Quantum vs. Classical, Qtrans, Equipartition, Internal Degrees of Freedom

5.62 Lecture #10: Quantum vs. Classical. qtrans. Equipartition. Internal Degrees of Freedom. GOAL: Calculate average translational energy via quantum and classical descriptions and compare results. QUANTUM DESCRIPTION — Calculate –ε x — average kinetic energy in x-direction εx = ∞ ε x P ( ε x ) dε x ∫0 P ( ε x ) = ( πkT) ⎛ 1 ⎞ εx = ⎜ ⎝ πkT ⎟⎠ 1/2 ⎛ 1 ⎞ εx = ⎜ ⎝ πkT ⎟⎠ 1/2 −1/2 −1/2 −ε x kT εx e ∞ ∫0 −ε x ε1/2 x e kT dε x kT ( πkT)1/2 = 1 kT 2 2 1 This is not an accident. It is our first glimpse of “equipartition” of energy, 2kT into each independent degree of freedom. Integral table for two useful integrals ∫ ∞ ∫ ∞ 0 0 x1/2 e− ax dx = π1/2 2a 3/2 x −1/2 e− ax dx = ( π a ) 1/2 ( dimension: length 3/2 ) ( dimension: length1/2 ) Since the translational energies in each dimension are uncorrelated, average total energy is the sum of the average energy in each direction. Also, as a consequence of the separability of the Hamiltonian with respect to x, y, z coordinates, total E is sum of Ex + E y + E z. –ε = –ε + –ε + –ε x y z 1 2kT 1 1 3 + 2kT + 2kT = 2kT 1 Each degree of freedom contributes 2kT to total energy. Agrees with result from ensemble average 5.62 Spring 2008 Lecture #10, Page 2 3 ⎛ ∂ ln Q trans ⎞ ε = E = kT2 ⎜ = ∑ Pα E α = kT ⎟⎠ ⎝ ∂T 2 N,V α CLASSICAL MECHANICAL DESCRIPTION — calculate –ε State of a molecule is described by p momentum, q position. The energy of ~ ~ N–molecule system is ε(p 3N, q 3N), which is a continuous function of 6N variables. ~ ~ q trans = ∑ e − ε i kT ∫∫ e let's assume this by analogy ⇒ i quantum =V∫ ∫∫∫ ~ ~ dp dq ~ ~ classical — integral over 3 momentum and 3 position coordinates for each particle 2 q trans,cl = − ε( p,q )/kT 2 2 ↓= p x + p y + p z dq x dq y dq z ∫∫∫ dp x dp y dp z e−p dp x e−p x /2mkT ∫ 2 dp y e −p 2y /2mkT ∫ 2 / (2mkT) 2 dp z e−p z /2mkT 2 3/2 ⎡ ∞ ⎤3 = V⎢ ∫ dp x e p x /2mkT ⎥ = V (2πmkT) ⎣ −∞ ⎦ [qtrans,cl. should be dimensionless, but this has units m3l6t–3 = h3] The need to divide by a factor of h3 was recognized by Boltzmann even before Planck “invented” h. It is impossible for a trajectory in phase space to intersect itself. [Why? Because the classical Hamiltonian is a function of q,p. If the system is at q(t1), p(t1) at t1 then the H tells us the values of q and p at t1 + δt. If the system returns to q(t1), p(t1) at t2, then it must move to q(t1 + δt), p(t1 + δt) at t2 + δt.] The ergodic hypothesis requires trajectories to visit “each cell” of phase space of volume h3N, not “each location” in phase space. This avoids the problem of self-intersecting trajectories. 3/2 ⎛ 2πmkT ⎞ V , we see that we need a factor of h3 in the denominator Comparing to q trans,qm = ⎜ ⎝ h 2 ⎟⎠ of qtrans,cl.. In fact, it turns out that to construct a partition function from a classical Hamiltonian, this h–3 factor is required, so our assumption above must be corrected: q trans,cl = Q trans,cl 1 ∫∫ e −ε classical 3 h kT dp dq ~ (dimensionless) ~ qN 1 ⎡ −ε = trans,cl = ⎣ ∫ ∫ e classical 3N N! N!h kT N VN ⎛ 2πmkT ⎞ ⎤ dp dq ⎦ = ⎜ ⎟ N! ⎝ h 2 ⎠ 3N/2 revised 2/29/08 8:52 AM 5.62 Spring 2008 Lecture #10, Page 3 CONCLUSION: Classical and quantum descriptions of translational degrees of freedom yield consistent results. QM is operating in classical limit because energies of the translational states are so closely spaced that they can be approximated as continuous, as in classical mechanics. kT = 0.6 kcal mol–1 at 300K ∆ε = 10 –20 kcal mol–1 (particle in box) 3 The average molecular translational energy, 2 kT, is independent of the kind of molecule, or more precisely, independent of the energy level spacings because these spacings are ~ 1020 times smaller than kT. Spacings don't change much from He, to C6H6, to DNA. If T were decreased to – 10–19 K, then He, C6H6, and DNA would have different Etrans. CLASSICAL EQUIPARTITION OF ENERGY PRINCIPLE 1 Each squared momentum or position term in the energy of a particle contributes 2 kT to the average energy; according to this principle, translation along each coordinate axis contributes 1 1 2 kT; rotation about each principal axis contributes 2 kT; and each vibrational mode contributes 1 kT (2 kT each for kinetic and potential energy) to the average energy. −ε (p 3N ,q 3N ) kT g(ε)e P (ε) = ∫∫ −ε(p e ~ 3N ~ ,q 3N ) kT ~ ~ Classical Boltzmann distribution function d p 3N d q 3N N!h 3N ~ ~ g(ε) is the density of states, often denoted as ρ(ε). It has units (energy)–1. ∞ 1 1 1 3 ε = ∫ εP ( ε ) dε = εx + εy + εz = kT + kT + kT = kT 0 2 2 2 2 INTERNAL DEGREES OF FREEDOM — FACTORIZATION OF Q ATOMS — have one internal degree of freedom ELECTRONIC degree of freedom revised 2/29/08 8:52 AM 5.62 Spring 2008 Lecture #10, Page 4 MOLECULES — have other degrees of freedom ELECTRONIC, VIBRATION, AND ROTATION, each of which contributes to total energy and to other macroscopic properties. Nuclear hyperfine? [Nuclear spin degeneracy factors. LATER.] Internal energy adds to translational energy to get total energy ε = εtrans εint + quantum #'s N,M,L internal quantum #'s where εint = energy from internal degrees of freedom q = ∑ e−εi −ε +ε = ∑ e ( trans int ) kT kT i all molecular states all molecular states We do not have to start over. qtrans factors out. q = ( ∑ e −ε trans kT translational states q= qtrans • qint N q trans q int ) ( Q= N! Q = QtransQint Q trans = QN trans N! Q int = Q N int )( ∑ e−ε int kT ) internal states ← INTERNAL MOLECULAR PARTITION FUNCTION N ⎛q ⎞ = ⎜ trans ⎟ q N ⎝ N! ⎠ int CANONICAL PARTITION FUNCTION CANONICAL TRANSLATIONAL PARTITION FUNCTION CANONICAL INTERNAL PARTITION FUNCTION revised 2/29/08 8:52 AM 5.62 Spring 2008 Lecture #10, Page 5 NOTE: N! was included with Qtrans. This is because it's the translational motion that causes the positions of identical particles to be interchanged, requiring the factor of N! The internal motions don't interchange particles. Classically Qcl = Qtrans,cl Qint,cl Q trans,cl = q Ntrans,cl N! [∫ e − ε trans /kT dpdq] N = N!h 3 N Q int,cl = q Nint = ∫ dp 3N dq 3N e − εint /kT h 3N   CONTRIBUTION OF INTERNAL DEGREES OF FREEDOM TO MACROSCOPIC PROPERTIES ⎛ ∂ ln Q ⎞ E = kT2 ⎜ = ⎝ ∂T ⎟⎠ N,V ⎛ ∂ ln Q transQ int ⎞ kT2 ⎜ ⎟⎠ ⎝ ∂T N,V ⎛ ∂ ln Q trans ⎞ E = kT2 ⎜ ⎟ ⎝ ∂T ⎠ N,V + E= ⎛ ∂ ln Q int ⎞ kT2 ⎜ ⎝ ∂T ⎟⎠ N,V E trans E int (will equipartition be applicable here?) A = −kTln Q = −kTln Q transQ int = −kTln Q trans + −kTln Q int = A trans + Aint Likewise: S = Strans + Sint ⎛ ∂ln Q ⎞ p = kT ⎜ = p trans + p int = p trans ⎟ ⎝ ∂V ⎠N,T because no V dependence for internal coordinates But for pressure: Could use the Classical Mechanical approach to compute any average quantity, e.g. −ε q 3 , p 3 ∫  ∫ dq 3dp 3x q 3 , p 3 e (   ) x=    −ε  q 3 ,p 3 kT ∫  ∫ dq 3dp 3e (   )   ( ) kT revised 2/29/08 8:52 AM 5.62 Spring 2008 Lecture #10, Page 6 The factors of h–3 in numerator and denominator cancel. The Classical picture will prove extremely useful when there are inter-particle interactions. Non-Lecture Example of a classical mechanical calculation: particle bound in a deep well near a solid surface. ε β L V(z) α 0 0 0≤z≤α α≤z≤L εz = p 2z 2m εz = p 2z 2m + β pz = [2mε]1/2 pz = [2m(ε−β)]1/2 To compute qCl we need L ∞ ∫ = [ α + (L − α)e = [ α + (L − α)e −β kT q z = [ α + (L − α)e −β kT qz = 0 dz ∫ dp z e − ε(z,pz ) kT = −∞ α −∞ dp z ∫ dze − pz 2 2mkT 0 ∞ L −∞ α + ∫ dp z ∫ dze − pz 2 2mkT ] ∫ dp e ][ 2πmkT] ∞ −∞ −β kT ∫ ∞ − p 2z 2mkT z 1/2 1/2 ⎤ ] ⎡⎢⎣ 2πmkT h 2 ⎥⎦ The multiplicative contributions from the x and y directions do not have the pre-factor q Cl = [ α + (L − α)e −β kT ⎤ ] ⎡⎢⎣ 2πmkT 2 ⎥ h ⎦ 3/2 for kT  β ⎡ 2πmkT ⎤ q Cl = ⎢ ⎣ h 2 ⎥⎦ 3/2 for kT  β ⎡ 2πmkT ⎤ q Cl = ⎢ ⎣ h 2 ⎥⎦ 3/2 L2 L3 L2α revised 2/29/08 8:52 AM 5.62 Spring 2008 Lecture #10, Page 7 Of course this derivation is not quantitatively correct because it is necessary to assume that kT  β. It is also incorrect because the exact quantum mechanical energy level in the deep, nearsurface well is not at the bottom of the well. But this calculation reveals the qualitative change in qtrans between the low-T limit (all molecules are adsorbed, thus confined to a volume L2α) and the high-T limit (all molecules are desorbed). revised 2/29/08 8:52 AM