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Quantum Physics II Angular Momentum Particle on a Sphere Recommended Reading: Harris, Chapter 6 Angular Momentum Angular Momentum: L = m r v = r mv = r p O r L=r p m v p r Note that the angular momentum vector is perpendicular to both the vectors r and p Some Math: The vectors r and p can be expressed in terms of their components r xi y j zk p px i p y j p z k Then the angular momentum is defined by the cross product i L rp x px j y py k z i ypz zpy j xpz zpx k xpy - ypx pz (11) General expression for the angular momentum in Cartesian coordinates Angular Momentum If the particle is confined to the x-y plane, so the only non-zero component of its angular momentum is the z-component LZ. Lz = (r p)Z = xpy - ypx x and y are the coordinates of the particle, px and py are the x and y components of the momentum of the particle. We can now find the quantum mechanical operator corresponding to the angular momentum by using the operator definition for the momenta p L̂ Z x i y i i (12) y x φ Operate on the wavefunctions found above with this angular momentum operator L̂ Z ψ L̂ Z A expimL φ i A expimL φ (13) φ mL A expimL φ mL φ Eigenvalue equation. The wavefunction Aexp(imLx) corresponds to a state with with angular momentum mL Z component of Angular momentum is quantised in units of mL Rotation in three dimensions In this section we consider a particle that is free to move in three dimensions but always at a constant distance r from a fixed point O. In this way the motion of the particle is confined to the surface of a sphere. The potential V is constant and can be set equal to zero This model can be applied to the rotation of rigid molecules, and also to describe electron and nuclear spin (see later), but here we we introduce it as a first step towards obtaining a solution for the 3_D Schrodinger equation for the Hydrogen atom. O r Cartesian Coordinates If we consider the proton to be at the origin then the position of the electron relative to the proton is specified by three coordinates (x,y,z) z r x 2 y2 z 2 Because of the form of r solving the Schrodinger equation in Cartesian coordinates is very difficult. So we change to another coordinate system to specify the position of the electron relative to the proton. Spherical coordinates are the best for this problem. (x,y,z) r x (x,y) y Spherical Coordinates (r,,) Instead of the three coordinates (x,y,z), we can specify the position of the electron by three Spherical coordinates (r,,). z r range 0 r = radius, = polar angle range 0 = azimuthal angle range 0 2 y y = r sin sin x x = r sin cos Transformation from Spherical (r,,) to Cartesian (x,y,z) Coordinates We can transform form spherical to cartesian coordinates and vice-versa using the following equations (x,y,z) (r,,) (r,,) (x,y,z) x r sinθ cos φ y r sinθ sinφ z r cosθ r x2 y2 z 2 2 2 z2 x y z 1 θ cos cos1 r y φ tan1 x and the volume element dV, is given by dV dxdydz r 2 sinθdrdθdφ z Schrodinger Equation in Spherical Coordinates The time independent Schrodinger equation in three-dimensions is 2 2 ψr, θ, φ Uψr, θ, φ Eψr, θ, φ 2m We must also express the Laplacian operator 2 in polar coordinates (see additional notes) the result is that 1 2 1 1 2 2 sinθ r 2 2 2 θ sin θ φ r r r sinθ θ therefore the Schrodinger equation in 3D polar coordinates is 1 2 1 1 2 2m r 2 E Uψ 0 sinθ 2 2 2 θ sin θ φ r r r sinθ θ this looks terrible but it in not much different from a particle in a 3d box or the SHO equation. Particle confined to a Sphere We now return to the particle confined to move on the surface of a sphere. This is really a 2-dimensional problem, since we can specify the position of the particle with just two coordinates, (, ). Since r is constant 1 2 r r r 0 2 r and the Schrodinger equation becomes 2 1 1 2 Eψ sinθ 2 sinθ θ 2 2 θ sin θ φ 2mr 2 2 Which we can write as L̂ Eψ 2I 1 1 2 L̂ sinθ 2 2 θ sin θ φ sinθ θ 2 where is the square of the angular momentum operator and I = mr2 is the Moment of Inertia of the particle about the origin. We now want to find the eigenvalues and eigenvectors of the L2 operator. This will give us the allowed values of the angular momentum. Particle confined to a Sphere The Schrodinger equation is with 2 2 L̂ Eψ 2I (1) 1 1 2 L̂ sinθ 2 2 θ sin θ φ sinθ θ 2 Yet again we use the method of separation of variables. We try a solution of the form , where () is a function of only and () is a function of only, I.e. these are one dimensional functions. When we substitute into equation (1) above and rearrange, we get 2 2IE 0 sinθ 2 2 2 θ sin θ φ sinθ θ Multiplication by sin2 and division across by gives: sinθ 1 2 2IE 2 sinθ 2 2 sin 0 θ θ φ The two variables and can now be separated by moving the term in to the right hand side of the equation sinθ 2IE 1 2 2 sinθ 2 sin - 2 θ θ φ (2) only depends on only depends on Since and can be varied independently, this equation can be satisfied only when both sides of the equation are equal to a constant. Set this constant equal to ml. Then sinθ 2IE 1 2 2 2 sinθ 2 sin ml - 2 θ θ φ We have already seen the equation on the right-hand side, (particle on a ring). 1 2 2 2 ml m2 (3) l 0 2 φ2 φ 1 expiml with ml = 0, ±1, ±2... solutions (eigenfunctions) are 2 ml is known as the azimuthal quantum number. The equation on the left hand side for becomes after rearranging, m2 1 2IE l 0 sinθ 2 2 sinθ θ θ sin putting 2 IE 2 and multiplying across by gives 2 1 ml 0 sinθ sinθ θ θ sin2 (4) we now have another differential equation to solve. This one is a little more complicated than those encountered previously. Solutions can be found for all values of but most of these diverge and therefore are nt physically acceptable solutions. It can be shown that non-divergent (physically acceptable) solutions only exist when l (l 1) wherel 0, 1, 2, 3,... and ml 0, 1, 2, 3,... l so if l = 0 then ml = 0; l =1 then ml = -1, 0, +1; l =2 then ml = -2, -1, 0, +1, +2; l =3 then ml = -3, -2, -1, 0, +1, +2, +3 etc. inserting l (l 1) and 1 2 2 ml φ 2 into eqn (4) gives 1 d d 1 d2 l l 1 sinθ 2 2 sinθ dθ dθ sin d multiply across by d d d2 l l 1 sinθ 2 2 sinθ dθ dθ sin d (5) Recall that (, ) = ()() then we can write equation (5) as L̂2 , l l 1 , (6) This is an eigenvalue equation for the square of the angular momentum operator, L2. From this equation we see that, since l is an integer the the angular momentum is quantised. So we have found another quantum number l. What about the solutions to the differential equation for (equation 4) ? 2 1 ml 0 sinθ 2 sinθ θ θ sin Associated Legendre Polynomials This is a difficult equation to solve, it is called the The Associated Legendre Equation. We use the same technique that we used for the Simple Harmonic Oscillator Equation, that is we look for a Series Solutions The solutions are a called the Associated Legendre Functions, Pl,mL(cos ) they are polynomials that depend on the angle and the two quantum numbers l and mL. For each allowed set of quantum numbers (l, mL) there is a solution Pl,ml(). The first few are given in the following table l 0 1 1 2 2 2 ml Pl,mL(cos ) 0 1 0 cos 1 sin 0 (3cos2 - 1)/2 1 3cos sin 2 3sin2 l 3 3 3 3 What do these functions look like ? ml 0 1 2 3 Pl,mL(cos ) (5cos3 - 3cos)/2 3sin(5cos2 -1)/2 15cossin2 15sin3 Polar Plots to visualise the () wavefunctions P1,0 l = 1, mL = 0 z x-y plane P1,1 l = 1, mL = 1 P2,0 l = 2, mL = 0 P2,1 l = 2, mL = 2 P2,1 l = 2, mL = 1 P3,0 P3, 2 l = 3, mL = 0 l = 3, mL = 2 P3,1 P3, 3 l = 3, mL = 1 l = 3, mL = 3 We now have the solutions to both parts of the wave function, , The solutions to the - dependent part () are given by 1 expiml 2 with ml = 0, ±1, ±2... While the - dependent solutions () are the associated Legendre polynomials Pl,ml(cos ). Putting this all together we see that the wavefunctions for a particle confined to move on the surface of a sphere are 1 , expiml Pl,m cosθ l 2 The product of these two functions are usually combined into a single function called a Spherical Harmonic, denoted Yl,ml(, ), where l and ml are the two quantum numbers that determine the particular form that the wavefunction takes. Note that the Spherical Harmonics depend on two variables and . With this notation the eigenvalue equation for L2 becomes L̂2 Yl ,ml , l l 1Yl ,ml , The first few Spherical Harmonics Yl,ml l ml 0 0 2l 1 l -m! Pl,ml cos eiml 4 l m! Yl ,ml , Definition Yl ,ml , 1 4 l ml 2 0 2 1 l ml 1 0 1 1 Yl ,ml , 3 cos 4 3 sin e i 8 2 2 Yl ,ml , 5 3 cos2 1 16 15 cos sin e i 8 15 sin2 e 2i 32 These can be visualised using the applets on the following websites http://www.falstad.com/qmrotator/ http://www.bpreid.com/applets/poasDemo.html http://www.uniovi.es/qcg/harmonics/harmonics.html These images show the Real part of the first few Spherical Harmonics l = 1: ml = -1, l = 2: ml = -2, -1, l=0 ml = 0 0, 0, +1 +1, +2 Energies of the Wavefunctions Going back to equation 1 we have 2 2 L̂ Eψ 2I But we know that the wavefunctions for a particle confined to a sphere are the spherical harmonics so this equation becomes: 2 2 2 L̂ Yl ,m l l 1Yl ,m EYl ,m 2I 2I This is an eigenvalue equation and we see that the allowed energies of a particle confined to the surface of a sphere and rotating about the originare quantised according to the equation 2 2 E l l 1 l l 1 2 2I 2mr where l = 0, 1, 2, 3, ... Note that the energy is independent of the value of ml, so there will be 2 l + 1 states with the same energy, that is, each energy level is 2 l + 1 degenerate. Examples: l = 0, degeneracy = 1: l = 1, degeneracy = 3 l = 2, degeneracy = 5 ... Angular Momentum and Spatial Quantisation The equation relating rotational energy and angular momentum is 1 2 L2 L2 E I 2 2 2I 2 mr We have also seen that the angular momentum is quantised and its eigenvalues are given by L̂2 Yl ,ml , l l 1Yl ,ml , from this we see that the magnitude of the angular momentum is limited to the following values L̂ L̂2 l l 1 where l = 0, 1, 2, 3, ... We have also seen that for a particle confined to rotate in the xy-plane its z-component of angular momentum is also quantised according to the equation L̂ z Yl ,ml , ml Yl ,ml , So that component of the angular momentum along the z-axis s quantised according to the equation L z ml where ml = 0, ±1, ±2, ± 3, ... ± l l=2 Putting this together we see that the quantised angular momentum for a given value of l can be represented by a vector of length proportional to l l 1 and orientated so that the projection of the vector on the z-axis is equal to ml z ml = 2 z ml = 1 l = 1, ml = 1 ml = 0 l = 1, ml = 0 l = 0, ml = 0 l = 1, ml = -1 ml = -1 ml = -2 for a given value of the quantum number l there are 2l +1 directions in which the angular momentum vector can point. If all directions in space are equivalent, then the direction we define for the z-axis is arbitrary and the 2l +1 orientations that the rotating particle can adopt all have the same energy, that is, they are degenerate. In the presence of an externally applied electric or magnetic field , however, the z-axis is determined by the direction of this field and the orientation of the angular momentum with respect to this axis will affect the energy, that is, the external field can remove the degeneracy (Zeeman efect, see later) So far we have only discussed the component of angular momentum along the z-axis, what about the components in the x and y directions? The Uncertainty Principle forbids complete knowledge of the orientation of the angular momentum vector. So if the component in the z-direction is known then the other two components remain undetermined. This situation is represented by the following diagram: one end of the angular momentum vector is considered to be at the apex of the appropriate cone and the other end then lies anywhere along the circular cross section at the top of the cone, so its projection can lie anywhere in the xy-plane Figure depicts this for l = 2 c ml = 2h/2 c ml = h/2c ml = 0 ml = -h/2 ml = -2h/2 Application: Rotational Spectroscopy of Diatomic Molecules r m2 r m1 Centre of mass The motion of the two masses m1 and m2 is mathematically equivalent to the rotation of a single particle of mass about a fixed point with the distance between the particle and fixed point being equal to the bond length r. It is therefore possible to treat the rotation of a diatomic molecule as the motion of a particle of mass confined to the surface of a sphere m1m2 reduced mass m1 m2 In rotational spectroscopy it is customary to use J as the angular quantum number instead of l. With this modification the allowed rotational energy levels are 2 2 EJ J(J 1) J(J 1) 2 2 I 2 r Where I = r2 is the moment of inertia of the molecule. For a given rotational energy there will be (2J + 1) possible orientations of the molecule in space, so each rotational state is (2J +1) -fold degenerate . The molecule can absorb or emit a photon and make a transition to a higher or lower rotational state. However there is a selection rule that states that J can only change by 1 in a transition. So the change in rotational energy if the molecule absorbs a photon is given by 2 2 (J 1)(J 2) J(J 1) (J 1) E E J1 E J 2I I Example In the absorption spectrum of 12C16O the rotational transition from the J = 12 to the J = 13 rotational state causes absorption of radiation at a wavenumber of 50.2 cm-1. (a) Calculate the energy change involved in the transition (b) the moment of inertia and (c) the bond length of CO (a) E(photon) h hc hc~ 9.98 1022 J (b) This energy is equal to the difference in rotational energy between the J = 12 and J = 13 states 2 E E13 E12 13 I So the moment of inertia is 2 I 13 1.45x10- 46 kg.m2 r 2 E The reduced mass is Which gives a bond length 12x16 amu 1.14x10-26 kg 12 16 r I 1.13x10-10 m rendered images of the Spherical Harmonics websites on the following http://www.uniovi.es/qcg/harmonics/harmonics.html http://odin.math.nau.edu/~jws/dpgraph/Yellm.html