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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 |
¯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 atleast C1,1) in order to define 2nd derivatives ... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
+ 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(x) + c(x + h · el)u(x + h · el) − c(x)u... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
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 in
Cα(Ω) (and not in Cα( ¯Ω)!).
Putting the above facts together we now see that both sid... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
∆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 remarked above ˜w necessarily equals Dlg
by definition.
Second, we sh... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
. 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 leading to the boundary.... | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
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
without this assumption (exercise). We needed it... | 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 |
(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
(cid:8)
(cid:6)
(cid:9)
... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
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 solution, then Z(λ) ≤ ZIP for λ ≥ 0.
• The fun... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
K, j ∈ J .
3.4 Example
minimize
subject to
3x1 − x2
x1 − x2 ≥ −1
5
−x1 + 2x2 ≤
3x1 + 2x2 ≥
3
6x1 + x2 ≤ 15
x1,x2 ≥ 0
x1, x2 ∈ Z.
Slide 14
• Relax x1 − x2 ≥ −1
• X = {(1, 0), (2, 0), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2), (1, 3), (2, 3)} .
(cid:11)
• Z(λ) = min(x1,x2)∈X 3x1 − x2 + λ(−1 − x1 + x2) ,
(c... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
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 |
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 is linear, then we can use superposition to simplify the
analysis. For a lin... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
1v
=
IsR1
And so
1i
Is=
Vs
R
2
=
i
2
How much current is going through the voltage source Vs?
Another example:
For the following circuit let’s calculate the node voltage v.
R1
v
Vs
R2
Is
Nodal analysis gives:
or
v
Vs
−
R
1
+
Is
−
v
R
2
=
0
v
=
2
R
R
R
1
+
2
Vs
+
1 2
R R
R
R
1
2
+
Is
(1.6)
(1.7)
(1.8)
(1.9)
(... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
the
box as indicated. Our communication with the circuit is via the port A-B. This is a single
port network regardless of its internal complexity.
R4
Vn
In
R3
i
A
+
v
-
B
If we apply a voltage v across the terminals A-B as indicated we can in turn measure the
resulting current i . If we do this for a number of diff... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
circuit.
This circuit is known as the Thevenin Equivalent Circuit.
Since we are dealing with linear circuits, application of the principle of superposition
results in the following expression for the current i and voltage v relation.
i m v
0
=
+
m V
j
j
+
∑
j
∑
b I
j
j
j
(1.17)
Where
jV and
the coefficients
jI... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
venin’s theorem
6kΩ
12 V
2kΩ
6kΩ
1kΩ
+
vo
-
The 1kΩ resistor is the load. Remove it and compute the open circuit voltage Voc or
VTh.
6kΩ
12 V
2kΩ
6kΩ
+
Voc
-
Voc is 6V. Do you see why?
Now let’s find the Thevenin equivalent resistance RTh.
6.071/22.071 Spring 2006. Chaniotakis and Cory
9
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
R
R
+
RTh R
=
13
+
R
24
(1.22)
(1.23)
(1.24)
The open circuit voltage across terminals A-B is equal to
6.071/22.071 Spring 2006. Chaniotakis and Cory
11
+
Vs
-
R1
R2
A
vA
B
vB
R3
R4
VTh
=
vA vB
−
=
Vs
⎛
⎜
⎝
R
3
1
R
R
+
3
−
R
4
2
R
+
4
R
⎞
⎟
⎠
(1.25)
And we have obtained the equiva... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
Vs
-
R1
R3
A
+
vA
-
R2
Ru
B
+
vB
-
The voltage vu is given by
Where,
vu
=
vA vB
−
(1.26)
6.071/22.071 Spring 2006. Chaniotakis and Cory
13
And
And vu becomes:
vA Vs
=
vB Vs
=
R
3
1
R
R
+
3
Ru
2
+
Ru
R
vu Vs
=
⎛
⎜
⎝
R
3
1
R
R
+
3
−
Ru
2
+
Ru
⎞
⎟
⎠
R
(1.27)
(1.28)
(1.29)
A typical us... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
replaced by an equivalent circuit consisting of a current
source In in parallel with a resistor Rn. The current In is equal to the short circuit current
through the terminals of the port and the resistance Rn is equal to the open circuit voltage
Voc divided by the short circuit current In.
The Norton equivalent cir... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
able to obtain the solution by
simplifying the circuit.
First, let’s perform the transformation of the part of the circuit contained within the
dotted rectangle indicated below:
3 V
i
3 Ω
6 Ω
6 Ω
3 Ω
2 A
The transformation from the Thevenin circuit indicated above to its Norton equivalent
gives
0.5 A
i
3 Ω... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
R
R
3
2
1
4
+
+
+
(1.38)
Next we calculate the short circuit current
Is
R3
Vs
R1
R4
Isc
R2
X
Y
6.071/22.071 Spring 2006. Chaniotakis and Cory
18
Resistor R2 does not affect the calculation and so the corresponding circuit is
Is
R3
Vs
R1
R4
Isc
X
Y
By applying the mesh method we have
Isc
=
Vs
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
6.071/22.071 Spring 2006. Chaniotakis and Cory
20
The curve has a maximum which occurs at RL=RTh.
Figure 3.
In order to show that the maximum occurs at RL=RTh we differentiate Eq. (1.42) with
respect to RL and then set the result equal to zero.
and
dP
dRL
=
VTh
2
(
⎡
⎢
⎣
dP
dRL... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
Th.
VTh Vs
=
⎛
⎜
⎝
3
R
1
R
R
+
3
−
4
R
2
R
+
4
⎞
⎟
⎠
R
RTh
=
1 3
R R
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. Chaniot... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
the circuit for the calculation of the equivalent resistance
is
RTh
Rs
attenuator
a
Rp
Reff
b
The effective resistance is the parallel combination of Rp with Rs+RTh.
Reff
=
=
Rp
RTh Rs
(
+
RTh Rs Rp
(
+
RTh Rs Rp
+
) //
)
+
(1.50)
6.071/22.071 Spring 2006. Chaniotakis and Cory
23
Which ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
io=1.35 A, vo=10 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... | 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 |
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 obtained by acting with (A
I) on the physical state (... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
� 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 one usually squares the expression in (1.2) so that... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 follows. Consider the one-
dimensional vector subspace UΨ g... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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
The uncertainty principle is an inequality that is sat... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 Hermitian. For this first note that the commutator of two Hermitian operators is anti-Hermitian:
[A, B]† = (AB)†
(BA)† = B†A† ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 two states:
Note that by the definition (1.2) of uncertainty,
f
|
g
|
) ≡
) ≡
(A
(B ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 above we find a nice expression for the imaginary part of
f
(
g
|
:
)
f
|
For the real part the expression is not tha... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 any rate, the term is positive or zero so it can be
dropped while preserving the inequality. This is often done, thus giving the celebrated form (2.14)
that w... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
= 0 ,
= β
f
(
f
(
g
(
f
|
f
|
f
|
f
|
g
|
+
)
)
)
)
)
More explicitly this requires
g
|
)
= iλ
f
|
,
)
R .
λ
∈
Saturation Condition:
(B
B
− (
Ψ
I)
|
)
)
= iλ (A
Ψ
I)
A
|
)
− (
.
)
(2.30)
(2.31)
This must be viewed as a condition for Ψ, given any two operators A and B. Moreover, note that
and
A
(
)
equation... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
The problem is time. Time is not an operator in quantum mechanics, it is a parameter,
a real number used to describe the way systems change. Unless we define Δt in a precise way we
cannot hope for a well-defined uncertainty relation.
We can try a rough, heuristic definition, in order to illustrate the spirit of the ine... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
classical, the above identity is something electrical engineers are well aware of.
It
represents a limit on the ability to ascertain accurately the frequency of a wave that is observed for
a limited amount of time. This becomes quantum mechanical if we speak of a single photon, whose
energy is E = 1ω. Then ΔE = 1Δω... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
spin-1/2 particle, the operator
+
|
. Such
)(−|
6
[H, Q] Ψ
.
(3.36)
operator Q can easily have time-dependent expectation values, but the time dependence originates
from the time dependence of the states, not from the operator Q itself.
To explore the meaning of [H, Q] we begin by computing th... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
important result. Each time you see [H, Q] you should think ‘time derivative of
classical mechanics one usually looks for conserved quantities, that is, functions of the dynamical vari
ables that are time independent. In quantum mechanics a conserved operator is one whose expectation
value is time independent. An op... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
are roughly of the same size.
for “appreciable” change in
Q
(
(cid:12)
(cid:12)
(cid:12)
dt
)
Q
(
. This is certainly so when
)
Q
(
)
In terms of ΔtQ the uncertainty inequality reads
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
ΔtQ
≡
ΔQ
d�Q)
dt
ΔHΔtQ
1
2
.
≥
7
(3.43)
... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
now, it is not, if the Hamiltonian is a time-independent operator. Indeed, if H is time
independent, we can use H and H 2 for Q in (3.39) so that
d
dt
d
dt
H
(
H 2
(
)
)
=
=
i
1
i
1
[H, H] = 0 ,
(cid:11)
[H, H 2] = 0 .
(cid:11)
\
\
It then follows that
d
dt
(ΔH)2 =
d
dt
showing that ΔH is a constant. So... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
is a very tiny split: δE = 5.88
×
difference δE resulting in a wavelength of 21.1cm and a frequency ν = 1420.405751786(30)MHz. The
eleven significant digits of this frequency attest to the sharpness of the emission line. The issue of
uncertainty arises because the excited state of the hyperfine splitting has a lifetime ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
so useful in astronomy. For decays with much
uncertainty ΔE. But ΔE/δE
×
∼
3
shorter lifetimes there can be an observable broadening of an emission line due to the energy-time
uncertainty principle.
4
Lower bounds for ground state energies
You may recall that the variational principle could be used to find upper... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
(4.50)
(4.51)
for any state of the system. We should note however, that for the ground state (or any bound state)
= 0 so that in fact
p
(
)
From the inequality
A2
(
2 we have
A
)
) ≥ (
p 2
(
)gs = (Δp)2
gs
,
x 4
(
x 2
) ≥ (
2 .
)
Moreover, just like for momentum above, (Δx)2 =
x2
(
x
) − (
2 leads to
)
so t... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
Hgs
(
)
f (Δxgs) but we don’t know the value of Δxgs. As a result, we can only be
is greater than or equal to the lowest value the function f (Δxgs) can take.
If we knew the value of Δxgs we would immediately know that
)gs is bigger than the value taken
by the right-hand side. This would be quite nice, since we wan... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
2/3
(cid:17)
.
(4.61)
This is the final lower bound for the ground state energy. It is actually not too bad, for the ground
state instead of the prefactor 0.4724, we have 0.668.
10
5 Diagonalization of Operators
When we have operators we wish to understand, it can be useful to find a basis on the v... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
envector. But not even in complex vector spaces all operators have enough
eigenvectors to span the space. Those operators cannot be diagonalized. The simplest example of
such operator is provided by the two-by-two matrix
0 1
0 0
(
)
.
The only eigenvalue of this matrix is λ = 0 and the associated eigenvector is
... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
.64)
(5.65)
where the matrix Aij is the representation of A in the original v-basis. The operator T 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 wh... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
��
.
Ank
.
(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 unit... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 because both U ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
−
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 spectral theorem. It states that a matrix i... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 the orthonormal basis
−
×
. We know there is at least one eigenvalue λ1 with a non-zero
n
|
) o... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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. Moreover
j
(
1
M1|
)
|
j
= λ1(
1
)
|
= λ1δi,j ,
j
M1|
1
|
(
where we used M †
= λ∗
1
1|
1
)
equations that M1, in the original... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
matrix. By the induction
1)-by-(n
1)
−
′
ˆU =
1
0
. .
.
0
0
. . . 0
′
U
.
(6.81)
It follows that Uˆ †M1Uˆ = Uˆ †U M U1Uˆ = (U1Uˆ )†M (U1Uˆ ) is diagonal, proving the desired result. D.
†
1
Of course this theorem implies that Hermitian and unitary operators are unitarily diagonalizable.
... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
T -invariant subspace of
dimension dk
1 times. An eigenvalue λk is degenerate if dk > 1.
1 spanned by eigenvectors with eigenvalue λk:
It follows that
≥
≥
|
By the spectral theorem Uk has a basis comprised by dk orthonormal eigenvectors (u1
Note that while the addition of eigenvectors with different eigenvalues ... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
u
(m)
dm
) .
(6.84)
The matrix T is manifestly diagonal in this basis because each vector above is an eigenvector of T and
is orthogonal to all others. The matrix representation of T reads
T = diag λ1, . . . , λ1 ,
(cid:0)
d1 times
v
(cid:1)
dm times
v
. . .
, λm, . . . , λm
(6.85)
"
This is is clear beca... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
of eigenvectors of T with each of them orthogonal to the rest. Moreover,
(cid:0)
(cid:1)
the first d1 vectors are in U1, the second d2 vectors are in U2 and so on and so forth. More explicitly,
for example, within Uk
(k)
Vku
i
(k)
, T (Vku ) = λk
j
(k)
Vku , Vku
i
(k)
j
= λk(
)
(k)
u , u
i
(k)
j
)
(
= λkδij
(6.87... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
. Since arbitrary linear operators S
and T on a complex vector space cannot be diagonalized, the vanishing of [S, T ] does not guarantee
simultaneous diagonalization. But if the operators are Hermitian it does, as we show now.
Theorem. If S and T are commuting Hermitian operators they can be simultaneously diagonali... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
ui is also an eigenvector of S, this time with eigenvalue ωi. Thus any eigenvector of T is
also an eigenvector of S, showing that these operators are simultaneously diagonalizable.
Now consider case (ii). Since T has degeneracies, as explained in the previous section, we have a
decomposition of V in T -invariant sub... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
Uk are also S-invariant subspaces! To show this let u
the vector Su. We have
T (Su) = S(T u) = λkSu
→
Su
∈
Uk .
Uk and examine
∈
(7.93)
We use the subspaces Uk and the basis (7.91) to organize the matrix representation of S in blocks.
It follows that this matrix must have block-diagonal form since each subsp... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
.
D
Remarks:
1. Note that the above proof gives an algorithmic way to produce the common list of eigenvectors.
One diagonalizes one of the matrices and constructs the second matrix in the basis of eigenvectors
of the first. These second matrix is block diagonal, where the blocks are organized by the
degeneracies i... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
block. This is repeated for each block, until we get a common basis of eigenvectors.
3. An inductive algorithm is clear. If we know how to simultaneously diagonalize n commuting
Hermitian operators we can diagonalize n + 1 of them, call them S1, . . . Sn+1, as follows. We
diagonalize S1 and then consider the remaini... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
can be uniquely distinguished by labeling it with
the corresponding eigenvalue of S. The physical quantity associated with the observable can be used
to distinguish the various eigenstates. Moreover, these eigenstates provide an orthonormal basis for
the full vector space. In this case the operator S provides a “com... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
can
find a basis of V comprised by simultaneous eigenvectors of the operators. These states can be labeled
by two observables, namely, the two eigenvalues. If we are lucky, the basis eigenstates in each of the
S-invariant subspaces of dimension higher than one can be organized into T eigenstates of different
eigenval... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
quantum system there is a complete
set of commuting observables, for otherwise there is no physical way to distinguish the various states
that span the vector space. So in any physical problem we are urged to find such complete set, and
we must include operators in such set until all degeneracies are broken. A CSCO n... | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
need another observable that can
be used to distinguish these states. There are several options, as you will discuss in the homework.
19
MIT OpenCourseWare
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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-2013/005979fa741c3ea2e0430456b70caf93_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 |
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
)
(
P1 − P2
ρ0l
(5.3)
page 2
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
where Eqn. 5.3 is the s... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
analogy
Mechanics: Mobility
analogy
Acoustics: Impedance
sound pressure p(t)
volume velocity u(t)
30 Sept -2004
page 3
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
analogy
Acoustics: Mobility
analogy
volume velocity u(t)
sound pressure p(t)
In all of the above analogies, pow... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
equations
that relate v(t) (or e(t)) to i(t).
30 Sept -2004
page 4
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
Figure 5.2
Electric elements and their
mechanical and acoustic counter-
parts in the “Impedance analogy”
From Kinsler, Frey, Coppens, &
Sanders, Fundamentals of Acoustics ,
3rd E... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
Resistor
Resistor
Mass
Inductor
Inertance
V vs F
Spring
I vs E
U vs P
Capacitor
Compliance
V(ω) = jωCM F(ω)
I(ω) = jωCE E(ω)
U (ω) = jωCA P(ω)
Damper
1
RM
V (ω) =
F(ω)
Resistor
1
RE
I(ω) =
E(ω)
Resistor
1
RA
U (ω) =
P(ω)
V (ω) =
Mass
1
jωLM
F (ω )
Inductor
1
jωLE
Inertance
1
jωL A
I(ω) =
E (ω )
U... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
Note that the acoustic mass is equivalent to the mass of the air in the enclosed
element divided by the square of the cross-sectional area of the element. Also since
some small volume of the medium on either end of the tube is also entrained with the
media inside the tube, the “acoustic” length is usually somewhat l... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
)
P(jω)
U(jω)
Another structure that may be well
approximated by an acoustic
compliance.
C =
Volume
Adiabatic Bulk modulus
U = jωCAP
The variations in sound pressure within an enclosed air volume generally occur about
the steady-state atmospheric pressure, the ground potential in acoustics. Therefore,
one termina... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
; a = 0.1
Fig. 5.3 Relative Particle velocity amplitude as a
function of radial position in a small pipe of radius
a = 0.1 cm, at frequency f = 200 Hz.
After Kinsler and Frey, 1950; p. 238
⎛
(
− 0.1−r
⎜
δ
⎝
The velocity profile in Figure 5.3 varies as v(r) =1− e
1/ 2]
[
)
δ = η/ ρ0ω(
constant”
m-2 for air at ST... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
ecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
There is a catch, however, in that
l is effectively infinite
p(t)
u(t)
P0
U1(jω) RA
P1(jω)
this lumped element always has
one end coupled to ground and
therefore can only be used to
either terminate acoustic circuits
or be placed in parallel with other
el... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
=
1
jωLR
=
+
1
jω0.8a
2πa2
+
1
RR
1
ρ0c
2πa2
Note that the radiation mass is equivalent to the addition of a tube of radius a and
length 0.8a to the end of the pipe. This is the end correction!!
Range of applicability of acoustic circuit theory.
E.
1. Pressure and volume-velocity ranges consistent with “linear acou... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
the impedance of the three series elements.
30 Sept -2004
page 16
Lecture 7
Acoustics of Speech & Hearing
6.551 - HST 714J
Z IN ω( )= jωL A + RA +
1
jωCA
B. An Electrical Analog of the Acoustic Circuit Description
In Electrical circuits the wires that connect the ideal elements are perfect conductors.
... | https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/00610bbb35656bb99990aa06a6b78205_lec_7_2004.pdf |
MIT OpenCourseWare
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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 |
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
Lecture 2 Ver 2.0
More on Document Dis... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
‘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:
Read file
Make word list
Count frequencies
Sort in order
Once vectors D1,D2 are obtained:
Compute the angle
2
Lectur... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
down the complexity and
scale up the efficiency of the Insertion Sort routine.
Figure 1: Divide/Conquer/Combine Paradigm
3
input array of size nALRsortsortL’R’mergesorted array A2 arrays of size n/22 sorted arrays of size n/2sorted array of size n Lecture 2 Ver 2.0
More on Document Distance
6.006 Spring 2008
Figure... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
Document Distance
6.006 Spring 2008
Figure 3: Efficiency of Running Time for Problem of size n is of order Θ(n lg(n))
Question: When is Merge Sort (in Python) 2n lg(n) better than Insertion Sort (in C)
0.01n2?
Aside: Note the 20X constant factor difference between Insertion Sort written in Python
and that written in ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/0084579211f6874fb95258a10a701ae3_lec2.pdf |
MIT OpenCourseWare
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6.046J / 18.410J Design and Analysis of Algorithms
Spring 2015
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MIT OpenCourseWare
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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 |
∫ 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 potential
b
i
∫
E ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
�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
_ ∂Φ
+ i
z
∂z
∫ E ds = Φ ... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
Markus Zahn
Lecture 4
Page 3 of 6
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
... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
x
+
∂Φ
y
∂
_
i
y
+
∂Φ
z
∂
_
⎤
i
⎥
z
⎦
=
2
∂ Φ
2x
∂
+
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 Charg... | https://ocw.mit.edu/courses/6-641-electromagnetic-fields-forces-and-motion-spring-2005/00b2fd4ae919656537f533c1aa421721_lecture4.pdf |
T (
)P =
=
⎡
⎢
N
⎢∑ qn
rn
−r
=
1
ε
4π 0 ⎢n 1
⎢
⎣
'
λ r dl
'
⎛
⎜
⎝
⎞
⎟
⎠
r − r'
+ ∫
L
'
⎛
s ⎜
⎝
⎞
'
σ r da
⎟
⎠
+ ∫
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 Mot... | 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 |
√
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:19)
µ0 − µ1
σ/ n
¯
X > µ... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
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 calculations!
Example
3
7.2 Inferences on Small Samples
If n < 30, we often need t... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
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 = P χ2
n−1,1−α/2
≤
⇑
(*1)
(n−1)S2
σ2
≤ χ2
(cid:19)
n−1,α/2
and remember χ2 =
(n − 1)S2
,
σ2
⇑
(*2)
and we want:
1 − α = P (L ≤ σ2 ≤ U ) for some L a... | https://ocw.mit.edu/courses/15-075j-statistical-thinking-and-data-analysis-fall-2011/00da8087b863814d7d150a63912fa9da_MIT15_075JF11_chpt07.pdf |
= 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
σ2
0
or χ2 < χ2
n−1,1−α/2.
First we calculate the probability to be as extreme in either direction:
5
PU = P (χ2 ≥ χ2)
n−1
or
PL = P (χ2 ≤ χ2)
n−1
... | 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 |
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 + u3)
= �v + �u.
�
We... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
+ �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 =
−→
P R.
Proof. Let us consider the result of adding
P + (
−→
P Q +
−→
QR... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
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 called the standard basis. Any vector can be written uniquely
as a linear combination of these vectors,
�v = (v1, v2, v3) = v1ˆı + v2jˆ+ v3k.
We can us... | https://ocw.mit.edu/courses/18-022-calculus-of-several-variables-fall-2010/00eb8d3551c3c991805c0cf42970ee47_MIT18_022F10_l_1.pdf |
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