MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Statistical Physics I Spring Term Solutions to Problem Set #1


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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Deprtment Sttisticl Physics I Spring Term 03 Problem : Doping Semiconductor Solutions to Problem Set # ) Mentlly integrte the function p(x) given in the figure. The result rises from zero t decresing rte, jumps discontinuously by 0. t x = d, then continues to rise symptoticlly towrd the vlue. This behvior is sketched below. b) < x > = 0.8 x p(x) dx = x exp( x/l) dx + 0. l } 0 0 {{}} l = 0.8 l + 0. d xδ(x d) dx {{} d c) < > = x 0.8 x p(x) dx = x exp( x/l) dx + 0. x δ(x d) dx l } 0 {{}} 0 {{} l 3 d =.6 l + 0. d Vr(x) < (x < x >) > = < x > < x > = (.6 l + 0. d ) (0.64 l ld d ) = 0.96 l 0.3 ld 0.6 d
2 d) < exp( x/s) > = exp( x/s) p(x) dx = 0.8 l exp( x/s) exp( x/l) dx + 0. exp( x/s)δ(x d) dx } 0 {{ } 0 } {{ } (/s + /l) exp( d/s) = ( ) exp( d/s) + l/s Check to see tht this result is physiclly resonble. Note tht if the skin depth s is much less thn the distnce d, the impurities on the grin boundry do not contribute to the surfce impednce. Similrly, if the skin depth is much less thn the chrcteristic diffusion distnce l, the impurity contribution to the surfce impednce is gretly reduced.
3 Problem : A Peculir Probbility Density ) = p(x) dx = 0 dx b + x = dξ = (π/b) b } 0 + ξ {{ } π/ = (b/π) b) P (x) = x b x p(x ) dx = dx π b + x b = π [ x rctn(x /b) b = rctn(x/b) + π c) < x >= 0 by symmetry. p(x) is n even function nd x is odd. d) p(x) flls to hlf its vlue t x = ±b. e) b x < x >= dx π b + x However the limit of x /(b + x ) s x ± is unity, so this integrl diverges. Neither the men squre nor the Vrince of this distribution exist. 3
4 Problem 3: Visulizing the Probbility Density for Clssicl Hrmonic Oscilltor ) First find the velocity s function of time by tking the derivtive of the displcement with respect to time. ẋ(t) = d [x 0 sin(ωt + φ)] dt = ωx 0 cos(ωt + φ) But we don t wnt the velocity s function of t, we wnt it s function of the position x. And, we don t ctully need the velocity itself, we wnt the speed (the mgnitude of the velocity). Becuse of this we do not hve to worry bout losing the sign of the velocity when we work with its squre. ẋ (t) = (ωx 0 ) cos (ωt + φ) = (ωx ) [ sin 0 (ωt + φ)] = (ωx 0 ) [ (x(t)/x 0 ) ] Finlly, the speed is computed s the squre root of the squre of the velocity. ẋ(t) = ω(x 0 x (t)) / for x(t) x 0 b) We re told tht the probbility density for finding n oscilltor t x is proportionl to the the time given oscilltor spends ner x, nd tht this time is inversely proportionl to its speed t tht point. Expressed mthemticlly this becomes p(x) ẋ(t) = C(x 0 x ) / for x < x 0 where C is proportionlity constnt which we cn find by normlizing p(x). x0 p(x)dx = C (x 0 x ) / dx x 0 x0 dx/x 0 = C let x/x 0 y 0 (x/x 0) dy = C 0 y }} = πc {{ π/ = by normliztion The lst two lines imply tht C = /π. We cn now write (nd plot) the finl result. 4
5 p(x) = ( πx 0 (x/x0 ) ) x < x 0 = 0 x > x 0 As check of the result, note tht the re of the shded rectngle is equl to /π. The re is dimensionless, s it should be, nd is resonble frction of the nticipted totl re under p(x), tht is. c) The sketch of p(x) is shown bove. By inspection the most probble vlue of x is ±x 0 nd the lest probble ccessible vlue of x is zero. The men vlue of x is zero by symmetry. It is the divergence of p(x) t the turning points tht gives rise to the pprent imge of the pencil t these points in your experiment. COMMENTS If n oscilltor oscilltes bck nd forth with some fixed frequency, why is this p(x) independent of time? The reson is tht we did not know the strting time (or equivlently the phse φ) so we used n pproch which effectively verged over ll possible strting times. This wshed out the time dependence nd left timeindependent probbility. If we hd known the phse, or equivlently the position nd velocity t some given time, then the process would hve been deterministic. In tht cse p(x) would be delt function centered t vlue of x which oscillted bck nd forth between x 0 nd +x 0. Those of you who hve lredy hd course in quntum mechnics my wnt to compre the clssicl result you found bove with the result for quntum hrmonic oscilltor in n energy eigenstte with high vlue of the quntum number n nd the sme totl energy. Will this probbility be time dependent? No. Recll why the energy eigensttes of potentil re lso clled sttionry sttes. 5
6 Problem 4: Quntized Angulr Momentum ) Using the expression for the normliztion of probbility density, long with expressions for the men nd the men squre, we cn write three seprte equtions relting the individul probbilities. p( ) + p(0) + p() = p( ) + 0 p(0) + p() = < L x > = 3 x p( ) + 0 p(0) + p() = < L > = 3 We now hve three simple liner equtions in three unknowns. The lst two cn be simplified nd solved for two of our unknowns. p( ) + p() = p() = 3 p( ) + p() = p( ) = 3 6 Substitute these results into the first eqution to find the lst unknown. + p(0) + = p(0) = 6 3 b) 6
7 Problem 5: A Coherent Stte of Quntum Hrmonic Oscilltor [ iωt i x αx Ψ( r, t) = (πx ) /4 cos 0 exp (αxx 0 sin t α ω ω x t x 0 ) 0 sin ω ) ( ( x 0 0 t ) ] ) First note tht the given wvefunction hs the form Ψ = exp[ib + c] = exp[ib] exp[c] where, b nd c re rel. Thus the squre of the mgnitude of the wvefunction is simply exp[c] nd finding the probbility density is not lgebriclly difficult. (x αx 0 cos ωt) p(x, t) = Ψ(x, t) = exp[ ] πx x 0 0 b) By inspection we see tht this is Gussin with time dependent men < x >= αx 0 cos ωt nd time independent stndrd devition σ = x 0. c) p(x, t) involves time independent pulse shpe, Gussin, whose center oscilltes hrmoniclly between αx 0 nd αx 0 with rdin frequency ω. t=/ T t=3/4 T t=/4 T t=0 αx 0 0 αx 0 x Those lredy fmilir with quntum mechnics will recognize this s coherent stte of the hrmonic oscilltor, stte whose behvior is closest to the clssicl behvior. It is not n energy eigenstte since p(x) depends on t. It should be compred with clssicl hrmonic oscilltor with known phse φ nd the sme mximum excursion: x = αx 0 cos ωt. In this deterministic clssicl cse p(x, t) is given by p(x, t) = δ(x αx 0 cos ωt). The coherent stte is good representtion of the quntum behvior of the electromgnetic field of lser well bove the threshold for oscilltion. 7
8 Problem 6: BoseEinstein Sttistics We re given the discrete probbility density p(n) = ( ) n n = 0,,, ) First we find the men of n. < n >= np(n) = ( ) n n } {{ S } The sum S cn be found by mnipulting the normliztion sum. p(n) = ( ) n = ( ) n must = Rerrnging the lst two terms gives the sum of geometric series: n =. But note wht hppens when we tke the derivtive of this result with respect to the prmeter. d n = n n = n n = S d d ( ) lso = = d ( ) Equting the two results gives the vlue of the sum we need, S = /( ), nd llows us to finish the computtion of the men of n: < n >=. c) To find the vrince we first need the men of the squre of n. < n >= n p(n) = ( ) n n }{{} S 8
9 Now try the sme trick used bove, but on the sum S. d d }{{} n n = n n = S n n = Then nd S d ( ) lso = = + d ( ) ( ) 3 ( ) < n > = [ ] ( ) ( ) ( ) + = + ( ) 3 ( ) = < n > + < n >, Vrince = < n > < n > =< n > + < n > = < n > (+ < n >). This is greter thn the vrince for Poisson, < n >, by fctor + < n >. c) p(x) = ( = f(x) δ(x ) n δ(x n) n) Try f(x) = Ce x/φ, then f(x = n) = Ce n/φ = C(e /φ ) n = ( ) n. This tells us tht C = nd exp( /φ) =. We cn invert the expression found bove for < n > to give s function of < n >: =< n > /(+ < n >). ( < n > ) /φ = ln = ln + < n > /φ = ( < n > + ) ( ) ln = ln + < n > < n > 9
10 Recll tht for smll x one hs the expnsion ln( + x) = x x / Therefore in the limit < n > >>, /φ / < n > which implies φ < n >. 0
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