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Translational Motion Free Motion in One Dimension

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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