# Translational Motion Free Motion in One Dimension by sanmelody

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```									                     Translational Motion -
Free Motion in One Dimension

• A particle of mass m free to move in one dimension ( along entire
x-axis) with zero potential energy
• Ek = k2ћ2/2m
– Ek is the kinetic energy determined from the Schrödinger equation
– ћ = h/2
• All values of the energy, Ek, are permitted
– k can take on any value for a particle free to move along entire x-axis
• The translational energy of a free particle is not quantized

Chapter 8                                    1
A Particle in a Box

• A particle of mass m confined between two walls at x=0 and x=L
• The potential energy is zero inside the box but rises to infinity at
the walls – an infinite square well
• k(x) = C coskx + D sinkx
– This is the general solution of the Schrödinger equation for the
particle inside the box
– Ek = k2ћ2/2m
– ћ = h/2

Chapter 8                              2
Particle in a Box – Acceptable Solutions

•   There is no chance of finding the particle outside the box.
–  is zero outside the box
•   The amplitude of wavefunction must be zero at each end of the box (
must be continuous)
•   Boundary conditions: k(0) = 0 and k(L)=0
•   k(0) = C
– sin0 =0 and cos0 =1
•   Thus, C must be zero
•   k = D sinkx
•   k(L) = D sinkL = 0
•   sinkL must be zero
– D cannot be zero (this would mean that =0 for all x-values which is
impossible)

Chapter 8                               3
Particle in a Box – Possible Wavefunctions

• sin kL = 0
• kL = n       n = 1, 2, 3, …….. (any integer)
– n=0 is ruled out because k would be zero
• k = n/L
• n(x) = D sin (nx/L)       n = 1, 2, 3, ….
– These are the possible wavefunctions
• En = n2h2/8mL2       n = 1, 2, 3, ……
• The energy is quantized
• To satisfy the boundary conditions, only certain wavefunctions are
possible
– Each wavefunction corresponds to a certain energy
– The energy can only have discrete values

Chapter 8                         4
Particle in a Box –
Normalization of the Wavefunctions

• n(x) = D sin (nx/L)
•  * dx = 1
– Integrated from x=0 to x=L
– The particle must be found inside the box
• D = (2/L)½
• n(x) = (2/L)½ sin (nx/L)
• En = n2h2/8mL2        n = 1, 2, 3, ……..
• A quantum number, n, is used to label energies and wavefunctions
– A quantum number is an integer (or in some cases a half-integer) that
labels the state of the system

Chapter 8                                5
Particle in a Box –
Properties of the Wavefunctions

• All wavefunctions for the particle in the box have the same
amplitude but different wavelengths
• Each wavefunction is a standing wave that has zero amplitude at
each end of the box
– For n=2, the wavefunction has one complete cycle in the box
– The number of nodes increases as n increases
• The wavefunction is not an eigenfunction of the linear momentum
operator.
• However, each wavefunction is a superposition of functions with
definite linear momenta
– p = nh/2L or -nh/2L
– particles can travel in opposite directions – equal probability for a
particle traveling to the right or to the left

Chapter 8                                  6
Particle in a Box –
Properties of the Solutions

• E1 = h2/8mL2
– E1 is the energy of the lowest energy state (n=1)
– E1 is called the zero-point energy
• En+1 – En = (2n +1)h2/8mL2
– The energy separation between energy levels (n+1) and n
– The energy separation between adjacent levels increases as n
increases
– The energy separation decreases as length of container increases
– The energy separation is very small when the container has
macroscopic dimensions

Chapter 8                              7
Particle in a Box –
The Probability Density

• The probability density for a particle in a box is given by:
• * = 2(x) = 2/L sin2(nx/L)
• For a given wavefunction, n, we can by integration calculate the
probability that the particle is found in a given region of the box
• 2(x) becomes more uniform as n increases
– The quantum mechanical result corresponds to the classical
prediction at high quantum numbers ( the correspondence principle)

Chapter 8                                 8
Orthogonality

• Two wavefunctions are orthogonal if the integral of their product
is zero
• n and m are orthogonal if  n*m d = 0
– Integrated over all space
• Wavefunctions corresponding to different energies are orthogonal
– For example n=1 and n=3 are orthogonal
– Orthogonality is important in chemical bonding and spectroscopy

Chapter 8                             9
Harmonic Oscillator

• A mass connected to a spring is a good example of a Harmonic
oscillator (1degree of freedom). We define the displacement of the
mass from the equilibrium position, Re, by the displacement
coordinate, x
•       x = R - Re
• Assume the force acting on the particle for small displacements is
given by Hooke’s Law: F = -kx where k is the force constant
• A negative sign is used because this is a restoring force. It acts in
the opposite direction to the displacement
• The force is related to the potential energy by: F = -dV/dx
• Thus, dV/dx = kx

Chapter 8                             10
Harmonic Oscillator

• The potential energy is given by:
• V(x) = ∫ kx dx = ½kx2
• The Hamiltonian in the displacement coordinate is:
H = px2 + V(x)

• The Schrodinger equation can be written:
• [(-ћ2/2m)(d2/dx2) + kx2/2]ψ = Eψ

• The reduced mass, µ, of a molecule AB is given by
•  µ = mAmB /(mA + mB)

Chapter 8                11
Energy Levels of a Harmonic Oscillator

• Quantization of energy levels arises from the boundary conditions:
• The oscillator will not be found with infinitely large compressions
or extensions ( ψ = 0 at x = ±∞ )
• The permitted energy levels are:
• Ev = (v + ½) hω          v = 0, 1, 2, 3,…..
• ω (the angular frequency) is given by:
• ω = (k/m)½ or ω = (k/µ)½ (k is the force constant and µ the
reduced mass)
• ω = 2πυ υ is the vibrational frequency (periods per second)
• Therefore, Ev = (v + ½) hυ v= 0, 1, 2, 3, ……..

Chapter 8                           12
Energy Levels of Harmonic Oscillator

• The smallest permitted value of v is zero
• The corresponding energy, E0 , is called the zero-point energy
• E0 = ½hω = ½hυ
• An oscillating particle is always fluctuating around its equilibrium
position according to quantum mechanics. Classical mechanics
would allow the particle to be perfectly still
• The separation between adjacent energy levels is given by:
• ΔE = Ev+1 – Ev = hυ
• Same separation for all values of v

Chapter 8                            13
Wavefunctions for the Harmonic Oscillator

•   Form of wavefunctions (Solutions of Schrodinger equation):
•   ψ(x) = N (polynomial in x) (bell-shaped Gaussian Function)
•   N is a normalization constant
•   Gaussian function is of form exp(-x2)
•   The precise form of the wavefunctions are:
•   ψv(x) = Nv Hv(y) exp(-y2/2)
•   Here, y = x/α and α = (h2/mk)1/4
•   Hv(y) is a Hermite polynomial
•   For ground state (v=0), H0(y) = 1
•   Thus, ψ0(x) = N0 exp(-x2/2α2)
•   ψ0(x) and ψ02(x) are bell-shaped Gaussian functions

Chapter 8                         14
Wavefunctions and Probabilities for an Harmonic
Oscillator

• The wavefunction , ψv , has v nodes
• Wavefunctions extend beyond the classical turning points
(tunnelling – typical quantum mechanical results)
• For v=0 (the ground state), the probability density is the highest at
x=0 (Classical probability: max at the turning points for any
energy)
• For large v-values, the highest probability is near the classical
turning points (The Correspondence Principle)

Chapter 8                            15
Rotational Motion -
Angular Momentum

• Particle of mass m rotating in a circular path of radius r
• J = I·
• J is the angular momentum of the particle
– J can be represented by a vector perpendicular to the rotational plane
• I is the moment of inertia
– I = mr2
•  is the angular velocity in radians per second

Chapter 8                               16
Rotation in Two Dimensions

• Assume the particle is rotating in the xy-plane and that the
potential energy, V, is zero
• E = p2/2m
– E is the total energy (kinetic energy only)
– p is the linear momentum
• Jz = ±pr
– rotation in opposite directions give angular momenta of opposite sign
• E = Jz2/2I
– I = mr2

Chapter 8                               17
Quantization of Rotational Motion

• The wavefunction describing the particle must reproduce itself on
successive circuits
• Requirement: 2r = ml
– ml is an integer
–  is the wavelength of the particle
• Using the de Broglie relation (  = h/p ), it follows:
• Jz = ћml (ml = 0, ±1, ±2, ±3, …….. )
– the angular momentum is quantized
– positive values of ml correspond to rotation in clockwise direction
– negative values of ml correspond to rotation in counter-clockwise
direction

Chapter 8                                18
Rotation in Two Dimensions -
Quantization of Energy

• E = Jz2/2I = ml2ћ2/2I
– ћ = h/2
– ml = 0, ±1, ±2, ±3, ……
• Each ml-value gives a certain discrete energy
– the energy is quantized

Chapter 8         19
Rotation in Three Dimensions

• Particle free to move on the surface of a sphere of radius r
• Requirement: Wavefunction should match as a path is traced over
the poles as well as around the equator of the sphere
• A second cyclic boundary condition
• A second quantum number is needed
• Quantum numbers: l and ml
• Possible values: l = 0, 1, 2, 3, ………
•                  ml = l, l-1, .., 0, …, -l+1, -l
• For a given l-value, there are (2l+1) permitted values of ml

Chapter 8                         20
Wavefunctions and Energies for
Rotational Motion

• The normalized wavefunctions, Yl,m(,), are called spherical
harmonics
• E = l(l+1) ћ2/2I
– l = 0, 1, 2, 3, ……
– ћ = h/2
• The energy, E, is quantized
• E is independent on ml
– there are (2l+1) wavefunctions foreach l-value. The energy level
corresponding to a particular l-value is (2l+1) degenerate

Chapter 8                              21
Angular Momentum

• J =[l(l+1)]½ћ
• J is the magnitude of the angular momentum
• z-component of angular momentum = mlh
– ml = l, l-1,…,0, ….., -l
– orientation of a rotating body is quantized - space quantization
• From the l-value, one can calculate the orbital angular momentum
of an electron orbiting around the nucleus
• One can also calculate the z-component of the orbital angular
momentum ( mlћ )
– ml is called the orbital magnetic quantum number

Chapter 8                              22
Electron Spin

• Electrons behave as if each is spinning around its own axis
• The electron spin quantum number, s = ½
• The magnitude of the spin angular momentum = [s(s+1)]½ ћ
– ћ = h/2
• The electron can have two spin directions, ms = +½ or -½
– ms is called the spin magnetic quantum number
• The z-component of the spin angular momentum is msћ

Chapter 8                       23

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