Solucionario Del Miller

Solutions to Selected Problems for Quantum Mechanics for Scientists and Engineers

David A. B. Miller
Stanford University

© Cambridge University Press 2008

Introduction
Selected problems in the book Quantum Mechanics for Scientists and Engineers (Cambridge University Press, 2008) are marked with an asterisk (*), and solutions to these problems are collected here so that students canhave access to additional worked examples.

David A. B. Miller Stanford, California March 2008

2.6.1

2.6.1
The normalized wavefunctions for the various different levels in the potential well are

ψ n ( z) =

2 ⎛ nπ z ⎞ sin ⎜ ⎟ Lz ⎝ Lz ⎠

The lowest energy state is n = 1, and we are given Lz = 1 nm. The probability of finding the electron between 0.1 and 0.2 nm from one side of thewell is, using nanometer units for distance,
P = ∫ ψ 1 ( z ) dz = ∫ 2sin 2 (π z )dz
2 0.1 0.1 0.2 0.2

= ∫ [1 − cos(2π z ) ] dz
0.1

0.2

= 0.1 − ∫ cos(2π z )dz
0.1

0.2

1 [sin(2π × 0.2) − sin(2π × 0.1)] 2π = 0.042 = 0.1 − (Note: For computation purposes, remember that the argument of the sine is in radians and not degrees. For example, when we say sin( π ) = 0, it is implicit herethat we mean π radians.)

2.8.1

2.8.1
The wave incident from the left on the infinite barrier will be reflected completely because of the boundary condition that the wavefunction must be zero at the edge of, and everywhere inside of, the infinite barrier. So if the barrier is located at x = 0

ψ ( x) = 0
2mo E
2

(x > 0)

Now, for an electron of energy E, which here is 1 eV, we knowthat it will have a wavevector
k= = 5.12 × 109 m -1

The general solution for a wave on the left of the barrier is a sum of a forward and a backward wave each with this magnitude of wavevector, with amplitudes A and B, respectively; that is

ψ ( x) = A exp(ikx) + B exp(−ikx) (x < 0)
Knowing from our boundary condition that the wave must be zero at the boundary at x = 0 , A + B = 0 ⇒ A = −B ⇒ψ ( x) = A(exp(ikx) − exp(−ikx)) = 2iA sin(ikx) The probability density for finding the electron at any given position is (x < 0)

Thus, the wave function on the left hand side of the infinite barrier is a standing wave.

ψ ( x) = 0 ψ ( x) = 4 A sin 2 (kx)
which has a period π / k .
2 2

2

( x > 0) ( x < 0)

The period of the standing wave shown in the graph is therefore ~ 6.1Angstroms.

Probability density

6.1 Å x=0

x

(The amplitude of the standing wave is 4 A , but A here has to remain as an arbitrary number. We cannot actually normalize such an infinite plane wave, though this problem can be resolved for any actual situation, for example by considering a wavepacket or pulse rather than just an idealized plane wave.)

2

2.8.3

2.8.3
For E = 1.5 eV andVo = 1 eV, the incoming particle/wave from the left will be partly reflected and partly transmitted at the barrier. We write the general form of the wavefunctions on both sides of the barrier

ψ left ( z ) = C exp(ik L z ) + D exp(−ik L z ) ψ right ( z ) = F exp(ik R z )

i.e., the sum of the incident and reflected waves i.e., the transmitted wave

(Note that we do not have a backwardpropagating wave on the right hand side because there is no reflection beyond the barrier.) Here
kL = 2mE
2

= 6.27 ×109 m -1 and k R =

2m( E − Vo )
2

= 3.62 × 109 m-1

Now applying boundary conditions (a) the continuity of the wavefunction at z=0 (barrier edge): (C + D) = F (b) continuity of the derivative of the wavefunction at z=0: (C − D)k L = k R F Adding and subtracting, we get D = kL− kR 2k L C C and F = kL + kR (k L + k R )

The absolute phase of any one of these wave components is arbitrary because it does not affect any measurable result, including the probability density (we are always free to choose such an overall phase factor). If we choose that phase such that C is real, then our algebra becomes particularly simple and, from the above equations, D and F are also...