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)

# E [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)

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 h^{2 }/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

20

gives

2a

Both sides are positive, so squaring preserves the inequality

h2

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

8ma2

C—l. Find the length of the vector v

Use Equation C.5:

# = lvl -( 1

— [22 -k (— l) 2 + 32

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,

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:

= LlxVx + LCyVy + uzVz

—3

Then use Equation C.6,

lullvl cose = 3

3

cos 0 =0.327

(1 1) ^{1}/^{2 }(9 + 1 + 4) ^{1 2 }o = 109^{0}

4—13. Defining

## ¯ + + 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

The Postulates and General Principles of Quantum Mechanics

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

= (Lynx y — ihizi

= —ihLyLz —

– ihiy Lz

= Lz – +ihLyiz

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,

^{2}

Similarly, F, = 0 because

and L^{2 }, L z = 0 because

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.

## ix, 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

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* 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 gÄ*f* dx g( a )*f* dx

g (Ä — ( a ) ) * f* dx

Chapter 4

Indeed, Ä — ( a ) is Hermitian.

If B is also Hermitian, then

f *fig dx

It follows that

f*Ä*g dx + gÄ*f* dx + | f* B* g* dx |

g (Ä + i) * f* dx

Thus, the sum of two Hermitian operators is Hermitian.

4-22. Show that

1/4 —x2/2

= ) 1/4xe¯x2/2

l/fl(x) (4/1T

x2/2

IP2(x) =

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

First show that the functions are normalized, that is, | t/f*nl/fn dx |

e ^{x }dx

dx

dx

1/4

2

1/2

2 —x dx

2

1/2

00

— I dx

1/2

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

dx

1 4 1/4 2 dx = xe—x d.v

IT 1/4

-o

1

IPO* 1/12 dx = —1/4

The Postulates and General Principles of Quantum Mechanics

1/11* 1//2 dx = | dx |

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 )):

| 1) 1 2) I n)

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 ) =

n

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

Show that c = ( n). Show that ( d) I can be written as

Now show that if ( I is normalized, then

and use this result to argue that

is a unit operator.

Since the summation variable is arbitrary,

1 4 )

then

n

Using this relation,

I n) n

Similarly,

173

The Postulates and General Principles of Quantum Mechanics

ml

^{m }l n )

mn

n

Furthermore,

Combining the expressions for 1 ) and ( I and using the normalization relation,

1 4 )

n

which can be written as

( 4 1 4 ) = ( 4

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

—i Emt/ ^{h }+ 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 ih3 ^{1}V/öt are equal.

üW(.v, t) =

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

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

-iEn -i Ent/h

= ih Cm Von (X)h e

-i Ent/h

iE,nt/h

+

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

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 e-iE2t/h | 2rx | |

V(x, t) | e¯iElt/h • — TX + Slll | ||

a Now show that it is normalized: a | a | a | a |

I 1/21/2 2r.v

W* (x, t) dx = eiElt/h sin eiE2t/h sm

1 | a 2rrx | ||||

sin | –—- dx + | sin — sin — dx | |||

a | o | a | 0 | a a |

o 0 aa

x | I a | 1/2 | i Elt/h Sin a | 1 a | 1/2 | 2rx i E2t/h sin |

2nx a 2 2nx dx

—— sin clx + sin ———

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, | I t) dx = | a | 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)

# E [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)

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 h^{2 }/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

20

gives

2a

Both sides are positive, so squaring preserves the inequality

h2

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

8ma2

C—l. Find the length of the vector v

Use Equation C.5:

# = lvl -( 1

— [22 -k (— l) 2 + 32

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,

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:

= LlxVx + LCyVy + uzVz

—3

Then use Equation C.6,

lullvl cose = 3

3

cos 0 =0.327

(1 1) ^{1}/^{2 }(9 + 1 + 4) ^{1 2 }o = 109^{0}

4—13. Defining

## ¯ + + 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

The Postulates and General Principles of Quantum Mechanics

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

= (Lynx y — ihizi

= —ihLyLz —

– ihiy Lz

= Lz – +ihLyiz

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,

^{2}

Similarly, F, = 0 because

and L^{2 }, L z = 0 because

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.

## ix, 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

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* 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 gÄ*f* dx g( a )*f* dx

g (Ä — ( a ) ) * f* dx

Chapter 4

Indeed, Ä — ( a ) is Hermitian.

If B is also Hermitian, then

f *fig dx

It follows that

f*Ä*g dx + gÄ*f* dx + | f* B* g* dx |

g (Ä + i) * f* dx

Thus, the sum of two Hermitian operators is Hermitian.

4-22. Show that

1/4 —x2/2

= ) 1/4xe¯x2/2

l/fl(x) (4/1T

x2/2

IP2(x) =

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

First show that the functions are normalized, that is, | t/f*nl/fn dx |

e ^{x }dx

dx

dx

1/4

2

1/2

2 —x dx

2

1/2

00

— I dx

1/2

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

dx

1 4 1/4 2 dx = xe—x d.v

IT 1/4

-o

1

IPO* 1/12 dx = —1/4

The Postulates and General Principles of Quantum Mechanics

1/11* 1//2 dx = | dx |

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 )):

| 1) 1 2) I n)

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 ) =

n

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

Show that c = ( n). Show that ( d) I can be written as

Now show that if ( I is normalized, then

and use this result to argue that

is a unit operator.

Since the summation variable is arbitrary,

1 4 )

then

n

Using this relation,

I n) n

Similarly,

173

The Postulates and General Principles of Quantum Mechanics

ml

^{m }l n )

mn

n

Furthermore,

Combining the expressions for 1 ) and ( I and using the normalization relation,

1 4 )

n

which can be written as

( 4 1 4 ) = ( 4

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

—i Emt/ ^{h }+ 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 ih3 ^{1}V/öt are equal.

üW(.v, t) =

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

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

-iEn -i Ent/h

= ih Cm Von (X)h e

-i Ent/h

iE,nt/h

+

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

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 e-iE2t/h | 2rx | |

V(x, t) | e¯iElt/h • — TX + Slll | ||

a Now show that it is normalized: a | a | a | a |

I 1/21/2 2r.v

W* (x, t) dx = eiElt/h sin eiE2t/h sm

1 | a 2rrx | ||||

sin | –—- dx + | sin — sin — dx | |||

a | o | a | 0 | a a |

o 0 aa

x | I a | 1/2 | i Elt/h Sin a | 1 a | 1/2 | 2rx i E2t/h sin |

2nx a 2 2nx dx

—— sin clx + sin ———

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, | I t) dx = | a | 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)

# E [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)

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 h^{2 }/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

20

gives

2a

Both sides are positive, so squaring preserves the inequality

h2

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

8ma2

C—l. Find the length of the vector v

Use Equation C.5:

# = lvl -( 1

— [22 -k (— l) 2 + 32

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,

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:

= LlxVx + LCyVy + uzVz

—3

Then use Equation C.6,

lullvl cose = 3

3

cos 0 =0.327

(1 1) ^{1}/^{2 }(9 + 1 + 4) ^{1 2 }o = 109^{0}

4—13. Defining

## ¯ + + 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

The Postulates and General Principles of Quantum Mechanics

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

= (Lynx y — ihizi

= —ihLyLz —

– ihiy Lz

= Lz – +ihLyiz

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,

^{2}

Similarly, F, = 0 because

and L^{2 }, L z = 0 because

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.

## ix, 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

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* 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 gÄ*f* dx g( a )*f* dx

g (Ä — ( a ) ) * f* dx

Chapter 4

Indeed, Ä — ( a ) is Hermitian.

If B is also Hermitian, then

f *fig dx

It follows that

f*Ä*g dx + gÄ*f* dx + | f* B* g* dx |

g (Ä + i) * f* dx

Thus, the sum of two Hermitian operators is Hermitian.

4-22. Show that

1/4 —x2/2

= ) 1/4xe¯x2/2

l/fl(x) (4/1T

x2/2

IP2(x) =

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

First show that the functions are normalized, that is, | t/f*nl/fn dx |

e ^{x }dx

dx

dx

1/4

2

1/2

2 —x dx

2

1/2

00

— I dx

1/2

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

dx

1 4 1/4 2 dx = xe—x d.v

IT 1/4

-o

1

IPO* 1/12 dx = —1/4

The Postulates and General Principles of Quantum Mechanics

1/11* 1//2 dx = | dx |

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 )):

| 1) 1 2) I n)

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 ) =

n

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

Show that c = ( n). Show that ( d) I can be written as

Now show that if ( I is normalized, then

and use this result to argue that

is a unit operator.

Since the summation variable is arbitrary,

1 4 )

then

n

Using this relation,

I n) n

Similarly,

173

The Postulates and General Principles of Quantum Mechanics

ml

^{m }l n )

mn

n

Furthermore,

Combining the expressions for 1 ) and ( I and using the normalization relation,

1 4 )

n

which can be written as

( 4 1 4 ) = ( 4

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

—i Emt/ ^{h }+ 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 ih3 ^{1}V/öt are equal.

üW(.v, t) =

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

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

-iEn -i Ent/h

= ih Cm Von (X)h e

-i Ent/h

iE,nt/h

+

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

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 e-iE2t/h | 2rx | |

V(x, t) | e¯iElt/h • — TX + Slll | ||

a Now show that it is normalized: a | a | a | a |

I 1/21/2 2r.v

W* (x, t) dx = eiElt/h sin eiE2t/h sm

1 | a 2rrx | ||||

sin | –—- dx + | sin — sin — dx | |||

a | o | a | 0 | a a |

o 0 aa

x | I a | 1/2 | i Elt/h Sin a | 1 a | 1/2 | 2rx i E2t/h sin |

2nx a 2 2nx dx

—— sin clx + sin ———

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, | I t) dx = | a | a |

a | 2 | 2 |