteed to be sparse because the vectors ω1, . . . , ωM obtained from the
Varshamov-Gilbert Lemma may themselves not be sparse. To overcome this
limitation, we need a sparse version of the Varhsamov-Gilbert lemma.
Lemma 5.14 (Sparse Var... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
(k) are equally likely under this distribution and
therefore, ω is uniformly distributed on C0(k). Observe that
ωj = ωk : ρ(ωj, ωk) < k
P
I
∃
(cid:0)
(cid:1)
=
1
d
k x
(cid:0) (cid:1)
1
d
k x
(cid:0) (cid:1)
= M IP
≤
(cid:0)
ωj = x : ρ(ωj, x) <
k
2
IP
d
∃
(cid:0)
M
0,1
X
∈{
}
|0=k
x
|
IP
ωj = x : ρ(ωj, x) <
(cid:0)
d
j... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
2k
k ≤ d
1) ≤ d
i
1
l=1 Z
−
Qi =
(i
P
−
−
k
−
l
since k
d/2.
≤
Next we apply a Chernoff bound to get that for any s > 0,
IP ω = x0 : ρ(ω, x0) <
k
2
≤
k
IP
Zi >
k
= IE exp s
Zi
sk
e− 2
k
2
i=1
(cid:0) X
The above MGF can be controlled by induction on k as follows:
i=1
(cid:0) X
h
(cid:1)
(cid:1)
(cid:0)
(cid:1)i
k
IE exp... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
(cid:0)
sk
− 2
k
2
k
2
−
−
(cid:1)
= exp log M + k log 2
(cid:17)
lo
g(1 + )
(cid:16)
log M + k log 2
log(1 + )
d
2k
d
2k
(cid:17)
e
xp log M
log(1 + )
e
xp
≤
≤
< 1 .
d
2k
(cid:17)
(cid:17)
(for d
8k)
≥
(cid:16)
(cid:16)
(cid:16)
k
4
k
4
−
d
2k
log M < log(1 + )
If we take M such that
Apply the sparse Varshamov-Gilbert... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
�
(cid:0)
k
θ
|0≤
|
ˆ
θ
θ
2
|2 ≥
−
α2σ2
64n
It implies the following corollary.
k log(1 + )
1
d
2k ≥ 2 −
2α .
(cid:1)
Corollary 5.15. Recall that
of IRd. The minimax rate of estimation over
model is φ(
least squares estimator θls
IR denotes the set of all k-sparse vectors
B0(k) in the Gaussian sequence
B0(k)) = σ k log... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
Theorem 5.11:
θj|1 = R for j = 1, . . . , M . We can check the conditions of
(i)
θj −
|
θ 2
k|2 =
R2
k2
ρ(ωj, ωk)
R2
≥ 2k ≥
4R min
R
8
, β2σ
log(ed/√n)
8n
(ii)
θj −
|
2
θk|2 ≤
2R2
k
≤
4Rβσ
r
log(ed/√n)
n
(cid:0)
≤
2ασ2
n
log(M ) ,
.
(cid:1)
for β small enough if d
Applying now Theorem 5.11 yields
≥
Ck for some constant... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
to zero for small R. Note that if
R, we have
θ∗
|
|1 ≤
0
|
−
θ∗
|
2
2 =
θ∗
|
2
2 ≤ |
|
θ∗
|
1 = R2 .
2
5.5. Application to the Gaussian sequence model
119
2
Remark 5.17. Note that the inequality
1 appears to be quite loose.
Nevertheless, it is tight up to a multiplicative constant for the vectors of the
σ log d ,
form... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
C/εd .
≤
2ε
|
θi −
θj| ≥
θ1, . . . , θN }
(b) Show that for any x
θi|2 ≤
2ε.
−
x
|
∈ B2(0, 1), there exists i = 1, . . . , N such that
(c) Use (b) to conclude that there exists a constant C′ > 0 such that N
C′/εd .
≥
Problem 5.4. Show that the rate φ = σ2d/n is the minimax rate of estimation
over:
(a) The Euclidean Bal... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
, NJ, third edition, 2008. With
an appendix on the life and work of Paul Erdo˝s.
Dennis S. Bernstein. Matrix mathematics. Princeton University
Press, Princeton, NJ, second edition, 2009. Theory, facts, and
formulas.
Patrick Billingsley. Probability and measure. Wiley Series in
Probability and Mathematical Statistics. J... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
istical estimation when p is much larger than n. Ann. Statist.,
35(6):2313–2351, 2007.
T. Tony Cai and Harrison H. Zhou. Minimax estimation of large
covariance matrices under ℓ1-norm. Statist. Sinica, 22(4):1319–
1349, 2012.
T. Tony Cai, Cun-Hui Zhang, and Harrison H. Zhou. Opti-
mal rates of convergence for covariance... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
2001. Data mining, inference, and
prediction.
I. A. Ibragimov and R. Z. Hasminski˘ı. Statistical estimation,
volume 16 of Applications of Mathematics. Springer-Verlag, New
York, 1981. Asymptotic theory, Translated from the Russian by
Samuel Kotz.
[Joh11]
Iain M. Johnstone. Gaussian estimation: Sequence and wavelet
mode... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
let. Adaptive density estimation using the block-
wise Stein method. Bernoulli, 12(2):351–370, 2006.
Philippe Rigollet and Alexandre Tsybakov. Exponential screen-
ing and optimal rates of sparse estimation. Ann. Statist.,
39(2):731–771, 2011.
Jun Shao. Mathematical statistics. Springer Texts in Statistics.
Springer-Ver... | https://ocw.mit.edu/courses/18-s997-high-dimensional-statistics-spring-2015/619e4ae252f1b26cbe0f7a29d5932978_MIT18_S997S15_CourseNotes.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.102 Introduction to Functional Analysis
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
18
LECTURE NOTES FOR 18.102, SPRING 2009
Lecture 4. Thursday, 12 Feb
I talked about step functions, then the covering le... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/61a26bd09e9c192d3da2e538d1a35f52_MIT18_102s09_lec04.pdf |
a sum
(4.4)
N
�
ciχ[ai,bi)
f =
i=1
of multiples of the characteristic functions of our intervals. Note that such a ‘pre
sentation’ is not unique but can be made so by demanding that the intervals be
disjoint and ‘maximal’ – so f is does not take the same value on two intervals with
a common endpoint.
Now, a c... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/61a26bd09e9c192d3da2e538d1a35f52_MIT18_102s09_lec04.pdf |
fine a
Lebesgue integrable function, however we need to do some work to flesh out the
definition.
�
LECTURE NOTES FOR 18.102, SPRING 2009
19
Definition 3. A function g : R −→ C is Lebesgue integrable if there exists an
absolutely summable sequence of step functions fn, i.e. satisfying
(4.7)
such that
(4.8)
f (x)... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/61a26bd09e9c192d3da2e538d1a35f52_MIT18_102s09_lec04.pdf |
i =�
j = ⇒
N
�
(bi − ai) ≤ (b − a).
i=1
On the other hand
(4.10)
[a, b) ⊂
Ci = ⇒
N
N
�
(bi − ai) ≥ (b − a).
i=1
i=1
You can prove this by inserting division points etc.
Now, what we want is the same thing for a countable collection of intervals.
Proposition 2. If Ci = [ai, bi), i ∈ N, is a countable colle... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/61a26bd09e9c192d3da2e538d1a35f52_MIT18_102s09_lec04.pdf |
20
LECTURE NOTES FOR 18.102, SPRING 2009
Now, by Heine-Borel – the compactness of closed bounded intervals – a finite subcol
lection of these open intervals covers [a, b − δ) so (4.10) does apply to the semi-open
intervals and shows that for some finite N (hence including the finite subcollection)
(4.14)
N
�
i=1
(... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/61a26bd09e9c192d3da2e538d1a35f52_MIT18_102s09_lec04.pdf |
that
fn is a decreasing (meaning non-increasing) se
quence. So there are only two possibilities, it converges to 0, as we claim, or it
converges to some positive value. This means that there is some δ > such that
�
�
fn > δ for all n, so we just need to show that this is not so.
Given an � > 0 consider the sets
(... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/61a26bd09e9c192d3da2e538d1a35f52_MIT18_102s09_lec04.pdf |
(Bj ) < �.
j≥N
Dividing the integral for fk, k ≥ N, into the part over SN and the rest we see that
(4.23)
fk ≤ (b − a)� + �A.
�
The first estimate comes for the fact the fact that fk ≤ � on SN , and the second
that the total lengths of the remaining intervals is no more than � (and the function
is no bigger than... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/61a26bd09e9c192d3da2e538d1a35f52_MIT18_102s09_lec04.pdf |
WAVE MECHANICS
B. Zwiebach
September 13, 2013
Contents
1 The Schr¨odinger equation
2 Stationary Solutions
3 Properties of energy eigenstates in one dimension
4 The nature of the spectrum
5 Variational Principle
6 Position and momentum
1 The Schr¨odinger equation
1
4
10
12
18
22
In classical mechanics th... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
complex: if it were real, the right-hand side of (1.2) would be real while the left-hand side
would be imaginary, due to the explicit factor of i.
∈
Let us make two important remarks:
1
1. The Schr¨odinger equation is a first order differential equation in time. This means that if
we prescribe the wavefuncti... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
define the probability density P (x, t), also denoted as ρ(x, t), as the norm-squared of
the wavefunction:
P (x, t) = ρ(x, t)
Ψ∗(x, t)Ψ(x, t) =
≡
Ψ(x, t)
|
2 .
|
(1.4)
This probability density so defined is positive. The physical interpretation of the wavefunction
arises because we declare that
P (x, t) dx is th... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
wave-
function remains normalized for all times. Proving this is a good exercise:
2
Exercise 1. Show that the Schr¨odinger equation implies that the norm of the wavefunction
does not change in time:
d ∞
Z
dt −∞
Ψ(x, t)
dx
|
|
2 = 0 .
(1.8)
You will have to use both the Schr¨odinger equation and its com... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
This equation applied to a fixed volume V implies that the rate of change of the enclosed charge
QV (t) is only due to the flux of Ji across the surface S that bounds the volume:
dQV
dt
(t) =
dia .
i
J
·
−
i
S
(1.11)
Make sure you know how to get this equation from (1.10)! While the probability current in
more... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
the rate at which probability flows out at the
right boundary of the interval.
It is sometimes easier to work with wavefunctions that are not normalized. The normaliza
tion can be perfomed if needed. We will thus refer to wavefunctions in general without assuming
normalization, otherwise we will call them normalized... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
�(x, t) ,
(2.1)
where we have introduced the Hamiltonian operator Hˆ :
~
ˆH
2 ∂2
2m ∂x2
ˆH is an operator in the sense that it acts on functions of x and t to give functions of x and t:
it acts on the space of complex functions, a space that contains wavefunctions. Note that V (x)
acts just by multiplication. N... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
energy E been a complex number E = E0−
would not drop out:
iΓ, with E0 and Γ real, the time dependence
P (x, t) = Ψ∗(x, t) Ψ(x, t) = e
i(E∗ −E)t/
~
+iE∗t/
~
ψ ∗(x) e
−2Γt/
~
ψ∗(x)ψ(x) = e
= e
−iEt/
~
ψ(x)
ψ(x)
|
2 .
|
This kind of state is not acceptable: the normalization cannot be preserved in time.
Let us ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
~
2 d2
2m dx2
−
(cid:16)
+ V (x) ψ(x) = E ψ(x) .
(cid:17)
(2.7)
(2.8)
(2.9)
(2.10)
Note that the derivatives along x need not be denoted as partial derivatives since the functions
they act on have no other argument except x. Using primes to denote derivatives with respect
to the argument, the above equation is... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
and, of course, finding
those solutions for each E.
A solution ψ(x) associated with an energy E is called an energy eigenstate of energy E.
The set of all allowed values of E is called the spectrum of the Hamiltonian Hˆ . A degeneracy
in the spectrum occurs when there is more than one solution ψ(x) for a given value... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
In the spectrum of a Hamiltonian, localized energy eigenstates are particularly important.
This motivates the definition:
An energy eigenstate ψ(x) is a bound state if ψ(x)
0 when
x
| → ∞
|
.
→
(2.13)
Since a normalizable eigenstate must have a wavefunction that vanishes as
state is just a normalizable eigenstat... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
that for rather general potentials the Hˆ
eigenstates ψn(x) can be chosen to be orthonormal. What does it mean for two functions to
be orthogonal? Orthogonal vectors have a vanishing dot product, where the dot product is a
(clever) rule to obtain a single number from two vectors. For two functions f1 and f2 an inner ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
expanded as a superposition of energy eigenstates. Namely, there exist
complex numbers bn such that
ψ(x) =
∞
L
n=1
bn ψn(x) ,
C .
bn ∈
(2.17)
This is a very powerful statement: it means that if the energy eigenstates are known, the general
solution of the Schr¨odinger equation is known. Indeed assume that the... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
be emphasized that the superposition of stationary states is generally not a sta
tionary state. The expansion coefficients bn used above can be calculated explicitly if we know
the energy eigenstates. Indeed using (2.16) and (2.17) a one-line computation (do it!) gives
bn
=
∞
Z
−∞
dx ψ ∗ (x)ψ(x) .
n
(2.21)
A curi... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
� , x)δ(x ′
x0) = K(x0, x) .
−
(2.24)
We therefore conclude that K(x ′ , x) = δ(x
thus find
∞
x ′ ) (recall that δ(x) = δ(
x)). Back in (2.22) we
−
−
Completeness:
∗ (x ′ )ψn(x) = δ(x
ψn
−
x ′ ) .
L
n=1
(2.25)
Let us compare the completeness relation above with the orthonormality relation (2.16). In
the... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
on a normalized state Ψ by
∞
Aˆ
(
)Ψ(t)
≡
Z
−∞
dx Ψ∗(x, t)(AˆΨ(x, t)) .
(2.26)
What happens when we take the operator to be Hˆ ? Using (2.20) twice, we get
ˆH
)Ψ(t) =
(
=
=
=
so that we get
∞
Z
−∞
dx Ψ∗(x, t)( ˆHΨ(x, t))
∞
Z
−∞
L
n,n ′
dx b∗
n e iEnt/ ψ∗
n(x) bn ′ e −iE ′
~
~
nt/ ˆHψn ′ (x)
nb... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
is normalizable, then the wavefunction
Ψ(x, t)
dx Ψ∗Ψ
V
J
(2.29)
is normalized. We can thus use this normalized wavefunction in the definition on
the expectation value is given by
Aˆ
)
(
to find
Aˆ
(
)Ψ(t)
≡
∞
−∞ dx Ψ∗(x, t)(AˆΨ(x, t))
J
R
dx Ψ∗(x, t)Ψ(x, t)
.
(2.30)
This formula can be used for any norm... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
.
With this the Schr¨odinger equation (3.1) becomes
ψ ′′ + (
E − V
(x))ψ = 0 .
(3.1)
(3.2)
(3.3)
We are now ready to consider a basic result:
two or more bound states for any given energy.
in a one-dimensional potential there cannot be
Theorem 1. There is no degeneracy for bound states in one-dimensional poten... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
(3.7)
The constant can be evaluated by examining the left-hand side for
that ψ1 →
assumed in (2.12). It follows that the left-hand side vanishes as
We thus have
. We then have
0, since they are bound states, while the derivatives are bounded, as
and therefore c = 0.
0 and ψ2 →
| → ∞
| → ∞
x
|
|
x
ψ2ψ ′
1 =
... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
of the above equation gives
ψ ′′ + (
E − V
(x))ψ = 0 ,
(ψ∗) ′′ + (
E − V
(x))ψ∗ = 0 .
(3.10)
(3.11)
So ψ∗ if different from ψ defines a degenerate solution. By superposition we can then get two
real (degenerate) solutions
(ψ + ψ∗) , ψim ≡
ψr ≡
These are, of course, the real and imaginary parts of ψ.
1
2
1
2i
ψ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
is an even function of x: V (
x) = V (x) the eigenstates can be
−
x.
→ −
Proof. Again, we begin with our main equation
ψ ′′ (x) + (
E − V
(x))ψ(x) = 0 .
(3.14)
Recall that primes denote here derivative with respect to the argument, so ψ ′′ (x) means the
function “second-derivative-of-ψ” evaluated at x. Similarly ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
indeed ϕ(x) = ψ(
x) provides a degenerate solution to the Schr¨odinger equation:
−
d2
dx2
ϕ(x) + (
E − V
(x))ϕ(x) = 0 .
(3.18)
Equipped with the degenerate solutions ψ(x) and ψ(
antisymmetric (a) combinations that are, respectively, even and odd under x
−
x) we can now form symmetric (s) and
x:
→ −
ψs(x)
1
(ψ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
odd under x
→ −
x.
±
−
→ −
4 The nature of the spectrum
Consider the time-independent Schr¨odinger equation written as
ψ ′′ =
2m
~
2 (E
−
−
V (x)) ψ .
(4.1)
We always have that ψ(x) is continuous, otherwise ψ ′′ has singularities worse than delta func
tions and we would require potentials V (x) that are wors... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
hard wall at x = a. In such a case, the
wavefunction will vanish for x
a from the left,
≥
and will vanish for x > a. Thus ψ ′ is discontinuous at the wall.
a. The slope ψ ′ will be finite as x
→
In conclusion
Both ψ and ψ ′ are continuous unless the potential has delta functions
or hard walls in which cases ψ ′... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
and k determined if E is
known. Finally to the right we have a solution α4 exp(
κx) since the wavefunction must
. So we got four (real) unknown constants αi, i = 1, 2, 3, 4. Since ψ and
vanish as x
cψ are the same solution we can scale the solution and thus we only have three unknown
constants to determine. There ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
�
→
→ ∞
→
±∞
freedom is accounted in. We also have two boundary conditions at the interface. So we
can expect a solution. Indeed there should be a solution for each value of the energy. The
spectrum here is continuous and non-degenerate.
(c) Two constants are needed here in each of the three regions: they multiply s... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
the figure we indicate the type of spectrum for
energies in the various intervals defined: E > V+, then V− < E < V+, then V0 < E < V− and
finally E < V0.
Figure 2: A generic potential and the type of spectrum for various energy ranges.
A node in a wavefunction is a point x0 where ψ(x0) = 0 (a zero of ψ) and ψ ′(x0) = ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
�4(x) with
three nodes at x1, x2, and x3 and zeroes at x =
. For ψ5 there must be a node
w1 in (
−∞
(x1, x2) and so on until a last node w4 ∈
Example: Potential with five delta functions. We will discuss the bound states of the
Schr¨odinger equation with potential
, x1], a node w2 ∈
and x =
(x3,
−∞
∞
∞
).
−
2... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
)ψ(x) = Eψ(x) ,
(4.4)
and integrate this equation from a
to zero. By doing this we will get one out of the five delta functions to fire. We find
ǫ to a + ǫ, where ǫ is a small value that we will take down
−
~
2
a+ǫ
−
2m
Z
a−ǫ
dx
d2ψ
dx2
+
a+ǫ
Z
a−ǫ
dxV (x)ψ(x) = E
Z
a+ǫ
a−ǫ
dx ψ(x) .
(4.5)
The first term in... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
= 0 .
2m
This implies that the discontinuity Δψ ′ of ψ ′ is given by
−
−
−
)
(
Δψ ′ (a)
ψ ′ (a +)
≡
−
ψ ′ (a −) =
2m
2 (
~
V0a) ψ(a) .
−
(4.7)
(4.8)
The discontinuity of ψ ′ at the position of the delta function is proportional to the value of ψ at
this point. The constant of proportionality is linear on th... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
a and b have the same sign and it will
have exactly one zero if a and b have opposite signs.
Figure 5: Plots of ae −κx and beκx with a, b > 0. This can be used to show that any linear superposition
of these two functions can at most have one zero.
Let us then make the following remarks:
17
2a (no... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
argued that such function can at
most have one zero.
3. Zeroes appear at x = 0 for all the antisymmetric bound states.
In those cases, there
a, but this
cannot be another zero in the interval [
is presumably not generic. There are at most five bound states because the maximum
number of nodes is four; one in betwee... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
out the ground state wavefunction.
For this purpose, consider an arbitrary normalized wavefunction ψ(ix):
dix ψ∗(ix)ψ(ix) = 1 .
Z
(5.12)
By arbitrary we mean a wavefunction that need not satisfy the time-independent Schr¨odinger
equation, a wavefunction that need not be an energy eigenstate. Then we claim the grou... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
energies En are ordered as
E2 ≤
Of course Hψn = Enψn. Since the energy eigenstates are complete, any trial wavefunction can
be expanded in terms of them (see (2.18)):
Egs = E1 ≤
E3 ≤
(5.14)
. . . .
ˆ
ψ(ix) =
∞
L
n=1
bn ψn(ix) .
(5.15)
Such a ψ is not an energy eigenstate in general. The normalization cond... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
function
ψ(x)
√
N
with N =
Z
dixψ∗(ix)ψ(ix) ,
(5.19)
is normalized and can be used in (5.13). We therefore find that
Egs ≤
dix ψ∗(ix) Hψ(ix)
ˆ
dix ψ∗(ix)ψ(ix)
≡ F
J
J
[ψ] .
(5.20)
This formula can be used for trial wavefunctions that are not normalized. We also introduced
[ψ]. A functional is a machine that... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
a one-dimensional problem with the delta function potential
Hˆ
(
(
In this problem the ground state energy is calculable exactly and one has
V (x) =
α δ(x) , α > 0 .
−
Egs =
mα2
~
2 2
−
.
(5.21)
(5.22)
So this problem is just for illustration. Consider an unnormalized gaussian trial wavefunction,
with a real... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
the final expression above provides an upper bound for
the ground state energy, and the best upper bound is the lowest one. We thus have that the
ground state energy satisfies
Egs ≤
Minβ
(cid:16)
~
β2 2
4m
β
α .
π
− √
(cid:17)
The minimum is easily found
β =
2mα
~
2
π
√
→
Egs ≤ −
mα2
~
2
π
=
Comparing wi... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
than the ground state
F
F
1We use the integrals
due−u
2
= √
π and
J
J
duu2 e
2
−u
1
= √
2
π.
21
6 Position and momentum
In quantum mechanics the position operator ˆx and the momentum operator ˆp do not commute.
They satisfy the commutation relation
[ˆ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
.
(6.29)
The N + 1 component column vector summarizes the values of the wavefunction at equally
separated points. N is some kind of regulator: a precise description requires N
0.
→ ∞
→
Associated with the description (6.29) the operator ˆx can be viewed as the (N + 1)
(N + 1)
d... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
��
(6.31)
2The time dependence is irrelevant to the present discussion, which applies without changes to time-
dependent wavefunctions Ψ(x, t).
22
which is indeed the representation of xψ(x). Given our definition of the action of ˆx, expectation
values in normalized states are naturally defined by
xˆ
)... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
i ∂x
This is to say that acting on a wavefunction we have
≡
pˆ
~
(6.35)
(6.36)
p ψ(x)
ˆ
=
~
dψ
i dx
.
Note that the commutation relation (6.27) is satisfied by the above definitions, as we can check
acting on any wavefuntion:
[ˆx , pˆ]ψ(x) = (ˆxpˆ
pˆxˆ)ψ(x)
−
~
−
x
= ˆ
xˆ
= ˆp ψ(x)
~
dψ
i dx
dψ
x
i dx ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
∂ e
~
i ∂x 2π
√
~
ipx/
e
= p
√
~
2π
= p ψp(x) .
(6.39)
So ψp(x) is a momentum eigenstate with momentum eigenvalue p. It is a plane wave.
The so-called momentum representation is mathematically described by Fourier transforms.
The Fourier transform ψ˜(p) of ψ(x) is defined by
˜
ψ(p)
∞
≡
Z
−∞
dx
−ipx/
~
e
√
~... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
are just two different representations of the same state:
ψ(x)
←→
˜
ψ(p) .
(6.42)
The arrow above is implemented by Fourier Transformation. Calculate now the action of
on (6.41)
d
i dx
~
d
i dx
ψ(x) =
~
d
i dx
∞
dp
Z
−∞
~
ipx/
e
√
~
2π
∞
˜
ψ(p) =
dp
Z
−∞
~
ipx/
e
√
2π
~
˜
p ψ(p) .
(6.43)
In the la... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
�x, pˆ] = i acting on momentum space wavefunctions.
~
25
25
MIT OpenCourseWare
http://ocw.mit.edu
8.05 Quantum Physics II
Fall 2013
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.726 Algebraic Geometry
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
18.726: Algebraic Geometry (K.S. Kedlaya, MIT, Spring 2009)
Divisors on curves and Riemann-Roch (updated 31 Mar 09)
We continue the discus... | https://ocw.mit.edu/courses/18-726-algebraic-geometry-spring-2009/61d8135f20b057ebeee0f387e612985a_MIT18_726s09_lec15_divisors2.pdf |
to cohomology.
Theorem (Riemann-Roch). There exists a nonnegative integer g = g(X) with the following
property. For any divisor D and any canonical divisor K,
l(D) − l(K − D) = deg(D) + 1 − g.
Corollary. The integer g in Riemann-Roch can be identified as
g = l(K) = dimk �(X, �X/k).
Proof. Take D = 0. Then l(D) = 1... | https://ocw.mit.edu/courses/18-726-algebraic-geometry-spring-2009/61d8135f20b057ebeee0f387e612985a_MIT18_726s09_lec15_divisors2.pdf |
2g − 2, then deg(K − D) < 0. In this case, (f ) + K − D has negative degree
and so cannot be effective, so l(K − D) = 0 no matter what.
Corollary. For g ∼ 2, for any divisor D of degree at least 2g − 1, the complete linear system
associated to D defines a closed immersion of D into a projective space.
2 The canonical... | https://ocw.mit.edu/courses/18-726-algebraic-geometry-spring-2009/61d8135f20b057ebeee0f387e612985a_MIT18_726s09_lec15_divisors2.pdf |
discussed in the problem set, so I’ll only sketch the
general argument. Put D = (P ) + (Q) for P, Q ≤ X(k) not necessarily distinct. We need to
check whether we always have
l(K − D) = l(K) − 2 = g − 2.
2
By Riemann-Roch,
l(K − D) = l(D) + g − 3
so we have an embedding if and only if l(D) = 0 for any effective D ... | https://ocw.mit.edu/courses/18-726-algebraic-geometry-spring-2009/61d8135f20b057ebeee0f387e612985a_MIT18_726s09_lec15_divisors2.pdf |
) is separable). The ramification divisor of f is defined as
R = �
length(�X/Y )P (P ),
P �X(k)
where as usual �X/Y is the module of K¨ahler differentials.
Proposition. We have
KX � f �KY + R.
Proof. (Compare Hartshorne Proposition IV.2.3.) Note that
0 � f ��Y /k � �X/k � �X/Y � 0
is exact; this follows from prope... | https://ocw.mit.edu/courses/18-726-algebraic-geometry-spring-2009/61d8135f20b057ebeee0f387e612985a_MIT18_726s09_lec15_divisors2.pdf |
Namely, put Q = f (P ), and pick t ≤ k(Y ) which generates mY,Q; then f �(t) generates
m
for some nonnegative integer e. We call e = eP the ramification index of P . Then
e
X,P
length(�X/Y )P ∼ eP − 1,
with equality if and only if f is tamely ramified, i.e., eP is not divisible by the characteristic
of k.
In case ... | https://ocw.mit.edu/courses/18-726-algebraic-geometry-spring-2009/61d8135f20b057ebeee0f387e612985a_MIT18_726s09_lec15_divisors2.pdf |
Lectures 15 & 16
Local Area Networks
Eytan Modiano
Eytan Modiano
Slide 1
Carrier Sense Multiple Access (CSMA)
•
In certain situations nodes can hear each other by listening to the channel
- “Carrier Sensing”
• CSMA: Polite version of Aloha
– Nodes listen to the channel before they start transmission
Channel idl... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
(duration = β)
Eytan Modiano
Slide 4
�Analysis of CSMA
• Let the state of the system be the number of backlogged nodes
• Let the state transition times be the end of idle slots
– Let T(n) = average amount of time between state transitions when the system is
in state n
T(n) = β + (1 - e-λβ (1-qr)n)
When qr is s... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
performance of CSMA
• Unslotted CSMA will have slightly lower throughput due to
increased probability of collision
• Unslotted CSMA has a smaller effective value of β than slotted
CSMA
– Essentially β becomes average instead of maximum propagation
delay
Eytan Modiano
Slide 8
CSMA/CD and Ethernet
Two way cable
... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
Assume N users and that each attempts transmission during a
free “mini-slot” with probability p
– P includes new arrivals and retransmissions
P(i users attempt) =
N
Pi(1− P)N −i
i
P(exactly 1 attempt) = P(success) = NP(1 -P)N -1
To maximize P(success),
d
dp
[NP(1- P)N-1] = N(1 -P)N -1 − N(N − 1)... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
λ < 1/(1+4.4β)
• Compare to CSMA without CD where λ <
1
1 + 2β
Eytan Modiano
Slide 14
Notes on CSMA/CD
• Can be viewed as a reservation system where the mini-slots are
used for making reservations for data slots
•
In this case, Aloha is used for making reservations during the
mini-slots
• Once a users captures... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
Node replaces token on ring as soon as it is done transmitting the
packet
– Next node can use token after short propagation delay
• Release after reception
– Node releases token only after its own packet has returned to it
Serves as a simple acknowledgement mechanism
Eytan Modiano
Slide 18
PACKET TRANSMISSION
(... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
system
Eytan Modiano
Slide 21
Throughput analysis (non-exhaustive)
• Gated system with limited service - each node is limited to
sending one packet at a time
– When system is heavily loaded nodes are always busy and have a
packet to send
• Suppose each node transmits one packet and then releases the
token to the... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
(1 − λ( E[ X] + ( m + 1)v))
Eytan Modiano
Slide 24
Token ring issues
• Fairness: Can a node hold the token for a long time
– Solution: maximum token hold time
• Token failures: Tokens can be created or destroyed by noise
– Distributed solution:
Nodes are allowed to recognize the loss of a token and create a new... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
27
IMPLICIT TOKENS
• The idle tokens on a token bus can be replaced with silence
• The next node starts to transmit a packet after hearing the bus
•
•
become silent
If the next node has no packet, successive nodes start with
successively greater delay
If the bus propagation delay is much smaller than the time to... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
Res
Data
Res
Wait for Reser-
vation Interval
Wait for Assigned
Data Slot
• Satellite reservation system
– Use mini-slots to make reservation for longer data slots
– Mini-slot access can be inefficient (Aloha, TDMA, etc.)
• To a crude approximation, delay is 3/2 times the propagation delay
plus ideal queueing del... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
to packet transmission time
– GEO example: Dp = 0.5 sec, packet length = 1000 bits, R = 1Mbps
Latency = 500 => very high
– LEO example: Dp = 0.1 sec
Latency = 100 => still very high
– Over satellite channels data rate must be very low to be in a low
latency environment
• Low latency protocols
– CSMA, Polling, T... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/623c1e88221701c134811e228f8b7f11_Lectures15_16.pdf |
18.354J Nonlinear Dynamics II: Continuum Systems Lecture 20
Spring 2015
20 Classical aerofoil theory
We now know that through conformal mapping it is possible to transform a circular wing into
a more realistic shape, with the bonus of also getting the corresponding inviscid, irrotational
flow field. Let’s consider some... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/62405b4d167552f452b8df52513a0ef8_MIT18_354JS15_Ch20.pdf |
−2iθ
(cid:1)
(490)
(491)
At θ = 0 and θ = π we are in trouble because the velocities are infinite. Notably, however,
this problem can be removed at θ = 0 if the circulation is chosen so that the numerator
vanishes
(cid:2)
e−iθ u (ei(θ−α)
0
− e−i(θ−α)) − iΓ
2πR
(cid:3)
U − iV =
1 − e−2iθ
.
(492)
Thus for a finite velocity... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/62405b4d167552f452b8df52513a0ef8_MIT18_354JS15_Ch20.pdf |
find that
u − iv =
dW
dZ
=
dw/dz
dZ/dz
= u0
e−iα −
(cid:17)2
(cid:16) R+λ
z+λ
1 − R2
z2
− iΓ
2π(z+λ)
.
The value of Γ that makes the numerator zero at the trailing edge is
Γ = −4πu0(R + λ) sin α.
(497)
(498)
The flow is then smooth and free of singularities everywhere (because we have successfully
trapped the rogue singu... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/62405b4d167552f452b8df52513a0ef8_MIT18_354JS15_Ch20.pdf |
(cid:12)
.
Thus, we find
(cid:73) (cid:16)
p0 −
f = i
|v|2(cid:17)
ρ
2
dz = −i
ρ (cid:73)
2
|v|2 dz
Taking the complex conjugate, we have
¯f = fx − ify = i
(cid:73)
ρ
2
|v|2 dz¯.
96
(500)
(501)
(502)
(503)
(504)
Furthermore, since v is parallel to z on boundary (which is a stream line), we may use
0 = vxdy − vydx
(505)... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/62405b4d167552f452b8df52513a0ef8_MIT18_354JS15_Ch20.pdf |
�ow remains finite at |z| → ∞ and, in this case,
we can identify
a−2 + ...
z2
= a0 +
dw
dz
(508)
+
a0 = vx(∞) − ivy(∞).
(509)
In particular, if the wing moves along the x-axis and surrounding gas is at rest, then simply
a0 = vx(∞).
To obtain the physical meaning of a−1, we note that by virtue of the residues theorem26
a... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/62405b4d167552f452b8df52513a0ef8_MIT18_354JS15_Ch20.pdf |
a0 = ρΓvy(∞) + iρΓvx(∞).
(514)
Recall that FD = (cid:96)fx and FL = (cid:96)fy, this is indeed the generalization of our earlier results
for drag and lift on a cylinder, if we identify vy(∞) = 0 and vx(∞) = −u0. Note that the
results FD = 0 is again a manifestation of d’Alembert’s paradox (now for arbitrarily shaped
wi... | https://ocw.mit.edu/courses/18-354j-nonlinear-dynamics-ii-continuum-systems-spring-2015/62405b4d167552f452b8df52513a0ef8_MIT18_354JS15_Ch20.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.950 Differential Geometry
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
CHAPTER 4
Geometry of lengths and distances
1
Lecture 36
Let’s start by looking at standard Rn . Straight lines are distinguished by
... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
2 .
Definition 36.2. Let M ⊂ Rn+1 be a hypersurface. A smooth map γ : I →
M , where I ⊂ R is an interval, is called a geodesic if γ��(t) is perpendicular
to T Mγ(t) for all t.
Remember that γ�(t) ∈ T Mγ(t), essentially by definition of tangent space.
Geodesics are curves held to M by a constraint force.
Lemma 36.3. If... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
for all times.
Examples 36.7. (i) The nontrivial geodesics on Sn are just the great circles,
parametrized with arbitrary constant speed. More explicitly, take u, v ∈ Sn
which are orthogonal to each other, and write γ(t) = cos(αt)u + sin(αt)v,
where α ∈ R is any constant.
(ii) Take the infinite cylinder M = {x ∈ R3 ... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
Particular solutions are where x2 is constant, or where x1 is constant at a
value where l1
2(t)2 = 0,
� (x1) = 0.
2(t) = 0.
1(t)c�
c�
1
Consider a hypersurface M ⊂ Rn+1, but where now Rn+1 carries the Minkowski
inner product. We assume that M is space-like, which means that the re
striction of �·, ·�M in to T My ... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
become straight line segments (their speed, obviously, is not
constant). In the parametrization by the Poincar´e ball model, they become
circle segments which intersect the boundary of ball perpendicularly (on, in
the limiting case, a line segment through the center of our ball).
�
Lecture 38
This lecture covers t... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
and any interval [a, b], there is a geodesic γ : [a, b] M
which is an absolute minimizer of the energy.
→
This provides a practical way of finding geodesics numerically, by applying
some minimization method to the energy functional.
Now consider a partial parametrization f : U Rn+1 of M , and its associ
ated first ... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
ecture 39
Let M ⊂ Rn+1 be a hypersurface. The length of a path γ : [a, b] → M is
� b
L(γ) =
�γ�(t)� dt.
a
Define the distance dist(p, q) = inf γ L(γ), where the infimum is taken over
all paths from p to q.
Lemma 39.1. If M is a connected hypersurface, then (M, dist) is a metric
space. By this we mean that it sat... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
Then, for any
two points p, q there is a path γ connecting them, such that L(γ) = dist(p, q).
In other words, the infimum in the definition of distance is always attained.
Given a parametrization f : U M with first fundamental form I, one can
define the lengths of paths c : [
→
U to be equal to the length of their
im... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
ollary 39.9. Any holomorphic function h : U
→
increasing for the hyperbolic metric: dist(h(p), h(q)) ≤
dist(p, q).
U is distance-non
Lecture 40
Let (X, d) be a metric space. This means that X is a set, and d : X ×X → R
a function satisfying the three axioms from the last lecture. In particular,
this allows one to... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
odesics with the same starting point is
d(γ1(t), γ2(t)) = α arctanh(1/ tanh(t)).
for some constant α, which is a convex function.
Example 40.5. Any metrized tree is nonnegatively curved in the sense of
Busemann.
Example 40.6. A combinatorial surface in R3 is Busemann if and only if it
is topologically simply-conn... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
with com
parison points x�, y�, we have dist(x, y) ≤ dist(x�, y�).
All examples listed above are in fact CAT (which implies Busemann). There
are also important local versions of all the notions in this lecture, where the
conditions are assumed to hold only locally (“for every point x ∈ X there
exists an open subse... | https://ocw.mit.edu/courses/18-950-differential-geometry-fall-2008/624f231226f7c15ddce6ba6dfad2db64_ch4_revised.pdf |
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