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NMR signal will be proportional to the difference in occupation number ofthe levels, which in turn is proportional 1/T.4.6 Localised one-dimensional harmonic oscillatorsFrom quantum mechanics we have that the energy levels of a harmonicoscillator are given byεn (n 12 )!ω , n 0,1,2 We can now easily write down the partition functionZ e εn / k BT en ( n 12 ) ! ω / kB T e ( n 12 )θ /Tn 12 θ /T e e (n n 12 )θ /T Z0 Ztermnwhere we have introduced the scale temperature θ !ω /k B .In this case we can explicitly sum the partition function, it is a geometricalseries.Z e ( n 12 )θ /Tn e θ /2T e nθ /T n()e θ /2T 1 e θ /T e 2θ /T e θ /2T1 θ /T θ /T1 ee 1The internal energy isE NNk θ ln Z 1N!ω 2 N!ω !ω /kBT 12 NkBθ θ /T B βe 1e 1The first term is the zero-point energy. At hightemperatures the internal energy isE NkBTa result that we derived in an earlier chapter bycounting the number of quadratic terms in the expression for the total energy.However, the quantum mechanical model also gives the behaviour at lowtemperature.θ /T ) eθ /T(dEThe heat capacity is C . NkB2dTeθ /T 12()It starts from zero at T 0 and saturates asexpected for high temperatures at Nk B . Againquantum mechanics describes correctly thebehaviour at low temperatures. We can see theneed for a quantum mechanical description as for low temperatures both theinternal energy and the heat capacity contain Planck’s constant.We can finally compute the entropy from S Fwhere F Nk BT ln Z . ��37

eθ /Tθ /T Exercise. Show that S Nk B ln θ /T θ /T e 1 e 1 Exercise. Show that S 0 when T 0 and S Nk B lnTkT Nk B ln B whenθ!ωT ��Note that also in the high temperature limit the entropy contains Planck’sconstant. Entropy is fundamentally a quantum mechanical quantity. On theother hand the energy and heat capacity can be described classically at hightemperatures.Albert Einstein used the above model with one-dimensional harmonicoscillators in the so-called Einstein model to explain the properties of solids atlow temperatures. The model is qualitatively correct but with closercomparison with experiment it has the wrong temperature dependence.Experiments show that the heat capacity at low temperatures must be3proportional to T . We will in a later chapter present the Debye model that hasa correct T dependence. However, we can already now explain that small heatcapacities of graphite and diamond. As we have seen, the scale temperaturedetermines the transition between the quantum mechanical and classicalregions in temperature, in this case θ !ω /k B . For a harmonic oscillator wehave that the frequency ω k/m where k is the spring constant of the forcebetween the atoms in the solid. Diamond is a very hard solid, thus we canexpect that the spring constants in diamond are very large. This implies thatthe scale temperature is high, about 500 K for diamond. This means that theheat capacity is quite far away from its asymptotic value at room temperatureand that the heat capacity is much below the classical Dulong-Petit value.Even more interesting is the situation with graphite. Themolecular structure of graphite is such that is consists oflayers of carbon atoms with each layer being a very rigidhexagonal lattice. These layers are stacked on top ofeach other and quite loosely connected. This is one of the reasons for graphitebeing used in pencils; the layers are easily ripped off and get attached to thepaper. This is also the reason for graphite being used as a lubricant, thedifferent layers in the graphite can slide relative one another with littlefriction. This structure means that the chemical bindings (the springconstants) are strong in two dimensions, in the hexagonal layer, and weakbetween these layers. We then have two different scale temperatures, a high38

one for vibrations in the layer and a low one for vibrations between the layers.Actually only the vibrations between the layers are activated at roomtemperature. When we count quadratic terms in the total energy we shouldonly count these latter vibrations. This implies that the number of quadraticvibration terms only is 1/3 of the number in a solid where the vibrations canhappen in three dimensions and the heat capacity thus only becomes 1/3 ofthe normal, something that agrees perfectly with experiment.4.7. A note about the partition function for systems withdegenerate energy levelsWe have earlier noted that we should sum over all possible states in thepartition function. This means that if an energy level is degenerate I has to becounted several times, as many times as the degeneracy or the multiplicity ofthe degeneration. Mathematically we write thisZ gi e ε i /k BTiwhere gi is the multiplicity of energy level i. We will use this way of writingvery often when we in the next chapter will treat the statistical description ofgases where we will use so-called group distributions.Exercise problems. Chapter 41. Compute the internal energy of a two-level system using E N ln Zand βcheck that you get the same result as we got in the lecture.2. Assume that we could have a stable equilibrium for a two-level systemwhere we had more particles in the upper level than in the lower one. Whatcan you say about the temperature of such a system? In reality such systemsare not stable but can be realised quasi-stably in for instance a laser bypumping particles to the upper level using external energy. Such systems aresaid to have an inverted population.3. Compute the internal energy and heat capacity for a system with one nondegenerate level with energi 0, two degenerate levels with energy ε and onenondegenerate level with energy 2ε. Hint: Chose a suitable zero-point energy.4. Below what temperature do you have deviations that are larger than 5%from Curie’s law in an ideal electronic spin 1/2 system?5. Magnetisation of Pt nuclei is often used as a thermometer for lowtemperatures. The external magnetic field is 10 mT and the magnetic momentof a Pt nucleus is 0.60 nuclear magnetons. Estimate the usable temperatureinterval of the thermometer if we assume a) that a magnetisation less than1/10 000 of the maximal one cannot be measured with precision and b) thatdeviations from Curie’s law that are larger than 5% are unacceptable.Assume that we use NMR technique to detects the split in the energy levels.What is in this case the NMR frequency?39

6. Experimental results frommeasurements of the specificheat capacity of gadolinium areshown in the figure. At lowtemperatures you see a bump inthe curve caused by a few lowlying energy levels. Explain thisand estimate the distancebetween these levels. Give youranswer in electron volts. Youhave to motivate your answerbut you don’t have to makedetailed computations.Cp[mJ / mol]20001000500200540101520T [K]

5. The ideal gas5.1 Density of statesTo begin with we will consider gases with non-relativistic massive particles.We will need the density of state in energy.We now study bas particles in a “box” with dimensions LxLxL.If we solve the Schödinger equation for such a particle we get wave functionsof the typeϕ ( x, y, z ) Ae ikx x e y e ikz zWe now use so-called periodic boundary conditions where we demand thatϕ x nx L, y n y L, z n y L ϕ ( x, y, z )ik y()This impliesk x nx2π2π2π, ky ny, k z nzLLL2πbetween theLlattice points. Each lattice point corresponds to a state of the gas particle. In anelementary cube we then have exactly one state. The volume of an elementaryIn k-space, these k-values form a cubic lattice with a distance( 2π ) . The density of states in k-space is then 2π cube is evidently L V3( 2π )1/3 V3V( 2π )3You always start with the number of states in an infinitesimal cube in k-spaceand the transform step by step to ε-space:V( 2π )3d3k ( 2π )34π 2 p 2mε V 4π m ( 2m)p dp !3!3( 2π )31/22V( 2π )spherical1 3 symmetryd p !3V3 ε 1/2 dε g ( ε ) dεwhere V is the normalisation volume. The factor 4π comes from that weintegrate over the uninteresting space angles. We have also exploited that themomentum p !k and the relation (in this case) between energy andp2.2mThe relation between energy and momentum/wave vector is called adispersion relation.momentum ε V (2m)This gives the density of states in energy g(ε) 23(2π ) !413 /2ε1/ 2

This expression must be corrected with a spin factor, 2 for spin 1/2 particlesand 3 for (massive) spin 1 particles. The density of states is such thatg ( ε ) dε is the number of states in the energy interval [ ε , ε dε ] .Note 1. It is easy to realise that our box doesn’t have to be cubic. Actually outderivation works for a volume of arbitrary form, thermodynamical propertiesdo not depend on the form of the container.Note 2. We could have used the usual condition when you solve theSchödinger equation in a box, namely that we have standing waves betweenthe walls. This gives a final result that is identical to that we got here but themethod with periodic boundary conditions often is simpler to use.5.2 Group distributionsFirst we observe that the number of states in a gas in a macroscopic volume isenormous and that the energy levels are very close to each other. To handlethis situation we introduce something called group distributions. See the figure!We group such that the number of levels within a group i, gi , is very large,and such that the number of particles within the group, ni , also is large, butsuch that the average distances Δε i between the levels in the different groupsstill is very small. In practice it turns out that you can gi of order 10 and stillhave Δε i 10 9 k B T . It also turns out that it is not critical for the final resulthow we do the grouping.105.3 Identical particlesWe now want to count the number of microstates in a certain distribution. Wethen have to take into account an important fact: particles in a gas arefundamentally identical, that is it is impossible to tell them apart. We cannotdecide which particle occupies a certain state; we can only decide the numberof particles in a certain state. The other important fact is that we have twokinds of particles in nature, fermions and bosons. Fermions follow the Pauliprinciple that says that two fermions cannot be in the same state. Fermionsalways have half integer spin; examples are electrons, protons, and neutrons.Bosons do not follow the Pauli principle; an arbitrary number of bosons can bein the same state. Bosons have integer spin; examples of bosons are photons,helium-4 atoms, and deuterons. This means that we have to count microstates42

differently for fermions and bosons. For fermions we will get Fermi-Dirac(FD) statistics for bosons we will get Bose-Einstein (BE)-statistics.5.4. Counting microstates for fermionsConsider the level group i where we have gi possible states and ni particles tooccupy these states. As we deal with fermions we can have at most oneparticle in each state. Consequently we have gi ni empty states. We then canarrange filled and empty states ingi !ni ! ( gi ni )!ways. The total number of microstates for all levels then isgi !ΩFD i ni ! ( gi ni )!5.5. Counting microstates for bosonsAgain consider level group i. We have gi possible states and ni particles todistribute in these states. We can now have several particles in each statesomething that complicates the counting. However, we can get around thisproblem using a trick. Consider the states as slots with walls in between. Ifthere are gi states there are gi 1 walls between them. Together with the niparticles we now have gi ni 1 objects to handle, gi 1 of one kind (thewalls) and ni of the other kind (the particles) and we can arrange them in( gi ni 1)! ( gi ni )!ni ! ( gi 1)!n i! g i !ways. The total number of microstates for all level is then ΩBE i( gi ni )!ni ! gi !5.6. Dilute gasesNow suppose that we have a dilute gas. By this we mean that ni gi (butstill ni 1 ). This is a very common situation if we for instance consider air atnormal temperature and pressure. In the fermion case we then can make theapproximationg ( g 1) ( gi ni 1) gin igi ! i i ni ! ( gi ni )!ni !ni !In the boson case we have( gi ni )! ( gi ni )( gi ni 1) ( g i 1) g ni ini ! gi !ni !ni !that is we get the same result. This “classical limit” gives us the so-calledMaxwell-Boltzmann (MB) statistics where we havegin iΩ MB ni !i43

5.6. Distributions in thermodynamical equilibrium5.6.1. FermionsWe now copy the procedure that we used for distinguishable particles inchapter 3. We define the entropyS kB ln Ωand try to maximise the entropy given the constraintsN ni and E ε iniiithat is the number of particles and the internal energy are conserved.We maximise the function f gi ln gi ni ln ni ( gi ni ) ln ( gi ni ) α ni N β ε ini E i i iwhere we have used Stirling’s approximation and introduced Lagrangianmultipliers. We compute all partial derivatives with respect to ni and putthem to zero and get after some manipulationgni α βε iie 1It is here useful to define the filling factor, the ratio between the number ofparticles and the number of accessible states in this leveln1fi i α βεigi e 1As the levels are very close we can id we want write this as a continuousdistribution1fFD ( ε ) α βεe 1()5.6.2 BosonsWe repeat the procedure for the boson case and get with a similarcomputation1fBE (ε ) α βεe ��Exercise. Show this by doing the detailed ��––––5.6.3 Dilute gasesFinally, for dilute gases we have1f MB (ε) α βεeExercise. Show ––44

5.7. SummaryAs before β 1/k BT . α can in principle be determined from the condition ni N but it turns out that you can explicitly solve the resulting equationionly in the MB case, we will return to this in the next chapter. If we for theαmoment introduce e B , we can summarize our results in the following way 1 ε / kBT(FD) Be 1 1f (ε ) ε / kBT(MB) Be 0 ε / k1T Be B 1 (BE)As we will see the different kinds of particle system will have very differentphysical properties.45

6. Maxwell-Boltzmann gases6.1. The partition functionWe start by rewriting the filling factor1fi ε i /k BT Ae ε i /k BTBeIn this case we can easily compute A (i.e. implicitly the value of α). We haveN ni gi fi A gi eii ε /k BTiorfi withN ε i /k BTeZZ gi e ε i /k BTithe partition function. This is very similar to what we had before in chapter 3.We now consider a monoatomic gas where the gas atoms have spin zero asfor instance in helium-4. The partition function can be written as an integral asthe energy levels are close3/ 2 ε i / k BTV (2m) 1/ 2 ε / kB TZ gie g(ε)dε e ε dε e ε / kBT 23( 2π ) ! 0i0We put x2 ε /kBT and get 2V ( 2m)3/2Z 2kTx2 e x dx() B32π!0where we have “extracted the physics” from the integral that is now just anumber. The integral has the numerical value π 1/2 /4 and we have finally3/21/2V ( 2m)3/2 πZ 2kT(B ) 42π!3We can now check if our gas can be considered as dilute when for instance thetemperature is 5 K. If we insert physical values (at normal pressure we havethat N /V is one mol of particles per 20 litres) we find f N /Z 0.1 , stillquite small. The quantity N /Z is sometimes called the degeneracy parameter.At room temperature this parameter is extremely small which shows that wethen safely can describe normal gases using MB statistics. For further use wecan also note that Z V .3/246

6.2. Velocity distributionsIn many cases in kinetic gas theory we are interested in the velocitydistribution of the molecules of the gas. We then need the density of states inspeed that we easily deriveVV 1 3d3 k d p 33 3(2π )( 2π ) !V(2π )3p mv1 2V 12p dp 4π mv) mdv323 (!2π !that givesg( v) we then haveV 12mv) m23 (2π !N ε( v ) /k BTeZIf we insert our expression for the density of states and the earlier derivedexpression for the partition function we get m 3/2 2 mv 2 /2k TB v en(v) dv 4πN dv , 2πkBT the number of particles with speed in the interval [ v, v dv ] .Maxwell derived this relation was by long before quantum mechanics wasknown. We also see that the result does not contain Planck’s constant thatmeans that the result is “classical” and consequently be derived without usingquantum mechanics. We also know that the result cannot be expected to bevalid for very dense or cold gases when quantum phenomena start to appear.The distribution n(v) as a function of v, starts from zero, increases as thespeed increases, has a maximum and then decreases exponentially. You cancompute some interesting representative speed for the distribution:n(v) dv g (v) dv f (v) g( v)dv1. The speed for which n ( v ) has a maximum. Maximum occurs when k BT 1/2 k BT 1/2 1.4 vmax 2 m m Exercise. Show this!2. The average speed is v v n( v)dv0 n(v)dv1/2 kBT 1/28 kBT 1.6 π m m 0Exercise. Show this!47

3. The RMS (root mean square) speed is 2 1/2 v n( v)dv k BT 1/2 kBT 1/2 0 2 1.7 v 3 m m n( v)dv 0 From the last result we find that13ε m v2 kBT ,22a result that we got in chapter 3 using the equipartition theorem: a freeparticle has three quadratic terms in the total energy, each one correspondingto an (average) energy of 12 kBT .6.3. Internal energy and heat capacity ln Z 3 2 NkBT βdE 3From which follows CV NkdT 2 BHere, we could have used the result from the last section where we computedthe average energy of one molecule.We use E N6.4. Entropy and free energy and moreWe useginiS kB ln Ω kB ln i ni !()kB ( ni ln gi ni ln ni ni ) kB ni ln ( gi /ni )i 1 iikB ni ( ln Z ln N ε i / kBT 1) iNkB ln Z NkB ln N NkB E/T NkB ln Z kB ln N ! E/TWe note that this expression is similar to what we had before fordistinguishable particles only we also have the term kB ln N! . The extra termcompensates for that we count to many permutations if we havedistinguishable particles.For the free energy we now haveF E TS NkBT ( ln Z ln N 1)Pressure is defined by F p NkB Tln Z Nk B Tln(const V ) NkB T /V V V VorpV NkB TWe have derived the equation of state for an ideal gas!48

Exercise problems. Chapter 61. What is the RMS speed for air molecules at room temperature? Why doesn’tthe moon have an atmosphere?2. Consider a model of a HCl molecule (chloric acid). You may assume thatthe molecule c

zero quit fast, like T 2e θ/T. At high temperatures the heat capacity also goes to zero like T 2. This behaviour is typical for a two-level system and is called a Schottky anomaly. That the heat capacity goes to zero as the temperature goes to zero is universal for any system. As we have seen this is required