Physics 701 - Solutions 3

Department of Physics Due Feb 17, 2010University of New Hampshire

Reading: Please read 2.1, 2.2 and 2.3 in Griffiths.

1 General statements about solutions to the Schrodinger

Equation

[10] Problem 2.3 in Griffiths, reworded see also problems 2.1 and 2.2 in theextra section.By explicitly solving the Schrodinger equation for the infinite square well potential,show that there are no acceptable solutions to the time-independent Schrodingerequation for E = 0 and E < 0. Since V = 0, the implication is that there are nosolutions when E V everywhere.Solution: When E = 0 the time-independent Schrodinger equation becomes:

d2

dx2(x) = 0 (1)

which can be integrated twice to result in a solution:

(x) = A+Bx (2)

where A and B are constants. The boundary conditions require (0) = 0, so wemust have A = 0 and (a) = 0, so Ba = 0 and thus B = 0. This results in a wavefunction psi(x) = 0, which cannot be normalized, hence E = 0 is not an allowedenergy.

For E < 0, we solve the time-independent Schrodinger equation to get:

h22m

d2

dx2(x) = E(x)

d2

dx2(x) = k2(x), with k =

2mEh

, real (3)

The solution for this differential equation is then:

(x) = Ae+kx +Bekx (4)

1

The boundary conditions result in: (x) = 0 Ae+0 +Be0 = 0 B = A and(a) = 0 A (e+ka eka) = 0. We thus get either A = 0, but then (x) = 0everywhere, or we get e+ka eka = 0 +ka = ka k = 0, but then again(x) = 0, which cannot be normalized.

We conclude that E 0 is not allowed for this system.

2 Infinite Square Well Problems

(1) [10] Problem 2.4Calculate x, x2, p, p2, x and p, for the nth stationary state (energyeigenstate) of the infinite square well. Check that the uncertainty principleis satisfied for all n. Which state comes the closest to the uncertainty limit?For each of your answers, check to make sure you have the correct dimension(units).Solution: We use the general solution for a stationary state:

n(x) =

2

asin(npiax)

(5)

We now compute:

x = n |x|n = a

0

nxndx =2

a

a0

x sin2(npiax)dx (6)

The function we integrate over is odd, but the limits are not symmetric, so wecannot immediate make any immediate conclusions. Use an integral table,or the trig identity: sin2 = 1

2 1

2cos(2) and then integrate by parts on the

x cos(2npiax) term using u = x and dv = cos(2npi

ax), so v = a

2npisin(2npi

ax). We

get:

x = a

0

(x

a xa

cos(2npi

ax)

)dx

=

[1

a

1

2x2]a

0

1a

a0

a

2npisin(

2npi

ax)dx+

[x

2npisin(

2npi

ax)

]a0

=a

2+

[a

4n2pi2cos(

2npi

ax)

]a0

=a

2+

a

4n2pi2(1 cos(2npi))

=a

2(7)

2

The result has units of length and is between 0 and a.The integral for x2is quite similar, except that you have to integrate by parts twice:

x2

=nx2n = a

0

nx2ndx =

2

a

a0

x2 sin2(npiax)dx

=

a0

(x2

a x

2

acos(

2npi

ax)

)dx

=

[x3

3a

]a0

+

a0

2x

a

a

2npisin(

2npi

ax)dx

[x2

npisin(

2npi

ax)

]a0

=a2

3 a

0

1

npi

a2npi

cos(2npi

ax) +

[ ax2n2pi2

cos(2npi

ax)

]a0

=a2

3+

ax

2n2pi2

[a

2npisin(

2npi

ax)

]a0

a2

2n2pi2cos(2npi)

=a2

3 a

2

2n2pi2= a2

(1

3 1

2n2pi2

)(8)

This result has units of length squared, as expected, and is larger than x2for all n. Now we know we can do p the fast way:

p = m tx = 0 (9)

The units here are not important since the result is zero.

What about p2? Since we know the energy E of the wave functions we cantake a shortcut:

E | = H | E =

p22m + V = 12m p2+ V

p2

= 2m (E V )(10)

In our case the potential is zero, so

p2

= 2m E = n2pi2h2

a2(11)

3

The long, but reliable, method for finding the same is:

p2

= h2 2a

a0

sin(npi

ax)

d2

dx2sin(

npi

ax)dx

=2h2a

a0

sin(npi

ax)n2pi2a2

sin(npi

ax)dx

=n2h2pi2

a2

a0

(x)(x)dx

=n2pi2h2

a2(12)

To check the units we reduce everything to dimensions of combinations ofmass, length and time. Since h has units of Joules times seconds, h/a hasunits J s/m = N m s/m = kg m/s2 s = kg m/s, which are indeed theunits for momentum (remember classically p = mv). Note that the integralthat needed to be evaluated is just the normalization integral.

We then have:

x = a

1

3 1

2n2pi2 1

4= a

1

12 1

2n2pi2=a

2

1

3 2n2pi2

p =npih

a

xp =npih

a

a

2

1

3 2n2pi2

=h

2

n2pi2

3 2 (13)

The uncertainty principle holds for all n and the ground state comes closestto the uncertainty limit.

(2) Problem 2.5A particle in the infinite square well of width a has as its initial wave functionan even mixture of the first two energy eigenstates:

(x, 0) = A [1(x) + 2(x)] (14)

where n(x) are the properly normalized solutions of the time-independentSchrodinger equation for the infinite square well potential.

a. [5] Normalize (x, 0) without performing any actual integral.Solution: We know the n wave functions are orthonormal (since they

4

are solutions to an eigen equation) so:

n |m =

n(x)m(x)dx = nm (15)

then:

| = AA (1|+ 2|) (|1+ |2)= |A|2 (1 |1 + 1 |2 + 2 |1 + 2 |2)= |A|2 2 = 1 A = 1

2(16)

In integral notation:

| =

(x, 0)(x, 0)dx

= AA

(1(x) + 2(x)) (1(x) + 2(x)) dx

= |A|2

(1(x)1(x) + 2(x)1(x) +

1(x)2(x) +

2(x)2(x)) dx

= |A|2 2 = 1 A = 12

(17)

b. [5] Find (x, t) and |(x, t)|2. Express the latter as a sinusoidal func-tion of time. To simplify your result, let = pi2h/2ma2.Solution: Normally (see problem 2.7) you need to do an integral to getthe coefficients of the wave function expansion:

|(t) =n

cn |n eiEnt/h

(x, t) =n

cnn(x)eiEnt/h (18)

where the coefficients are given by:

cn = n |(0) =

n(x)(x, 0)dx (19)

But in this case, you can just match the cn to equation 14 and imme-diately write: c1 = c2 = A and all other cn are zero. Setting =

pi2h2ma2

,

5

and using the energy levels En = n2h we can thus write:

(x, t) =121(x)e

iE1t/h +121(x)e

iE1t/h

=

1

asin(piax)eit +

1

asin

(2pi

ax

)ei4t (20)

When we compute ||2 we need to be careful about the cross terms,which do not drop out (until you integrate over them). We get:

|(x, t)|2 = 1a

(sin(piax)eit + sin

(2pi

ax

)ei4t

)(

sin(piax)eit + sin

(2pi

ax

)ei4t

)=

1

a

{sin2

(piax)

+ sin2(

2pi

ax

)+ sin

(piax)

sin

(2pi

ax

)(ei3t + ei3t

)}=

1

a

{sin2

(piax)

+ sin2(

2pi

ax

)+ 2 sin

(piax)

sin

(2pi

ax

)cos(3t)

}(21)

c. [5] Compute x and find the angular frequency and amplitude of theoscillation.Solution:

x = a

0

x |(x, t)|2 dx

=

a0

1

a

{x sin2

(piax)

+ x sin2(

2pi

ax

)+ 2x sin

(piax)

sin

(2pi

ax

)cos(3t)

}(22)

The first two integrals we have done already, together they equate to a2.

The last integral requires: a0

x sin(piax)

sin

(2pi

ax

)dx = a

0

x

[cos(piax) cos

(3pi

ax

)]= 16a

2

9pi2

(23)

6

So:

x = a2 16a

9pi2cos(3t) (24)

The angular frequency is 3 = 3pi2h

2ma2, the amplitude is 16a

9pi2 a

20.3603...

so less than a2.

d. [5] Compute p. (No integration needed! If you want to, do the integraland very this.)Solution: We can just take the derivative with respect to time of theprevious answer:

p = m xx

=16ma

3pi2sin(3t)

=8h

3asin(3t) (25)

e. [5] If you measured the energy of this particle, what values might youget and what is the probability of getting these values? Compute theexpectation value H and comment on the result. (How does it com-pare with E1 and E2).

Solution: A measurement of the energy would results in either E1 =pi2h2

2ma2

or in E2 =2pi2h2

ma2with equal probability: P1 = P2 =

12. The expectation

value for E is then:

E =n

|cn|En = 12

(1 + 4)pi2h2

2ma2=

5pi2h2

4ma2(26)

This result is right in between E1 and E2, as you would expect for anequal weighted average.

(3) [10] Problem 2.6Although the overall phase (embedded in the constant A) of the wave func-tion is of no physical significance, the relative phase of the coefficients inequation 2.17, does matter. For example, suppose we change the relativephase in the previous problem to:

(x, 0) = A[1(x) + e

i2(x)]

(27)

7

where is a constant. Find (x, t), |(x, t)|2 and x and compare youranswer with what you got before. Study the special cases of = 0, = pi/2and = pi.Solution: The exponential term drops out of the normalization integral, so

we still have A =

12. Now:

(x, t) =

1

asin(piax)eit +

1

aei sin

(2pi

ax

)ei4t

|(x, t)|2 = 1a

{sin2

(piax)

+ sin2(

2pi

ax

)+ sin

(piax)

sin

(2pi

ax

)(ei3t+i + ei3ti

)}=

1

a

{sin2

(piax)

+ sin2(

2pi

ax

)+ 2 sin

(piax)

sin

(2pi

ax

)cos(3t )

}(28)

So then:

x = a2 16a

9pi2cos(3t )

(29)

For = 0 we get our previous result. For = pi/2 we get x = a2

16a9pi2

sin(3t), so x starts at a2

instead of a2 16a

9pi2.

For = pi we get x = a2

+ 16a9pi2

cos(3t), so x starts at a2

+ 16a9pi2

instead ofa2 16a

9pi2.

3 Extra Problems

This extra problem section is for those of you who want to dig a little more deeplyinto the material. These problems are a bit more difficult. Sometimes they maybe a bit too advanced to expect everyone to do them. Thus, they are not required,but I encourage you to think about them, and then check the solutions when theybecome available.

(1) Problem 2.1 in Griffiths.

8

a. Suppose E is not real, then the solution to the time dependence of(x, t) has an energy term E = Er + iEi where now Er and Ei are bothreal. We then get:

|(t) = | ei(Er+iEi)t/h = | eEit/heiErt/hbar (30)The normalization then leads to:

| = e2Eit/hbar | = 1 (31)Since | = 1 for all time t, we must have e2Eit/hbar = 1 Ei = 0.

b. If satisfies the time independent Schrodinger equation: h22m

2

x2 +

V = E, then, noting that V , E and m are real, the complex conju-gate gives: h2

2m2

x2 + V = E so that both and satisfy the

Schrodinger equation for the same energy E. Then any linear combina-tion of and also satisfy the Schrodinger equation (with the sameenergy E) Check for yourself by inserting A = c1 + c2

. Thus thecombinations (+) and i(), both of which are real, also satisfythe Schrodinger equation. So from any complex solution we can alwaysconstruct two real solutions.

c. If (x) satisfies the time independent Schrodinger equation, then chang-ing variables x x and noting that 2/(x)2 = 2/x2, we get:

h2

2m

2

x2(x) + V (x)(x) = E(x) (32)

Now if V (x) = V (x), in other words, the potential is an even function,then (x) is a solution to the same Schrodinger equation. Thus thecombinations = (x) (x) are also solutions, where + is evenand is odd. For any particular energy level, in one dimension, oneor the other will be zero.

(2) Problem 2.2 in Griffiths. Writing the Schrodinger equation as:

d2

dx2 =

2m

h2[V (x) E] (33)

you can see that if E < Vmin then and always have the same sign. If

is positive (negative) then is also positive (negative). This means that always curves away from the x-axis. But we must have a normalizable

9

solution, which is not possible if continues to curve away from the x-axis.To get to curve back so that it approaches zero as x goes to infinity, it needsto turn over, which requires a negative second derivative for a positive and visa versa. We conclude that for E < Vmin we cannot get a normalizablesolution.

Total number of points: 55

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