Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

11
Problem 3-2 In Hilbert Space finctions are square integrable, the inner product must exist, and in particular, < f f >= a b Abs@f HxLD 2 x; f HxL = x Ν on the interval 0 to 1 f HxL = x Ν 0 1 Hx Ν L 2 x Set::write : Tag Times in f x is Protected. x Ν IfBRe@ΝD >- 1 2 , 1 1 + 2 Ν , IntegrateBx 2 Ν , 8x, 0, 1<, Assumptions Re@ΝD £- 1 2 FF AssumingARe@ΝD >- 1 2 , 0 1 Hx Ν L 2 xE 1 1 + 2 Ν AssumingARe@ΝD <- 1 2 , 0 1 Hx Ν L 2 xE Integrate::idiv : Integral of x 2 Ν does not converge on 80, 1<. 0 1 x 2 Ν x AssumingARe@ΝD =- 1 2 , 0 1 Hx Ν L 2 xE 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 , IntegrateBx 2 Ν , 8x, 0, 1<, Assumptions fi- 1 2 &&Re@ΝD £- 1 2 FF if Ν<- 1 2 , solution blows up, and if Ν=- 1 2 , 0 1 Hx Ν L 2 x = 0 1 x 2 Ν x = 0 1 x -1 x = 0 1 x x Printed by Mathematica for Students

Transcript of Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

Page 1: 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.

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

,

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Page 2: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

= 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

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

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Page 3: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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Page 4: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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Page 5: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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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

-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

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Page 7: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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Page 8: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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Page 9: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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Page 10: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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Page 11: Griffiths Ch 3 Selected Solutions in Quantum Mechanics Prob 2,5,7,10,11,12,21,22,27

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

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