translation vibration rotation 2. Quantum theory: techniques and applications 2.1.Translational motion 2.1.1 Particle in a box 2.1.2 Tunnelling 2.2. Vibrational motion 2.2.1 The energy levels 2.2.2 The wavefunctions 2.3. Rotational motion 2.3.1 Rotation in 2 dimensions 2.3.2 Rotation in 3 dimensions 2.3.3 Spin translation vibration rotation The energy in a molecule is stored as molecular vibration, rotation and translation.

2.1 The translational motion For a free particle (V=0) travelling in one dimension, the Schrödinger equation has a general solution k, where k is a value characteristic of the energy (eigenvalue) of the particle Ek. k= Aeikx + Be-ikx For a free particle, all the values of k, i.e. all the energies are possible: there is no quantization 2.1.1 Particle in a box Particle of mass m is confined in an infinite square well. Between the walls: V=0 and the solution of the SE is the same as for a free particle. k= C sinkx + D coskx NB: with D= (A+B) C= i(A-B) A. boundary condition (BC): The difference with the free particle is that the wavefunction of a confined particle must satisfy certain constraints, called boundary conditions, at certain locations.  BC1: k(0)=0  k (0) = C 0 + D 1=0  D=0  after BC1: k= C sinkx

 BC2: k(L)=0  k (L) = C sinkL =0  absurd solution: C=0, it gives k(x)=0 and |k(x)|2=0… the particle is not in the box!  physical solution: kL= n with n=1,2,… (n0 is also absurd) The wavefunction n(x) of a particle in an infinite square well is now labeled with “n” instead of k. Because of the boundary conditions, the particle can only have particular energies En: B. Normalization: Let’s find the value of the constant C such that the wavefunction is normalized.

C. Properties of the solutions  The solutions are labeled with n, called “quantum number”. This is an integer that specifies the energetic state of the system. In order to fit into the cavity, n(x) must have specific wavelength characterized by the quantum number. With an increase of n, n(x) has a shorter wavelength (more nodes) and a higher average curvature  the kinetic energy of the particle increases.

 The probability density to find the particle at a position x in the box is More n increases, more uniform is 2n(x): we approach the case of a ball bouncing between two walls, for which there is no preferred position between the two walls.  The classical mechanics emerges from the quantum mechanics as high quantum numbers are reached.  The zero-point energy: because n>0, the lowest energy is not zero but E1=h2/(8mL2). That follows the Uncertainty Principle: if the location of the particle is not completely indefinite (in the well), hence the momentum p cannot be precisely zero and E0.  The energy level separation E= increases with n. E decreases with the size L of the cavity  for a molecule in gas phase free to move in a laboratory-sized vessel, L is huge and E is negligible: the translational energy of a molecule in gas phase is not quantized and can be described in classical physics.

2.1.2 Tunnelling X=0 X=L If the energy E of the particle is below a finite barrier of potential V, the wavefunction of the particle is non-zero inside the barrier and outside the barrier.  there is certain probability to find the particle outside the barrier, even though according to classical mechanics the particle has insufficient energy to escape: this effect is called “tunnelling”.  Let’s try to estimate the transmission probability of the particle through the barrier.  For x<0: the wavefunction is that of a free particle: (x<0)= Aeikx + Be-ikx with kħ=(2mE)1/2.  For 0<x<L: the wavefunction must be solution of the SE for a particle in a constant potential V. The general solutions are (0<x<L)= Ceqx + De-qx with qħ=[2m(V-E)]1/2. NB: here, the two exponentials are real!  For x>L: V=0, it’s like for a free particle: (x>L)= A’eikx + B’e-ikx with kħ=(2mE)1/2.

x  Direction of the transmitted wave is (Left Right): B’=0 since B’e-ikx is a wave travelling in the (Right  Left) direction.  The wave function must be continuous at the edges of the barrier (for x=0 and L): (1) For x=0: (x<0)= (0<x<L)= Aeik0 + Be-ik0 = Ceq0 + De-q0  A+B = C+D (2) For x=L: (0<x<L)= (x>L)= CeqL + De-qL= A’eikL  The derivative of the wave function must be also continuous (see 1.4.B, no singular point): (3) For x=0: [(x<0)/x]0= [(0<x<L)/x]0 = ikA-ikB=qC-qD (4) For x=L: [(0<x<L)/x]L = [(x>L)/x]L = qCeqL- qDe-qL= ikA’eikL x X=0 X=L

2.2 The vibrational motion Classical mechanics Quantum mechanics A particle undergoes harmonic motion if it experiences a restoring force proportional to its displacement Eigenvalues:  Energy separation : constant = ħω  Zero-point energy : E(=0)=½ ħω  classical limit : for a huge mass m, ω is small and the energy levels form a continuum

A. The form of the wavefunctions N is the normalization constant NB: < x >= 0  the oscillator is equally likely to be found on either side of x=0, like a classical oscillator.

we also know that <E>= <T> + <V> B. The virial theorem In a 1-dimensional problem with a potential V(x)= xn, the expectation values of the kinetic energy <T> and the potential energy <V> verify the following equality: 2 <T> = n <V> ; with the total energy: <E>= <T> + <V>  The harmonic oscillator, V=½kx2, is a special case of the virial theorem since n=2 and we have seen that <T> = <V> we also know that <E>= <T> + <V>

C. Quantum behavior of the oscillator The probability to find an oscillator (in its ground state: =0) beyond the turning point xtp (the classical limit), is: xtp xtp Classical behavior Quantum behavior -xtp xtp Quantum behavior Classical behavior  In the harmonic approximation, a diatomic molecule in the vibration state = 0 has a probability of 8% to be stretched (and 8% to be compressed) beyond its classical limit. These tunnelling probabilities are independent of the force constant and the mass of the oscillator.  Classical limit: for huge  (the case of macroscopic object), P 0

2.3.1 Rotation in 2 dimensions 2.3 The rotational motion  2.3.1 Rotation in 2 dimensions  Classical mechanics: The angular momentum Jz = pr The moment of inertia I= mr2  In quantum mechanics: not all the values of Jz are permitted, and therefore the rotational energy is quantized. Where does this quantization come from? No physical meaning  The wavelength  of the wavefunction () cannot have any value. When  increases beyond 2, we must have ()= (+2), such that the wavefunction is single-valued: |()|2 is then meaningful.  The wavelength  should fit to the circumference 2r of the circle. The allowed wavelengths are = 2r/ml ; where ml is an integer that is the quantum number for rotation.

 Schrödinger equation: A. Schrödinger equation for rotation in 2D Go to cylindrical coordinates: x= r cos; y= r sin  Schrödinger equation:  The normalized general solutions: have to fulfill the cyclic boundary condition ()= (+2): 2ml = an even integer  ml = 0, 1, 2, 3, ...  The eigenvalues are given by NB: With ml2, the energy does not depend on the sense of rotation

Plots of the real part of the wavefunction ()  For an increasing ml, the real part of the wavefunction has more nodes  the wavelength decreases and consequently, the momentum of the particle that travels round the ring increases (de Broglie relation): p=h/  The probability density to find the particle in  is a constant: |()|2=1/2  knowing the angular momentum precisely eliminates the possibility of specifying the particle’s location: the operator position and angular momentum do not commute: uncertainty principle. Plots of the real part of the wavefunction ()

cylindrical coordinates: B. The angular momentum operator lz Classical mechanics Correspondence principles (chap 11) cylindrical coordinates: x= r cos; y= r sin Quantum mechanics ux, uy, uz are unitary vectors What are the eigenfunctions and eigenvalues of lz? Let’s apply lz to the wavefunctions that are solutions of the Schrödinger equation: Vector representation of angular momentum: the magnitude of the angular momentum is represented by the length of the vector, and the orientation of the motion in space by the orientation of the vector  The solutions of the Schrödinger equation, eigenfunctions of the Hamiltonian operator, are also eigenfunctions of the angular momentum operator lz: H and lz are commutable: the energy and the angular momentum can be known simultaneously.  ml() is an eigenfunction of the angular momentum operator lz and corresponds to an angular momentum of mlħ.

2.3.2 Rotation in 3 dimensions x y z r 2.3.2 Rotation in 3 dimensions A particle of mass m free to travel (V=0) over a sphere of radius r. spherical coordinates: x= r sincos; y= r sin sin; z= r cos r = 0 is the Legendrian  The Schrödinger equation is : Since I = mr2, we can write: with

 We consider that (,) can be separated in 2 independent functions:  the Hamiltonian can be separated in 2 parts  the SE is divided into 2 equations + ml2 -ml2 At the moment, ml2 is just introduced as an arbitrary constant The solutions  should also fulfill the cyclic boundary condition: ()=(+2); because of that another quantum number “l” appears and is linked to ml. Plm(cos ) is a polynomial called the associated Legendre functions. Nlm is the normalization constant. Same as for the rotation in 2-D with l = 0, 1, 2, 3,… |m|l

 The normalized functions lm(,)=Ylm(,) are called spherical Harmonics The figure represents the amplitude of the spherical harmonics at different points on the spherical surface. Note that the number of nodal lines (where lm(,)=0) increases as the value of l increases: a higher angular momentum implies higher kinetic energy.  From the solution of the SE, the energy is restricted to:  The energy is quantized and is independent of ml. Because there are (2l+1) different wavefunctions (one for each value of ml) that correspond to the same energy, the energetic level characterized by “l” is called “(2l+1)-fold degenerate”.

Spherical harmonics ml = 0: a path around the vertical z-axis of the sphere does not cut through any nodes. For those functions, the kinetic energy arises from the motion parallel to the equator because the curvature is the greatest in that direction. http://www.sci.gu.edu.au/research/laserP/livejava/spher_harm.html http://mathworld.wolfram.com/SphericalHarmonic.html

Vector representation of the angular momentum  The comparison between the classical energy E=J2/2I and the previous expression for E, shows that the angular momentum J is quantized and has the values ( length of the vector): J={l(l+1)}1/2 ħ ; l= 0, 1, 2,...  As for the rotation in 2-D, the z-component Jz is also quantized, but with the quantum number ml ( orientation of the vector J): Jz= ml ħ ; ml= l, l-1, …, -l  For a particle having a certain energy (e.g. characterized by l=2), the plane of rotation can only take a discrete range of orientations (characterized by one of the 2l+1 values ml)  The orientation of a rotating body is quantized

Cone representation of the angular momentum While J2 and Jz commute, Jz and Jx (or Jy) do not commute  Jz and Jx (or Jy) cannot be measured accurately and simultaneously  If Jz is known precisely, Jx and Jy are completely unknown: representation with a cone is more realistic than a simple vector. It means that once the orientation of the rotation plane is known, Jx and Jy can take any value.

2.3.2 Spin of a particle The spin “s” of a particle is an angular momentum characterizing the rotation (the spinning) of the particle around its own axis.  The wavefunction of the particle has to satisfy specific boundary conditions for this motion (not the same as for the 3D-rotation). It follows that this spin angular momentum is characterized by two quantum numbers:  s (in place of l) > 0 and   the magnitude of the spin angular momentum: {s(s+1)}1/2ħ  ms  |s|  the projection of the spin angular momentum on the z-axis: msħ NB: In this course the spin is introduced as such. But in the Relativistic Quantum Field Theory, the spin appears naturally from the mathematics.  Electrons: s = ½  the magnitude of the spin angular momentum is 0.8666 ħ. The spins may lie in 2s+1= 2 different orientations (see figure). The orientation for ms= -½, called  and noted ; the orientation for ms= -½ is called  and noted .  Photons: s = 1  the angular momentum is 21/2 ħ