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/ ESD.07J Statistical Thinking and Data Analysis
Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
1. Vectors in R2 and R3
Definition 1.1. A vector �v ∈ R3 is a 3-tuple of real numbers (v1, v2, v3).
Hopefully the reader can well imagine the definition of a vector in
R2 .
Example 1.2. (1, 1, 0) and (
√
2, π, 1/e) are vectors in R3 .
Definition 1.3. The zero vector in R3 , denoted �0, is the vector
(0, 0, 0). If �v... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
0 + �v = �0.
(2) �u + (�v + w� ) = (�u + �v) + w� .
(3) �u + �v = �v + �u.
·
(4) λ (µ �v) = (λµ) �v.
(5) (λ + µ) �v = λ �v + µ �v.
·
(6) λ (�u + �v) = λ �u + λ �v.
·
·
·
·
·
·
·
Proof. We check (3). If �u = (u1, u2, u3) and �v = (v1, v2, v3), then
�u + �v = (u1 + v1, u2 + v2, u3 + v3)
= (v1 + u1, v2 + u2, v3... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
get another
point Q. If P = (p1, p2, p3) and �v = (v1, v2, v3) then
Q = P + �v = (p1 + v1, p2 + v2, p3 + v3).
(q1, q2, q3), then there is a unique vector
−→
P Q, such that Q
=
If Q =
P + �v, namely
−→
P Q = (q1 − p1, q2 − p2, q3 − p3).
Lemma 1.6. Let P , Q and R be three points in R3 .
Then
−→
P Q +
−→
QR = ... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
effect of two forces is represented by the vector sum. Sim
ilarly we can use vectors to measure both velocity and acceleration.
The equation
F� = m�a,
is the vector form of Newton’s famous equation.
Note that R3 comes with three standard unit vectors
ˆı = (1, 0, 0)
jˆ = (0, 1, 0)
and
kˆ = (0, 0, 1),
which are ... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
We can always rewrite
this as,
(x, y, z) = (p1, p2, p3) + t(v1, v2, v3) = (p1 + tv1, p2 + tv2, p3 + tv3).
Writing these equations out in coordinates, we get
x = p1 + tv1
y = p2 + tv2
and
z = p3 + tv3.
Example 1.7. If P = (1, −2, 3) and Q = (1, 0, −1), then �v = (0, 2, −4)
and a general point of the line contai... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
Where does the line
(x, y, z) = (1 − t, 2 − 3t, 2t + 1)
intersect the plane
We must have
2x − 3y + z = 6?
2(1 − t) − 3(2 − 3t) + (2t + 1) = 6.
Solving for t we get
9t − 3 = 6,
so that t = 1. The line intersects the plane at the point
(x, y, z) = (0, −1, 3).
3
Example 1.10. A cycloid is the path traced in th... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
and at time t = 0, the angle
3π/2. So we have
−→
P Q = (a cos(3π/2 − t), a sin(3π/2 − t)) = (−a sin t, −a cos t)
Putting all of this together, we have
(x, y) = (at − a sin t, a − a cos t).
4
MIT OpenCourseWare
http://ocw.mit.edu
18.022 Calculus of Several Variables
Fall 2010
For information about citing these... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
LECTURE 4
Broken circuits, modular elements, and supersolvability
This lecture is concerned primarily with matroids and geometric lattices. Since
the intersection lattice of a central arrangement is a geometric lattice, all our results
can be applied to arrangements.
4.1. Broken circuits
y in L, we have seen (Th... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
, 124. With respect to the second ordering O� the circuits are 123, 145, 2345,
and the broken circuits are 12, 14, 234.
It is clear that the broken circuit complex BC(M ) is an abstract simplicial
BC(M ). In Figure 1 we
BC(M ) and U
T , then U
complex, i.e., if T
≤
∗
≤
1
5
3
5
2
4
2
4
3
1
Figure 1. Tw... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
5, f1(�) = 8, f2(�) = 4.
Note, moreover, that
ψM (t) = t3
5t2 + 8t
4.
−
In order to generalize this observation to arbitrary matroids, we need to introduce
a fair amount of machinery, much of it of interest for its own sake. First we give
a fundamental formula, known as Philip Hall’s theorem, for the M¨obius func... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
0, 1) = ci,
�
ˆ 1
0, ˆ . Define
}
�(P �) to be the set of chains of P �, so �(P �) is an abstract simplicial complex. The
reduced Euler characteristic of a simplicial complex � is defined by
Note. Let P be a finite poset with ˆ
1, and let P � = P
0 and ˆ
ζ)k (ˆ ˆ
ζ)i(ˆ
− {
−
−
−
−
where fi is the number of i-dimensional... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
later.
Now let P be a finite graded poset with ˆ
0 and ˆ
1. Let
(x, y) : x � y in P
,
}
{
the set of (directed) edges of the Hasse diagram of P .
E(P ) =
Definition 4.11. An E-labeling of P is a map ϕ : E(P )
P then there exists a unique saturated chain
∃
P such that if x < y in
C : x = x0 � x1 � x1 �
satisfying... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
.g., the first has a chain whose edge labels are not all different,
while every maximal chain label of Figure 2(c) is a permutation of
i
}
{
−
Theorem 4.11. Let ϕ be an E-labeling of P , and let x
the M¨obius function of P . Then (
decreasing saturated chains from x to y, i.e.,
−
.
}
y in P . Let µ denote
→
1)rk(x... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
0 � x1 � � xn = ˆ
< yj−1 < ˆ
1 in P such that
1 such
→
→
−
· · ·
ϕ(xi−1 , xi) > ϕ(xi , xi+1)
i
S, 1
i n.
To prove the claim, let ˆ = y0 < y1 <
1
j
i
0
· · ·
1. By the definition of E-labeling, there exists a unique refinement
≤
⊆
< yj−1
→ →
ˆ
< yj = 1 with rk(yi) = ai for
→
→
−
ˆ
0 = y0 = x0 � x1 �
� x... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
· · ·
Now for S
[n
∗
−
1] define
(28)
λP (S) =
(
T →S
�
1)#(S−T )κP (T ).
−
The function λP is called the flag h-vector of P . A simple Inclusion-Exclusion
argument gives
(29)
κP (S) =
λP (T ),
T →S
�
−
∗
[n
for all S
the number of maximal chains ˆ
if and only if i
decreasing maximal chains ˆ
1]. It fol... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
now
�
We come to the main result of this subsection, a combinatorial interpretation
< yi = ˆ
· · ·
of the coefficients of the characteristic polynomial ψM (t) for any matroid M .
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
45
5
5
4
5
4 2
3
1
4
1
5
1
4
2
2
3
2
1
3
2... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
ϕ(x, y) = max
{
⇒
xi = y
.
}
As an example, Figure 3 shows the lattice of flats of the matroid M of Figure 1
with the edge labeling (30).
Claim 1. Define ϕ : E(L(M ))
P by
∃
ϕ(x, y) = m + 1
−
Then ϕ is an E-labeling.
˜ϕ(x, y).
To prove this claim, we need to show that for all x < y in L(M ) there is a
unique s... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
· · ·
· · ·
→
⊂
⊂
→
⇔→
⇒
46
R. STANLEY, HYPERPLANE ARRANGEMENTS
By the induction hypothesis there exists a unique saturated chain y1 � y2 �
� yk = y satisfying ˜
ϕ(y1, y2),
˜
ϕ(yk−1, yk). Since ˜
ϕ(y0, y1) = j > ˜
ϕ(y1, y2)
· · ·
the proof of Claim 1 follows by induction.
· · ·
⊂
⊂
Claim 2. The bro... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
ait for 1
1
→
j. Now for any circuit
is a broken circuit, with 1
i1 < < ij
· · ·
u1, . . . , uh}
{
r
}
∅ {
and any
h we have
k. Let B
t
→
→
→
x
⇒
u1
⇒
u2
⇒ · · · ⇒
uh = u1
⇒ · · · ⇒
ui−1
⇒
ui+1
⇒ · · · ⇒
uh.
i
→ →
Thus
⇒
zi1
zij−1
zi2 ⇒ · · · ⇒
z = zi1 ⇒
xr = yij , contradicting the maximality of ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
, so T
To complete the proof of the theorem, note that we have shown that fi−1(BC(M ))
0 = y0 � y1 � � yi such that ˜
· · ·
is the number of chains C : ˆ
ϕ(C) is strictly increas-
ing, or equivalently, ϕ(C) is strictly decreasing. Since ϕ is an E-labeling, the proof
�
follows from Theorem 4.11.
Corollary 4.6. The ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
(t) at t = 1). See Exercise 21 for more information on λ(M ).
1)rank(M )−1λ(M ) = ψ�
2, where (
−
(b) [to be inserted]
As an example of the applicability of our results on matroids and geometric
lattices to arrangements, we have the following purely combinatorial description of
the number of regions of a real cen... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
Now suppose
y) = 1 + rk(y), while rk(a) = 1 and
a
⇔→
�
rk(a
y) = rk(ˆ0) = 0, so again (31) holds.
By semimodularity, rk(a
y = a and a
y. Then a
y.
y
→
⇒
∈
⇒
∈
(c) Suppose that rk(L) = 3. All elements of rank 0, 1, or 3 are modular by
(a) and (b). Suppose that rk(x) = 2. Then x is modular if and only i... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
in the vector space Fn(q).
Figure 4 shows the Hasse diagrams of B2(3) and B3(2).
q
≤
Note that for x, y
Bn(q) we have x
= x + y
(subspace sum). Clearly Bn(q) is atomic: every vector space is the join
(sum) of its one-dimensional subspaces. Moreover, Bn(q) is graded of rank
n, with rank function given by rk(x) =... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
p < L if p
1 adjoined. It
is an immediate consequence of the axioms that when R is finite, L(R)
is a modular geometric lattice of rank 3. It is an open (and probably
intractable) problem to classify all finite projective planes. Now let P and
Q be posets and define their direct product (or cartesian product ) to be
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
rangement Bn.
Proposition 4.9. A partition β
Γn is a modular element of Γn if
≤
and only if β has at most one nonsingleton block. Hence the number of
modular elements of Γn is 2n
n.
−
Proof. If all blocks of β are singletons, then β = ˆ0, which is modular by
(a). Assume that β has the block A with r > 1 elemen... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
1,
1, so β is
k + j
−
−
−
with #B1 > 1 and #B2 > 1.
β
|
|
= π = k
|
|
b, . . . , B3, . . . , Bk}
β
β
∈
⇒
−
π =
π =
a, b, B1
{
B1
{
a, B2
B2, B3, . . . , Bl}
∅
Hence rk(β) + rk(π) = rk(β
∈
−
β
⊆ |
∈
π = k + 2
|
β
|
π = k
⇒ |
1.
−
⊆
π) + rk(β
π), so β is not modular.
⇒
�
∈
y = ˆ
0 and x
In a fini... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
rk(y) = rk(x
y).
∈
⇒
(b) (transitivity of modularity) If x is a modular element of L and y is modular
in the interval [ˆ0, x], then y is a modular element of L.
(c) If x and y are modular elements of L, then x
y is also modular.
∈
The next result, known as the modular element factorization theorem [16], is
our p... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
[ˆ0, x] is modular, so ψx(t) is divisible by ψa(t) = t
1. From this it
is immediate (e.g., because the characteristic polynomial ψG(t) of any geometric
lattice G of rank n begins xn
, where a is the number of atoms of G) that
ψx(t) = (t
5). On the other hand, since y is
1)(t
modular, ψM (t) is divisible by ψy(t), and w... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
is
µ(v)µ(y)(v
y) =
⇒
µ(v)µ(y)
πs
v⊇z
�
y≥z=ˆ
0
s∗v∞y
�
v⊇z
�
y≥z=ˆ
0
=
πs
µ(v)µ(y)
s
�
v⊇s,v⊇z
�
y⊇s,y≥z=ˆ0
=
πs
s
�
µ(v)
v⊇s≥z
�
�
ˆ0,s�z
��
�
�
⎡
⎡
⎡
⎡
⎡
⎢
⎤
⎥
⎥
⎥
⎥
⎥
�
�
⎤
⎤
µ(y)
�
y⊇s
�
y≥z=ˆ0
⎥
⎥
�
⎡
⎡
⎢
=
πs
s≥z=ˆ0
�
= πˆ0.
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
�
µ(y)
y⊇s
�
y≥z=ˆ0 (redundant)
�
ˆ0,s
��
�
�
�
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
rk(z
Proof of Claim 1. Clearly z
y)
rk(v). Since z is modular we have
(v
⇒
∈
v, so it suffices to show that rk(z
(v
∈
⇒
⊂
(v
∈
⇒
y)) = rk(z) + rk(v
= rk(z) + rk(v
y)
y)
⇒
⇒
−
−
rk(z
y)
(rk(z) + rk(y)
⇒
−
y))
rk(z
∈
0
= rk(v
y)
rk(y)
⇒
(rk(v) + rk(y)
−
→
−
rk(v
∈
0
)
y)
−
�
��
�
= ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
⇒
vy
tn−rk(v)−rk(y)
= t
n−rk(v∞y)
.
⇒
∃
∃
Now v
tute µ(x)x
∃
preserved. We thus obtain
⇒
y is just vy in the M¨obius algebra A(L). Hence if we further substi
µ(x)tn−rk(x) in the left-hand side of (33), then the product will be
µ(x)t
n−rk(x)
=
x⊆L
�
�L (t)
as desired.
�
��
�
⎤
⎥
⎥
⎥
⎥
⎥
�
v⊇z
�
�
µ(v)tr... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
the arrangement in K n
with equations
×
K (where y denotes the last coordinate)
L1(x) = a1y, . . . , Lm(x) = amy, y = 0.
LECTURE 4. BROKEN CIRCUITS AND MODULAR ELEMENTS
53
Let H0 denote the hyperplane y = 0. It is easy to see by elementary linear algebra
that
L(A) ∪= L(cA)
... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
A) ∪= L(A)
x
L(A) : x
.
H0}
⊂
≤
− {
4.3. Supersolvable lattices
For some geometric lattices L, there are “enough” modular elements to give a
factorization of ψL(t) into linear factors.
Definition 4.13. A geometric lattice L is supersolvable if there exists a modular
maximal chain, i.e., a maximal chain ˆ
1 suc... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
be a supersolvable geometric lattice of rank n, with modular
maximal chain ˆ
1. Let T denote the set of atoms of L, and
set
0 = x0 � x1 � � xn = ˆ
· · ·
(34)
Then ψL(t) = (t
e1)(t
e2)
· · ·
−
−
a
ei = #
{
≤
(t
T : a
xi, a
xi−1}
.
⇔→
→
en).
−
Proof. Since xn−1 is modular, we have
xn−1 = ˆ
0
y
∈
y
≤
√
T ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
the boolean algebra of rank n. By Exam
ple 4.9(d) every element of Bn is modular. Hence Bn is supersolvable.
Clearly each ei = 1, so ψBn (t) = (t
1)n .
(b) Let L = Bn(q), the lattice of subspaces of Fq . By Example 4.9(e) every
element of Bn(q) is modular, so Bn(q) is supersolvable. If
denotes the
number of j-dim... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
� βn−1 = ˆ
only if it has the form ˆ
1, where βi for i > 0 has
· · ·
exactly one nonsingleton block Bi (necessarily with i + 1 elements), with
n−1 = [n]. In particular, Γn is supersolvable and has
B1
exactly n!/2 modular chains for n > 1. The atoms covered by βi are the
partitions with one nonsingleton block
Bi.... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
is a more general class of geometric
lattices L than the supersolvable ones for which ψL(t) factors into linear factors
(over Z). There is a profound such generalization due to Terao [22] when L is an
intersection poset of a linear arrangement A in K n . Write K[x] = K[x1, . . . , xn]
and define
T(A) =
(p1, . . . ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
n + 1, elements that are linearly independent over K[x]. We say that A
is a free arrangement if T(A) is a free K[x]-module, i.e., there exist Q1, . . . , Qn
≤
T(A) can be uniquely written in the form
T(A) such that every element Q
K[x]. It is easy to see that if T(A) is free,
+ qnQn, where qi
Q = q1Q1 +
· · ·
t... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
to supersolvability, we can try to characterize the supersolvable prop
erty for various classes of geometric lattices. Let us consider the case of the bond
lattice LG of the graph G. A graph H with at least one edge is doubly connected if
it is connected and remains connected upon the removal of any vertex (and all ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
osition 4.12. Let G be a doubly connected graph, and let β =
be a
coatom of the bond lattice LG, where #A #B. Then β is a modular element of
→
LG if and only if #A = 1, say A =
, and the neighborhood N (v) (the set of
}
vertices adjacent to v) forms a clique (i.e., any two distinct vertices of N (v) are
adjacent... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
− {
a, b
− {
}
}
}}
rk(π) = rk(β) = n
2, rk(π
β) = n
1, rk(π
β) = n
4.
−
∈
−
⇒
−
Hence β is not modular.
Conversely, let β =
E(G).
It is then straightforward to show (Exercise 8) that β is modular, completing the
�
proof.
As an immediate consequence of Propositions 4.10(b) and 4.12 we obtain a
. Assume that ... | https://ocw.mit.edu/courses/18-315-combinatorial-theory-hyperplane-arrangements-fall-2004/0111284e101e01b31f3e8d8abb074975_lec4.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1: Integral Form of Maxwell’s Equations ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
2 farads/meter
ε 0 ≈
10-9
36π
1
ε µ 0 0
free space)
c =
3≈
× 108 meters/second (Speed of electromagnetic waves in
4. Gauss’ Law for Magnetic Field
(cid:118)∫ µ 0H da = 0
i
S
In free space:
B = µ H0
magnetic
flux
density
(Teslas)
magnetic
field
intensity
(amperes/meter)
5. Conservation of Charge
Take... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
Motion
Prof. Markus Zahn
Lecture 1
Page 3 of 6
3
Mg
⎡2π ε0r
q = ⎢
l
⎣
2
1
⎤
⎥
⎦
III. Faraday Cage
J da = i = -
i
∫(cid:118)
S
d
dt
∫ ρ dV = -
d
dt
(
) =
-q
dq
dt
∫
idt = q
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 1
Page 4 of 6
IV. Boundary Conditions
1... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
�
⎣
H a - H b
⎤ = 0
⎦
(cid:118)
∇ × H = J ⇒ ∫ H i ds = J i da
∫
C
S
H ds - Hatds = Kds
bt
H - Hat = K
bt
n × ⎡
⎣
H a - H b
⎤ = K
⎦
∂ρ
∇ i J + = 0
∂t
5. Conservation of Charge Boundary Condition
i(cid:118)∫
d
∫
J da + ρdV = 0
dt V
S
n i ⎣
⎡ J a - Jb ⎦
⎤ +
∂
t
∂
σ = 0
s
6.641, Electromagnetic Field... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2009/01706aa68260e3c0295d85ca927c0870_MIT6_641s09_lec01.pdf |
18.433 Combinatorial Optimization
NP-completeness
November 18
Lecturer: Santosh Vempala
Up to now, we have found many efficient algorithms for problems in Matchings, Flows,
Linear Programs, and Convex Programming. All of these are polynomial-time algorithms.
But there are also problems for which we have found no polynomi... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
such that every clause in F is equal
1?
4. Longest Cycles
Given a graph G = (V, E), find the longest cycle.
5. Cliques
A clique is a complete subgraph. Given a graph G = (V, E), find a clique of maximum
cardinality (vertices).
As different as these examples might seem, they have two main properties in common:
A) None of t... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
Is there a clique of size ≤ k? To find the optimal size
k∗, again we do a binary search. We then consider the graph with vi and all its neighbors.
If the optimum in this graph remains the same, then save that vertex, then we can delete
all other vertices. Else, delete vi because it is not in the max clique.
2 P and NP
2... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
to another problem B in NP, then B is
also NP-complete.
2.3 Examples of Reduction
SAT is NP-complete (we will not prove this in class).
1. ILP is NP-complete Let’s take the following SAT problem and see if it can be solved
by an ILP.
F = (x1 ∨ x2 ∨ ... ∨ ¯xi) ∧ (x4 ∨ ¯x5) ∧ ... ∧ (xa ∨ xb ∨ ... ∨ xc)
This SAT problem c... | https://ocw.mit.edu/courses/18-433-combinatorial-optimization-fall-2003/0174bd8254153827d2ccaf345a0fd0e4_l20.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.727 Topics in Algebraic Geometry: Algebraic Surfaces
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ALGEBRAIC SURFACES, LECTURE 5
LECTURES: ABHINAV KUMAR
1. Examples
→
(1) If S ⊂ Pn, p ∈ S, then projection ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
of
3 independent conics passing through three point p, q, r (non-collinear).
Generally, 2 conics passing through p, q, r would have a unique 4th point
of intersection, gives the birational map.
z
x
y
(3) Linear systems of cubics: let p1, . . . , pr be r distinct points in the plane
(r ≤ 6) in general position, i.... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
+ zn−1 = 0.
E6 x2 + y3 + z4 = 0.
1
2
LECTURES: ABHINAV KUMAR
E6 x2 + y3 + yz3 = 0.
E8 x2 + y3 + z5 = 0.
If you resolve these, you get the corresponding Dynkin diagrams for the dual
graph of the exceptional curves.
Theorem 1 (Artin Contraction). A connected set of curves {Ci} on a surface
Y is the exceptional l... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
) is called the generic curve of the pencil π.
→
Definition 3. A smooth morphism X
over B if the fibers are all isomorphic to P1 .
Theorem 2 (Noether-Tsen). Let π : X ��� B be a pencil of curves s.t. the
generic curve has arithmetic genus zero. Then X is birational to P1 × B (and
the generic fiber of π is ∼
B is call... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
+ xmem we see that f can be expressed as a homogeneous polynomial of
· · ·
degree md in mn variables. Since d < n, md < m, we have a nontrivial solution
�
NL/K (F (x)) = 0 = F (x) = 0.
⇒
ALGEBRAIC SURFACES, LECTURE 5
3
Proposition 1. Let k be algebraically closed. Then k(T ) (purely transcendental
extension in... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
isT s
, Yns) = f (
s
� �
Y1j T j ,
Y2j T j , . . . ,
Ynj T j )
�
This has degree sd + µ in T . Write it as
j=0
(4) φ = φ0(Y10,
· · ·
, Yns) + T φ1(Y10,
· · ·
, Yns) + + T sd+µφsd+µ(Y10,
· · ·
· · ·
, Yns)
, Yns. Since
i.e. have ds+µ+1 homogeneous polynomials φj of degree d in Y10,
n > d, for large enough... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
Blowing up X enough times, we get φ : X �
B
completing π φ. Note that this does not change the generic fiber. By assumption,
k(B) is algebraically closed in k(X). We see Xη = (η φ)(η) is geometrically
◦
(Xη) ∼ k(η)(t) for t an
integral, and therefore is k(η)-isomorphic to P1
k(η). So k
=
�
independent variable ove... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
0). By
the base change theorem, we see that H 1(π−1(b�), Oπ−1(b�)) = 0 for b� in a neigh
borhood V of B, and π∗OX ⊗ k(b) → H 0(π−1(b�), Oπ−1(b�)) is an isomorphism for
b� ∈ V .
(6)
π−1(b) ∼
= P1
= ⇒ dim H 0(π−1(b�), Oπ−1(b�)) = 1
so π∗OX is locally free of rank 1, i.e. is OB. Thus, k(B) is alg. closed in k(X),
a... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
π� are P1, and
∼
= OFb� (1), we have dim k(b�)H 0(OX (D) ⊗ k(b�)) = 2 for b� ∈ U .
OX � (D) ⊗ OFb�
Again applying the base change theorem, we have E = π∗(OX � (D)) a locally free
O+U -module of rank 2 and the canonical homomorphism
→
→
·
◦
(7)
π∗OX � (b) ⊗ k(b�) → H 0(OX (D) ⊗ OFb� )
is an isomorphism for b� ∈ U ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/01767d4bef386f92c482c478e832fd2c_lect5.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.306 Advanced Partial Differential Equations with Applications
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Stability of Numerical Schemes for PDE’s.
Rodolfo R. Rosales
.
(cid:3)
MIT, Friday February 12, 1999.
A... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
. . . .
7
3.1 Corrected scheme, cosine initial data with 55 points. . . . . . . . . . . . . . . . . . .
13
3.2 Corrected scheme, cosine initial data with 190 points.
. . . . . . . . . . . . . . . . .
14
(cid:3)
MIT, Department of Mathematics, room 2-337, Cambridge, MA 02139.
1
Stability of Numerical Schemes for PDE’s.... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
as follows
n
j
x
= x
+ n(cid:1)x and t
= j(cid:1)t :
(1.3)
n
j
0
Here (cid:1)x and (cid:1)t are some \small" positive constants and x
is arbitrary. Next replace the function
0
u = u(x; t) of the continuum variables x and t by a discrete double sequence fu
g, where
n
j
j
u
= u(x
; t
) :
(1.4)
n
n
j
Finally, introduce th... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
t
)
j
+1
j
j
j
j
j
j
j
+1
+1
(cid:1)t
j
j
u
= u
+ (cid:1)t v
and v
= v
+
u
(cid:0) 2u
+ u
;
(1.7)
n
n
n
n
n
+1
n
1
n
n
(cid:0)
2
((cid:1)x)
2
(cid:16)
(cid:17)
where the errors should be of size O((cid:1)t; ((cid:1)x)
), that is: small.
Upon implementation one quickly discovers that this algorithm is disastrously bad. ... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
time of period 2 (a standing wave). For the numerical
solution we take (cid:1)x = 2 (cid:1)t = 2=N (for some \large" N ) and x
= (cid:0)1 in (1.3). Then we im-
0
plement (1.7) for 1 (cid:20) n (cid:20) N (the periodicity of the solution means that the indexes n + N and n
are equivalent) and solve the equations over one... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
reasonable numerical
scheme ought to give a better approximation when we do this. Figure 1.2 shows the result of in-
creasing N to N = 57 (a rather smal l increase). The new approximation is not only not better; it
is a disaster. By time t (cid:25) 2, O(1) grid scale (i.e. wavelength = 2 (cid:1)x) oscil lations appear ... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
scale on this (cid:12)gure!
Stability of Numerical Schemes for PDE’s.
5
MIT, Friday February 12, 1999 | Rosales.
Final ly, we point out that if (instead of increasing N ) we compute for longer times, the same e(cid:11)ect
of large amplitude grid scale oscil lations arising (which grow exponential ly in time) i... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
0) (2n + 1), with n integer. These
initial values are not smooth | as were the ones in the prior example. There is a smal l corner in
u
(x), whenever x is an odd integer (in particular for x = (cid:6)1). This is because at these points there
0
is a cut-o(cid:11) from a Gaussian centered at x (cid:0) 1 to one centered a... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
=0.02.
Figure 1.4: Solution of (1.5) with initial data (1.9) using (1.7) with 100 points in
the space grid and a = 10. To avoid an over-dense graph not all the points in the
numerical grid are plotted (enough points to show all the relevant details are kept).
In the second calculation, we take a smal ler value a = 6. T... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
1.9) using (1.7) with 100 points in
the space grid and a = 6. To avoid an over-dense graph not all the points in the
numerical grid are plotted (enough points to show all the relevant details are kept).
The next section gives a detailed explanation of why this is happening.
2 von Neumann stability analysis for PDE’s.
I... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
j
j
ikn
j
j
ikn
u
= U G
e
and v
= V G
e
;
(2.1)
n
n
where U , V , G and k are constants (with k real). Generally double sequences like this will be solu-
tions provided G, U and V are restricted by some functional relations of the form G = G(k ; (cid:1)x; (cid:1)t),
U = U (k ; (cid:1)x; (cid:1)t) and V = V (k ; (cid:1)... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
.1 and 1.2), the Fourier modes (2.1) must also satisfy the periodicity condition.
That is, one must have (cid:21) = T =‘, where ‘ is an integer and T is the period in space. Since in this
case one would normal ly take (cid:1)x = T =N , where N is a large natural number, the acceptable values
for k end up restricted to ... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
cid:1)x (grid size oscillations, with period 2 in n: the
‘
solution alternates between two values on the grid). To see this recal l that k
and k
give the same
‘
‘
N
+
Fourier mode in (2.1). Thus the mode (N (cid:0) ‘) has the same wavelength as the mode (cid:0)‘, i.e. T =‘.
This means that, after ‘ = N=2 the wavelength... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
0 or ‘ (cid:25) N in
(2.3)) should be treated "accurately" by the scheme. By this we mean that their time evolution
(given by the factors G
in (2.1)) should be as close as possible to the one provided by the PDE the
j
scheme approximates. This is what consistency is al l about.
Consider now the special case of the algo... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
Stability of Numerical Schemes for PDE’s.
10
MIT, Friday February 12, 1999 | Rosales.
Notice that the maximum ampli(cid:12)cation for the scheme (1.7) occurs | as follows from (2.5)
| for k = (cid:25) . This corresponds to ‘ = N=2 in (2.3), i.e.: grid size oscillations with (cid:21) = 2 (cid:1)x.
In this case
where (ci... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
to a superposition of just three modes in (2.1), with k = k
, k = k
and
0
1
k = k
(cid:24) k
in (2.3). Thus, the exact solution for the scheme equations is rather simple
1
1
N
(cid:0)
(cid:0)
and has the form
j
j
1
g
+ (cid:22)g
2n(cid:25)
g
(cid:0) (cid:22)g
2n(cid:25)
(cid:25)
j
j
j
j
u
=
1 (cid:0)
cos(
)
and v
=
^v ... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
This sum wil l general ly include all the modes,
in particular the highly ampli(cid:12)ed ones with grid size wavelengths. Consider then what would happen
with the solution of the scheme if we add to the initial data above
a smal l amount of the component
3
corresponding to the maximum ampli(cid:12)cation rate above in... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
:12)nal time t = 2 and
it is easy to check that
80
80
(cid:25)
Re(g
) = Re
1 + i sin(
)
(cid:25) 1:28 :
(
)
40
(cid:18)
(cid:19)
This agrees quite wel l with the (cid:25) 30% growth in the wave amplitude observed in (cid:12)gure 1.1.
2. Second, for N = 57, (2.8) gives A
(cid:25) 1:4 (cid:2) 10
for the (cid:12)nal time ... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
with
wavenumbers low multiples of k
= 2(cid:25)=N | which, indeed, are quite obvious in (cid:12)gure 1.3.
1
Remark 2.3 Now consider example 1.2, where N = 100 and 0 (cid:20) t (cid:20) 0:5. Then, for the time
t = 0:5, equation (2.8) gives A
(cid:25) 3:4 (cid:2) 10
:
2
7
In this case the initial data has components in a... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
present. However, now
their amplitudes and phases are al l correlated because they (mostly) are generated by the corner in
the initial data. Thus they interfere with each other in ways subtler than the mere beating observed in
the prior example; i.e.: the pattern of grid size oscil lations has a clear maximum near the ... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
:1)x
(cid:26)
(cid:18)
(cid:19)(cid:27)
Thus the exact evolution corresponds to a factor G given by
G
= exp
(cid:6)i
k
= 1 (cid:6) i
k + O
(
k)
:
(2.11)
exact
(cid:18)
(cid:19)
(cid:18)
(cid:19)
(cid:1)x
(cid:1)x
(cid:1)x
(cid:1)t
(cid:1)t
(cid:1)t
2
This should be compared with (2.10) above. It is clear then that (for... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
0.2
0
2
1.5
1
0.5
1
0.5
0
-0.5
Time t --- dt=1/N.
0
-1
Space x --- dx=2/N.
Figure 3.1: Solution of (1.5) with initial data (1.8) using the corrected scheme (3.1)
with 55 points in the space grid. To avoid an over-dense graph not all the points
in the numerical grid are plotted. However, enough points to show all the re... | https://ocw.mit.edu/courses/18-306-advanced-partial-differential-equations-with-applications-fall-2009/0177f1f3a5f5642edbc5a5ef7692ca80_MIT18_306f09_lec29_Num_Scheme_Stab.pdf |
Lecture 10: Solving the Time-Independent Schr¨odinger Equation
Contents
1 Stationary States
2 Solving for Energy Eigenstates
3 Free particle on a circle.
1 Stationary States
B. Zwiebach
March 14, 2016
1
3
6
Consider the Schr¨odinger equation for the wavefunction Ψ(x, t) with the assumption that the potential
energy V i... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
i(cid:126)
1 dg(t)
g(t) dt
=
1
ψ(x)
ˆ
Hψ(x) .
(1.4)
The left side is a function of only t, while the right side is a function of only x (a time dependent
potential would have spoiled this). The only way the two sides can equal each other for all values of t
and x is for both sides to be equal to a constant E with units... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
= EΨ(x, t) ,
(1.10)
since the time dependent function in Ψ cancels out.
We have noted that the energy E must be real. If it was not we would also have trouble normalizing
the stationary state consisten
tly. The normalization condition for Ψ, if E is not real, would give
(cid:90)
∗
1 = dx Ψ(cid:3)
= ei(E∗(cid:0)E)t/(cid... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
, t) = E,
∗
(1.13)
Since the stationary state is an eigenstate of H, the uncertainty ∆H of the Hamiltonian in a stationary
state is zero.
ˆ
ˆ
There are two important observations on stationary states:
2
(1) The expectation value of any time-independent operator Q on a stationary state Ψ is time-
ˆ
independent:
(cid:90... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
�2 are H eigenstates with energies E1 and E2, respectively. Consider a Hermitian
operator Q. With the system in state (1.15), its expectation value is
ˆ
ˆ
(cid:90) 1
(cid:90) ∞
(cid:0)1
−∞
1
∞
(cid:90)
(cid:0)1
−∞
1
∞
(cid:90)
hQiΨ =
(cid:105)
(cid:104)
=
=
(cid:0)1
−∞
dx Ψ(cid:3)
∗
ˆ
(x, t)QΨ(x, t)
(cid:0) (cid:3) iE1... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
possible time dependence arising from the cross terms. The first two terms are
simple time-independent expectation values. Using the hermitically of Q in the last term we
then get
ˆ
hQiΨ = jc1j2hQi
(cid:105)
|
(cid:104)
| (cid:104)
(cid:105)
2
ψ1 + jc2j hQiψ2
(cid:105)
| (cid:104)
|
(cid:0)E2)t/(cid:126) (cid:90)
1
∞
−
... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
ian operator.
(cid:54)=
(cid:104)
(cid:105)
2 Solving for Energy Eigenstates
We will now study solutions to the time-independent Schr¨odinger equation
ˆHψ(x) = E ψ(x).
3
(2.19)
6
ˆ
For a given Hamiltonian H we are interested in finding the eigenstates ψ and the eigenvalues E,
which happen to be the corresponding energi... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
that become plus infinity beyond certain points. These points
represent hard walls.
We want to understand general properties of ψ and the behavior of ψ at points where the potential
V (x) may have discontinuities or other singularities. We claim: we must have a continuous
wavefunction. If ψ is discontinuous then ψ0 cont... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
x ! a from the left, and will vanish for x > a.
Thus ψ0 is discontinuous at the wall.
→
≥
(cid:48)
(cid:48)
In the first two cases ψ0 is continuous, and in the second two it can have a finite discontinuity. In
(cid:48)
conclusion
Both ψ and ψ0 are continuous unless the potential has delta functions
or hard walls in which... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
ψ(x) .
(2.25)
By definition, the discontinu
side:
Back in (2.25) we then have
ity in the
derivative of ψ at x = a is the limit as (cid:15) ! 0 of the left-hand
→
∆a
(cid:18) dψ (cid:19)
dx
(cid:17) lim
≡
(cid:15)!0
→
(cid:16) dψ
(cid:12)
(cid:12)
(cid:12)
dx a+(cid:15)
(cid:12)
−
dψ
dx
(cid:12)
(cid:12)
(cid:12)
(cid:12... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
2)
(cid:12)
dx x0
(cid:12)
2m (cid:90)
(cid:0) (cid:126)
−
x0
x
(E
(cid:0) V (x0))dx0.
(cid:48)
(cid:48)
−
(2.29)
Note that the integral on the right is a bounded
a + (cid:15). Since the first term on the right-hand side is a constant we find
function of x. We now integrate again from a (cid:0) (cid:15) to
−
ψ(a + (cid:1... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
[0, L] with the endpoints identified.
It is perhaps clearer mathematically to think of the circle as the full real line x with the identification
∈
x (cid:24) x + L ,
∼
(3.1)
which means that two points whose coordinates are related in this way are to be considered the same
point. If follows that we have the periodicity ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
)2
(cid:104)(cid:16)
2m
∗
ψ(cid:3)
dψ
dx
(cid:17)(cid:12)
(cid:12)
(cid:12)
x=L
(cid:16)
− ∗
(cid:0) (cid:3)
ψ
dψ
dx
(cid:17)(cid:12)
(cid:12)
(cid:12)
x=0
(cid:105)
+
(cid:126)2
(cid:90)
2m
0
L dψ 2
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
dx
dx = E .
(3.6)
Since ψ(x) and its derivatives are p
are left wi... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
cid:126)k. Using k2 the differential equation becomes the familiar
E =
(cid:126)2k2
2m
,
d2ψ
dx
2 (cid:0)k2ψ .
−
=
(3.8)
(3.9)
(3.10)
(3.11)
We could write the general solution in terms of sines and cosines of kx, but let’s use complex expo-
nentials:
ψ(x) (cid:24) eikx.
∼
(3.12)
This solves the differential equation and... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
energies are
(cid:126)24π2n2
2mL2
There are infinitely many energy eigenstates. We have degenerate states because En is just a
−
function of jnj and thus the same for n and (cid:0)n. Indeed ψn and ψ(cid:0)n both have energy En. The only
nondegenerate eigenstate is ψ0 = p1 , which is a constant wavefunction with zero ene... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
cid:0) ψ(cid:0)n (cid:24) sin(knx) .
While these are real energy eigenstates, they are not momentum eigenstates. Only our exponentials
are simultaneous eigenstates of both H and pˆ.
(3.21)
ˆ
The energy eigenstates ψn are automatically orthonormal since they are pˆ eigenstates with no
degeneracies (and as you recall eig... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/01b09433137651bedbbfe4f039b0c5be_MIT8_04S16_LecNotes10.pdf |
8. Geometric problems
Convex Optimization — Boyd & Vandenberghe
extremal volume ellipsoids
centering
classification
placement and facility location
•
•
•
•
8–1
Minimum volume ellipsoid around a set
L¨owner-John ellipsoid of a set C: minimum volume ellipsoid
s.t. C
parametrize
as
=
v
Av + b
E
≤
is proport... | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
∈
Rn
⊆
maximize
subject to
log det B
supkuk2≤1 IC(Bu + d)
0
≤
(where IC(x) = 0 for x
C and IC(x) =
for x
C)
�∈
∞
∈
convex, but evaluating the constraint can be hard (for general C)
polyhedron
T x
ai
x
{
|
bi, i = 1, . . . , m
}
:
≤
maximize
subject to
log det B
Bai
T d
�2 + ai
�
(constraint follows fr... | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
transformations
Geometric problems
8–5
Analytic center of a set of inequalities
the analytic center of set of convex inequalities and linear equations
fi(x)
≤
0,
i = 1, . . . , m,
F x = g
is defined as the optimal point of
m
minimize
i=1
subject to F x = g
− �
log(
fi(x))
−
•
•
more easily computed than ... | https://ocw.mit.edu/courses/6-079-introduction-to-convex-optimization-fall-2009/01bc28bea61ac801e76e9bcdf7971cdb_MIT6_079F09_lec08.pdf |
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