A few principles of classical and quantum mechanics

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1 A few principles of classical and quantum mechanics The classical approach: In classical mechanics, we usually (but not exclusively) solve Newton s nd law of motion relating the acceleration a of the system of mass m to the external forces F: F ma with a dv dt d dt x The kinetic energy K of the system expresses the energy involved in the motion of the system. Hence, K depends on the momentum p (or speed v = p/m) by: K mv p m The potential energy V of a system expresses the energy associated with the position of the system within a force field F. Hence, V depends on the position x and is related to the force field by: F dv / dx The total energy E of a system is the sum of its kinetic energy K and of its potential energy V: E K V

2 The quantum mechanical approach: In quantum mechanics, we need to solve the appropriate Schrödinger equation: Ĥ E where Ĥ is the Hamiltonian operator and is the wavefunction describing the state of our system at energy E. We shall see that only certain wavefunctions and their corresponding energies are acceptable. Hence, quantization of energies is a natural consequence of the equation and the conditions imposed on it (boundary conditions). For a single particle of mass m, the Hamiltonian operator takes a simple form: Ĥ m d dx V(x) Very importantly, we notice that the first term defining Ĥ can be related to the kinetic energy of the particle. It follows that the momentum operator is defined by: Pˆ i d dx

3 Translational motion The classical approach: The classical description of free motion (V = 0) for a particle of mass m and momentum p in one dimension is given by: E p m Furthermore, its position is perfectly defined at all times: F 0 a 0 v v x v t 0 0 x0 The quantum approach: We solve the Schrödinger equation: Ĥ E, with Ĥ d d hence: E m dx m dx The general solutions are (indexed with k): k Ae ikx Be ikx with: E k k m All values of k are permitted. It follows that translational motion of a free particle is not quantized. independent of x position of the particle is completely unpredictable. This is consistent with Heisenberg s uncertainty principle since the momentum is certain: k is p x / p 0 hence x

The classical approach: Vibrational motion (Characteristic of the harmonic oscillator) A particle undergoes harmonic motion if it experiences a restoring force proportional to its displacement (no

4 The classical approach: Vibrational motion (Characteristic of the harmonic oscillator) A particle undergoes harmonic motion if it experiences a restoring force proportional to its displacement (no dissipation): F kx (k = force constant) Potential energy is derived from force field: dv F V kx dx We solve the problem with Newton s law: d x F ma m kx 0 dt This is a well known differential equation whose solutions are: x(t) A cos0t with k 0 m Total energy of the system: E mv kx ka We notice that the total energy of a given vibration (or harmonic oscillation is a constant, independent of time (hence of frequency = 0 ). Classically, as expected, the total energy can take any value (since A can take any value ).

5 The quantum approach: Schrödinger s equation: m d dx kx E This is a standard differential equation whose solutions are known (see later). We solve for E n by forcing the wavefunctions 0 at x (boundary conditions): n E n n with: Separation in energy for two adjacent levels: En E n for any given level n. Therefore, the energy levels form a ladder of spacing. k m and: n = 0,,, Zero-point energy: E 0 The detailed solutions for the wavefunctions have the form: n x N n H n (y)e y / x with : y ; mk / 4 N n = normalized constant H n (y) = Hermite polynomial y e / = Gaussian function

6 Expressions for the ground state and the st excited state: / x / y / x 0 0 e x N ye N x e N x Probability of locating the ground state particle in position x: x / 0 0 e N x

7 An example of application: Atoms vibrate to one another in molecules with the bond acting like a spring. Consider an X-H chemical bond, where a heavy X atom forms a stationary anchor for the very light H atom. That is, only the H atom moves, vibrating as a simple harmonic oscillator. The force constant of a typical X-H chemical bond is around 500 Nm - ; for example, 56.3 Nm - for the H 35 Cl bond. Because the mass of the proton is about kg, using 500 Nm - we get = s - (5.4 0 THz). It follows that the separation of adjacent levels is = J (about 0.36 ev). This energy separation corresponds to 34 kj mol -, which is chemically significant. The zero-point energy of this molecular oscillator is about J, or 0.8 ev, or 7 kj mol -. The excitation of the vibration of the bond from one level to the level immediately above requires J. Therefore, if it is caused by a photon, the excitation requires radiation of frequency = E/h = 86 THz and the wavelength is = c/ = 3.5 m. The associated wavenumber is about = / = 900 cm -. It follows that transitions between adjacent vibrational energy levels of molecules are stimulated by or emit infrared radiation.

8 The classical approach: Rotational motion The rotational motion of a particle of mass m around a central point at a fixed distance r (radius) is described by its angular momentum, J (a vector!). Its magnitude, J = J is given by: J = I = angular velocity (rad/s). I = moment of inertia (kg m ). For a point of mass m moving in a circle of radius r, the moment of inertia about the axis of rotation is given by: I = mr Note the analogous roles of m and I, of v and, of p and J in the translational and rotational cases, respectively, with = v/r and J = pr. Its rotation energy is - as function of : E I J - as function of J: E I (V = 0). Classically, we expect the rotational energy to take any arbitrary value.

angular momentum about the z-axis is: Jz = ±hr/.

9 The quantum approach: The treatment is broken down in two parts: motion in D and motion in 3D. ) Rotation in D: a particle on a ring We recall from classical mechanics: J z E I Because Jz = ±pr, and, from the de Broglie relation, p = h/, the angular momentum about the z-axis is: Jz = ±hr/. For meaningful sense, the only allowed wavelengths are: r ml ml = 0, ±, ±, When: ml 0 Hence, the angular momentum is quantized, that is: J z ml ml 0,,,... ml > 0 (< 0) corresponds to CW (CCW) rotation around the zaxis. It follows that the energy is limited to the values: Jz ml E I I ml 0,,,...

Formally we solve Schrödinger s equation (in D, V = 0): ˆ H m x y ˆ E H x r cos( ) Transformation to cylindrical coordinates: y r sin( ) r r r x y r r with r 0 ˆ Hence: H I mr Schrödinger s

10 Formally we solve Schrödinger s equation (in D, V = 0): ˆ H m x y ˆ E H x r cos( ) Transformation to cylindrical coordinates: y r sin( ) r r r x y r r with r 0 ˆ Hence: H I mr Schrödinger s equation becomes: d d EI Normalized general solutions are: m l ( ) e im l / with ml EI /. Cyclic boundary conditions: ( ) ( ). By substitution it becomes: m l ( ) m l ( ) m l m Hence: l ml = 0, ±, ±,

The ground state ( ml 0 ): 0 ( ) / / has the same value at all points on the circle. We arrive at the following conclusions: () energy quantized (ml) and independent of sense of rotation ( ml ).

11 The ground state ( ml 0 ): 0 ( ) / / has the same value at all points on the circle. We arrive at the following conclusions: () energy quantized (ml) and independent of sense of rotation ( ml ). () levels doubly degenerate for ml >0 (except for ml = 0). (3) wavefunctions have increasing # of nodes with increasing Jz. (4) decreases as ml increases. * (5) probability density is m /, l ml independent of ml and of hence the location of the particle is completely indefinite (since Jz or p are precisely known!) ) Rotation in 3D: a particle on a sphere A second cyclic boundary condition (the wavefunction should match as a path is traced over the poles as well as round the equator of a sphere of radius r) will introduce a second quantum number, l. Again, we solve Schrödinger s equation (in 3D, V = 0) with: ˆ H E m m where = laplacian (for convenience) x y z

x r cos( ) sin( ) Transformation to spherical coordinates: y r sin( ) sin( ) x r cos( ) r r r r r with sin ( ) sin sin( ) Schrödinger s equation becomes: EI Separation of variables, and substitution

12 x r cos( ) sin( ) Transformation to spherical coordinates: y r sin( ) sin( ) x r cos( ) r r r r r with sin ( ) sin sin( ) Schrödinger s equation becomes: EI Separation of variables, and substitution in Schrödinger s equation leads to two equations: d m l d sin( ) d d EI sin( ) sin ( ) ml d d The first is exactly what we found for the D case; hence it has the solutions described earlier. The second is much more complex to solve. Cyclic boundary conditions on result in the introduction of a second quantum number, l, identifying acceptable solutions and restricting the values of ml by the value of l. Quantization: l = 0,,, and ml = l, l-,, -l+, -l. l is named the orbital angular momentum quantum number. ml is named the magnetic quantum number.

Energy is: E l (l ), l = 0,,, independent of ml, hence I a level with quantum number l is (l + )-fold degenerate. Normalized wavefunctions Yl,m l (, ) = spherical harmonics.

13 Energy is: E l (l ), l = 0,,, independent of ml, hence I a level with quantum number l is (l + )-fold degenerate. Normalized wavefunctions Yl,m l (, ) = spherical harmonics. It follows that: J l (l ) l = 0,,, And we saw that: Jz ml, ml = l, l-,, -l Note that: () Number of angular nodes increase with l. () No angular nodes around z-axis for ml = 0. (3) For a given l, the probability of migrating towards the x-y plane increases with ml.

14 An example of application: Under some circumstances, the particle on a sphere is a reasonable model to describe the rotation of diatomic molecules. Consider the rotation of a H7I molecule: because the large difference in atomic masses, it is appropriate to picture the H atom as orbiting a stationary I atom at a distance r = 60 pm, the equilibrium bond distance. The moment of inertia of HI is then: I = mhr = kg m It follows that:.97 0 J 0.80 mev I This energy corresponds to 78. Jmol-. The table below reports a few values for the first four levels: () rotational energy levels (in mev), () the degeneracies of the levels, and (3) the magnitudes of the angular momentum of the molecule: l 0 3 () () (3) 0 / 6/ / It follows from our calculation that the l = 0 and l = levels are separated by E = J =.69 mev. A transition between these two rotational levels of the molecule can be brought about by the emission or absorption of a photon with a frequency given by the Bohr frequency condition: E Hz 39.5 GHz h Radiation with this frequency belongs to the microwave region of the EMR spectrum, so microwave spectroscopy is a convenient method for the study of molecular rotations.

Lecture 5: Harmonic oscillator, Morse Oscillator, 1D Rigid Rotor

Lecture 5: Harmonic oscillator, Morse Oscillator, 1D Rigid Rotor It turns out that the boundary condition of the wavefunction going to zero at infinity is sufficient to quantize the value of energy that