763620SS STATISTICAL PHYSICS Solutions 2 Autumn 2012

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1 763620SS STATISTICAL PHYSICS Solutions 2 Autumn Continuous Random Walk Conside a continuous one-dimensional andom walk. Let w(s i ds i be the pobability that the length of the i th displacement is between s i and s i + ds i. Assume that the displacements ae independent on each othe and obey the same distibution w(s. a Show that the pobability of finding the total displacement (afte steps x s i between x and x + dx is P(xdx [ ( w(s 1 w(s 2 w(s δ x ] s i dx ds 1 ds 2 ds, whee δ(x is the Diac delta function and the integations go fom to. Solution: Let s take a set of displacements s,..., that sum to a fixed x. As the displacements ae independent of each othe, the common pobability is a staight poduct of the pobabilities fo the individual displacements: w(s 1 ds 1 w(s 2 ds 2 w(s ds. But as we ae not inteested only the one set of displacements s,..., that sum to a x, but all of them, we integate (i.e. sum ove all values of s,..., that sum to x, fomally: P (xdx... w(s 1 w(s 2 w(s ds 1 ds 2 ds dx }{{} x n s i [ ( ]... w(s 1 w(s 2 w(s δ x s i dx ds 1 ds 2 ds. b Show that P(x 1 dke ikx Q (k, whee Q(k dse iks w(s. Solution: Hee, the key thing is to expess the delta function by using the hint given in the Ex. 2: ( δ x s i 1 e ik(x si dk 1 e ikx e iks i e iks2 e iks dk. 1

2 Then, we can a bit make eaangements in the P (xdx P (xdx 1 e ikx e iks 1 w(s 1 ds 1 e iks 2 w(s 2 ds e ikx Q (kdk dx. e iks w(s ds 2. Continuous Random Walk - Discete Steps Assume that a paticle is popagating along a discete one-dimensional andom walk with the pobability density w(s pδ(s l+qδ(s+l, whee p (q is the pobability of taking a step of length l to the ight (left. Use the esults of the pevious poblem to show that the paticle can be found only at locations x (2n l, n 0, 1, 2,...,. Show that the pobability of finding the paticle at those locations is P (2n! n!( n! pn q n. (You might need the binomial expansion (x+y! n!( n! xn y n and the elation δ(x 1 dke ikx. Solution: Fist, we wite the explicit expession fo Q(k Q(k ds e iks w(s Then the pobability P (x P (xdx 1 e ( ikx pe ikl + qe ikl 1 dk dx n0 n0 ( n p n q n 1 ds e iks [pδ(s l + qδ(s + l] pe isl + qe isl. e ikx e ik(2n l dk dx! n!( n! pn q n δ(x (2n ldx, e ikx n0 ( (pe ikl n ( qe ikl n dk dx n whee we can ead that the paticle can be found only at the locations x (2n l, n 0, 1, 2,...,, with the pob. P (2n! n!( n! pn q n. 2

3 3. H-theoem - Appoach to Equilibium Assume that P is the pobability of finding the system in the micostate. The tansition pobability between any two micostates obeys W s W s. a Wite down the maste equation govening the time evolution of P. Solution: The maste equation means the equation fo the time evolution of P, i.e,. We intepet that the W s means the pobability pe time unit fo tansition fom the state to the state s. Thus P can incease by tansitions fom othe states to the state and decease by the tansitions fom the the state to othe states, fomally witten: s P s W s s P W s s W s (P s P. b Show that d P 0, whee the equality holds when P P s fo all and s. Intepet this esult! Solution:. ow, P means the expectation value fo P, that is P P P. Ou task is now to fom the time evolution equation fo it: d P ( dp 1 P + P ( P + 1. P ow, we can make estimates. Fist of all, fo P 1, ( P and P 0. These imply that d d P ( P + 1 0, that is, d P / 0. And if P P s fo all the states, we see that s W s (P s P 0, which implies that also d P / 0. Intepetation: If the pobabilities P ae not equal with each othe (i.e. P P s fo all possible pais s,, then the system is in a state of highe ode (H P is highe, entopy S ( H is lowe than in equilibium. This state tends to change so that H deceases, i.e. d P / < 0, until the equilibium is eached, whee P P s fo all the state s and. In othe wods, the entopy becomes maximized. 3

4 4. Spin System Conside an isolated system of vey weakly inteacting localized spin- 1 paticles with 2 a magnetic moment µ pointing eithe paallel o anti-paallel to an applied magnetic field B. a What is the total numbe of states Ω(E lying in enegy ange between E and E + δe, whee δe is small compaed to E but δe µb? Solution: The enegy of a 1-spin in the magnetic field is E µ B. Thus, if n 2 p (n a is the numbe of spins that ae paallel (anti-paallel with the magnetic field, then we can fom a set of equations fo the total numbe paticles and the total enegy E: { { n a + n p E (n a n p µb na E + 2 n p E +. 2 With fixed E, thee numbe of micos states is ( (E! n p n p!n a!! ( (! E 2. E! 2 The possible enegy states ae spaced by. By assuming that is sufficiently lage, we can estimate that the total numbe of states lying in the enegy ange [E, E+δE] is the numbe of micostates at the enegy E multiplied with the numbe of possible enegy states within δe. In othe wods, we assume that (E changes vey slowly within δe. Then we wite fo the total numbe of states: Ω(E (!! ( δe!. b Wite down the expession fo Ω(E. Use Stiling s fomula to simplify. Solution:The Stiling s fomula is n! n n n. By applying it we can wite: Ω(E δe ( + ( ( 2 + E ( 2 + E. c Use Gaussian appoximation fo Ω(E when E µb. Solution: The total numbe of states has maximum at E 0 and aound it it behaves smoothly as Gaussian ( e x2. Thus, we make Taylo expansion of Ω(E at E 0: Ω(E Ω(E 0 + d Ω(E 0 de E + 1 d 2 Ω(E 0 E 2 + O 3 2 de 2 4

5 By foming the deivatives of the expession fo Ω(E in (b, we get that Ω(E δe d Ω(E 1 [ ( de [ d 2 Ω(E 1 1 de 2 ( 2 which implies that Ω(E 2 + δe 1 Ω(E δe 2 exp ( E 2(µB 2 E2,. ( E2 2µ 2 B E ] ], 5. Appoach to Themal Equilibium Assume that two systems with diffeent values of β ae bought into themal contact. Show that the system with highe value of β will absob heat fom the othe until the two β values ae the same. Solution: In a themal contact, the total enegy E (0 E 1 + E 2 is fixed but the two subsystems can exchange enegy. Fo the combined system, the total numbe of micostates is Ω (0 Ω 1 (E 1 Ω(E 2 Ω 1 (E 1 Ω 2 (E 0 E 1. In the non-equalibium, we know that the entopy tends to incease (the second law of themodynamics. Let s study thus the entopy S k B Ω (0 k B Ω 1 (E 1 + k B Ω 2 (E 2 : ds d(k B Ω (0 ( Ω1 (E 1 E 1 k B E 1 t + Ω 1(E 2 E 2 E 1 E 2 E 1 t k B (β 1 β 2 E 1 t Thus, if β 1 β 2 then E 1 / t 0, i.e. the system with highe value of β will absob heat fom the othe system until the two β values ae the same. 0. 5