Electricity and Magnetism II Griffiths Chapter 7 Maxwell’s Equations Clicker Questions 7.1.
Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27
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Transcript of Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27
Problem 3-2
In Hilbert Space finctions are square integrable,
the inner product must exist, and in particular, < f f > = àa
b
Abs@f HxLD2âx;
f HxL = xΝon the interval 0 to 1
f HxL = xΝ
à0
1HxΝL2âx
Set::write : Tag Times in f x is Protected.
xΝ
IfBRe@ΝD > -
1
2,
1
1 + 2 Ν
, IntegrateBx2 Ν, 8x, 0, 1<, Assumptions ® Re@ΝD £ -
1
2FF
AssumingARe@ΝD > -1
2, à
0
1HxΝL2âxE
1
1 + 2 Ν
AssumingARe@ΝD < -1
2, à
0
1HxΝL2âxE
Integrate::idiv : Integral of x2 Ν
does not converge on 80, 1<.
à0
1
x2 Νâx
AssumingARe@ΝD = -1
2, à
0
1HxΝL2âxE
Set::write : Tag Re in Re@ΝD is Protected.
$Assumptions::bass : -
1
2
is not a well-formed assumption.
Integrate::bass : -
1
2
is not a well-formed assumption.
IfBRe@ΝD > -
1
2,
1
1 + 2 Ν
, IntegrateBx2 Ν, 8x, 0, 1<, Assumptions ® -
1
2&& Re@ΝD £ -
1
2FF
if Ν < -1
2
, solution blows up,
and if Ν = -1
2
, à0
1HxΝL2âx = à
0
1
x2 Ν
âx = à0
1
x-1
âx = à0
1 âx
x
,
> , ,Printed by Mathematica for Students
= ln HxL0
1= ln H1L - ln H0L = ¥, and it blows up.
Therefore Ν > -1
2
, and then the inner product of f with itself =1
1 + 2 Ν,
otherwise the integral does not converge;
3-2 b.
In Hilbert Space, àa
b
f HxL2âx < ¥, and when Ν =
1
2
,
this is true, so f HxL is in Hilbert space ,
à0
1Kx1
2 O2
âx
1
2
In Hilbert Space, àa
b
f HxL2âx < ¥, and when Ν =
1
2
, this is true,
so f HxL is in Hilbert space ; as is x f HxL; but the derivative of f Hx0L is1
2 x
and
àa
b 1
2 x
2
âx =1
4à
0
1 âx
x
which is not less than ¥.
à0
1
x Kx1
2 O2
âx
1
3
DAx1
2 , xE1
2 x
à0
1 1
2 x
2
âx
Integrate::idiv : Integral of
1
x
does not converge on 80, 1<.
à0
1 1
4 xâx
3-5. a. Find Hermitian conjugate of x, p, and â
âx
2 Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb
Printed by Mathematica for Students
3-5. a. Find Hermitian conjugate of x, p, and â
âx
Q` †
= H Q` TL*
so
for x, using braket notation,
X f xg^ = à-¥
¥
f* Hx gL âx
= à-¥
¥HxfL* HgL âx
= Xxf g\ ® therefore x†
= x
ä†
= äT*
= ä*
= -ä
for ä, using braket notation,
X f ä g_ = à-¥
¥
f* Hä gL âx
= à-¥
¥Hä fL* HgL âx
= X-ä f g_ ® therefore ä†
= -ä
forâ
âx
, using braket notation,
[ fâ
âx
g_ = à-¥
¥
f*
â
âx
g âx so this must be integrated by parts :
= @f*gD-¥
¥- à
-¥
¥ â f
âx
*
g âx and the first term goes to zero
= - à-¥
¥ â f
âx
*
g âx
therefore [ fâ
âx
g_ = - [ â f
âx
g_ ®â
âx
†
= -â
âx
3-5. b. Find Hermitian conjugate of a+
Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb 3
Printed by Mathematica for Students
a+ =1
2 Ñ m Ω
H-ä p + m Ω xLSince the operators p and x are Hermitian,
p†
= p and x†
= x, and from a above, ä†
= -ä,
a+†
=1
2 Ñ m Ω
Hä p + m Ω xL
3-5.
c. Show HQRL† = R† Q† (using Q`
= Q, and R` = R for brevity)
Yf ¡HQRL†g] = à f
* HQRL†g âx =
à f* IQ†
R†gM âx = à IQ†
f*M IR†
gM âx = à IR†Q
†f
*M HgL âx = à IHR QL†f
*M HgL âx
therefore HQRL†= R
†Q
†
3-7. a.
Q f = q f since q is an eigenfunction of Q and
Q g = q g
so if m = f + g then
Q m = Q f + Q g = qf + qg
so Q m = q Hf + gL and m is an eigenfunction of Q with eigenvalue q
b.
f HxL = ãx
g HxL = ã-x
operate on both functions with operatorâ
2
âx2
â2
âx2
ãx
= ãx
4 Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb
Printed by Mathematica for Students
â2
âx2
ã-x
= ã-x
so both f and g are eigenfunctions of operatorâ
2
âx2, and the eigenvalue is 1 for both;
ãx
+ ã-x
= 2 Cosh@xD;ã
x- ã
-x= 2 Sinh@xD;
these are orthogonal to each other;
3-10.
Ground state wave function :
Ψ1 =2
a
SinB n Π x
a
F where n = 1
momentum operator p = -ä Ñ
â
âx
Question : is b Ψ1 = -ä Ñ
â
âx
Ψ1?
-ä Ñ
â
âx
2
a
SinB Π x
a
F =
-ä Ñ
2
a
Π
a
CosB Π x
a
F which can never equal a constant b Ψ1;
so the ground state wave function is not an eigenfunction of the momentum operator
DB 2
aSinB Π x
aF , xF
21
a
32
Π CosB Π x
aF
3-11.
Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb 5
Printed by Mathematica for Students
Find the momentum space wave function F Hp, tLfor a particle in the ground state of the harmonic oscillator.
Ground state Ψ0 Hx, tL =m Ω
Π Ñ
1
4
ã-
m Ω
2 Ñ
x2
ã-
ä Ω t
2
F Hp, tL =1
2 Π Ñ
à-¥
¥
ã-ä p xÑ
Ψ Hx, tL âx
=1
2 Π Ñ
à-¥
¥
ã-ä p xÑ
m Ω
Π Ñ
1
4
ã-
m Ω
2 Ñ
x2
ã-
ä Ω t
2 âx
=1
2 Π Ñ
m Ω
Π Ñ
1
4 à-¥
¥
ã-ä p xÑ
ã-
m Ω
2 Ñ
x2
ã-
ä Ω t
2 âx
F Hp, tL =1
Hm Ω Π ÑL 1
4
ã-
1
2ä t Ω-
p2
2 m Ω Ñ
à-¥
¥
ã-ä p xÑ
ã-
m Ω
2 Ñ
x2
ã-
ä Ω t
2 âx
IfBReB m Ω
Ñ
F > 0,ã
-1
2ä t Ω-
p2
2 m Ω Ñ 2 Π
m Ω
Ñ
,
IntegrateBã-
2 ä p x+m x2 Ω+ä t Ω Ñ
2 Ñ , 8x, -¥, ¥<, Assumptions ® ReB m Ω
Ñ
F £ 0FF
AssumingBReB m Ω
Ñ
F > 0,1
2 Π Ñ
m Ω
Π Ñ
1
4ã
-ä Ω t
2 à-¥
¥
ã-ä p xÑ
ã-
m Ω
2 Ñ
x2
âxF FullSimplify
ã-
1
2ä t Ω-
p2
2 m Ω Ñ
Π14 I m Ω
Ñ
M14Ñ
3-11. b
3-12.
Given :
Ψ HxL =1
2 Π Ñ
à-¥
¥
ãä p xÑ
F HpL â p
6 Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb
Printed by Mathematica for Students
Ψ* HxL =
1
2 Π Ñ
à-¥
¥
ã-ä q xÑ
F* HqL â q
F HpL =1
2 Π Ñ
à-¥
¥
ã-ä p xÑ
Ψ HxL â x
F* HpL =
1
2 Π Ñ
à-¥
¥
ãä p x'Ñ
Ψ* Hx'L â x'
1
2 Π Ñà
-¥
¥
ãä p x'Ñ
ã-ä p xÑ
â p = ∆ Hx - x'L
Substitute the expression for F HpL and F* HpL into the right hand side of the equation, then
à-¥
¥
F* HpL -
Ñ
ä
¶
¶ pF HpL â p =
à-¥
¥ 1
2 Π Ñ
à-¥
¥
ãä p x'Ñ
Ψ* Hx'L â x' -
Ñ
ä
¶
¶ p
1
2 Π Ñ
à-¥
¥
ã-ä p xÑ
Ψ HxL â x â p
Now move the derivative inside the x integral to get
à-¥
¥
F* HpL -
Ñ
ä
¶
¶ pF HpL â p =
1
2 Π Ñà
-¥
¥ Kà-¥
¥
ãä p x'Ñ
Ψ* Hx'L â x' O Kà
-¥
¥
x ã-ä p xÑ
Ψ HxL â xO â p
Now move all the p dependence to interior and obtain
à-¥
¥
F* HpL -
Ñ
ä
¶
¶ pF HpL â p =
à-¥
¥
à-¥
¥
Ψ* Hx'L Ψ HxL 1
2 Π Ñà
-¥
¥
ãä p x'Ñ
ã-ä p xÑ
â p x â x â x'
The expression in parentheses is the Dirac delta function. By
integrating over x' we get what we expect,
à-¥
¥
F* HpL -
Ñ
ä
¶
¶ pF HpL â p = à
-¥
¥
à-¥
¥
Ψ* Hx'L Ψ HxL ∆ Hx - x'L x â x â x' =
à-¥
¥
Ψ* HxL Ψ HxL x â x = à
-¥
¥
Ψ* HxL x Ψ HxL â x = < x >
3-21a. A matrix is said to be idempotent if A2
= A;
the projection operator is defined as P`
= a] Za ;
so P` 2
f^ = P`
P`
f^ = P`
a^ Xa f\
Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb 7
Printed by Mathematica for Students
and we know that a bra acting on a function results in a scalar,
c , so Xa f\ can be shifted, and leaves P`
acting on a^ :
P` 2
f^ = Xa f\ Y P`
a] = Xa f\ a^Xa a\ where we know Xa a\ = 1
so P` 2
f^ = Xa f\ a^ = a^ Za f = P`
f
so P` 2
f^ = P`
f and the projection operator is idempotent.
3-21b. Find the eigenvalues, and characterize the eigenfunctions.
we have two equations :
P` 2
Λ = f2
Λ
and P`
Λ = f Λ where Λ is the eigenvalue
so since P`
= P` 2
then f - f2
= 0 so f Hf - 1 L = 0;
the two solutions are 1 and 0;
3-22.
Αket = ä 1 \ - 2 2 \ - ä 3 \Β ket = ä 1 \ - 2 3 \
Syntax::sntxf : "ä" cannot be followed by " 1 \ ".
Syntax::sntxi : Incomplete expression; more input is needed.
so
abra = -ä < 1 -2 < 2 +ä < 3
Βbra = -ä < 1 -2 < 3
X Α Β > = H-ä < 1 - 2 < 2 + ä < 3 L Hä 1 > +2 3 >L= -ä HäL < 1 1 > + HäL H2L < 3 3 >
= 1 + 2 ä
X Α Β > =
8 Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb
Printed by Mathematica for Students
aket = 8ä, -2, -ä<Βket = 8ä, 0, 2<abra = 8-ä, -2, ä<Βbra = 8-ä, 0, 2<
8ä, -2, -ä<8ä, 0, 2<8-ä, -2, ä<8-ä, 0, 2<abra Βket
81, 0, 2 ä<Βbra aket
81, 0, -2 ä<so
< a Β > = < Β a >;
Aop = a > < Β ;
A11 = < 1 a > < Β 1 > = HäL H-äL = 1;
A12 = < 1 a > < Β 2 > = HäL H0 L = 0;
A13 = < 1 a > < Β 3 > = HäL H2L = 2 ä;
A21 = < 2 a > < Β 1 > = H-2L H-äL = 2 ä;
A22 = < 2 a > < Β 2 > = H-2L H0L = 0;
A23 = < 2 a > < Β 3 > = H-2L H2L = -4;
A31 = < 3 a > < Β 1 > = H-äL H-äL = -1;
A32 = < 3 a > < Β 2 > = H-äL H0L = 0;
A33 = < 3 a > < Β 3 > = H-äL H2L = -2 ä;
so Am = 881, 0, 2 ä<, 82 ä, 0, -4<, 8-1, 0, -2 ä<<Am = 881, 0, 2 ä<, 82 ä, 0, -4<, 8-1, 0, -2 ä<<;
% MatrixForm
1 0 2 ä
2 ä 0 -4
-1 0 -2 ä
Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb 9
Printed by Mathematica for Students
Transpose@AmD;
% MatrixForm
1 2 ä -1
0 0 0
2 ä -4 -2 ä
Am daggar also requires conjugate, but i already see that Am is not equal to Am daggar and it is not hermitian.
3-27.
Ψ1 =H3 Φ1 + 4 Φ2L
5
Ψ2 =H4 Φ1 - 3 Φ2L
5
Set
A Ψ1 = a1 Ψ1
A Ψ2 = a2 Ψ2
B Φ1 = b1 Φ1
B Φ2 = b2 Φ2
Ψ Hx, tL = âcn Ψn
A Ψn = an Ψn
measure A
âan cn2
where cn2
is the probability that a
particle in state Ψ will be in state Ψn at a time in the future ;
The value obtained is a1, an eigenvalue, so Ψ1 is now = a1;
measure B
the state of the system is Ψ1 =3
5Φ1 +
4
5Φ2
since Φ1 and Φ2 are both eigenstates of B, either b1 or b2 could occur
Probability of b1 =3
5
2
=9
25
Probability of b2 =4
5
2
=16
25
10 Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb
Printed by Mathematica for Students
Put Φ1 and Φ2 in terms of Ψ1 and Ψ2
Φ1 =3
5Ψ1 +
4
5Ψ2
Φ2 =4
5Ψ1 -
3
5Ψ2
to get probability of a1, look at Ψ1
Prob@a1D =9
25
3
5
2
+16
25
4
5
2
= 0.5392
9
25
3
5
2
+16
25
4
5
2
337
625
NB 9
25
3
5
2
+16
25
4
5
2
F0.5392
Griffiths Ch 3 Selected Solutions in Quantum Mechanics prob 2,5,7,10,11,12,21,22,27.nb 11
Printed by Mathematica for Students