Quantum Mechanics Representation by Linear Hermitian Operator Questions

3—10. Prove that if V(x) is a solution to the Schrödinger equation, then any constant times V(x) is also a solution.

If l/f(x) is a solution of the Schrödinger equation, then fid’ (x) = E V (x). The Hamiltonian operator, H, is a linear operator, so if c is a constant,

ficdf(x) = cfiV(x)

= cEV(x)

word image 375E [cV(x)]

where the last step follows because both c and E are numbers whose multiplication commutes. This shows that ct/J (x) is an eigenfunction of H with eigenvalue equal to E, and it follows that Cl/J(x) is also a solution to the Schrödinger equation.

3-28. Consider a particle of mass m in a one-dimensional box of length a. Its average energy is given by

1

(E) word image 376

2m

Because (p) = 0, (p ) = a 2 , wherea can be called the uncertainty in p. Using the uncertainty principle, show that the energy must be at least as large as h2 /8ma because ax, the uncertainty in x, cannot be larger than a.

Starting with Equation 3.43 and the inequality ax a,

The Schrödinger Equation and a Particle in a Box

word image 377 20

gives

2a word image 378

word image 379 Both sides are positive, so squaring preserves the inequality

h2

4a2 — Since (p2) = 0 2 and (E) (p 2)/2m, then

8ma2

C—l. Find the length of the vector v

Use Equation C.5:

word image 380word image 381word image 382word image 383word image 384word image 385word image 386word image 387word image 388word image 390= lvl -( 1word image 391

— [22 -k (— l) 2 + 32 word image 393

14

C—4. Show that the vectors u = 2 i — 4 j — 2 k and v = 3 i + 4 j — 5 k are orthogonal.

Find the scalar product of the two vectors using Equation C.9,

word image 396

word image 398

Since u • v = O, the two vectors are orthogonal. (See Problem C—3.)

C—6. Find the angle between the two vectors u = —i 2 j k and v j+2k

First find the scalar product using Equation C.9:

word image 399 = LlxVx + LCyVy + uzVz

word image 402 word image 404 —3

Then use Equation C.6,

lullvl cose = 3

word image 406 3

cos 0 =0.327

(1 1) 1/2 (9 + 1 + 4) 1 2 o = 1090

4—13. Defining

word image 408¯ + + L 2

z

show that L 2 commutes with each component separately. What does this result tell you about the ability to measure the square of the total angular momentum and its components simultaneously?

First consider the commutator L 2 L word image 410

The Postulates and General Principles of Quantum Mechanics

word image 412

Since L2x commutes with ix, the commutator F, is zero. Using, for example, ihL: — word image 414 ix, , the other two commutators in the last expression can be written as

word image 416

word image 418

= (Lynx y — ihizi word image 420

= —ihLyLz — word image 422

word image 423

word image 425 – ihiy Lz

= Lz – +ihLyiz

word image 428

where the commutation relations in Problem 4—12 have been applied. Combining the three comutation relations between the square of the operator of each angular momentum component and Lx,

word image 430

2

word image 431

Similarly, F, = 0 because

word image 433

word image 435

and L2 , L z = 0 because

word image 438

word image 439

Since L2 commutes with the operator of each angular momentum component, the corresponding observable quantities for these operators have simultaneously well-defined values. It is therefore possible to measure the square of angular momentum and one of its components simultaneously to arbitrary precision.

4—16. Can the position and kinetic energy of an electron be measured simultaneously to arbitrary precision? (See Problem 4-42.)

Consider first whether the kinetic energy and the x position of an electron can be measured simultaneously to arbitrary precision.

word image 441ix, x + iy,x + iz, x

The commutator Tx, x can be readily evaluated using the results from Problem 4—1 la:

The Postulates and General Principles of Quantum Mechanics

word image 443 2m Ox 2 ‘ h2 0 m Ox

The commutators T , x and Tz, x are both zero since Ty and Tz contain partial derivatives that do not involve x. Combining the three commutators,

Similarly, T, y # 0 and T, z #0. Therefore, it is not possible to measure the position and kinetic energy of any electron simultaneously to arbitrary precision.

4—19. Show that if A is Hermitian, then Ä — ( a ) is Hermitian. Show that the sum of two Hermitian operators is Hermitian.

To show that Ä — ( a ) is Hermitian, show that the two expressions f* (A — a ) gdx

and fX g Ä — ( a ) ) * f* dx are equal.

  

f* word image 448 gdx

f*Äg dx

f* ( a )g dx

Since A is Hermitian, it follows that f*Äg dx gÄ*f* dx

Furthermore, ( a ) is real; thus, ( a ) * = ( a ). Finally, the order of multiplication among functions is irrelevant. Thus,

f* (Ä — ( a ) ) g dx word image 449 gÄ*f* dx g( a )*f* dx

word image 450 g (Ä — ( a ) ) * f* dx

Chapter 4

Indeed, Ä — ( a ) is Hermitian.

If B is also Hermitian, then

f *fig dx word image 453

It follows that

word image 456

word image 457

f*Ä*g dx +

gÄ*f* dx +

f* B* g* dx

word image 461

word image 464 g (Ä + i) * f* dx

Thus, the sum of two Hermitian operators is Hermitian.

4-22. Show that

1/4 —x2/2

word image 466

= ) 1/4xe¯x2/2

l/fl(x) (4/1T

word image 468 x2/2

IP2(x) = word image 470

are orthonormal over the interval —00 < x < 00.

First show that the functions are normalized, that is,

t/f*nl/fn dx

word image 474 e x dx

word image 477 dx

word image 478 word image 480 word image 483 word image 484 word image 487 dx

word image 490

1/4

word image 493

2

word image 495

1/2

word image 497 2 —x dx

2

word image 499

1/2

00

— I dx

word image 502

1/2

word image 507

Now show that the functions are orthogonal, that is, for m # n,

word image 510

dx

word image 511 1 4 1/4 2 dx = word image 513 xe—x d.v

IT 1/4

-o

word image 517 1

IPO* 1/12 dx = —1/4

The Postulates and General Principles of Quantum Mechanics

1/11* 1//2 dx = word image 5202 — 1

dx

word image 523 4 1/4

2x —x dx

The first and third integrals are simply evaluated as 0 because they involve odd functions integrated over an interval symmetric about x = 0.

4—28. A general state function, expressed in the form of a ket vector 1 ) , can be wntten as a superposition of the eigenstates I l), | 2)of an operator A with eigenvalues al, ’12, . . . (in other words, ÅI n ) = anl n )):

word image 526 | 1) word image 528 1 2) word image 532 I n)

word image 535 word image 538 word image 543 word image 545 n

Show that cn = ( n 1 ). This quantity is called the amplitude of measuring an if a measurement of A is made in the state 1 ). The probability of obtaining an is c*ncn. Show that I ) can be written as

1 4 ) = word image 549

n

Similarly, the corresponding bra vector of I ) can be written in terms Of the COrresponding bra vectors of the I n) as

word image 552

Show that c word image 556 = ( word image 558 n). Show that ( d) I can be written as

word image 564

Now show that if ( word image 566 I is normalized, then

word image 568

and use this result to argue that

word image 573

is a unit operator.

Since the summation variable is arbitrary,

1 4 ) word image 575

then

word image 579

word image 580

n

Using this relation,

word image 581 I n) n

word image 584

word image 589

Similarly,

173

The Postulates and General Principles of Quantum Mechanics

word image 590 ml

word image 594m l n )

word image 598

mn

n

Furthermore,

word image 600

word image 602

word image 605 word image 609 word image 613 word image 614 Combining the expressions for 1 ) and ( I and using the normalization relation,

word image 617

word image 621 1 4 )

word image 623

n

which can be written as

( 4 1 4 ) = ( 4 word image 627

n

For the equality to hold,

— 1

n

4—32. In this chapter, we learned that if Vn is an eigenfunction ofthe time-independent Schrödinger equation, then

Wn(x, t) = —iEnt/h

Show that if (x) and (x) are both stationary states of , then the state

word image 630 —i Emt/ h + word image 635 iEnt/h satisfies the time-dependent Schrödinger equation.

If (x) and (x) are both stationary states of h, then

fit//n(x) = Enl/fn(x)

To show that V(x, t) satisfies the time-dependent Schrödinger equation, show that word image 639 ih3 1V/öt are equal.

üW(.v, t) = word image 643

= cm Em (x)e iEmt/h + c E (x)e iEnt/h

word image 647 ih-— = ih—i E,nt/h + Cn t/fn—-e-iEnt/h at

word image 651 -iEn -i Ent/h

= ih Cm Von (X)h e

-i Ent/h

word image 655

iE,nt/h

+

word image 657

Indeed, V(x, t) satisfies the time-dependent Schrödinger equation.

word image 661 word image 664 4—33. Show that V(x, t) given by Equation 4.78 is normalized.

The wave function is given by Equation 4.78:

1 1/2

1 1/2

word image 665 e-iE2t/h

2rx

V(x, t)

e¯iElt/h • — TX +

Slll

a

Now show that it is normalized:

a

a

a

a

word image 670 I 1/21/2 2r.v

W* (x, t) dx = word image 674 eiElt/h sin eiE2t/h sm word image 678

1

    

a

2rrx

word image 680

 

sin

–—- dx + word image 684 word image 688

word image 690

sin — sin — dx

a

o

 

a

0

a a

word image 693 o 0 aa

x

I

word image 698

a

1/2

word image 701 i Elt/h word image 703

Sin a

1

a

1/2

word image 707

word image 710

2rx

i E2t/h sin word image 714

2nx a 2 2nx dx

—— sin word image 719 clx + sin ———

word image 722

o

a 0 a

As shown in Problem 3—23,

a nit x mrx a

sinsin ¯ dx = — (S

0 a a 2 nm

Thus,

V* (x, word image 724 word image 727 word image 731 word image 734 word image 740 word image 743

I

t) dx = word image 746

a

word image 748

a

 

a

2

2

3—10. Prove that if V(x) is a solution to the Schrödinger equation, then any constant times V(x) is also a solution.

If l/f(x) is a solution of the Schrödinger equation, then fid’ (x) = E V (x). The Hamiltonian operator, H, is a linear operator, so if c is a constant,

ficdf(x) = cfiV(x)

= cEV(x)

word image 389E [cV(x)]

where the last step follows because both c and E are numbers whose multiplication commutes. This shows that ct/J (x) is an eigenfunction of H with eigenvalue equal to E, and it follows that Cl/J(x) is also a solution to the Schrödinger equation.

3-28. Consider a particle of mass m in a one-dimensional box of length a. Its average energy is given by

1

(E) word image 392

2m

Because (p) = 0, (p ) = a 2 , wherea can be called the uncertainty in p. Using the uncertainty principle, show that the energy must be at least as large as h2 /8ma because ax, the uncertainty in x, cannot be larger than a.

Starting with Equation 3.43 and the inequality ax a,

The Schrödinger Equation and a Particle in a Box

word image 394 20

gives

2a word image 395

word image 397 Both sides are positive, so squaring preserves the inequality

h2

4a2 — Since (p2) = 0 2 and (E) (p 2)/2m, then

8ma2

C—l. Find the length of the vector v

Use Equation C.5:

word image 400word image 401word image 403word image 405word image 407word image 409word image 411word image 413word image 415word image 417= lvl -( 1word image 419

— [22 -k (— l) 2 + 32 word image 421

14

C—4. Show that the vectors u = 2 i — 4 j — 2 k and v = 3 i + 4 j — 5 k are orthogonal.

Find the scalar product of the two vectors using Equation C.9,

word image 424

word image 426

Since u • v = O, the two vectors are orthogonal. (See Problem C—3.)

C—6. Find the angle between the two vectors u = —i 2 j k and v j+2k

First find the scalar product using Equation C.9:

word image 427 = LlxVx + LCyVy + uzVz

word image 429 word image 432 —3

Then use Equation C.6,

lullvl cose = 3

word image 434 3

cos 0 =0.327

(1 1) 1/2 (9 + 1 + 4) 1 2 o = 1090

4—13. Defining

word image 436¯ + + L 2

z

show that L 2 commutes with each component separately. What does this result tell you about the ability to measure the square of the total angular momentum and its components simultaneously?

First consider the commutator L 2 L word image 437

The Postulates and General Principles of Quantum Mechanics

word image 440

Since L2x commutes with ix, the commutator F, is zero. Using, for example, ihL: — word image 442 ix, , the other two commutators in the last expression can be written as

word image 444

word image 445

= (Lynx y — ihizi word image 447

= —ihLyLz — word image 449

word image 452

word image 454 – ihiy Lz

= Lz – +ihLyiz

word image 455

where the commutation relations in Problem 4—12 have been applied. Combining the three comutation relations between the square of the operator of each angular momentum component and Lx,

word image 459

2

word image 462

Similarly, F, = 0 because

word image 465

word image 467

and L2 , L z = 0 because

word image 469

word image 472

Since L2 commutes with the operator of each angular momentum component, the corresponding observable quantities for these operators have simultaneously well-defined values. It is therefore possible to measure the square of angular momentum and one of its components simultaneously to arbitrary precision.

4—16. Can the position and kinetic energy of an electron be measured simultaneously to arbitrary precision? (See Problem 4-42.)

Consider first whether the kinetic energy and the x position of an electron can be measured simultaneously to arbitrary precision.

word image 475ix, x + iy,x + iz, x

The commutator Tx, x can be readily evaluated using the results from Problem 4—1 la:

The Postulates and General Principles of Quantum Mechanics

word image 477 2m Ox 2 ‘ h2 0 m Ox

The commutators T , x and Tz, x are both zero since Ty and Tz contain partial derivatives that do not involve x. Combining the three commutators,

Similarly, T, y # 0 and T, z #0. Therefore, it is not possible to measure the position and kinetic energy of any electron simultaneously to arbitrary precision.

4—19. Show that if A is Hermitian, then Ä — ( a ) is Hermitian. Show that the sum of two Hermitian operators is Hermitian.

To show that Ä — ( a ) is Hermitian, show that the two expressions f* (A — a ) gdx

and fX g Ä — ( a ) ) * f* dx are equal.

  

f* word image 479 gdx

f*Äg dx

f* ( a )g dx

Since A is Hermitian, it follows that f*Äg dx gÄ*f* dx

Furthermore, ( a ) is real; thus, ( a ) * = ( a ). Finally, the order of multiplication among functions is irrelevant. Thus,

f* (Ä — ( a ) ) g dx word image 482 gÄ*f* dx g( a )*f* dx

word image 483 g (Ä — ( a ) ) * f* dx

Chapter 4

Indeed, Ä — ( a ) is Hermitian.

If B is also Hermitian, then

f *fig dx word image 485

It follows that

word image 488

word image 489

f*Ä*g dx +

gÄ*f* dx +

f* B* g* dx

word image 493

word image 495 g (Ä + i) * f* dx

Thus, the sum of two Hermitian operators is Hermitian.

4-22. Show that

1/4 —x2/2

word image 496

= ) 1/4xe¯x2/2

l/fl(x) (4/1T

word image 499 x2/2

IP2(x) = word image 501

are orthonormal over the interval —00 < x < 00.

First show that the functions are normalized, that is,

t/f*nl/fn dx

word image 505 e x dx

word image 510 dx

word image 511 word image 514 word image 518 word image 522 word image 525 dx

word image 530

1/4

word image 534

2

word image 537

1/2

word image 540 2 —x dx

2

word image 544

1/2

00

— I dx

word image 547

1/2

word image 550

Now show that the functions are orthogonal, that is, for m # n,

word image 554

dx

word image 556 1 4 1/4 2 dx = word image 560 xe—x d.v

IT 1/4

-o

word image 563 1

IPO* 1/12 dx = —1/4

The Postulates and General Principles of Quantum Mechanics

1/11* 1//2 dx = word image 5672 — 1

dx

word image 570 4 1/4

2x —x dx

The first and third integrals are simply evaluated as 0 because they involve odd functions integrated over an interval symmetric about x = 0.

4—28. A general state function, expressed in the form of a ket vector 1 ) , can be wntten as a superposition of the eigenstates I l), | 2)of an operator A with eigenvalues al, ’12, . . . (in other words, ÅI n ) = anl n )):

word image 572 | 1) word image 574 1 2) word image 578 I n)

word image 580 word image 582 word image 586 word image 588 n

Show that cn = ( n 1 ). This quantity is called the amplitude of measuring an if a measurement of A is made in the state 1 ). The probability of obtaining an is c*ncn. Show that I ) can be written as

1 4 ) = word image 592

n

Similarly, the corresponding bra vector of I ) can be written in terms Of the COrresponding bra vectors of the I n) as

word image 595

Show that c word image 597 = ( word image 600 n). Show that ( d) I can be written as

word image 603

Now show that if ( word image 607 I is normalized, then

word image 610

and use this result to argue that

word image 613

is a unit operator.

Since the summation variable is arbitrary,

1 4 ) word image 614

then

word image 616

word image 620

n

Using this relation,

word image 623 I n) n

word image 629

word image 632

Similarly,

173

The Postulates and General Principles of Quantum Mechanics

word image 635 ml

word image 639m l n )

word image 642

mn

n

Furthermore,

word image 646

word image 651

word image 656 word image 658 word image 662 word image 664 Combining the expressions for 1 ) and ( I and using the normalization relation,

word image 667

word image 671 1 4 )

word image 676

n

which can be written as

( 4 1 4 ) = ( 4 word image 677

n

For the equality to hold,

— 1

n

4—32. In this chapter, we learned that if Vn is an eigenfunction ofthe time-independent Schrödinger equation, then

Wn(x, t) = —iEnt/h

Show that if (x) and (x) are both stationary states of , then the state

word image 682 —i Emt/ h + word image 685 iEnt/h satisfies the time-dependent Schrödinger equation.

If (x) and (x) are both stationary states of h, then

fit//n(x) = Enl/fn(x)

To show that V(x, t) satisfies the time-dependent Schrödinger equation, show that word image 688 ih3 1V/öt are equal.

üW(.v, t) = word image 692

= cm Em (x)e iEmt/h + c E (x)e iEnt/h

word image 695 ih-— = ih—i E,nt/h + Cn t/fn—-e-iEnt/h at

word image 700 -iEn -i Ent/h

= ih Cm Von (X)h e

-i Ent/h

word image 702

iE,nt/h

+

word image 704

Indeed, V(x, t) satisfies the time-dependent Schrödinger equation.

word image 708 word image 711 4—33. Show that V(x, t) given by Equation 4.78 is normalized.

The wave function is given by Equation 4.78:

1 1/2

1 1/2

word image 715 e-iE2t/h

2rx

V(x, t)

e¯iElt/h • — TX +

Slll

a

Now show that it is normalized:

a

a

a

a

word image 718 I 1/21/2 2r.v

W* (x, t) dx = word image 722 eiElt/h sin eiE2t/h sm word image 723

1

    

a

2rrx

word image 728

 

sin

–—- dx + word image 732 word image 735

word image 739

sin — sin — dx

a

o

 

a

0

a a

word image 743 o 0 aa

x

I

word image 747

a

1/2

word image 750 i Elt/h word image 751

Sin a

1

a

1/2

word image 755

word image 761

2rx

i E2t/h sin word image 762

2nx a 2 2nx dx

—— sin word image 765 clx + sin ———

word image 768

o

a 0 a

As shown in Problem 3—23,

a nit x mrx a

sinsin ¯ dx = — (S

0 a a 2 nm

Thus,

V* (x, word image 771 word image 774 word image 776 word image 777 word image 780 word image 781

I

t) dx = word image 784

a

word image 786

a

 

a

2

2

3—10. Prove that if V(x) is a solution to the Schrödinger equation, then any constant times V(x) is also a solution.

If l/f(x) is a solution of the Schrödinger equation, then fid’ (x) = E V (x). The Hamiltonian operator, H, is a linear operator, so if c is a constant,

ficdf(x) = cfiV(x)

= cEV(x)

word image 446E [cV(x)]

where the last step follows because both c and E are numbers whose multiplication commutes. This shows that ct/J (x) is an eigenfunction of H with eigenvalue equal to E, and it follows that Cl/J(x) is also a solution to the Schrödinger equation.

3-28. Consider a particle of mass m in a one-dimensional box of length a. Its average energy is given by

1

(E) word image 449

2m

Because (p) = 0, (p ) = a 2 , wherea can be called the uncertainty in p. Using the uncertainty principle, show that the energy must be at least as large as h2 /8ma because ax, the uncertainty in x, cannot be larger than a.

Starting with Equation 3.43 and the inequality ax a,

The Schrödinger Equation and a Particle in a Box

word image 451 20

gives

2a word image 454

word image 455 Both sides are positive, so squaring preserves the inequality

h2

4a2 — Since (p2) = 0 2 and (E) (p 2)/2m, then

8ma2

C—l. Find the length of the vector v

Use Equation C.5:

word image 458word image 460word image 463word image 466word image 468word image 471word image 473word image 476word image 478word image 481= lvl -( 1word image 483

— [22 -k (— l) 2 + 32 word image 484

14

C—4. Show that the vectors u = 2 i — 4 j — 2 k and v = 3 i + 4 j — 5 k are orthogonal.

Find the scalar product of the two vectors using Equation C.9,

word image 486

word image 491

Since u • v = O, the two vectors are orthogonal. (See Problem C—3.)

C—6. Find the angle between the two vectors u = —i 2 j k and v j+2k

First find the scalar product using Equation C.9:

word image 492 = LlxVx + LCyVy + uzVz

word image 494 word image 498 —3

Then use Equation C.6,

lullvl cose = 3

word image 499 3

cos 0 =0.327

(1 1) 1/2 (9 + 1 + 4) 1 2 o = 1090

4—13. Defining

word image 501¯ + + L 2

z

show that L 2 commutes with each component separately. What does this result tell you about the ability to measure the square of the total angular momentum and its components simultaneously?

First consider the commutator L 2 L word image 504

The Postulates and General Principles of Quantum Mechanics

word image 509

Since L2x commutes with ix, the commutator F, is zero. Using, for example, ihL: — word image 512 ix, , the other two commutators in the last expression can be written as

word image 515

word image 517

= (Lynx y — ihizi word image 519

= —ihLyLz — word image 521

word image 525

word image 529 – ihiy Lz

= Lz – +ihLyiz

word image 533

where the commutation relations in Problem 4—12 have been applied. Combining the three comutation relations between the square of the operator of each angular momentum component and Lx,

word image 537

2

word image 541

Similarly, F, = 0 because

word image 544

word image 548

and L2 , L z = 0 because

word image 551

word image 555

Since L2 commutes with the operator of each angular momentum component, the corresponding observable quantities for these operators have simultaneously well-defined values. It is therefore possible to measure the square of angular momentum and one of its components simultaneously to arbitrary precision.

4—16. Can the position and kinetic energy of an electron be measured simultaneously to arbitrary precision? (See Problem 4-42.)

Consider first whether the kinetic energy and the x position of an electron can be measured simultaneously to arbitrary precision.

word image 560ix, x + iy,x + iz, x

The commutator Tx, x can be readily evaluated using the results from Problem 4—1 la:

The Postulates and General Principles of Quantum Mechanics

word image 562 2m Ox 2 ‘ h2 0 m Ox

The commutators T , x and Tz, x are both zero since Ty and Tz contain partial derivatives that do not involve x. Combining the three commutators,

Similarly, T, y # 0 and T, z #0. Therefore, it is not possible to measure the position and kinetic energy of any electron simultaneously to arbitrary precision.

4—19. Show that if A is Hermitian, then Ä — ( a ) is Hermitian. Show that the sum of two Hermitian operators is Hermitian.

To show that Ä — ( a ) is Hermitian, show that the two expressions f* (A — a ) gdx

and fX g Ä — ( a ) ) * f* dx are equal.

  

f* word image 565 gdx

f*Äg dx

f* ( a )g dx

Since A is Hermitian, it follows that f*Äg dx gÄ*f* dx

Furthermore, ( a ) is real; thus, ( a ) * = ( a ). Finally, the order of multiplication among functions is irrelevant. Thus,

f* (Ä — ( a ) ) g dx word image 570 gÄ*f* dx g( a )*f* dx

word image 571 g (Ä — ( a ) ) * f* dx

Chapter 4

Indeed, Ä — ( a ) is Hermitian.

If B is also Hermitian, then

f *fig dx word image 577

It follows that

word image 579

word image 580

f*Ä*g dx +

gÄ*f* dx +

f* B* g* dx

word image 581

word image 587 g (Ä + i) * f* dx

Thus, the sum of two Hermitian operators is Hermitian.

4-22. Show that

1/4 —x2/2

word image 588

= ) 1/4xe¯x2/2

l/fl(x) (4/1T

word image 590 x2/2

IP2(x) = word image 593

are orthonormal over the interval —00 < x < 00.

First show that the functions are normalized, that is,

t/f*nl/fn dx

word image 597 e x dx

word image 600 dx

word image 601 word image 606 word image 608 word image 611 word image 614 dx

word image 618

1/4

word image 621

2

word image 624

1/2

word image 628 2 —x dx

2

word image 631

1/2

00

— I dx

word image 637

1/2

word image 641

Now show that the functions are orthogonal, that is, for m # n,

word image 644

dx

word image 648 1 4 1/4 2 dx = word image 652 xe—x d.v

IT 1/4

-o

word image 654 1

IPO* 1/12 dx = —1/4

The Postulates and General Principles of Quantum Mechanics

1/11* 1//2 dx = word image 6592 — 1

dx

word image 663 4 1/4

2x —x dx

The first and third integrals are simply evaluated as 0 because they involve odd functions integrated over an interval symmetric about x = 0.

4—28. A general state function, expressed in the form of a ket vector 1 ) , can be wntten as a superposition of the eigenstates I l), | 2)of an operator A with eigenvalues al, ’12, . . . (in other words, ÅI n ) = anl n )):

word image 664 | 1) word image 665 1 2) word image 670 I n)

word image 675 word image 677 word image 679 word image 686 n

Show that cn = ( n 1 ). This quantity is called the amplitude of measuring an if a measurement of A is made in the state 1 ). The probability of obtaining an is c*ncn. Show that I ) can be written as

1 4 ) = word image 687

n

Similarly, the corresponding bra vector of I ) can be written in terms Of the COrresponding bra vectors of the I n) as

word image 691

Show that c word image 694 = ( word image 699 n). Show that ( d) I can be written as

word image 701

Now show that if ( word image 703 I is normalized, then

word image 709

and use this result to argue that

word image 711

is a unit operator.

Since the summation variable is arbitrary,

1 4 ) word image 717

then

word image 720

word image 722

n

Using this relation,

word image 725 I n) n

word image 728

word image 732

Similarly,

173

The Postulates and General Principles of Quantum Mechanics

word image 738 ml

word image 742m l n )

word image 743

mn

n

Furthermore,

word image 747

word image 750

word image 753 word image 756 word image 760 word image 762 Combining the expressions for 1 ) and ( I and using the normalization relation,

word image 766

word image 769 1 4 )

word image 771

n

which can be written as

( 4 1 4 ) = ( 4 word image 775

n

For the equality to hold,

— 1

n

4—32. In this chapter, we learned that if Vn is an eigenfunction ofthe time-independent Schrödinger equation, then

Wn(x, t) = —iEnt/h

Show that if (x) and (x) are both stationary states of , then the state

word image 776 —i Emt/ h + word image 779 iEnt/h satisfies the time-dependent Schrödinger equation.

If (x) and (x) are both stationary states of h, then

fit//n(x) = Enl/fn(x)

To show that V(x, t) satisfies the time-dependent Schrödinger equation, show that word image 780 ih3 1V/öt are equal.

üW(.v, t) = word image 783

= cm Em (x)e iEmt/h + c E (x)e iEnt/h

word image 785 ih-— = ih—i E,nt/h + Cn t/fn—-e-iEnt/h at

word image 787 -iEn -i Ent/h

= ih Cm Von (X)h e

-i Ent/h

word image 789

iE,nt/h

+

word image 791

Indeed, V(x, t) satisfies the time-dependent Schrödinger equation.

word image 793 word image 795 4—33. Show that V(x, t) given by Equation 4.78 is normalized.

The wave function is given by Equation 4.78:

1 1/2

1 1/2

word image 799 e-iE2t/h

2rx

V(x, t)

e¯iElt/h • — TX +

Slll

a

Now show that it is normalized:

a

a

a

a

word image 803 I 1/21/2 2r.v

W* (x, t) dx = word image 805 eiElt/h sin eiE2t/h sm word image 809

1

    

a

2rrx

word image 810

 

sin

–—- dx + word image 812 word image 814

word image 817

sin — sin — dx

a

o

 

a

0

a a

word image 820 o 0 aa

x

I

word image 822

a

1/2

word image 824 i Elt/h word image 827

Sin a

1

a

1/2

word image 830

word image 832

2rx

i E2t/h sin word image 834

2nx a 2 2nx dx

—— sin word image 837 clx + sin ———

word image 840

o

a 0 a

As shown in Problem 3—23,

a nit x mrx a

sinsin ¯ dx = — (S

0 a a 2 nm

Thus,

V* (x, word image 842 word image 846 word image 850 word image 852 word image 854 word image 857

I

t) dx = word image 858

a

word image 861

a

 

a

2

2

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