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The Quantum Mechanical Harmonic Oscillator

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The Quantum Mechanical Harmonic Oscillator The Quantum Mechanical Harmonic Oscillator Motivation: The quantum mechanical treatment of the harmonic oscillator serves as a good model for vibrations in diatomic molecules. The harmonic oscillator model accounts for the infrared spectrum a diatomic molecule. Using “normal mode analysis”, it also serves as the basis for vibrational analysis in larger molecules which are commonly used in chemistry today to interpret infrared spectra. Potential Energy Surface of a Typical Diatomic Molecule r ) r ( V m1 m2 bond breakage potential energy energy potential R0 (equilibrium bond length) internuclear distance, r The potential energy surface relates the geometry of a molecule (given by its nuclear coordinates) to the molecule’s potential energy. The Harmonic Oscillator With the harmonic oscillator, we approximate the true internuclear potential with a harmonic (parabolic) potential given by: 1 V (r) k(r R )2 2 0 E R0 is the equilibrium bond r length energy, energy, k is the force constant. R0 m1 m2 R0 and k are different for various diatomic molecule, i.e. internuclear distance, r HCl, N2, O2, … The harmonic oscillator is usually a good approximation for r values close to R0 (for example, “low” temperatures such as room temperature). The Classical Picture of the Harmonic Oscillator If we took our classical harmonic oscillator and stretched it to a length A at time zero, and let it go with zero velocity at time t = 0, then the harmonic oscillator would vibrate according to: r(t) Acost +A r(t) t classical turning points at r = A and r = A A The displacements of the masses never go beyond amplitude |A|. The total energy of the system can assume any value depending on how far we initially displace the spring: 1 E kA2 continuous energy 2 The system can have zero energy if the oscillator is not moving (A = 0). Reduction of the two-body problem to two one-body problems m r Consider the motion of the diatomic molecule 1 m2 in one dimension. The classical Hamiltonian function is given by: 1 2 1 2 1 H m x m x k(x x R )2 2 1 1 2 2 2 2 2 1 o x1 x2 Xcm r x21 x But wait!!! We can reduce this two-body problem to a one-body problem for the translational motion of the center of mass, Xcm, of the diatomic and the vibrational motion in terms of r, the distance between the nuclei. m1x1 m2 x2 X CM center of mass coordinate m1 m2 internuclear distance coordinate r x2 x1 m1x1 m2 x2 X CM r x2 x1 m1 m2 Using these two equations, we can also express x1 and x2 in terms of Xcm and r. (i.e., two equations and two unknowns) m2 m1 x1 X CM r x2 X CM r m1 m2 m1 m2 Use these relationships to derive the kinetic energy in terms of the center of mass coordinate and the bond distance coordinate (next tutorial!). The ultimate goal is to separate the translational motion from the vibrational motion. After some straightforward algebra, we end up with: 1122mm12 T m12 m XCM r 22mm12 The first term can be interpreted as the kinetic energy of translational motion of the whole molecule (m1 + m2) through space. The second term is the internal kinetic energy of the relative motion of the two nuclei, or the vibrational motion. To simplify things, we define m m the reduced mass: 1 2 m1 m2 The vibrational kinetic energy therefore reduces from a two body problem with masses m1 and m2 to a one-body problem with an ‘effective’ mass equal to . 1 T r2 Vib 2 Separation of Translational and Vibrational Motion It is reasonable to assume that the vibrational motion is independent of the motion of the center of mass as long as the molecule is isolated and does not interact with anything. The vibration of the molecule will then be unaffected by how fast or slow the center of mass is moving. We have already examined the quantum mechanical treatment of the translational motion in one dimension. This was the ‘free particle’. We now consider the vibrational motion of the m r molecule. The classical Hamiltonian function 1 m2 for vibrational motion only is: 1 1 H r2 k(r R )2 2 2 o where r is the internuclear distance and is the reduced mass. 1 1 H r2 k(r R )2 2 2 0 Let’s make a further simplification by defining a new coordinate, x: x r R0 x2 x1 Note: this new coordinate x (the displacement from the equilibrium internuclear separation) is not the same as the x1 and x2 coordinates used previously. In this way our classical Hamiltonian becomes: 1 1 H x 2 kx2 recall that 2 2 d dr(t) x (r(t) R ) r dt 0 dt so x2 r2 r displacement: x r R0 m1 m2 1 1 V (r) k(r R )2 V (x) kx2 2 0 2 V V potential energy, potential energy, potential energy, energy, potential R0 -x +x internuclear distance, r x = 0 When x is negative, the bond is compressed, and when x is positive, the bond is stretched. We have now reduced our vibrational motion of two bodies into the effective motion of a single particle. Before we look at solving for the quantum mechanical behavior, let’s relate what we’ve just done to the particle-in-a-box problem. 1 V (x) kx2 2 V(x) = 0 0< x < L V(x) = 0 x and x L V V = 0 V = V = L energy, potential x = 0 -x +x x = L x = 0 0 < x < L - < x < + The Quantum Mechanical Harmonic Oscillator Now solve for the wave function of the Harmonic Oscillator. We first need the Hamiltonian operator. We know the classical Hamiltonian is 1 1 H x 2 kx2 2 2 So our quantum mechanical Hamiltonian operator is given by 2 d2 1 Hˆ m m kx2 2 dx1 2 2 2 m1 m2 Notice the reduced mass is retained. The Schrödinger equation for the harmonic oscillator is therefore 2 d2 1 (x) kx2 (x) E (x) 2 dx2 2 We again have a ordinary differential equation. Rearranging into a more standard form, we get d2 2 1 (x) E kx2 (x) 0 x dx2 2 2 E – V is not constant in this case, but a function of x When E – V is variable, there is no general way of solving this type of differential equation. Each case must be studied individually, depending of the function. It turns out that we can use a power series method. This is a bit tedious and involves math we have not yet covered. For the moment, the solutions will only be presented and interpreted. The Quantum Mechanical Harmonic Oscillator 2 d2 1 (x) kx2 (x) E (x) 2 dx2 2 The Harmonic Oscillator Energies 1/ 2 k 1 En n 2 n = 0, 1, 2 ,3 …. m m 1 2 The energy levels are quantized: E0, E1, E2, E3, … m1 m2 The states are expressed in terms of the quantum number n. k is the force constant (characteristic of each diatomic bond). is the reduced mass defined by: This is usually rewritten as 1/ 2 k 1 En n 2 n = 0, 1, 2 ,3 …. where is classically interpreted as the vibration frequency in radians per second (the angular vibration frequency). Many textbooks also use: 1/ 2 1 k E h 1 = 0, 1, 2 ,3 …. n 2 2 2 where (“Nu”) is the frequency in cycles per second (Hz). where (“Upsilon”) is the harmonic oscillator quantum number. The Harmonic Oscillator Energies 1 n = 0, 1, 2 ,3 …. En n 2 • Energy levels are all equally spaced. n = 5 spacing n = 4 • Notice that the ground state in this case has n = 3 an index of n = 0, not n = 1 as with the particle E in a box. n = 2 • Again there is a zero-point energy with the n = 1 ground state having a positive non-zero energy: n = 0 E vibrational zero-point 0 2 energy The Harmonic Oscillator Wave Functions Solving the harmonic oscillator time-independent Schrödinger equation gives time-independent wave functions of the following form: x2 / 2 n (x) Nn Hn () e n = 0, 1, 2 ,3 …. 1/ 4 k 1 (normalization 2 1 x x x Nn 2n n! constant) 1/ 2 Hn () are Hermite polynomials, functions of “xi” = x 3 H0 () 1 H1() 2 H3 () 8 12 th th Hn () is an n order polynomial in and therefore an n order polynomial in x. The First Four Harmonic-Oscillator Wave Functions 1/ 4 α 2 ψ0 exp(αx / 2) π 1/ 4 4α3 2 ψ1 xexp(αx / 2) π k 1/ 4 α 2 2 ψ2 2x 1 exp(αx / 2) 4π 1/ 4 α3 3 2 ψ3 2x 3x exp(αx / 2) 9π 2 1/ 4 αx α3 3 2 ψ3 2x 3x e 9π αx 2 1/ 4 α 2 2 ψ2 2x 1 e 4π 2 1/ 4 αx 4α3 2 ψ1 xe π αx2 14/ α 2 ψ0 e π x = 0 x E Harmonic Oscillator Wave Functions 5 ψx4() 4 ψx3() 3 ψx2() 2 ψx1() 1 ψx0() 0 The harmonic oscillator wave functions are often superimposed on the potential. As n increases, the wave function ‘widens’ (i.e., it has more energy and can “stretch out more” than the classical oscillator). Verify that 0(x) is normalized. αx 2 1/ 4 α 2 ψ0 e π αx 2 Verify that 0(x) and 1(x) are orthogonal.
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