text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
Lecture 16
April 13th, 2004
Elliptic regularity
Hitherto we have always assumed our solutions already lie in the appropriate Ck,α space and
then showed estimates on their norms in those spaces. Now we will avoid this a priori assumption
and show that they do hold a posteriori. This is important for the consistency of o... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
0u = f ′ on all of Ω so
1
in particular on ¯B. By uniqueness on B therefore we have u| ¯B = v, and so u is C2,α smooth there.
As this is for any point and all balls we have u ∈ C2,α(Ω).
It is insightful to note at this point that these results are optimal under the above assumptions.
Indeed need C2 smoothness (or atle... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
u(x)
h
.
Namely we look at
∆hLu =
Lu(x + h · el) − Lu(x)
h
=
f (x + h · el) − f (x)
h
= ∆huf.
Note ∆hv(x) −→h→0Dlv(x) if v ∈ C1 (which we don’t know a priori in our case yet).
Expanding our equation in full gives
2
1
h h(aij(x + h · el) − aij(x) + aij(x))Dijuh − aij(x)Diju(x)
+bi(x + h · el)Diu(x + h · el) − bi(x)Diu(... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
) and not
just in C2(Ω). ⇔ Dijuh ∈ Cα(Ω) uniformly.
Remark. We take a moment to describe what we mean by uniformity. We say a function
gh = g(h, ·) : Ω → R is uniformly bounded in Cα wrt h when ∀Ω′ ⊂⊂ Ω exists c(Ω) such that
|gh|Cα(Ω′) ≤ c(Ω). Note this definition goes along with our local definition of a function being ... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
that’s how we define the first l-directional derivative at x). Now {∆hg}h>0
is family of uniformly bounded (in C0(A)) and equicontinuous functions (from the uniform H¨older
constant). So by the Arzel`a-Ascoli Theorem exists a sequence {∆hig}∞
i=1 converging to some
˜w ∈ Cα(A) in the Cβ(A) norm for any β < α. But as we re... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
on
Ω,
∂∂Ω.
with 0 < α < 1. Then u ∈ C2,α( ¯Ω).
4
Proof. Our previous results give u ∈ C2,α(Ω) and we seek to extend it to those points in ∂Ω. Note
that even though u = ϕ on ∂Ω and ϕ is C2,α there this does not give the same property for u. It
just gives that u is C2,α in directions tangent to ∂Ω, but not in directions... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
� ¯Ω)) for the induced Dirichlet
problem . Now our ˜u also solves it. So by uniqueness ˜u = v and ˜u has C2,α regularity as the
induced boundary portion, and by pulling back through C2,α diffeomorphisms we get that so does
u.
Remark. The assumption c ≤ 0 is not necessary although modifying the proof is non-trivial
witho... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
15.083J/6.859J Integer Optimization
Lecture 8: Duality I
Slide 1
Slide 2
Slide 3
Slide 4
1 Outline
• Duality from lift and project
• Lagrangean duality
2 Duality from lift and project
ZIP = max c�x
•
s.t. Ax = b
xi ∈ {0, 1}.
• {x ∈ (cid:3) | Ax = b, x ≥ 0} is bounded for all b.
n
• Without of loss of gen... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
2.
1
2.3 Lift-Project
(cid:2)
• Inequality form:
j∈N
Aj xj ≤ b
(cid:5)
• Multiply constraints with
for all S N to obtain using
i∈S
x
⊆
(cid:6) (cid:7) (cid:6) (cid:7)
xi +
Aj
Aj
i
(cid:7)
xi ≤ b
xi.
j∈S
i∈S
j /
∈S
i∈S∪{j}
i∈S
• Define yS =
(cid:5)
i∈S
xi, noting that yS ≥ 0 and setting y∅ = 1
(... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
solution to the dual problem, then
(cid:6) x ≤ u∅
c
(cid:6) b.
• (Strong duality) If the primal problem has an optimal solution, so does its
dual problem, and the respective optimal costs are equal.
2.6 Complementary slackness
x and u feasible solutions for primal and dual. Then, x and u are optimal solutions
if an... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
≥ 1
−3u4 + 9u∅ ≥ 5
0u1,4 + 9u1 + 3u4 ≥ 0
are all satisfied with equality.
3 Lagrangean duality
ZIP = min
c (cid:6) x
s.t. Ax ≥ b
Dx ≥ d
x ∈ Z n ,
(∗)
X = {x ∈ Z n | Dx ≥ d}.
3
Slide 10
Let λ ≥ 0.
Z(λ) = min c (cid:6) x + λ(cid:6)(b − Ax)
s.t. x ∈ X,
3.1 Weak duality
• If problem (*) has an optimal sol... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
be the extreme points and a extreme rays of
Slide 12
Slide 13
conv(X)
Z(λ) =
−∞,
⎧
⎪ ⎨
(cid:11)
⎪
⎩ min c x + λ (b − Ax ) ,
(cid:10) (cid:6) k
k
(cid:6)
k∈K
if (c (cid:6) − λ(cid:6)A)wj < 0,
for some j ∈ J,
otherwise.
•
•
(cid:10)
(cid:6) x
c
(b − Axk
(cid:11)
)
ZD = maximize min
k∈K
(c (cid:6) − ... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
(0, 2).
5
Z(λ)
4
2
− 1
3
5
3
λ
conv(X)
xIP
x2
3
xD
2
xLP
1
c
0
1
2
x1
6
MIT OpenCourseWare
http://ocw.mit.edu
15.083J / 6.859J Integer Programming and Combinatorial Optimization
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
Linear Circuits Analysis. Superposition, Thevenin /Norton Equivalent circuits
So far we have explored time-independent (resistive) elements that are also linear.
A time-independent elements is one for which we can plot an i/v curve. The current is
only a function of the voltage, it does not depend on the rate of cha... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
circuits, each of which has only one voltage source.
i1
Vs1
i2
R
R
Vs2
The total current is the sum of the currents in each circuit.
i
i
2
=
+
1
i
= +
Vs2
Vs
1
R
R
Vs Vs
2
1
+
R
=
(1.3)
Which is the same result obtained by the application of KVL around of the original
circuit.
If the circuit we are interested in i... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
shorted out (there is no voltage across R2), and therefore there is no
current through it. The current through R1 is Is, and so the voltage drop across R1 is,
6.071/22.071 Spring 2006. Chaniotakis and Cory
4
1v
=
IsR1
And so
1i
Is=
Vs
R
2
=
i
2
How much current is going throu... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
obtain the result given by Eq. (1.10).
More on the i-v characteristics of circuits.
As discussed during the last lecture, the i-v characteristic curve is a very good way to
represent a given circuit.
A circuit may contain a large number of elements and in many cases knowing the i-v
characteristics of the circuit i... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
Is
v
short circuit
current
6.071/22.071 Spring 2006. Chaniotakis and Cory
7
Thevenin Equivalent Circuits.
For linear systems the i-v curve is a straight line. In order to define it we need to identify
only two pints on it. Any two points would do, but perhaps the simplest ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
is as follows:
1. Calculate the equivalent resistance of the circuit (RTh) by setting all voltage and
current sources to zero
2. Calculate the open circuit voltage Voc also called the Thevenin voltage VTh
6.071/22.071 Spring 2006. Chaniotakis and Cory
8
The equivalent circ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
system is interface with it.
Therefore in order to characterize the network we must look the network characteristics
in the absence of RL.
6.071/22.071 Spring 2006. Chaniotakis and Cory
10
+
Vs
-
R1
R2
A
B
R3
R4
First lets calculate the equivalent resistance RTh. To do this we short the vo... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
Bridge Circuit as a measuring instrument.
Measuring small changes in large quantities – is one of the most common challenges in
measurement. If the quantity you are measuring has a maximum value, Vmax, and the
measurement device is set to have a dynamic range that covers 0 - Vmax, then the errors
will be a fraction... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
R ε
+
=
3
Under these simplifications,
vu Vs
=
=
Vs
⎛
⎜
⎝
⎛
⎜
⎝
R
3
R
R
1
+
3
R
3
R
R
1
+
3
−
−
R
Ru
Ru
2
+
⎞
⎟
⎠
R
3
ε
+
R
R
3
1
+
+
ε
(1.30)
(1.31)
⎞
⎟
⎠
As discussed above we are interested in the case where the variation in Ru is small, that is
in the case where
. Then the above expression may be approximated... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
RTh
the Thevenin and Norton circuits are equivalent
6.071/22.071 Spring 2006. Chaniotakis and Cory
15
RTh
i
Voc
A
B
+
v
-
(cid:8)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10)
Thevenin Circuit
i
+
v
-
In
RTh
(cid:8)(cid:11)(... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
side as indicated below:
0.5 A
i
3 Ω
6 Ω
6 Ω
3 Ω
2 A
The transformation from the Norton circuit indicated above to a Thevenin equivalent
gives
0.5 A
6 Ω
Which is the same as
0.5 A
i
i
3 Ω
3 Ω
6 Ω
6 Ω
6 Ω
6 Ω
6 V
6 V
By transforming the Thevenin circuit on the right with its Norton equivalent we hav... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
applying the mesh method we have
Isc
=
Vs
R
1
+
−
R
IsR
3
+
3
R
4
=
In
(1.39)
With the values for Rn and Isc given by Equations (1.38) and (1.39) the Norton
equivalent circuit is defined
In
Rn
X
Y
6.071/22.071 Spring 2006. Chaniotakis and Cory
19
Power Transfer.
In many cases an electronic syste... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
zero.
and
dP
dRL
=
VTh
2
(
⎡
⎢
⎣
dP
dRL
2
RTh RL
+
(
2
)
(
−
RTh RL
)
+
4
RL RTh RL
+
)
⎤
⎥
⎦
(1.43)
= → −
RL RTh
0
=
0
(1.44)
and so the maximum power occurs when the load resistance RL is equal to the Thevenin
equivalent resistance RTh.1
Condition for maximum power transfer:
RL RTh
=
The maximum power transfer... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
1
3
R
R
+
+
2 4
R R
2
4
R
R
+
The maximum power delivered to RL is
P
max
=
VTh
4
RTh
Vs
2
2
=
4
3
⎛
⎜
⎝
⎛
⎜
⎝
R
3
R
R
1
+
R R
1 3
R
R
1
3
+
−
+
2
R
4
R
R
2
+
R R
2 4
R
R
2
4
+
⎞
⎟
4
⎠
⎞
⎟
⎠
(1.47)
(1.48)
(1.49)
6.071/22.071 Spring 2006. Chaniotakis and Cory
22
In various applicat... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
Rs Rp
(
+
RTh Rs Rp
+
) //
)
+
(1.50)
6.071/22.071 Spring 2006. Chaniotakis and Cory
23
Which is constrained to be equal to RTh.
RTh
=
(
RTh Rs Rp
+
RTh Rs Rp
+
)
+
The second constraint gives
kVTh VTh
=
Rp
Rp RTh Rs
+
+
And so the constant k becomes:
k
=
Rp
Rp RTh Rs
+
+
By combining E... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
V)
+ vo -
io
4 Ω
3 Ω
3 Ω
24 V
1 Ω
2 A
3 Ω
12 V
6.071/22.071 Spring 2006. Chaniotakis and Cory
25
P4.
P5.
Find the Norton and the Thevenin equivalent circuit across terminals A-B of the
circuit. (Ans.
VTh
Rn = Ω ,
1.7
1.25
2.12
In
V
A
)
=
=
,
A
4 Ω
5 A
3 Ω
1 Ω
3 Ω
Calculat... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
UNCERTAINTY PRINCIPLE AND COMPATIBLE OBSERVABLES
B. Zwiebach
October 21, 2013
Contents
1 Uncertainty defined
2 The Uncertainty Principle
3 The Energy-Time uncertainty
4 Lower bounds for ground state energies
5 Diagonalization of Operators
6 The Spectral Theorem
7 Simultaneous Diagonalization of Hermitian Opera... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
should be a real number. In order to define such uncertainty we first recall that the expectation value
of A on the state Ψ, assumed to be normalized, is given by
A
)
(
=
Ψ
(
Ψ
A
|
|
)
=
Ψ, AΨ
(
)
.
(1.1)
is guaranteed to be real since A is Hermitian. We then define the uncertainty
as the norm of the vector obtai... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
λ =
.
A
)
(
(1.4)
Alternatively, if the state Ψ is an eigenstate, we now now that the eigenvalue if
state (A
I)Ψ vanishes and its norm is zero. We have therefore shown that
A
)
− (
and therefore the
A
)
(
The uncertainty ΔA(Ψ) vanishes if and only if Ψ is an eigenstate of A .
(1.5)
To compute the uncertainty on... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
))2 =
A2
(
2 .
A
)
) − (
(1.7)
(1.8)
(1.9)
Since the left-hand side is greater than or equal to zero, this incidentally shows that the expectation
value of A2 is larger than the expectation value of A, squared:
A2
(
2 .
A
)
) ≥ (
(1.10)
An interesting geometrical interpretation of the uncertainty goes as follo... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
)
Ψ⊥)
|
Ψ
A
|
) − (
A
Ψ
=
Ψ⊥)
|
,
)
)|
orthogonal to
Ψ
|
)
(1.12)
(1.13)
as is easily confirmed by taking the overlap with the bra Ψ. Since the norm of the above left-hand side
is the uncertainty, we confirm that ΔA =
, as claimed. These results are illustrated in Figure 1.
Ψ⊥|
|
2
The Uncertainty Principle
Th... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
is a real, non-negative number. For this to be consistent inequality, the right-hand
side must also be a real number that is not negative. Since the right-hand side appears squared, the
object inside the parenthesis must be real. This can only happen for all Ψ if the operator
1
2i
[A, B]
3
(2.15)
is Herm... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
1/2 we get
(Δx)2(Δp)2
1
2i
|
Ψ
≥ (
(cid:16)
Ψ
[ˆx, pˆ]
|
)
(cid:17)
2
.
(Δx)2(Δp)2
12
≥
4
→
Δx Δp
1
2
.
≥
(2.18)
(2.19)
We are interested in the proof of the uncertainty inequality for it gives the information that is
needed to find the conditions that lead to saturation.
Proof. We define the following... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
)
(2.23)
f
(
g
|
)
=
Ψ
(
AˇBˇ
|
Ψ
|
)
=
Ψ
(
(A
|
I)(B
A
)
− (
B
− (
Ψ
I)
|
)
)
=
Ψ
(
AB
|
Ψ
|
and since
and
f
|
)
g
|
)
go into each other as we exchange A and B,
g
(
f
|
)
=
Ψ
(
AˇBˇ
|
Ψ
|
)
=
Ψ
(
Ψ
BA
|
|
B
) − (
.
A
)
)(
4
A
B
) − (
)(
,
)
(2.24)
(2.25)
From the two equations ab... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
Ψ
(
(cid:16)
|
1
2
ˆA, ˆB
{
Ψ
}|
2
.
)
(cid:17)
(2.27)
(2.28)
This can be viewed as the most complete form of the uncertainty inequality. It turns out, however,
that the second term on the right hand side is seldom simple enough to be of use, and many times
it can be made equal to zero for certain states. At a... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
which requires β + β∗ = 0 or that the real part of β vanish. It follows that β must be purely imaginary.
So, β = iλ, with λ real, and therefore the uncertainty inequality will be saturated if and only if
(2.29)
= (β + β ∗ )
f
(
+ β ∗
= 0 ,
= β
f
(
f
(
g
(
f
|
f
|
f
|
f
|
g
|
+
)
)
)
)
)
More explicitly this requi... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
-Time uncertainty
A more subtle form of the uncertainty relation deals with energy and time. The inequality is sometimes
stated vaguely in the form ΔEΔt � 1. In here there is no problem in defining ΔE precisely, after
all we have the Hamiltonian operator, and its uncertainty ΔH is a perfect candidate for the ‘energy ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
typically fail to count full waves, because as the pulse gets started from
zero and later on dies off completely, the waveform will cease to follow the sinusoidal pattern. Thus
we expect an uncertainty ΔN � 1. Given the above relation, this implies an uncertainty Δω in the
value of the angular frequency
Δω T � 2π . ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
then have
ΔH ΔQ
≥
Ψ
1
2i
|
\
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:11)
(cid:12)
(cid:12)
(cid:12)
This starting point is interesting because the commutator [H, Q] encodes something very physical
about Q. Indeed, let us consider henceforth the case in which the operator Q hasno time dependence.
It coul... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
1
i1
d
dt
Q
(
)
=
=
=
(
i
1
(cid:16)
\
i
Ψ , (HQ
1
−
\
(cid:11)
QH)Ψ =
−
(cid:17)
Ψ , [H, Q]Ψ
(cid:11)
i
1
\
)
(3.38)
(cid:11)
\
where we used the Hermiticity of the Hamiltonian. We have thus arrived at
(cid:11)
d
dt
Q
(
)
=
i
1
\
[H, Q]
for time-independent Q .
(3.39)
(cid:11)
This is a very import... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
=
(cid:12)
(cid:12)
(cid:12)
(3.40)
,
for timeindependent Q .
(3.41)
This is a perfectly precise uncertainty inequality. The terms in it suggest a definition of a time ΔtQ
.
(3.42)
(cid:12)
(cid:12)
(cid:12)
to change by ΔQ if both ΔQ and the
This quantity has uni... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
time Δt⊥ that
it takes for a system to become orthogonal to itself. If we call the initial state Ψ(0), we call Δt⊥ the
smallest time for which
= 0. You will be able to show that
Ψ(0), Ψ(Δt⊥)
)
(
ΔH Δt⊥ ≥
h
4
.
(3.44)
The speed in which a state can turn orthogonal depends on the energy uncertainty, and in quant... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
3.46)
If H is time independent, the uncertainty ΔH is constant in time.
(3.47)
The concept of conservation of energy uncertainty can be used to understand some aspects of
atomic decays. Consider, for illustration the hyperfine transition in the hydrogen atom. Due to the
existence of proton spin and the electron spi... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
H
107sec, accurate to better than 1% ). This
years (= 3.4
lifetime can be viewed as the time that takes some observable of the electron-proton system to change
1014 sec, recalling that a year is about π
×
∼
×
significantly (its total spin angular momentum, perhaps) so by the uncertainty principle it must be
10−30... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
ground state energy of
1/2 to find rigorous lower
bounds for the ground state energy of one-dimensional Hamiltonians. This is best illustrated by an
certain systems. use below the uncertainty principle in the form ΔxΔp
≥
example.
Consider a particle in a one-dimensional quartic potential considered earlier
H =
2... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
2
(
x
) − (
2 leads to
)
so that
x 2
(
) ≥
(Δx)2 ,
x 4
(
) ≥
(Δx)4 ,
9
(4.52)
(4.53)
(4.54)
(4.55)
for the expectation value on arbitrary states. Therefore
H
(
)gs =
p2
(
2m
)gs
+ α
x 4
(
)gs
≥
(Δpgs)2
2m
+ α (Δxgs)4
From the uncertainty principle
Δxgs Δpgs
1
2
≥
→
Δpgs
≥
1
2Δxgs
.
(4.56)
(... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
would be quite nice, since we want the highest possible lower bound.
Since we don’t know the value of Δxgs, however, the only thing we can be sure of is that
)gs is
bigger than the lowest value that can be taken by the expression to the right of the inequality as we
H
(
H
(
vary Δxgs:
H
(
)gs
≥
MinΔx
(cid:16)
1... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
5 Diagonalization of Operators
When we have operators we wish to understand, it can be useful to find a basis on the vector space
for which the operators are represented by matrices that take a simple form. Diagonal matrices are
matrices where all non diagonal entries vanish.
matrix representing an operator is diago... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
by the two-by-two matrix
0 1
0 0
(
)
.
The only eigenvalue of this matrix is λ = 0 and the associated eigenvector is
(5.63)
1
0
( )
. Since a two
dimensional vector space cannot be spanned with one eigenvector, this matrix cannot be diagonalized.
Having seen that the question of diagonalization of an operator ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
is diagonalizable
u
if there is an operator A such that Tij (
{
) is diagonal.
}
There are two pictures of the diagonalization: One can consider the operator T and state that
its matrix representation is diagonal when referred to the u basis obtained by acting with A on the
original v basis. Alternatively, we can v... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
.
(5.67)
confirming that the k-th column of A is the k-th eigenvector of T .
While not all operators on complex vector spaces can be diagonalized, the situation is much
improved for Hermitian operators. Recall that T is Hermitian if T = T † . Hermitian operators can be
diagonalized, and so can unitary operators. Bu... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
operators. The proof is not harder
than the one for hermitian operators. An operator M is said to be normal if it commutes with its
adjoint:
M is normal :
[M †, M ] = 0 .
(6.69)
Hermitian operators are clearly normal. So are anti-hermitian operators (M † =
M is antihermitian).
Unitary operators U are normal beca... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
=
|
w , (M
(
−
λI)(M †
λ ∗ I)w
)
−
Since M and M † commute, so do the two factors in parenthesis and therefore
u
|
2 =
|
w , (M †
(
−
λ ∗ I)(M
λI)w
)
−
= 0
since (M
−
λI) kills w. It follows that u = 0 and therefore (6.70) holds.
(6.71)
(6.72)
(6.73)
D
We can now state our main theorem, called the spectr... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
M ] = U (D DM
†
M
DM D
† )U †
M
= 0 ,
−
Now let us prove that M provides a basis of orthonormal eigenvectors. The proof is by induction.
The result is clearly true for dim V = 1. We assume that it holds for (n
1)-dimensional vector spaces
and consider the case of n-dimensional V . Let M be an n
n matrix referred to... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
an orthonormal basis
, . . . ,
x1)
|
xN
)
|
(6.77)
(6.78)
(6.79)
(6.80)
†
1
1
= U M U1|
)
1
M1 is also normal and M1|
)
x1)
|
1
= λ1|
)
Let us now examine the explicit form of the matrix M1:
1
M1|
)
.
x1)
|
1
= λ1|
)
, so that
which says that the first column of M1 has zeroes in all entries except the first. M... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
M can be unitarily diagonalized so that U ′†M ′ U
unitary matrix U . The matrix U can be extended to an n-by-n unitary matrix Uˆ as follows
is diagonal for some (n
1)-by-(n
′
is a normal (n
−
−
′
′
′
1) matrix. By the induction
1)-by-(n
1)
−
′
ˆU =
1
0
. .
.
0
0
. . . 0
′
U
.
(6.... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
a description of the general situation that we may encounter when
diagonalizing a normal operator T . In general, we expect degeneracies in the eigenvalues so that each
- - - - - - - - - - - - -
eigenvalue λk is repeated dk
V has T -invariant subspaces of different dimensionalities. Let Uk denote the T -invariant sub... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
i=1
1 .
≥
(6.83)
m
All Ui subspaces are guaranteed to be orthogonal to each other. In fact the full list of eigenvectors is
a list of orthonormal vectors that form a basis for V
is conveniently ordered as follows:
(1)
(u1
(1)
, . . . , u ,
d1
. . . , u
(m)
1
, . . . , u
(m)
dm
) .
(6.84)
The matrix T i... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
4) can be changed considerably
without changing the matrix representation of T . Let Vk be a unitary operator on Uk, for each
k = 1, . . . , m. We claim that the following basis of eigenvectors leads to the same matrix T :
V1u
(1)
1
(1)
, . . . V1u ,
d1
. . . , Vmu
(m)
1
, . . . , Vmu
(m)
dm
.
(6.86)
Indeed... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
in which both the matrix representation of S and the matrix
representation of T are diagonal. It then follows that each vector in this basis is an eigenvector of S
and an eigenvector of T .
A necessary condition for simultaneous diagonalization is that the operators S and T commute.
Indeed, if they can be simultane... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
. . un) of eigenvectors of T with
different eigenvalues
T ui = λiui ,
i not summed , λi = λj for i
=
j .
(7.88)
We now want to understand what kind of vector is Sui. For this we act with T on it
T (Sui) = S(T ui) = S(λiui) = λi(S ui) ,
(7.89)
It follows that Sui is also an eigenvector of T with eigenvalue λi, th... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
)
T = diag λ1, . . . , λ1 ,
(cid:0)
d1 times
z
v
'
"
16
. . .
, λm, . . . , λm
in this basis.
dm times
z
v
'
"
(cid:1)
6
6
We also explained that the alternative orthonormal basis of V
(1)
V1u1
(1)
, . . . V1u ,
d1
. . . , Vmu
(m)
1
, . . . , Vmu
(m)
dm
.
(7.92)
(cid:1)
leads to the... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
along u .
i
Since S restricted to each S-invariant subspace Uk is hermitian we can find an orthonormal basis
of Uk in which the matrix S is diagonal. This new basis is unitarily related to the original basis
(u , . . . , u ) and thus takes the form (Vku , . . . , Vku ) with Vk a unitary operator in Uk. Note
that the... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
ent commuting Hermitian operators S1, S2
and S3, all of which have degenerate spectra, we would proceed as follows. We diagonalize S1
and fix a basis in which S1 is diagonal. In this basis we must find that S2 and S3 have exactly the
same block structure. The corresponding block matrices are simply the matrix represen... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
problem becomes one
of simultaneous diagonalization of n commuting Hermitian block matrices, which is assumed
known by the induction argument.
Corollary.
S1, . . . , Sn}
{
simultaneously diagonalized.
If
is a set of mutually commuting Hermitian operators they can all be
8
Complete Set of Commuting Observables
... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
S exhibits degeneracies in its spectrum.
This means that V has an S-invariant subspace of dimension d > 1, spanned by orthonormal eigen
vectors (u1, . . . , ud) all of which have S eigenvalue λ. This time, the eigenvalue of S does not allow us
to distinguish or to label uniquely the basis eigenstates of the invarian... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
T eigenvalues as well as S
eigenvalues we can label uniquely a basis of orthonormal states of V . In this case we say that S and
T form a CSCO.
We have now given enough motivation for a definition of a complete set of commuting observables.
Consider a set of commuting observables, namely, a set
of Hermitian operato... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
be unique.
Once we have a complete set of commuting observables, adding another observable causes no harm,
although it is not necessary. Also, if (S1, S2) form a CSCO, so will (S1 + S2, S1 −
the smallest set of operators.
S2). Ideally, we want
The first operator that is usually included in a CSCO is the Hamiltonian ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
Lecture 7: Lumped Elements
I. What is a lumped element?
Lumped elements are physical structures that act and move as a unit when
subjected to controlled forces. Imagine a two-dimensional block of lead on a one-
dimensional frictionless surface.
FORCE
mass=... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
x.
B. An example of a lumped acoustic element is a short open tube of moderate
diameter, where length l and radius a are <0.1 λ.
length l
u(t)
(t)
p
1
(t)
p
2
A SHORT
CIRCULAR TUBE
OF RADIUS
a
Under these circumstances particle velocity V and the sound pressures are simply
related by:
dV
dt
=
30 Sept -2004
)
(... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
) = through (t) across(t)
‘Across’ variable
‘Through’ variable
voltage e(t)
force f(t)
current i(t)
velocity v(t)
velocity v(t)
force f(t)
Electrics
Mechanics: Impedance
analogy
Mechanics: Mobility
analogy
Acoustics: Impedance
sound pressure p(t)
volume velocity u(t)
30 Sept -2004
page 3
... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
rical elements and
their constitutive
relations. The
orientation of the arrow
and the +/- signs
identifies the positive
reference direction for
each element. In this
figure the variable i is
current and v is voltage.
(From Siebert “Circuits,
Signals and System,
1986).
Note that R, C and L are the coefficie... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
v(t) =
Mass
1
∫
LM
f (t)
dt
Inductor
∫
1
LE
i(t) =
e(t)
dt
Inertance
u(t) =
1
L A
∫
p(t)
dt
page 5
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
In the Sinusoidal Steady State:
Mechanical
Electrical
Acoustical
Spring
Capacitor
Compliance
Damper
Resis... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
physical properties including the dimensions of structures, e.g. the
electrical resistance of a resistor depend on the dimensions and the resistivity of the
material from which it’s constructed.
A. Acoustic mass: units of kg/m4
An open ended tube with linear dimensions l and a <0.1 λ and S=πa2
l
circular tube
p1
u... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
t( )
dt
U = jωC A(P1 − P2 )
For a round, flat, “simply mounted” plate
C A =
πa6 7 + v
)
) 1 − v
(
(
16Et 3
,
where: a is the radius of the plate, v = 0.3 is Poisson’s ratio, E is the elastic constant
(Young’s modulus) of the material, and t is the thickness (Roark and Young, 1975, p.
362-3, Case 10a).
30 Sept -20... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
to the spatial derivative of the velocity and the fluid’s coefficient of shear viscosity η.
30 Sept -2004
page 10
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
The action of these forces results in a proportionality of pressure difference and
volume velocity that is analogous to an ... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
0.011 cm (Figure 5.3)
At 20 Hz δ=3.5x10-5 m = 0.035 cm
The effect of the viscous forces is insignificant when the radius of the tube is an order
of magnitude or more larger than the space constant and therefore we can ignore
viscosity for short tubes of moderate to large radius, 0.01λ < a <0.1λ.
30 Sept -2004
pag... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
within the tube) and a resistance (associated with
overcoming viscous drag at the stationary walls of the tube). Since the pressure
drop across the resistance and the mass elements add, we think of these as an R
and L in series.
(t)
p
1
length l
u(t)
An intermediate
tube
(t)
p
2
)
∆P = P2 − P1 = U jωρol S + R
(
3... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
714J
2. Frequency range limited by the assumption of “lumped” elements, i.e. the
dimensions of the structures need to be small compared to a wavelength:
a and l < 0.1 λ.
30 Sept -2004
page 15
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
VI. Circuit Descriptions of a Real Acoustic System
A Jug... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
of LE = LA, RE=RA, and CE=CA, then I1=U1, E1=P1 and
E2=P2.
C. A Mechanical Analog of the Acoustic Circuit Description
In Mechanical circuits the rods that attach ideal mechanical elements are rigid and
massless.
If the numerical values of LM = LA,
RM=RA, CM=CA, then V1=U1=U2, and
F=P1, then
F
V1
= jωLM + RM +
1... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.006 Introduction to Algorithms
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 2 Ver 2.0
More on Document Distance
6.006 Spring 2008
Lecture 2: More on the Document Distance
Problem
Lecture Overview
Toda... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
of size n
might be found to be T (n) = 4n2 − 2n + 2 µs. From an asymptotic standpoint, since n2
will dominate over the other terms as n grows large, we only care about the highest order
term. We ignore the constant coefficient preceding this highest order term as well because
we are interested in rate of growth.
1
... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
)) and T (n) = Ω(g(n))
Semantics: Read the Θ as ‘high order term is g(n)’
Document Distance so far: Review
To compute the ‘distance’ between 2 documents, perform the following operations:
Θ(n2)
+ op on list
Θ(n2)
double loop
insertion sort, double loop Θ(n2)
�
arccos D1·D2
�D1�∗�D2�
�
Θ(n)
For each of the 2 files... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
Insertion Sort with something faster because it takes
time Θ(n2) in the worst case. This will be accomplished with the Merge Sort improvement
which is discussed below.
Merge Sort
Merge Sort uses a divide/conquer/combine paradigm to scale down the complexity and
scale up the efficiency of the Insertion Sort routine. ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
� 0.1n lg(n) µs
The 20X constant factor difference comes about because Built in Sort is written in C while
Merge Sort is written in Python.
4
5473619234571269ij12345679inc jinc jinc iinc iinc iinc jinc iinc j(array L done)(array R done) Lecture 2 Ver 2.0
More on Document Distance
6.006 Spring 2008
Figure 3: Efficien... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.046J / 18.410J Design and Analysis of Algorithms
Spring 2015
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-046j-design-and-analysis-of-algorithms-spring-2015/009c51db8900141fb181971f2bb826f3_MIT6_046JS15_writtenlec1.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005
Please use the following citation format:
Markus Zahn, 6.641 Electromagnetic Fields, Forces, and Motion, Spring
2005. (Massachusetts Institute of Technology: MIT OpenCourseWare).
http://ocw.mit.edu (accessed MM DD, ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
b
a
b
b
i
(cid:118)∫ E ds = ∫ E ds + ∫
b
a
II
I
i
C
E ds
= 0 ⇒
i
∫ i
E ds
a
I(cid:11)(cid:9)(cid:11)
(cid:8) (cid:10)
Electromotive Force
(EMF)
= ∫ E ds
i
a
II
EMF between 2 points (a, b) independent of path
E field is conservative
− Φ r
( ref )
rref
∫
= E
r
i
ds
( )
Φ r
(cid:47) Scalar
electric... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
x, y, z )
=
∂Φ
∂x
∆x +
∂Φ
∂y
∆y +
∂Φ
∂z
∆z
= ⎢
_
i x +
∂Φ _
∂y
⎤
⎡ ∂Φ _
i z ⎥ i ∆r
⎣ ∂x
⎦
(cid:8)(cid:11)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10)
grad Φ = ∇Φ
∂Φ
∂z
i y +
_
∇ = i
x
_
+ i
y
∂
∂x
∂
∂ y
_ ∂
+ i
z
∂z
grad
_ ∂Φ
Φ = ∇Φ = i
x
∂x
_ ∂Φ
+ i
y
∂y
_... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
IV. Sample Problem
V xy
Φ (x, y ) = 0
2a
(Equipotential lines hyperbolas: xy=constant)
⎡ ∂Φ
E = −∇Φ = −
⎢
⎣
∂x x
_
i +
_
⎤
∂Φ
i
∂y y ⎥
⎦
=
_
_
⎞
−V0 ⎛
⎜ y i + x i ⎟
a2
⎠
⎝
y
x
Electric Field Lines [lines tangent to electric field]
dy
dx
=
Ey =
Ex
x
y
⇒ ydy = xdx
2
y
2
=
2
x
2
+ C
y2 − x2 =... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
∂
+
2
∂ Φ
2y
∂
+
2
∂ Φ
2z
∂
VI. Coulomb Superposition Integral
1. Point Charge
E
r
= −
∂Φ
r
∂
=
q
2
ε
r
4
π
0
⇒ Φ =
q
ε
r
4
π
0
+
C
Take reference
)
Φ → ∞ = ⇒
(
r
0 C 0
=
Φ =
q
πε
r
0
4
2. Superposition of Charges
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Z... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
r
r'−
V
⎛
ρ r
⎜
⎝
r
+ ∫
S
⎞
'
⎟
⎠
r'−
⎤
'
dV
⎥
⎥
⎥
⎥
⎦
Short-hand notation
Φ ( )r = ∫
'
⎞
ρ r dV
⎟
⎠
⎛
⎜
⎝
'
V 4πε0 r − r'
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 4
Page 6 of 6 | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
Chapter 7 Notes - Inference for Single Samples
• You know already for a large sample, you can invoke the CLT so:
¯X ∼ N (µ, σ2).
Also for a large sample, you can replace an unknown σ by s.
• You know how to do a hypothesis test for the mean, either:
– calculate z-statistic
z = √
x¯ − µ0
σ/ n
and compare it wi... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
�
(cid:18)
π(µ1) = Φ −zα + √
µ0 − µ1
σ/ n
�
(cid:19)
.
The alternative hypothesis only makes sense when µ1 < µ0. As µ1 increases (and gets
closer to a µ0), what happens to π(µ1)?
• For 2-sided tests,
(cid:18)
π(µ1) = P
�
¯
X < µ0 − zα/2
�
(cid:18)
�
(cid:18)
+ P
(cid:19)
�
σ
√
µ = µ1
n
(cid:18)
�
�
(cid... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
:
Now solve that for n:
δ
−zα + √ = zβ .
σ/ n
n =
2
(zα + zβ)σ
δ
.
• For lower 1-sided, n is the same by symmetry.
• For 2-sided, turns out one of the two terms of π(µ1) can be ignored to get an approxi
mation:
n ≈
2
(zα/2 + zβ)σ
δ
.
Remember to round up to the next integer when doing sample-size calcul... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
/ n
by χ2 =
(n − 1)S2
σ2
(and test for σ2 not µ). Replace zα by χ2
n−1,1−α
and/or χ2
n−1,α.
Hypothesis tests on variance are not quite the same as on the mean. Let’s do some of the
computations to show you. First, we’ll compute the CI.
4
2-sided CI for σ2 . As usual, start with what we know:
(cid:18)
1 − α ... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
Similarly, 1-sided CI’s for σ2 are:
(n − 1)s2
χ2
n−1,α
≤ σ2
and
σ2 ≤
(n − 1)s2
.
χ2
n−1,1−α
Hypothesis tests on Variance (a chi-square test)
To test H0 : σ2
= σ2
0 vs H1 : σ2
= σ2
0 , we can either:
• Compute χ2 statistic:
and reject H0 when either χ2 > χ2
n−1,α/2
• Compute pvalue:
χ2 =
(n − 1)s2
σ... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.