Grand-Canonical Ensembles

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Grand-canonical ensembles As we know, we are at the point where we can deal with almost any classical problem (see below), but for quantum systems we still cannot deal with problems where the translational degrees of freedom are described quantum mechanically and particles can interchange their locations – in such cases we can write the expression for the canonical partition function, but because of the restriction on the occupation numbers we simply cannot calculate it! (see end of previous write-up). Even for classical systems, we do not know how to deal with problems where the number of particles is not fixed (open systems). For example, suppose we have a surface on which certain types of atoms can be adsorbed (trapped). The surface is in contact with a gas containing these atoms, and depending on conditions some will stick to the surface while other become free and go into the gas. Suppose we are interested only in the properties of the surface (for instance, the average number of trapped atoms as a function of temperature). Since the numbers of atoms on the surface varies, this is an open system and we still do not know how to solve this problem. So for these reasons we need to introduce grand-canonical ensembles. This will finally allow us to study quantum ideal gases (our main goal for this course). As we expect, the results we’ll obtain at high temperatures will agree with the classical predictions we already have, however, as we will see, the low-temperature quantum behavior is really interesting and worth the effort! Like we did for canonical ensembles, I’ll introduce the formalism for classical problems first, and then we’ll generalize to quantum systems. So consider an open system in contact with a thermal and particle reservoir.

system: E,V,N,T,p, µ

µ reservoir: ERRRRR , V , N ,T , p , R

Figure 1: Model of an open thermodynamic system. The system is in contact with a thermal and particle reservoir (i.e., both energy and particles get exchanged between the two, but E + ER = ET = const, N + NR = NT = const). The reservoir is assumed to be much bigger than the system, E ≪ ER,N ≪ NR. In equilibrium T = TR,µ = µR.

As we know (see thermo review), the macrostate of such a system will be characterized by its temperature T (equal to that of the reservoir), its volume V and its chemical potential µ (which will also be equal to that of the reservoir) – again, I am using a classical gas as a typical example. We call an ensemble of very many copies of our open system, all prepared in the same equilibrium macrostate T,V,µ, a grandcanonical ensemble. As always, our goal is to find out the relation- ships that hold at equilibrium between the macroscopic variables, i.e. in this case to find out how U,S,p, hNi,... depend on T,V,µ. Note that because the number of particles is no longer fixed, we can only speak about the ensemble average hNi (average number of particles in the container for

1 given values of T,V,µ) from now on.

1 Classical grand-canonical ensemble

As was the case for the canonical ensemble, our goal is to find the density of probability ρg.c.(N, q, p) to find the system in a given microstate – once we know this, we can compute any ensemble average and answer any question about the properties of the system. Note that since the number of microsystems (atoms or whatever may be the case) that are inside the system varies, we will specify N explicitly from now on: a microstate is characterized by how many microsystems are in the system in that microstate, N, and for each for these microsystems we need f generalized coordinates and f generalized momenta to describe its behavior, so we have a total of 2Nf microscopic variables q, p. So we need to figure out ρg.c.(N, q, p). We will follow the reasoning we used for canonical ensembles – have a quick look over that and you’ll see the parallels. In fact, I’m going to copy and paste text from there, making only the appropriate changes here and there. First, we reduce the problem to one that we already know how to solve, by noting that the total system = system + reservoir is isolated, and so we can use microcanonical statistics for it. The R R microstate of the total system is characterized by the generalized coordinates N, q, p; NR, q , p , where the latter describe the microsystems that are inside the reservoir. Clearly, N +NR = NT is the total number of microsystems in the total system, which is a constant. So in fact the total system’s R R microstate is characterized by: N, q, p; NT − N, q , p . R R R R Its Hamiltonian is HT (N, q, p; NR, q , p ) = H(q, p)+ HR(q , p ) – here I won’t write the dependence of the Hamiltonian on N explicitly, but we know it’s there: for example, in the kinetic energy we sum over the energies of the microsystems inside the system, and the number of terms in the sum is N, i.e. how many microsystems are contributing to the energy in that particular microstate. Based on the microcanonical ensemble results, we know that the density of probability to find the R R total system in the microstate N, q, p; NT − N, q , p is:

1 R R R R Ω (E ,δE ,V,V ,N ) if ET ≤HT (N, q, p; NT − N, q , p ) ≤ ET + δET ρmc(N, q, p; NT −N, q , p )= T T T R T ( 0 otherwise where ΩT (ET ,δET ,V,VR,NT ) is the multiplicity for the total system. This is the same as saying that the total probability to ﬁnd the system in a microstate with N microsystems that are located between q, q + dq and p, p + dp AND the reservoir to have the remaining NT − N microsystems located between qR, qR + dqR and pR, pR + dpR is:

R R dqdpdq dp R R ρmc(N, q, p; NT − N, q , p ). Nf+(NT −N)fR GN GNT −N h Note that even though the microsystems are the same, they may have different number of degrees of freedom in the system and the reservoir – think about the previous example with atoms adsorbed on a surface. It is very likely that we need different degrees of freedom to describe the state of an atom when on the surface (inside the system) than when it’s in the gas (the reservoir). However, the total number of atoms on the surface plus inside the gas is fixed, and that’s all that matters. We do not care/need to know what the reservoir is doing, all we want to know is the probability that the system is in a microstate which has N microsystems between q, q + dq and p, p + dp. To

2 ﬁnd that, we must simply sum up over all the possible reservoir microstates, while keeping the system in the desired microstate. Therefore R R dqdp dqdpdq dp R R ρg.c.(N, q, p) = ρmc(N, q, p; NT − N, q , p ), Nf Res Nf+(NT −N)fR GN h Z GN GNT −N h where the integral is over all the reservoir’s degrees of freedom. After simplifying, we have:

R R dq dp R R ρg.c.(N, q, p)= ρmc(N, q, p; NT − N, q , p ) Res (NT −N)fR Z GNT −N h 1 dqRdpR = N N f E ≤H(q,p)+H (qR,pR)≤E +δE ( T − ) R ΩT Z T R T T GNT −N h 1 since ρmc = when this condition is satisﬁed, and zero otherwise. However, we can rewrite the ΩT condition as: 1 dqRdpR q p ρg.c.(N, , )= N N f → E −H(q,p)≤H (qR,pR)≤E −H(q,p)+δE ( T − ) R ΩT Z T R T T GNT −N h

ΩR(ET −H(q, p),δET ,VR,NT − N) ρg.c.(N, q, p)= ΩT (ET ,δET ,V,VR,NT ) since the integral is by deﬁnition just the multiplicity for the reservoir to be in a macrostate of energy ET −H(q, p), with VR and NT −N microsystems. If you have a quick look at the canonical ensemble derivation, you’ll see that so far everything went very similarly, except that we kept track carefully of how many microsystems are where. Using the link between the entropy of the reservoir and its multiplicity SR = kB ln ΩR (because the reservoir is so big and insensitive to what the system does, all microcanonical formulae valid for an isolated system apply to the reservoir), we then have:

∂SR ∂SR SR(ET ,δET ,VR,NT )−H(q,p) −N SR(ET −H(q,p),δET ,VR,NT −N) ∂ER ∂NR k k ΩR(ET −H(q, p),δET ,VR,NT − N)= e B ≈ e B where I used the fact that the energy of the system H(q, p) ≪ ET and N ≪ NT and performed a Taylor expansion. So the appearance of the second derivative is where the difference between canonical and grandcanonical ensembles shows up. But we know that at equilibrium ∂S 1 1 ∂S µ µ R = = ; R = − R = − ∂ER TR T ∂NR TR T so we find the major result: 1 ρ (N, q, p)= e−β[H(q,p)−µN] (1) g.c. Z where Z is a constant (what we obtain when we collect all terms that do not depend on (N, q, p)), called the grand-canonical partition function. We can find its value from the normalization condition: ∞ dqdp ∞ dqdp q p −β[H(q,p)−µN] 1= Nf ρg.c.(N, , ) →Z(T,V,µ)= Nf e (2) GN h GN h NX=0 Z NX=0 Z

3 Note that here “sum over all microstates” means to sum over all possible numbers N of microsystems in the system, and for each N to “sum” over all possible locations/momenta of the microsystems. Of course, for classical systems we know that the “sum” over locations/momenta is really an integral, because these are continuous variables. You might argue that we should stop the sum over N at NT , but since NT is by definition much much bigger than the average number of microsystems in the system, it will turn out that we can use the upper limit to be infinity – it makes calculations easier and the “error” can be made arbitrarily small by making the reservoir bigger, i.e. NT → ∞. Note also that Z is a function of T (through the β in exponent), of V (the integrals over positions are restricted to the volume V ), and µ (again through the exponent). Now that we know the grandcanonical density of probability, we can calculate the internal energy ∞ dqdp 1 ∞ dqdp q p q p q p q p −β[H(q,p)−µN] U = hH( , )i = Nf ρg.c.(N, , )H( , )= Nf H( , )e GN h Z GN h NX=0 Z NX=0 Z Here we have to be a bit careful. We can’t simply use the trick with the derivative with respect to β, since this will bring down both H (which we want), but also µN (which we don’t want): ∂ − e−β[H(q,p)−µN] =[H(q, p) − µN] e−β[H(q,p)−µN] =6 H(q, p)e−β[H(q,p)−µN] ∂β So here’s what we do. Let me define: α = βµ and use this instead of µ as a variable, so that Z = Z(β,V,α) and 1 ρ (N, q, p)= e−βH(q,p)+αN g.c. Z Now, we have to pretend that α and β are independent variables, i.e. we “forget” for a bit what is definition of α, we pretend that it’s just some quantity totally unrelated to β. If this was true, then we could use: ∂ − e−βH(q,p)+αN = H(q, p)e−βH(q,p)+αN ∂β and we could then write: ∞ 1 ∂ dqdp −βH(q,p)+αN U = − Nf e → Z " ∂β # GN h NX=0 Z 1 ∂ ∂ U = − Z(β,V,α)= − ln Z(β,V,α) Z(β,V,α) ∂β ∂β So the point is to treat α as a variable independent of β while we take the derivative, and only after we’re done with taking the derivative to remember that α = βµ. This “forgetfulness” is very useful since it allows us to calculate another ensemble average very simply, namely: ∞ dqdp 1 ∞ dqdp q p −βH(q,p)+αN hNi = Nf ρg.c.(N, , )N = Nf Ne GN h Z(β,V,α) GN h NX=0 Z NX=0 Z Again, treating α and β as independent variables while we take derivatives, we have: ∂ e−βH(q,p)+αN = Ne−βH(q,p)+αN ∂α

4 so that we avoid doing the integrals and we find: 1 ∂ ∂ hNi = Z(β,V,α)= ln Z(β,V,α). Z(β,V,α) ∂α ∂α So we can easily also calculate the average number of microsystems in the system. We will look at some examples soon and you’ll see that doing this is simple in practice, it just requires a bit of attention when taking the derivatives. This approach can be extended easily to find (check!) that: 1 ∂2 hH2i = Z(β,V,α) Z(β,V,α) ∂β2 and 1 ∂2 hN 2i = Z(β,V,α). Z(β,V,α) ∂α2 Of course, we would need these quantities to calculate standard deviations. Now, this trick I described above, with using α and β, is what people usually do, and what is given in textbooks etc. All is needed is that while you take the derivatives, you treat α and β as independent variable. For reasons which escape me, some students think that this is too fishy and refuse to use this trick. So here is an alternate trick, which is almost as good (takes just a bit more work) and gives the precise same answers at the end of the day. Let’s look again at: ∞ dqdp 1 ∞ dqdp q p q p q p q p −β[H(q,p)−µN] U = hH( , )i = Nf ρg.c.(N, , )H( , )= Nf H( , )e GN h Z GN h NX=0 Z NX=0 Z Clearly we’d like to not have to do the integrals explicitly, so we have to get rid of the H somehow. If you do not like the trick with introducing α, then we can do this. First, introduce an x in front of the Hamiltonian, in the exponent, and ask that x be set to 1 at the end of the calculation, since clearly: 1 ∞ dqdp U = H(q, p)e−β[xH(q,p)−µN] Z G hNf N=0 Z N x=1 X

Now we can take the derivative with respect to x, so that we have: 1 1 ∂ U = − Z(x) Z " β ∂x # x=1 where ∞ dqdp −β[xH(q,p)−µN] Z(x)= Nf e GN h NX=0 Z can be quickly calculated (this is where the extra work comes in), just like you calculated Z = Z(x = 1) (it’s just a matter of tracking where the extra x goes in some gaussians). So, putting these together, and since we set x = 1 at the end, we have: 1 ∂ U = − ln Z(x) β ∂x x=1

and similarly 1 ∂2 hH2i = Z(x) β2 ∂x2 x=1

5 and here the only “trick” is to ﬁrst take all the derivatives, and then set x = 1 as the very last step. To calculate ensemble averages of N we do not need to introduce any new variable, since we already have µ there, so we can also write:

∞ 1 dqdp −βH(q,p)+βµN 1 1 ∂ 1 ∂ hNi = Nf Ne = Z = ln Z Z(β,V,α) GN h Z "β ∂µ # β ∂µ NX=0 Z and similarly ∞ 2 2 1 dqdp 2 −βH(q,p)+βµN 1 1 ∂ hN i = Nf N e = 2 2 Z Z(β,V,α) GN h Z β ∂ µ NX=0 Z If you think about it, taking these derivatives is just the same as taking derivatives with respect to α, in the previous notation. How about calculating other ensemble averages, for example the entropy? Using the expression of ρg.c. in the general Boltzmann formula, we find that: ∞ dqdp q p q p S = −kBhln ρi = −kB Nf ρg.c.(N, , )ln ρg.c.(N, , ) GN h NX=0 Z ∞ dqdp 1 −β[H(q,p)−µN] q p = −kB Nf e [− ln Z− βH( , )+ βµN] → GN h Z NX=0 Z S = kB ln Z + kBβU − kBβhNi since the first integral is related to the normalization condition, while the second and third are just the ensemble averages of H,N. From this we find that the grand-canonical potential is:

φ(T,V,µ)= U − TS − µhNi = −kBT ln Z(T,V,µ) (3) As you see, things again ﬁt very nicely together. If you remember, when we reviewed thermodynamics, we decided that for an open system we need to be able to compute the grand-canonical potential φ(T,V,µ), since then we can use

dφ = −SdT − pdV −hNidµ (4) to find S, p and hNi as its partial derivatives, which tells us how they depend on T,V,µ. This seems to give us an alternate way to find hNi, but in reality we find the same formula with the derivative from ln Z (not surprisingly – we should get the same result no matter how we go about it). There is also an alternative way to find U, from using U = φ + TS + µhNi, where all the terms on the right hand side are known once Z is calculated. Of course, this gives the same result as the tricks with derivatives, but it involves a bit more work. So let us summarize how we solve a classical grandcanonical ensemble problem, once T,V,µ,... (the macroscopic variables) are given to us. As always, we begin by identifying the number of degrees of freedom f per microsystem, and all needed generalized coordinates q, p that fully characterize a microstate that has N microsystems in the system, and then the Hamiltonian of the system H(q, p). Then we calculate the partition function from Eq. (2). Once we have it, we know φ = −kBT ln Z. Once we have φ, we can find ∂φ ∂φ ∂φ S = − ; p = − ; hNi = − ∂T !V,µ ∂V !T,µ ∂µ!T,V

6 We can also calculate the internal energy and hNi from ∂ ∂ U = − ln Z(β,V,α); hNi = ln Z(β,V,α); ∂β ∂α where α, β are treated as independent variables while we take the derivatives, after which we can set α = βµ. Similarly, we can calculate averages of hH2i, hN 2i, hHNi,..., but using the proper number of derivatives with respect to α and β. If you do not like this, use the trick with introducing the x, and then setting it to be 1 after all derivatives were taken. Any other ensemble averages are calculated starting from the definition of an ensemble average and the known density of probability for the grandcanonical ensemble, and by doing all integrals over positions/momenta and sum over N. One last thing. Remember that for the canonical partition function and non-interacting systems, we could use the factorization theorem to simplify the calculation. It turns out we can do a similar thing here. First, start with the general definition of the grand-canonical partition function: ∞ ∞ dqdp −β[H(q,p)−µN] βµN dqdp −βH(q,p) Z(T,V,µ)= Nf e = e Nf e GN h GN h NX=0 Z NX=0 Z Now we recognize that the integrals are simply the canonical partition function for a system with N particles, so: ∞ Z(T,V,µ)= eβµN Z(T,V,N) NX=0 So in fact, if we know how to calculate Z (which we do) there isn’t much left of the calculation. Let’s simplify even further. For non-interacting systems where particles can move and exchange positions (such as gases), we know from the factorization theorem that: 1 Z(T,V,N)= [z(T,V )]N N! Using this in the above sum, we find:

∞ 1 N Z(T,V,µ)= z(T,V )eβµ = exp eβµz(T,V ) N! NX=0 h i since the sum is just the expansion of the exponential function. Those of you with good memory will be delighted to learn that eβµ has its own special name, namely fugacity. For problems where the microsystems are distinguishable because they are located at different spatial location (crystal-type problems), Gibbs’ factor is 1 and: Z(T,N)=[z(T )]N and therefore: ∞ N 1 Z(T,V,µ)= eβµz(T ) = 1 − eβµz(T ) NX=0 h i if |eβµz(T )| < 1 (otherwise the geometric series is not convergent). Of course, we’ll find that this condition is generally satisfied. So the conclusion is that once we calculate z (just as we did it for canonical ensembles), we immediately have Z, i.e. dealing with a grandcanonical classical system really does not involve any more work/math that dealing with a canonical system – at least so far as classical problems are concerned.

7 1.1 Classical ideal gas Let’s check how this works for a classical ideal gas – our favorite classical model. Assume a system of volume V in contact with a thermal and simple atom reservoir with temperature and chemical potential T,µ. Let’s calculate the average number of particles in the container, hNi, their internal energy U and their pressure p. Because these are simple, non-interacting atoms, they have f = 3 degrees of freedom, and we can use factorization theorem, and GN = N!. Of course, we know that:

3 2 2 N 1 ~p 2πmkBT z −β 2m z(T,V )= 3 d~r d~pe = V 2 → Z(T,V,N)= h Z Z h ! N! (three identical gaussians plus one integral over the volume). Therefore (see formula above):

3 ∞ ∞ 1 N 2πmk T 2 Z = eβµN Z(T,V,N)= eβµz = exp eβµV B N!  h2 !  NX=0 NX=0 h i   and so: 3 2 βµ 2πmkBT φ(T,V,µ)= −kBT ln Z = −kBTe V h2 ! This is indeed an extensive quantity. It is maybe not so obvious that it has the right (energy) units, but you should be able to convince yourselves that that is true (remember that z is a number, and µ is an energy). To calculate hNi we have two alternatives, either as a partial derivative of φ with respect to µ (I’ll let you do this) or using the trick with α and β. Let’s do it by the second method. First, we replace βµ → α everywhere where they appear together. We then ﬁnd:

3 3 2πm 2 2πm 2 Z(β,V,α) = exp eαV → ln Z(β,V,α)= eαV  βh2 !  βh2 !   Now, assuming α and β to be independent variables, we have:

3 ∂ 2πm 2 hNi = ln Z(β,V,α)= eαV ∂α βh2 ! This looks a bit strange, but let’s not loose heart – this tells us how hNi depends upon T,V,µ (the variables which characterize the macrostate for the open system) and it is not something we’ve looked at before. Notice that you can extract how µ depends on T,V, hNi from this – you should do that and see that the result agrees with what we obtained for canonical ensembles, if we replace N →hNi. The internal energy is:

3 2 ∂ 3 α 2πm 3 U = − ln Z(β,V,α)= e V → U = hNikBT ∂β 2β βh2 ! 2 So yeeeii! It works!

8 Finally, we ﬁnd the pressure from the partial derivative of φ with respect to V :

3 2 ∂φ βµ 2πmkBT p = − = kBTe 2 → pV = hNikBT ∂V !T,µ h ! So, yeeeiii again! As you can see, we do obtain the same results we got using micro- and canonical ensemble formalisms for this problem, except here we must use hNi instead of N. Why is this, why do we get the same predictions even though now the number of particles if not fixed? Exactly like in the case of the equivalence between microcanonical and canonical formulations’ results, this is due to the fact that the systems are very large, hNi ∼ 1023. Unlike in a canonical ensemble, in a grandcanonical ensemble the number of particles is not fixed, and it could be anything, in principle. However, because the system is so big, one can show (just as we did for the energy of the canonical ensembles) that the probability of finding the system to have anything but hNi particles is an extremely small number (basically zero), i.e. it is overwhelmingly likely that the system always has precisely hNi atoms inside. One can show that the relative standard deviation

σ hN 2i−hNi2 1 N = ∼ → 0 hNi q hNi hNi q (I’ll probably give this to you as an assignment exercise). So, as I’ve said all along, for (large) thermodynamic systems we can use whichever formalism is most convenient to solve a problem. We will see more examples of classical problems to be solved by grandcanonical formalism (the kinds of problems where the number of particles can really change) in the next assignment. We will also consider problems where we have more than one species of atoms in the container – in that case, it is possible that one species can pass through the wall (system is open, from its point of view) and another species cannot go through the wall (system is closed from its point of view). In this case, we need to use a mixed description, using canonical formalism for the atoms whose number is conserved, and grandcanonical formalism for those whose number can vary. This probably sounds more complicated than it is in practice, as you’ll see. But now it is ﬁnally time to concentrate on quantum gas -type of problems, which we could not solve by any other formalism. We are ﬁnally able to study them.

2 Quantum grandcanonical ensembles

First, we need to figure out how to properly characterize microstates of the quantum system, since the number of microsystems is not fixed. If you think about it, we can’t just say that the microstates are eigenstates of the Hamiltonian (the way we did for quantum canonical systems), because after all even the Hamiltonian is not unique – microstates with different numbers of particles have different Hamiltonians! (number of terms, e.g. kinetic energy contributions, in the Hamiltonian changes in proportion to how many microsystems are in the microstate). So we have to be a bit careful how we go about characterizing all the microstates. What will simplify things tremendously is the fact that we will only deal with non-interacting systems. For interacting systems, one really needs to use more formal objects like density matrices, but we won’t go there. Let eα be the energies of all (discrete) energy levels, if there was a single microsystem in the system. We’ll always assume the energy to be zero if there is no microsystem in the system (empty

9 system). α are all the needed quantum numbers to describe these levels – degeneracies are very important! These energies (and the single-particle wavefunctions associated with them) are usually called single-particle energies or single-particle orbitals. To make things more clear, let’s look at some examples as we go along. First, a simple “crystal”- like example. Assume we have a surface with a total number NT of sites where simple atoms could be trapped. If a trap is empty, its energy is zero. If it catches an atom, its energy is lowered to −ǫ0 < 0. This surface is in contact with a gas of atoms (the reservoir), with known T and µ. The question could be, for instance, what is the average number hNi of atoms trapped on the surface. What are the single-particle orbitals, in this case? Well, if we have a single atom in the system= surface with traps, it must be trapped in some site or other, and the energy will be −ǫ0. We could use as “quantum number” an integer 1 ≤ n ≤ NT which tells us at which site is the atom trapped – so here we have a NT degenerate spectrum of single-particle states, all with the same energy −ǫ0. As a second example, let’s consider a quantum gas problem. Assume simple atoms with quantum dynamics in a cubic box of volume V = L3. Of course, we’ll generally want to know what are the properties of the quantum gas when the system is in contact with a thermal and particle reservoir with known T,µ. However, right now all we want to know, is what are the single particle levels. For this, we must find the spectrum (the eigenstates) when there is just one atom in the system. This we know how to do. In this case the Hamiltonian is: h¯2 d2 d2 d2 hˆ = − + + 2m dx2 dy2 dz2 ! and the eigenstates are: h2 e = (n2 + n2 + n2) nx,ny,nz 8mL2 x y z where nx,ny,nz =1, 2,... are strictly positive integers. So here three quantum number α = nx,ny,nz characterize the single-particle orbitals (if you don’t remember where this formula comes from, it is a simple generalization of the 1d case we solved when we looked at multiplicities, when we discussed microcanonical ensembles). Strictly speaking, atoms also have some spin S, so we should actually include a 4th quantum number α = nx,ny,nz,m where m = −S, −S +1,...S − 1,S is the spin projection. For any other problem we can figure out the single-particle orbitals (or states) similarly. Now let’s go back to our general description, where eα are the energies of all possible single-particle orbitals. What happens if there are more microsystems in the system? Well, let’s start with two. Because these are non-interacting microsystems, they occupy the same set of single-particle orbitals. So to characterize the state, we now need two pairs of quantum numbers, say α and α′, to tell us which two states are occupied. The energy is simply eα + eα′ . For example, if there is a second atom on the surface, it must also be trapped in some trap or other, just like the first one, so I could specify the state by saying atom 1 is trapped at site n while atom 2 is trapped at site n′. Of course,

′ ′ ′ the energy is −2ǫ0. Similarly, a second atom in the box will be in some eigenstates enx,ny,nz and the energy will be the sum of the two. The wavefunction for the total system is now:

Ψ(~r1, ~r2)= φα(~r1)φβ(~r2) where φα(~r) is the single-particle wavefunction associated with the single-particle state eα. Right? WRONG! What quantum mechanics tells us is that if the particles are identical, their wavefunction must be either symmetric (for so-called bosonic particles, i.e. whose spin S is an integer) or antisymmetric (for so-called fermionic particles, i.e. whose spin S is half-integer ) to exchanges of the two, i.e. 2 2 Ψ(~r1, ~r2)= ±Ψ(~r2, ~r1) →|Ψ(~r1, ~r2)| = |Ψ(~r2, ~r1)|

10 This is what indistinguishability really means. If the particles are truly identical, then there is no measurement whatsoever that we can perform to tell us which of the two particles is at ~r1 and which at ~r2, so if we exchanged their positions we should see no diﬀerence (same probability to ﬁnd them at those locations). So going back, it follows that for two fermions, the two-particle wavefunction must be: ΨF (~r1, ~r2)= φα(~r1)φβ(~r2) − φα(~r2)φβ(~r1) while for bosons, we must have:

ΨB(~r1, ~r2)= φα(~r1)φβ(~r2)+ φα(~r2)φβ(~r1) (there is actually an overall normalization factor, but that is just a constant that has no relevance for our discussion). This immediately tells us that we cannot have two fermions occupy the same single-electron orbitals, i.e. α =6 β always. If α = β we find ΨF (~r1, ~r2) = 0, which is not allowed (wavefunctions must normalize to 1). This is known as Pauli’s exclusion principle. There is no such restriction for bosons, there we can have any number of bosons occupying the same single-particle state. If we now look at these two-particle wavefunctions, we see that it’s just as likely that particle 1 is in state α and 2 in β, as it is to have 1 in state β and 2 in state α. Therefore, it makes much more sense to characterize this state by saying that there is a particle in state α and a particle in state β, and not attempt anymore to say which is which – they’re identical and either one could be in either state with equal probability. We can rephrase this by saying that of all single-particle orbitals, only α and β are occupied, while all other ones are empty. In fact, we can define an occupation number which is an integer nα associated to each single-particle level. For empty levels nα = 0, while for occupied levels nα counts how many particles are in that particular orbital. For bosons, nα =0, 1, 2, 3,... – could be any number between 0 and infinity. For fermions, nα =0 or 1! Because of the exclusion principle, we cannot have more than 1 fermion occupying a state. This is the difference between fermions and bosons. It might not look like much, but as we will see, it will lead to extraordinarily different behavior of fermions vs. bosons at low temperatures, i.e. where quantum behavior comes into play. We can now generalize. For any number of microsystems (particles) present in the system, we can specify a possible microstate by giving the occupation numbers for all the single-particle orbitals (if there is an infinite number of orbitals, such as for a particle in a box, lots and lots of occupations numbers will be zero; but we still have to list all – infinite number – of them). So the microstate is now specified through the values {nα} of all occupation numbers in that microstate. Allowing all numbers {nα} to take all their possible values will generate all possible microstates for all possible numbers of microsystems. Of course, the total number of microsystems (particles) in the microstate must be

N{nα} = nα α X i.e. we go through all levels and sum how many particles are occupying each of them – clearly this sum gives the total number of particles in that microstate. The energy of this microstate is:

E{nα} = nαeα α X again we go through all levels, for each one which is occupied (nα =6 0) we add how many particles are in that level, nα, times the energy of each one of them, eα. Again, I think this should be quite

11 an obvious equation for non-interacting systems. For interacting systems, things are much more difficult, because the energy is no longer the sum of single-particle energies. This was the hard part. Now that we know how we describe the microstates, and how many particles are in a microstate and what is their energy, we’re practically done. Following the same reasoning as for classical ensembles, we find that the probability to be in a microstate is 1 p = e−β(Eµstate−µNµstate) µstate Z As in the canonical case, the only difference is that classical degrees of freedom are continuous, so we can only talk about density of probability to be in a microstate. For quantum systems, microstates are discrete so we can talk about probability to be in a microstate. The reason the formula above holds is because nowhere in its derivation did we have to assume that the energy is continuous, so things proceed similarly if the energy is discrete. Of course, the grand-canonical partition function Z must be such that the normalization condition holds: −β(Eµstate−µNµstate) pµstate =1 →Z = e µstates µstates X X In terms of occupation numbers which characterize the microstates, these formulae become:

1 −β(E −µN ) 1 −β( n e −µ n ) 1 −β n (e −µ) p = e {nα} {nα} = e α α α α α = e α α α {nα} Z Z Z P P P To ﬁnd Z we must sum over all microstates. Since the allowed values for the occupation numbers are diﬀerent, let’s do this separately for the two cases, For a fermionic system, each nα =0 or 1. As a result:

1 −β n (e −µ) Z = e α α α F   α n Y Xα=0 P   Throughout the remainder of the course, I’ll use this shorthand notation:

1 1 1 ≡ ···   α n =0 n =0 n =0 Y Xα αX1 αX2   where there is a sum for each single particle orbital (if there’s an infinite number of them, there’s an infinite number of sums). The short-hand notation simply says that for each possible α, there’s a sum sign in the product. This is actually very simple to calculate, because the exponential factorizes in terms each of which depend on a single occupation number nα. So we can group each exponential with its sum, and find:

1 Z = e−βnα(eα−µ) F   α n Y Xα=0   1 −βnα(eα−µ) −β(eα−µ) But now each sum is trivial, nα=0 e =1+ e , and so: P −β(eα−µ) ZF = 1+ e α Y Just to make sure you followed this, let us do this for the case of the atoms trapped on the surface (we’ll study the quantum gases in detail starting next lecture). In this case, we decided that we have

12 a ﬁnite number of single-particle levels indexed by the integer α → n, 1 ≤ n ≤ NT which tells us at which site is the single atom trapped. Let’s assume that the atoms are fermionic, i.e. there can be at most 1 atom in any trap (you might want to ask me some questions here ....). The microstate is now described by the occupation numbers n1,...,nNT where ni is zero if trap i is empty and 1 if trap 1 is occupied by an atom. NT NT The number of atoms in the microstate is N = i=1 ni, and the energy is E = i=1(−ǫ0)ni = NT −ǫ0N, and E − µN =(−ǫ0 − µ) i=1 ni. In this case:P P 1 1 P1 1 1 1 −β(E−µN) −β(−ǫ0−µ)(n1+···+nN ) ZF = ··· e = ··· e T n =0 n =0 n =0 n =0 n =0 n =0 X1 X2 NXT X1 X2 XT

1 1 1 N −β(−ǫ −µ)n −β(−ǫ −µ)n −β(−ǫ −µ)n −β(−ǫ −µ) T = e 0 1 e 0 2 ··· e 0 NT = 1+ e 0       n =0 n =0 n =0 X1 X2 NXT h i       In this case, there is a finite number of single particle orbitals, and each contributes the same since −β(eα−µ) they all have the same energy −ǫ0. In the general case, each orbital contributes 1 + e , and we must multiply over all the orbitals. That’s precisely what the general formula for ZF means. All the other formulae we have derived for classical grand-canonical system hold unchanged, in particular: −β(eα−µ) φF = −kBT ln Z = −kBT ln 1+ e α X h i For different fermionic systems we’ll have different energies ǫα and number of levels α, but this formula will always hold. Let us do the same for a bosonic system. In that case:

∞ −β n (e −µ) Z = e α α α B   α n Y Xα=0 P   since occupation numbers can now be anything. Again we can factorize the product:

∞ Z = e−βnα(eα−µ) B   α n Y Xα=0   Each sum is an inﬁnite geometric series. Note that we must have:

e−β(ǫα−µ) ≤ 1 in order for each of these series to be convergent. Since β > 0, it follows that for a bosonic system, we must always have µ ≤ eα for all single particle levels and therefore

for bosons, we must have: µ ≤ eGS where eGS is the energy of the single-particle ground-state. This restriction will turn out to have important consequences. For fermions we have no restrictions for the chemical potential. ∞ n If the restriction µ ≤ eGS is satisﬁed, then each geometric series n=0 x =1/(1−x) is convergent, and we ﬁnd: 1 P ZB = −β(eα−µ) α 1 − e Y

13 and −β(eα−µ) φB = −kBT ln ZB =+kBT ln 1 − e α X h i In fact, because of the similarities of the formulae, we can group together the results for both fermions and bosons and write: ±1 Z = 1 ± e−β(eα−µ) α Y and −β(eα−µ) φ = −kBT ln Z = ∓kBT ln 1 ± e α X where the upper sign is for fermions, and the lower sign is for boson systems. From partial derivatives of φ we can calculate S,p, hNi, as usual, since dφ = −SdT − pdV −hNidµ. We can also use the tricks with α and β to ﬁnd U and hNi. Let’s remember them, and check that they still hold. First, we replace βµ → α everywhere this product appear. In terms of α and β, we have: 1 p = e−βEµstate+αNµstate µstate Z where from normalization, Z(α,β,...)= e−βEµstate+αNµstate µstates X By deﬁnition,

1 −βEµstate+αNµstate 1 ∂ ∂ U = pµstateEµstate = Eµstatee = − Z(α,β,...)= − ln Z(α,β,...) µstates Z µstates Z ∂β ∂β X X if, while taking the derivative, we pretend that α and β are independent variables. Similarly,

1 −βEµstate+αNµstate 1 ∂ ∂ hNi = pµstateNµstate = Nµstatee = Z(α,β,...)= ln Z(α,β,...) µstates Z µstates Z ∂α ∂α X X So indeed, we have precisely the same formulae as before. This is because the only diﬀerence is what is meant by µstates. For classical systems that implies a sum over N and many integrals over all classical degreesP of freedom; for quantum systems, this is a sum over all possible occupation numbers. But nothing in the derivation depended on such details. For our quantum system, after replacing βµ → α, we have:

ln Z(α,β,...)= ± ln 1 ± e−βeγ +α γ X (again, upper sign for fermions, lower sign for bosons). I prefer to call the quantum numbers γ this time, since α = βµ is now taken. If we take the derivatives, we ﬁnd that:

−βeγ +α ∂ ∓eγe 1 U = − ln Z(α,β,...)= −(±) = eγ −βeγ +α β(eγ −µ) ∂β γ 1 ± e γ e ± 1 X X and ∂ e−βeγ +α 1 hNi = ln Z(α,β,...)=(±) = −βeγ +α β(eγ −µ) ∂α γ 1 ± e γ e ± 1 X X (we can go back to α → βµ after we took the derivatives. Results are generally in terms of µ.

14 But, on the other hand, using the expressions for number of particles and energy of a microstate in terms of occupation numbers, we have:

hNi = h nγi = hnγi γ γ X X and U = h eγnγi = eγhnγi γ γ X X Comparing these with the equations above, we see that we must have the average occupation number of level γ to be: 1 hnγi = eβ(eγ −µ) ± 1 These are extremely important results, which will come up time and time again. So let’s discuss them separately, in some detail. For fermions, the average occupation number of a level with energy ǫγ is: 1 hnγi = eβ(eγ −µ) +1

This is called the Fermi-Dirac distribution. Let’s analyze it a bit. First, since eβ(eγ −µ) ≥ 0 no matter what values β,µ,eγ have, it is clear that always 0 ≤hnγi≤ 1. This makes perfect sense. In any microstate, nγ can only be 0 or 1, so its average must be a number between 0 and 1! β(eγ −µ) At T = 0 → β → ∞, we see that if eγ < µ → e → 0 and so hnγi → 1. In other words, levels whose energy is below the chemical potential µ are certainly occupied at T = 0. β(eγ −µ) If, however, eγ > µ → e → ∞ and so hnγi → 0. Therefore, levels whose energy is above the chemical potential µ are certainly empty at T = 0. If the temperature is low, but not zero, the occupation number will be somewhat changed for levels whose energy is within about kBT of µ (see ﬁgure below). But levels with eγ − µ ≪−kBT are still certainly occupied, and levels with eγ − µ ≫ kBT are still certainly empty. It’s just within kBT of µ the transition from occupied to empty is no longer abrupt, instead the average occupation numbers continuously go from 1 to 0.

Fermi−Dirac

1 T=0

T > 0

µ eGS e

Figure 2: Fermi-Dirac distribution, showing the average occupation number of a level of energy e, as a function of e. There are no levels below eGS. At T = 0, all levels between eGS and µ are fully occupied, hni = 1, while all levels above µ are completely empty, hni = 0. At ﬁnite-T, the average occupation numbers for energies roughly in the interval [µ − kBT,µ + kBT ] are changed, and the decrease is now gradual.

15 For bosons, the average occupation number of a level with energy ǫγ is: 1 hnγi = eβ(eγ −µ) − 1 This is called the Bose-Einstein distribution. Let’s analyze it a bit. First, remember that for bosons we must always have µ ≤ eGS → eγ − µ ≥ 0. Now we see that this is very necessary, since with this restriction eβ(eγ −µ) ≥ 1 and the average occupation numbers are positive! They must be positive – the average of any number whose only allowed values are 0, 1, 2,... cannot be negative. Unlike for fermions, however, we see that an average occupation number could be anything between 0 and inﬁnity. In fact, let us consider T = 0 behavior. Here we have two cases: (1) µ

−β(eγ −µ) hnγi≈ e ≪ 1 As we will show soon, in this limit we indeed ﬁnd agreement with the expected classical results.