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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 our discussion.
Precisely what we would like to show is —
A priori regularity.
Let u ∈ C2(Ω) be a solution of Lu = f and assume 0 < α < 1. We
do not assume c(x) ≤ 0 but we do assume all the other assumptions on L in the previous Theorem
hold. If f ∈ Cα(Ω) then u ∈ C2,α(Ω)
• Here we mean the Cα norm is locally bounded, i.e for every point exists a neighborhood where
the Cα-norm is bounded. Had we written Cα( ¯Ω) we would mean a global bound on sup
x,y
|f (x) − f (y)|
|x − y|α
(as in the footnote if Lecture 14).
• This result will | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
,y
|f (x) − f (y)|
|x − y|α
(as in the footnote if Lecture 14).
• This result will allow us to assume in previous theorems only C2 regularity on (candidate)
solutions instead of assuming C2,α regularity.
Proof. Let u be a solution as above. Since the Theorem is local in nature we take any point in Ω
and look at a ball B centered there contained in Ω. We then consider the Dirichlet problem
L0v = f ′
= u
v
on
on
B,
∂B.
where L0 := L − c(x) and f ′(x) := f (x) − c(x) · u(x). This Dirichlet problem is on a ball, with
”c ≤ 0”, uniform elliptic and with coefficients in Cα. Therefore we have uniqueness and existence
of a solution v in C2,α(B) ∩ C0( ¯B). But u satisfies Lu = f or equivalently L0u = f ′ on all of Ω so
1
in particular on | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
u satisfies Lu = f or equivalently L0u = 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 atleast C1,1) in order to define 2nd derivatives wrt L! If one takes
u in a larger function space, i.e weaker regularity of u, and defines Lu = f in a weak sense then
need more regularity on coefficients of L! Under the assumption of Cα continuity on the coefficients
indeed we are in an optimal situation.
Higher a priori regularity.
Let u ∈ C2(Ω) be a solution of Lu = f and 0 < α < 1. We
do not assume c(x) ≤ 0 but we assume uniformly elliptic and that all | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
Lu = f and 0 < α < 1. We
do not assume c(x) ≤ 0 but we assume uniformly elliptic and that all coefficients are in Ck,α. If
f ∈ Ck,α then u ∈ Ck+2,α. If f ∈ C∞ then u ∈ C∞.
Proof. k = 0 was the previous Theorem.
The case k = 1. The proof relies in an elegant way on our previous results with the combination
of the new idea of using difference quotients. We would like to differentiate the u three times and
prove we get a Cα function. Differentiating the equation Lu = f once would serve our purpose but
it can not be done na¨ively as it would involve 3 derivatives of u and we only know that u has two.
To circumvent this hurdle we will take two derivatives of the difference quotients of u, which we
define by (let e1, . . . , en denote the unit vectors in Rn)
∆hu :=
u(x + h · el | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
��ne by (let e1, . . . , en denote the unit vectors in Rn)
∆hu :=
u(x + h · el) − u(x)
h
=: .
uh(x) − 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(x) + c(x + h · el)u(x + h · el) − c(x)u(x)i
= ∆haijDijuh − aijDij∆hu + ∆hbiDiuh + biDi∆hu + ∆hc | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
∆haijDijuh − aijDij∆hu + ∆hbiDiuh + biDi∆hu + ∆hc · uh + c · ∆hu = ∆hf.
or succintly
L∆hu = f ′
:= ∆hf − ∆haij · Dijuh − ∆hbi · Diuh − ∆hc · uh
where uh := u(x + h · e1).
We now analyse the regularity of the terms. f ∈ C1,α so so is ∆hf , but not (bounded) uniformly
wrt h (i.e C1,α norm of ∆hf may go to ∞ as h decreases). On the otherhand ∆hf ∈ Cα(Ω)
uniformly wrt h (∀h > 0): ∆huf = f (x+h·el)−f (x)
h
= Dlf (¯x) for some ¯x in the interval, and rhs
has a uniform Cα bound as f ∈ C1,α on all Ω! (needed as ¯x can be arbitrary).
For the same reason ∆haij, ∆h | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
∈ C1,α on all Ω! (needed as ¯x can be arbitrary).
For the same reason ∆haij, ∆hbi, ∆hc ∈ Cα(Ω). By the k = 0 case we know u ∈ C2,α(Ω) 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 in
Cα(Ω) (and not in Cα( ¯Ω)!).
Putting the above facts together we now see that both sides of the equation L∆hu = f ′ are in
Cα(Ω). And they are also in Cα(Ω′) with rhs uniformly so with | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
�hu = f ′ are in
Cα(Ω). And they are also in Cα(Ω′) with rhs uniformly so with constant c(Ω′).
By the interior Schauder estimate, ∀Ω′′ ⊂⊂ Ω′ and for each h
||∆hu||C2,α(Ω′′) ≤ c(γ, Λ, Ω
′′
) ·
||∆hu||C0,Ω′
(cid:0)
(+)||f ′
||
Cα,Ω′
(
(cid:1)
) ≤ ˜c(γ, Λ, Ω
′′, Ω
′, Ω, ||u||C1(Ω),
which is independent of h! If we assume the Claim below taking the limit h → 0 we get Dlu ∈
C2,α(Ω′′), ∀l = 1, . . . , n u ∈ C3,α(Ω′′). ∀Ω′′ ⊂⊂ Ω′ ⊂⊂ Ω ⇔ u ∈ C3,α( | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
′′). ∀Ω′′ ⊂⊂ Ω′ ⊂⊂ Ω ⇔ u ∈ C3,α(Ω).
3
Claim.
||∆hg||Cα (A) ≤ c independently of h ⇔ Dlg ∈ Cα(A).
First we we show g ∈ C0,1(A). This is tantamount to the existence of limh→0 ∆hg(x) (since if it
exists it equals Dluγ(x) - 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 remarked above ˜w necessarily equals Dlg
by definition.
Second, we show g ∈ C1(A) (i.e such that derivative is continuous not | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
w necessarily equals Dlg
by definition.
Second, we show g ∈ C1(A) (i.e such that derivative is continuous not just bounded) and actually
∈ C1,α(A):
c ≥ ||∆hg||Cα (A) ≥ lim
h→0
∆hg(x) − ∆hg(y)
|x − y|α
=
Dlg(x) − Dlg(y)
|x − y|α
= |Dlg|Cα(A)
where we used that c is independent of h.
The case k ≥ 2. Let k = 2. By the k = 1 case we can legitimately take 3 derivatives as
u ∈ C3,α(Ω). One has
L(Dlu) = f ′
:= Dlf − Dlaij · Diju − Dlbi · Diu − Dlc · u
with Dlu, f ′ ∈ C1,α(Ω). So again by the k = 1 case we have now Dlu ∈ C3,α(Ω), hence u ∈ C4,α(Ω).
The instances k ≥ 3 are in the same spirit.
Boundary regularity | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
α(Ω), hence u ∈ C4,α(Ω).
The instances k ≥ 3 are in the same spirit.
Boundary regularity
Let Ω be a C2,α domain, i.e whose boundary is locally the graph of a C2,α function. Let L be
uniformly elliptic with Cα coefficients and c ≤ 0.
T heorem. Let f ∈ Cα(Ω), ϕ ∈ C2,α(∂Ω), u ∈ C2(Ω)∩C0( ¯Ω) satisfying
Lu = f
u = ϕ
on
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 | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
� 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.
The question is local: restrict attention to B(x0, R) ∩ ¯Ω for each x0 ∈ ∂Ω. We choose a C2,α
homeomorphism Ψ1 : Rn → Rn sending B(x0, R) ∩ ∂Ω to a portion of a (flat) hyperplane and
∂B(x0, R) ∩ Ω to the boundary of half a disc. We then choose another C2,α homeomorphism
Ψ2 : Rn → Rn sending the whole half disc into a disc (= a ball). Therefore Ψ2 ◦ Ψ1 maps our
original boundary portion into a portion of the boundary of a ball.
Similarly to previous computations of this sort we define the induced operator ˜L on the induced
domain Ψ2 ◦ Ψ1(B(x0, R) ∩ Ω) and define the induced functions ˜ | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
domain Ψ2 ◦ Ψ1(B(x0, R) ∩ Ω) and define the induced functions ˜u, ˜ϕ, ˜f and we get a new Dirichlet
problem with all norms of our original objects equivalent to those of our induced ones. Note that
still ˜c := c ◦ Ψ1
−1 ◦ Ψ2
−1 ≤ 0, therefore by our theory exists a unique solution v ∈ C2,α(Ψ2 ◦
Ψ1(B(x0, R) ∩ Ω) ∪ Ψ2 ◦ Ψ1(B(x0, R) ∩ ∂Ω)) ∩ C0(Ψ2 ◦ Ψ1(B(x0, R) ∩ ¯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. | https://ocw.mit.edu/courses/18-156-differential-analysis-spring-2004/001210200bdd9ab37d9c20bf897d605f_da5.pdf |
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 in order to be able to use our existence result.
But since we already assume a solution exists we may use some of our previous results which do
not need c ≤ 0 and which secure C2,α regularity upto the boundary.
5 | 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 generality xi + xi+n = 1 are included in Ax = b.
2.1 LP1
(cid:3)
(cid:2) (cid:2)
ZLP1 = max
(cid:4)
cj wS
(cid:4)
j∈S
S⊆N
(cid:3)
(cid:2)
s.t.
Aj − b wS = 0 ∀ S ⊆ N,
j∈S (cid:2 | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
∅ = 1.
Theorem: ZLP1 = ZLP2.
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
(cid:8)
(cid:6)
(cid:9)
Aj − b yS +
(cid:6)
Aj yS∪{j} ≤ 0.
j∈S
j /� | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
yS +
(cid:6)
Aj yS∪{j} ≤ 0.
j∈S
j /∈S
2.4 The dual problem
Slide 5
x
2
i = xi:
Slide 6
min u (cid:6)
∅b
(cid:6)
j} (Aj − b) + u∅
s.t. u{
(cid:8)
(cid:9)
(cid:6)
(cid:6)
(cid:6) Aj ≥ cj ∀ j ∈ N,
(cid:6)
u
S
Aj − b
+
(cid:6)
S\{j}Aj ≥ 0 ∀ S ⊆ N, |
u
S| ≥ 2.
j∈S
j∈S
2.5 Strong Duality
Suppose that the only feasible solution to Ax = 0, x ≥ 0 is the vector 0.
Slide 7
• (Weak duality) If x is a feasible solution to | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
0.
Slide 7
• (Weak duality) If x is a feasible solution to the primal problem and u is a
feasible 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 and only if
(cid:10)
(cid:6)
u
{j} (Aj − b) + u
(cid:6)
(cid:9)
(cid:11)
(cid:9)
(cid:7)
(cid:6)
∅Aj − cj xj = 0 ∀ j ∈ N,
Aj − b +
(cid:6)
S\{ | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
x3 = 0 and x4 = 1, the dual
constraints associated with the subsets S = {1}, {4}, {1, 4}
, u3,4 =
, u1 =
7
24
1
18
5
12
1
2
−9u1 + 3u∅ ≥ 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 | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
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 function Z(λ) is concave.
Slide 11
• Lagrangean dual
ZD = max Z(λ)
s.t. λ ≥ 0.
• ZD ≤ ZIP.
3.2 Characterization
ZD = min c�x
s.t. Ax ≥ b
x ∈ conv(X).
3.3 Proof outline
(cid:10)
(cid:11)
(cid:6) x + λ(cid:6)
(b − Ax) .
• Z(λ) = min c
(cid:11)
(cid:10)
• Z(λ) = minx∈conv(X) c (cid:6) x + λ(cid:6)(b − Ax) .
(cid:11)
(cid:10)
(cid:6) x + | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
6) x + λ(cid:6)(b − Ax) .
(cid:11)
(cid:10)
(cid:6) x + λ(cid:6)
• ZD = max min
.
(b − Ax)
c
x∈X
λ≥0 x∈conv(X)
• Let x k, k ∈ K, and wj , j ∈ J, 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 | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/0036ce1440b69bb15bb6080528bae4e8_MIT15_083JF09_lec08.pdf |
λ,
⎩
6 − 3λ,
• Z(λ) =
0 ≤ λ ≤ 5/3,
5/3 ≤ λ ≤ 3,
λ ≥ 3.
• λ ∗ = 5/3, and the optimal value is ZD = Z(5/3) = −1/3. For λ = 5/3, the
corresponding elements of X are (1, 0) and (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 |
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 change of the voltage.
We will see latter that capacitors and inductors are not time-independent elements. Time-
independent elements are often called resistive elements.
Note that we often have a time dependent signal applied to time independent elements.
This is fine, we only need to analyze the circuit characteristics at each instance in time.
We will explore this further in a few classes from now.
Linearity
A function f is linear if for any two inputs x1 and x2
(
f x1 + x2
)= f x1(
)+ f x2(
)
Resistive circuits are linear. That is if we take the set {xi} as the inputs to a circuit and
f({xi}) as the response of the circuit, then the above linear relationship holds. The
response may be for example the voltage at any node of the | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
f({xi}) as the response of the circuit, then the above linear relationship holds. The
response may be for example the voltage at any node of the circuit or the current through
any element.
Let’s explore the following example.
i
R
Vs1
Vs2
Vs Vs
1
+
2
−
iR 0
=
i
=
2
1Vs Vs
+
R
KVL for this circuit gives
Or
6.071/22.071 Spring 2006. Chaniotakis and Cory
(1.1)
(1.2)
1
And as we see the response of the circuit depends linearly on the voltages
.
A useful way of viewing linearity is to consider suppressing sources. A voltage source is
suppressed by setting the voltage to zero: that is by short circuiting the voltage source.
and
2Vs
1Vs
Consider again the simple circuit above. We could view it as the linear superposition of
two circuits, each of which has only one voltage source.
i1
Vs1
i2
R
R
Vs2
The | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
of
two 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 is linear, then we can use superposition to simplify the
analysis. For a linear circuit with multiple sources, suppress all but one source and
analyze the circuit. Repeat for all sources and add the results to find the total response
for the full circuit.
6.071/22.071 Spring 2006. Chaniotakis and Cory
2
Independent sources may be suppressed as follows:
Voltage sources:
Vs
+
v=Vs
-
Current sources:
suppress
short
+
v=0
-
i=Is
Is
suppress
i=0
open
6.071/22.071 Spring | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
suppress
short
+
v=0
-
i=Is
Is
suppress
i=0
open
6.071/22.071 Spring 2006. Chaniotakis and Cory
3
An example:
Consider the following example of a linear circuit with two sources. Let’s analyze the
circuit using superposition.
R2
i2
R1
i1
Vs
+
-
Is
First let’s suppress the current source and analyze the circuit with the voltage source
acting alone.
R2
i2v
R1
i1v
Vs
+
-
So, based on just the voltage source the currents through the resistors are:
1i v 0=
Vs
R
2
=
i v
2
(1.4)
(1.5)
Next we calculate the contribution of the current source acting alone
R1
i1i
+
v1
-
R2
i2i
Is
Notice that R2 is shorted out (there is no voltage across R2), and therefore there is no
current through it. The current through R1 is Is, and so | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
ed 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 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)
(1.10)
We notice that the answer given by Eq. (1.10) is | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
1.8)
(1.9)
(1.10)
We notice that the answer given by Eq. (1.10) is the sum of two terms: one due to the
voltage and the other due to the current.
Now we will solve the same problem using superposition
The voltage v will have a contribution v1 from the voltage source Vs and a contribution
v2 from the current source Is.
6.071/22.071 Spring 2006. Chaniotakis and Cory
5
Vs
And
R1
v1
R1
v2
R2
R2
Is
v
1
=
Vs
v
2
=
Is
2
R
R
R
1
+
2
R R
1 2
R
R
1
2
+
(1.11)
(1.12)
Adding voltages v1 and v2 we 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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
.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 is sufficient in order to understand its behavior and be able to
interconnect it with other circuits.
The following figure illustrates the general concept where a circuit is represented by 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 different voltages and then plot them on
the i-v space we obtain the i-v characteristic curve of the circuit.
For a general linear network the i-v characteristic curve is a linear function
+
i m v b
=
6.071/22.071 Spring 2006. Chaniotakis and Cory
(1.13)
6 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
i m v b
=
6.071/22.071 Spring 2006. Chaniotakis and Cory
(1.13)
6
Here are some examples of i-v characteristics
i
i
+
v
-
R
v
In general the i-v characteristic does not pass through the origin. This is shown by the
next circuit for which the current i and the voltage v are related by
iR Vs v
0
− =
+
or
i
=
v Vs
−
R
i
R
Vs
i
+
v
-
Vs
-Vs/R
v
(1.14)
(1.15)
Similarly, when a current source is connected in parallel with a resistor the i-v
relationship is
i
Is
= − +
v
R
i
open circuit
voltage
(1.16)
i
+
v
-
Is
R
-Is
RIs
v
short circuit
current
6.071/22.071 Spring 2006. Chaniotakis and Cory
7
The | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 are where the
line crosses the i and v axes.
These two points may be obtained by performing two simple measurements (or make two
simple calculations). With these two measurements we are able to replace the complex
network by a simple equivalent 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 are voltage and current sources in the circuit under investigation and
jb are functions of other circuit parameters such as resistances.
jm and | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
jI are voltage and current sources in the circuit under investigation and
jb are functions of other circuit parameters such as resistances.
jm and
And so for a general network we can write
Where
And
+
i m v b
=
m m=
0
b
=
∑
j
m V
j
j
+
Thevenin’s Theorem is stated as follows:
∑
b
jI
j
j
(1.18)
(1.19)
(1.20)
A linear one port network can be replaced by an equivalent circuit consisting of a voltage
source VTh in series with a resistor Rth. The voltage VTh is equal to the open circuit
voltage across the terminals of the port and the resistance RTh is equal to the open circuit
voltage VTh divided by the short circuit current Isc
The procedure to calculate the Thevenin Equivalent Circuit 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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 circuit is now
R4
Vn
In
R3
Original circuit
i
A
+
v
-
B
RTh
i
Voc
A
B
+
v
-
Equivalent circuit
If we short terminals A-B, the short circuit current Isc is
Isc
=
VTh
RTh
(1.21)
Example:
Find vo using Thevenin’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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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
6kΩ
2kΩ
6kΩ
RTh
And the Thevenin circuit is
RTh
6
k
= Ω
// 6
k
Ω + Ω = Ω
k
5
2
k
5k Ω
6 V
1k Ω
6 V
5k Ω
1k Ω
+
vo
-
And vo=1 Volt.
Another example:
Determine the Thevenin equivalent circuit seen by the resistor RL.
+
Vs
-
R1
R2
RL
R3
R4
Resistor RL is the load resistor and the balance of the 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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 voltage source
resulting in the circuit.
R1
R2
A
B
R3
R4
≡
A
R1
R3
R2
R4
B
The resistance seen by looking into port A-B is the parallel combination of
In series with the parallel combination
R
13
=
1 3
R R
1
3
R
R
+
R
24
=
2 4
R R
2
4
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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 equivalent circuit with the Thevenin resistance given by Eq.
(1.24) and the Thevenin voltage given by Eq. (1.25).
6.071/22.071 Spring 2006. Chaniotakis and Cory
12
The Wheatstone 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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 of Vmax. However, many measurable quantities only vary slightly, and
so it would be advantageous to make a difference measurement over the limited range ,
Vmax- Vmin. The Wheatstone bridge circuit accomplishes this.
+
Vs
-
R1
R2
A
+
B
-
vu
R3
Ru
The Wheatstone bridge is composed of three known resistors and one unknown, Ru, by
measuring either the voltage or the current across the center of the bridge the unknown
resistor can be determined. We will focus on the measurement of the voltage vu as
indicated in the above circuit.
The analysis can proceed by considering the two voltage dividers formed by resistor pairs
R1, R3 and R2, R4.
+
Vs
-
R1
R3
A
+
vA
-
R2
Ru
B
+
vB
-
The voltage vu is | 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 use of the Wheatstone bridge is to have R1=R2 and R3 ~ Ru. So let’s take
Ru R ε
+
=
3
Under these simplifications,
vu Vs
=
=
Vs
⎛
⎜
⎝
⎛ | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
=
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 as,
+(cid:19)
1R
ε
R
3
(cid:17)
vu Vs
ε
+
R
1
R
3
(1.32)
6.071/22.071 Spring 2006. Chaniotakis and Cory
14
The Norton equivalent circuit
A linear one port network can be | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
aniotakis and Cory
14
The Norton equivalent circuit
A linear one port network can be 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 circuit model is shown below:
In
Rn
i
+
v
-
By using KCL we derive the i-v relationship for this circuit.
or
i
+
In
−
v
Rn
=
0
i
=
v
Rn
−
In
For
0i =
(open circuit) the open circuit voltage is
And the short circuit current is
Voc
=
InRn
Isc
In=
(1.33)
(1.34)
(1.35)
(1.36)
If we choose
Rn RTh
=
and
In
=
Voc
RTh
the Thevenin and Norton circuits are equivalent
6.071/22. | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
n RTh
=
and
In
=
Voc
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)(cid:11)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:10)
Norton Circuit
We may use this equivalence to analyze circuits by performing the so called source
transformations (voltage to current or current to voltage).
For example let’s consider the following circuit for which we would like to calculate the
current i as indicated by using the source transformation method. | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
voltage).
For example let’s consider the following circuit for which we would like to calculate the
current i as indicated by using the source transformation method.
3 V
i
3 Ω
6 Ω
6 Ω
3 Ω
2 A
By performing the source transformations we will be 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
6 Ω
6 Ω
3 Ω
2 A
6.071/22.071 Spring 2006. Chaniotakis and Cory
16
Next let’s consider the Norton equivalent on the right 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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
�
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 have
i
0.5 A
6 Ω
6 Ω
6 Ω
1 A
And so from current division we obtain
i
=
1 3
⎛
⎜
3 2
⎝
⎞
⎟
⎠
=
1
2
A
(1.37)
6.071/22.071 Spring 2006. Chaniotakis and Cory
17
Another example: Find the Norton equivalent circuit at terminals X-Y.
Is
R3
Vs
R1
R4
R2
X
Y
First we calculate the equivalent resistance across terminals | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
X-Y.
Is
R3
Vs
R1
R4
R2
X
Y
First we calculate the equivalent resistance across terminals X-Y by setting all sources to
zero. The corresponding circuit is
R3
R1
R4
R2
X
Y
Rn
And Rn is
Rn
=
R R
2( 1
R
3
R
4)
+
+
R
R
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
R
1
+
−
R
IsR
3
+
3
R
4
=
In
(1.39) | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 system is designed to provide power to a load. The general
problem is depicted on Figure 1 where the load is represented by resistor RL.
linear
electronic
system
RL
Figure 1.
By considering the Thevenin equivalent circuit of the system seen by the load resistor we
can represent the problem by the circuit shown on Figure 2.
RTh
i
VTh
+
vL
-
RL
The power delivered to the load resistor RL is
Figure 2
The current i is given by
And the power becomes
2
P i RL
=
i
=
VTh
RTh RL
+
P
⎛
= ⎜ | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
power becomes
2
P i RL
=
i
=
VTh
RTh RL
+
P
⎛
= ⎜
⎝
VTh
RTh RL
+
2
⎞
⎟
⎠
RL
(1.40)
(1.41)
(1.42)
For our electronic system, the voltage VTh and resistance RTh are known. Therefore if we
vary RL and plot the power delivered to it as a function of RL we obtain the general
behavior shown on the plot of Figure 3.
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
2
RTh RL
+
(
2 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 transferred from the source to the load is
P
max
=
2
VTh
4
RTh
(1.45)
(1.46)
1 By taking the second derivative
2
d P
2
dRL
and setting RL=RTh we can easily show that
2
d P
2
dRL
< , thereby
0
the point RL=RTh corresponds to a maximum.
6.071/22.071 Spring 2006. Chaniotakis and Cory
21
Example: | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
/22.071 Spring 2006. Chaniotakis and Cory
21
Example:
For the Wheatstone bridge circuit below, calculate the maximum power delivered to
resistor RL.
+
Vs
-
R1
R2
RL
R3
R4
Previously we calculated the Thevenin equivalent circuit seen by resistor RL. The
Thevenin resistance is given by Equation (1.24) and the Thevenin voltage is given by
Equation (1.25). Therefore the system reduces to the following equivalent circuit
connected to resistor RL.
RTh
i
VTh
+
vL
-
RL
For convenience we repeat here the values for RTh and VTh.
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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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. Chaniotakis and Cory
22
In various applications we are interested in decreasing the voltage across a load resistor
by without changing the output resistance of the circuit seen by the load. In such a
sit | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
applications we are interested in decreasing the voltage across a load resistor
by without changing the output resistance of the circuit seen by the load. In such a
situation the power delivered to the load continues to have a maximum at the same
resistance. This circuit is called an attenuator and we will investigate a simple example to
illustrate the principle.
Consider the circuit shown of the following Figure.
RTh
Rs
VTh
Rp
attenuator
+
vo
-
a
b
RL
The network contained in the dotted rectangle is the attenuator circuit.
The constraints are as follows:
1. The equivalent resistance seen trough the port a-b is RTh
2. The voltage vo
k VTh
=
Determine the requirements on resistors Rs and Rp.
First let’s calculate the expression of the equivalent resistance seen across terminals a-b.
By shorting the voltage source 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 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
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 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 Equations (1.51) and (1.53) we obtain
And
Rs
=
1 k
−
k
R
Th
Rp
=
1
−
k
1
R
Th
The maximum power delivered to the load occurs at RTh and is equal to
Pmax
=
2
2
k VTh
4
RTh
(1.51)
(1.52)
(1.53)
(1.54) | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
VTh
4
RTh
(1.51)
(1.52)
(1.53)
(1.54)
(1.55)
(1.56)
6.071/22.071 Spring 2006. Chaniotakis and Cory
24
Representative Problems:
P1.
Find the voltage vo using superposition.
(Ans. 4.44 Volts)
vo
2 Ω
3 Ω
4 Ω
1 Ω
6 V
2 V
P2.
Calculate io and vo for the circuit below using superposition
(Ans. io=1.6 A, vo=3.3 V)
2 A
4 Ω
2 Ω
3 Ω
12 V
io
1 Ω
4 Ω
3 Ω
1 A
P3.
using superposition calculate vo and io as indicated in the circuit below
(Ans. io=1.35 A, vo=10 V)
+ vo -
io
4 Ω
3 Ω
3 Ω | 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 Ω
3 Ω
Calculate the value of the resistor R so that the maximum power is transferred to
the 5Ω resistor. (Ans. 10Ω)
R
B
24 V
5 Ω
12 V
10 Ω
P6. Determine the value of resistor R so that maximum power is delivered to it from
the circuit connected to it.
Vs
+
-
R1
R3
R2
R4 | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/004de24dd40e3938b0393a6a2bd91788_linear_crct_ana.pdf |
so that maximum power is delivered to it from
the circuit connected to it.
Vs
+
-
R1
R3
R2
R4
R
P7
The box in the following circuit represents a general electronic element.
Determine the relationship between the voltage across the element to the current
flowing through it as indicated.
R1
R2
i
Vs
R3
v
+
-
6.071/22.071 Spring 2006. Chaniotakis and Cory
26 | 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 Operators
8 Complete Set of Commuting Observables
1 Uncertainty defined
1
3
6
9
11
12
16
18
As we know, observables are associated to Hermitian operators. Given one such operator A we can
use it to measure some property of the physical system, as represented by a state Ψ.
If the state
is in an eigenstate of the operator A, | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
a state Ψ.
If the state
is in an eigenstate of the operator A, we have no uncertainty in the value of the observable, which
coincides with the eigenvalue corresponding to the eigenstate. We only have uncertainty in the value
of the observable if the physical state is not an eigenstate of A, but rather a superposition of various
eigenstates with different eigenvalues.
We want to define the uncertainty ΔA(Ψ) of the Hermitian operator A on the state Ψ. This
uncertainty should vanish if and only if the state is an eigenstate of A. The uncertainty, moreover,
should be a real number. In order to define such | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
The uncertainty, moreover,
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 obtained by acting with (A
I) on the physical state (I is the identity
A
)
− (
The expectation
A
)
(
operator):
ΔA(Ψ)
≡
A
I Ψ .
A
)
− (
(1.2)
1
The uncertainty, so defined is manifestly non | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
)
− (
(1.2)
1
The uncertainty, so defined is manifestly non-negative. If the uncertainty is zero, the vector inside the
norm is zero and therefore:
A
(cid:0)
and the last equation confirms that the state is indeed an eigenstate of A (note that
I Ψ = 0
A
)
ΔA(Ψ) = 0
Ψ ,
A
)
(
A Ψ =
− (
→
→
(cid:1)
(1.3)
is a number).
You should also note that
A
)
(
and forming the inner product with another Ψ we get
is indeed the eigenvalue, since taking the eigenvalue equation AΨ = λΨ
A
)
(
Ψ, AΨ
(
)
= λ
Ψ, Ψ
(
) | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
� = λΨ
A
)
(
Ψ, AΨ
(
)
= λ
Ψ, Ψ
(
)
= λ
→
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 one usually squares the expression in (1.2) so that
I Ψ
A
)
(cid:1)
A
)
(
(ΔA(Ψ))2 =
A
(cid:0)
\
I Ψ , A
A
)
− (
− (
(cid:1 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
Ψ))2 =
A
(cid:0)
\
I Ψ , A
A
)
− (
− (
(cid:1)
(cid:0)
)
(1.6)
Since the operator A is assumed to be Hermitian and consequently
I)† =
A
)
I, and therefore we can move the operator on the first entry onto the second one to find
A
)
is real, we have (A
− (
− (
A
While this is a reasonable form, we can simplify it further by expansion
(ΔA(Ψ))2 = Ψ , A
\
(cid:0)
2
I Ψ
A
)
(cid:1)
− (
.
)
The last two term combine and we find
(ΔA(Ψ))2 = Ψ , A2
\
(cid:0)
A +
A
2
)
( | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
(ΔA(Ψ))2 = Ψ , A2
\
(cid:0)
A +
A
2
)
(
−
2I Ψ
A
)
(
(cid:1)
.
)
(ΔA(Ψ))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 follows. Consider the one-
dimensional vector subspace UΨ generated by Ψ. Take the state AΨ and project it to the subspace
Ψ and the | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
�. Take the state AΨ and project it to the subspace
Ψ and the part of A Ψ in the orthogonal subspace U ⊥ is a vector
UΨ. The projection, we claim is
A
)
(
of norm equal to the uncertainty ΔA. Indeed the orthogonal projector PUΨ is
Ψ
PUΨ =
Ψ
|
Ψ
)(
,
|
2
(1.11)
Figure 1: A state Ψ and the one-dimensional subspace UΨ generated by it. The projection of AΨ to UΨ is
Ψ. The orthogonal complement Ψ⊥ is a vector whose norm is the uncertainty ΔA(Ψ).
A
)
(
so that
Ψ
Moreover, the vector A | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
is the uncertainty ΔA(Ψ).
A
)
(
so that
Ψ
Moreover, the vector A
|
)
Ψ
PUΨA
|
Ψ
)(
|
minus its projection must be a vector
Ψ
A
|
|
Ψ
|
=
=
Ψ
)(
)
)
.
A
)
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
The uncertainty principle is an inequality | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
1.
Ψ⊥|
|
2
The Uncertainty Principle
The uncertainty principle is an inequality that is satisfied by the product of the uncertainties of two
Hermitian operators that fail to commute. Since the uncertainty of an operator on any given physical
state is a number greater than or equal to zero, the product of uncertainties is also a real number
greater than or equal to zero. The uncertainty inequality often gives us a lower bound for this product.
When the two operators in question commute, the uncertainty inequality gives no information.
Let us state the uncertainty inequality. Consider two Hermitian operators A and B and a physical
state Ψ of the quantum system. Let ΔA and ΔB denote the uncertainties of A and | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
of the quantum system. Let ΔA and ΔB denote the uncertainties of A and B, respectively,
in the state Ψ. Then we have
(ΔA)2(ΔB)2
Ψ
1
2i
|
≥
\
[A, B] Ψ
2
.
(cid:12)
(cid:12)
(2.14)
The left hand side 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 Hermitian. For this first | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
[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†
A†B†
−
BA =
[A, B]
−
−
−
(2.16)
The presence of the i then makes the operator in (2.15) Hermitian. Note that the uncertainty inequality
can also be written as
where the bars on the right-hand side denote absolute value.
ΔA ΔB
≥
Ψ
1
2i
|
[A, B] Ψ
[
[
(cid:12)
(cid:12)
(cid:12)
\
(cid:12)
(cid:12)
.
(cid:11)(cid:12)
(cid:12)
(cid:12)
(2.17)
Before we | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
)
(cid:12)
.
(cid:11)(cid:12)
(cid:12)
(cid:12)
(2.17)
Before we prove the theorem, let’s do the canonical example! Substuting ˆx for A and ˆp for B
results in the position-momentum uncertainty relation you have certainly worked with:
Since [ˆx, pˆ]/(2i) = 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 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
(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
Ψ
I)
A
|
)
)
Ψ
I)
B
|
)
)
− (
− (
f
(
f
|
g
(
g
|
= (ΔA)2 ,
= (ΔB)2 .
)
)
.
(2.20)
(2.21)
The Schwarz inequality immediately furnishes us an inequality involving precisely the uncertainties
and therefore we have
f
(
f
|
g
)(
g
|
f
) ≥ |(
g
|
2 ,
)|
( | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
have
f
(
f
|
g
)(
g
|
f
) ≥ |(
g
|
2 ,
)|
(ΔA)2(ΔB)2
f
≥ |(
g
|
2 = (Re
f
(
)|
g
|
)2 + (Im
)
f
(
g
|
)2 .
)
Writing Aˇ = (A
I) and Bˇ = (B
A
)
− (
B
− (
I), we now begin to compute the right-hand side:
)
(2.22)
(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 | 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 that simple, so it is best to leave it as the anticommutator of
Ψ
[A, B]
|
|
g
) − (
) =
)
f
Im
(
(2.26)
Ψ
(
g
|
g
|
)
)
.
1
f
= (
2i | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
Im
(
(2.26)
Ψ
(
g
|
g
|
)
)
.
1
f
= (
2i
(
1
2i
the checked operators:
Back in (2.23) we get
f
Re
(
g
|
)
=
1
2
f
(
(
g
|
)
+
g
(
f
|
) =
)
1
2
Ψ
(
ˆA, ˆB
|{
Ψ
}|
)
(ΔA)2(ΔB)2
1
2i
Ψ
(
(cid:16)
|
≥
Ψ
[A, B]
|
)
(cid:17)
2
+
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, | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 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 we have now established.
Now that we have a proven the uncertainty inequality, we can ask: What are the conditions for this
inequality to be saturated? If the goal is to minimize uncertainties, under what conditions can we
achieve the minimum possible product of uncertainties? As the | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
, under what conditions can we
achieve the minimum possible product of uncertainties? As the proof shows, saturation is achieved
under two conditions:
1. The Schwarz inequality is saturated. For this we need
g
|
)
= β
f
|
)
where β
C.
∈
) = 0, so that the last term in (2.28) vanishes. This means that
g
f
2. Re(
)
|
(
f
= β
|
in Condition 2, we get
g
|
)
)
Using
f
(
g
|
)
+
g
(
f
|
)
= 0.
f
(
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 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 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. Taking the norm of both sides we get | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
, note that
and
A
(
)
equation. Taking the norm of both sides we get
B
(
)
are Ψ dependent. What is λ, physically? Well, the norm of λ is actually fixed by the
ΔB =
λ
|
ΔA
|
λ
→ |
=
|
ΔB
ΔA
.
(2.32)
The classic illustration of this saturation condition is worked out for the x, p uncertainty inequality
ΔxΔp
≥
1/2. You will find that gaussian wavefunctions satisfy the saturation condition.
5
3 The Energy-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 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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
uncertainty’. 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 inequality. Consider
a photon that is detected at some point in space, | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
spirit of the inequality. Consider
a photon that is detected at some point in space, as a passing oscillatory wave of exact duration
T . Even without quantum mechanical considerations we can ask the observer what was the angular
frequency ω of the pulse.
In order to answer our question the observer will attempt to count the
number N of complete oscillations of the waveform that went through. Of course, this number N is
given by T divided by the period 2π/ω of the wave:
N =
ω T
2π
.
(3.33)
The observer, however, will typically fail to count full waves, because as the pulse gets started from
zero and later on dies off completely, the | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 .
(3.34)
This is all still 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 |
photon, whose
energy is E = 1ω. Then ΔE = 1Δω, so that multiplying the above inequality by 1 we get
ΔE T � h .
(3.35)
In this uncertainty inequality T is the duration of the pulse. It is a reasonable relation but the presence
of � betrays its lack of full precision.
We can find a precise energy/Q-ness uncertainty inequality by applying the general uncertainty
inequality to the Hamiltonian H and another Hermitian operator Q, as did the distinguished Russian
physicists L. Mandelstam and Tamm shortly after the formulation of the uncertainty principle. We
would then have
ΔH ΔQ
≥
Ψ
1
2i
|
\
(cid:12)
(cid:12) | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 could be, for example some function of ˆx and ˆp, or for a 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. | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 the time-derivative of the expectation
value of Q:
d
dt
Q
(
)
=
d
dt
\
Ψ , QΨ =
(cid:11)
∂Ψ
∂t
(
, QΨ + Ψ , Q
) (
∂Ψ
∂t
)
(3.37)
where we did not have to differentiate Q as it is time-independent. At this point we can use the
Schr¨odinger equation to find
1
i1
HΨ , QΨ + Ψ , Q HΨ
(
Ψ , QHΨ
HΨ , QΨ
)
1
i1
d
dt
Q
( | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
�
(
Ψ , QHΨ
HΨ , QΨ
)
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 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 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 operator Q is conserved if it commutes with the Hamiltonian!
With this result, the inequality (3.36) can be simplified. Indeed, using (3.39) we have
Q
(
’. In
)
and therefore
[H, Q]
1
2i
(
(cid:12)
(cid:12)
(cid:12)
ΔH ΔQ
≥
)
(cid:12)
(cid:12)
(cid:12)
1
2
=
(cid:12)
(cid:12)
(cid:12)
Q
d
(
dt
)
1
2i
1
i
Q
d
(
dt
)
Q
d
(
dt
)
1
2
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
)
Q
d
(
dt
)
1
2
(cid:12)
(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 units of time. It is the time it would take
velocity d�Q) were time-independent. Since they are not necessarily so, we can view ΔtQ as the time
and ΔQ are roughly of the same size.
for “appreciable” change in
Q
(
(cid:12) | 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)
This is still a precise inequality, given that ΔtQ has a concrete definition in (3.42).
As you will consider in the homework, (3.41) can be used to derive an inequality for time Δt⊥ that
it takes for a system to become orthogonal | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
used to derive an inequality for 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 quantum
computation it plays a role in limiting the maximum possible speed of a computer for a fixed finite
energy.
The uncertainty relation involves ΔH.
It is natural to ask if this quantity is time dependent.
As we show | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
H.
It is natural to ask if this quantity is time dependent.
As we show 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 we have shown that
H 2
(
H
) − (
2 = 0 .
)
(cid:1)
(cid:0)
(3.45)
(3.46)
If | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
(
2 = 0 .
)
(cid:1)
(cid:0)
(3.45)
(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 spin, the ground state of hydrogen is fourfold degenerate,
corresponding to the four possible combinations of spins (up-up, up-down, down-up, down-down).
The magnetic interaction between the spins actually breaks this degeneracy and produces the so-
10−6ev (compare with about 13.6 ev
for the ground state energy). | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
produces the so-
10−6ev (compare with about 13.6 ev
for the ground state energy). For a hyperfine atomic transition, the emitted photon carries the energy
called “hyperfine” splitting. This 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 τH for decay to
11 million
the ground state and emission of a photon. This lifetime is extremely long, in fact τH
107sec, | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
emission of a photon. This lifetime is extremely long, in fact τH
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
10−30ev. of the original excited state of the
related to some energy uncertainty ΔE
2
1/τH
∼
×
≃
8
hydrogen atom. Once the decay takes place the atom goes to the fully stable ground state, without
any possible energy uncertainty. By the conservation of energy uncertainty, the photon must carry | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
stable ground state, without
any possible energy uncertainty. By the conservation of energy uncertainty, the photon must carry the
10−25, an absolutely infinitesimal effect on the photon. There is
no broadening of the 21 cm line! That’s one reason it is 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 bounds on ground state
energies. The uncertainty principle can be used to find lower bounds for | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
ground state
energies. The uncertainty principle can be used to find lower bounds for the 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
p
2m
+ α x4 ,
(4.48)
where α > 0 is a constant with units of energy over length to the fourth power. Our goal is to find
a lower bound for the ground state energy
)gs. Taking the ground state expectation value of the
p2 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
for the ground state energy
)gs. Taking the ground state expectation value of the
p2
(
2m
Hamiltonian we have
)gs =
)gs ,
x 4
(
(4.49)
H
(
H
(
)gs
+ α
Recalling that
we see that
(Δp)2 =
p 2
(
p
) − (
2 ,
)
p 2
(
) ≥
(Δp)2 ,
(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 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
Δp)2
gs
,
x 4
(
x 2
) ≥ (
2 .
)
Moreover, just like for momentum above, (Δx)2 =
x2
(
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)
(4.57)
Back to | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
�pgs
≥
1
2Δxgs
.
(4.56)
(4.57)
Back to the value of
H
(
)gs we get
H
(
12
8m(Δxgs)2
The quantity to the right of the inequality is a function of Δxgs. This function has been plotted in
Figure 2.
+ α (Δxgs)4 .
(4.58)
)gs
≥
Figure 2: We have that
certain that
Hgs
(
) ≥
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 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 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)
12
8m(Δx)2
+ α (Δx)4
.
(cid:17)
(4.59)
The minimization problem is straightforward. | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
α (Δx)4
.
(cid:17)
(4.59)
The minimization problem is straightforward. In fact
f (x) =
A
2
x
+ Bx4
is minimized for x =
2
2−1/3
1/3
A
B
(cid:16) (cid:17)
yielding f =
21/3 3
2
(A2B)1/3
.
(4.60)
Applied to (4.59) we obtaine
H
(
)gs
≥
12 α 2/3
21/3 3
√
8 m
(cid:16)
(cid:17)
≃
0.4724
(cid:16)
√
m
12 α 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 | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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 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 diagonal we say that the operator is diagonalizable.
If we can find a set of basis vectors for which the
If an operator T is diagonal in some basis (u1, . . . un) of the vector space V , | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
operator T is diagonal in some basis (u1, . . . un) of the vector space V , its matrix takes the
form diag (λ1, . . . λn), with constants λi, and we have
T u1 = λ1u1 ,
. . . , T un = λnun .
(5.62)
The basis vectors are recognized as eigenvectors with eigenvalues given by the diagonal elements. It
follows that a matrix is diagonalizable if and only if it possesses a set of eigenvectors that span the
vector space. Recall that all operators T on complex vector spaces have at least one eigenvalue and
thus at least a one eigenvector. But not even in complex vector spaces all operators have enough
eigenvectors to | https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf |
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