Quantum ideal gases: fermions

Transcription

1 Quantum ideal gases: fermions Any particle with half-integer spin is a fermion. In this notes, we will discuss the main features of the statistics of a gas of N free, non-interacting fermions of spin S (S = /, /,...). The prototypical example we will have in mind will be electrons in a metal, and the following discussion will be focused on them, although one would treat any other example of non-interacting fermions in a similar manner. As discussed in class, because we cannot solve the problem in the canonical formalism, we will solve it using the grand-canonical formalism, and adjust the value of µ such that the average number of particles in the system N = N, i.e. the specific number of particles that we would like to have in the system. Electrons have spin S =, so they are fermions. What should surprise you in the previous statements, is the assertion that we can treat them as if they are non-interacting. How can that be?! We know that electrons carry electric charge e and therefore they very strongly repel one another. How can we ignore these strong electric interactions and hope to get any sensible results? The answer comes from understanding their behavior in a metal, where it turns out that this is a good approximation (this approximation would never work if the electrons were in vacuum, for instance). The generic model of a metal is the following: consider atoms arranged on an ordered lattice. Typically, an atom has a few completely filled shells, and one which is only partially filled. The electrons from the filled shells are called core electrons. They are very happy together (a filled shell is a very stable state) and basically do not participate in any properties (we will discuss more about them when we study semiconductors, later on). The electrons from the last, only partially filled shell, are called valence electrons. These are unhappy, and ready to participate in any arrangements that might lead to a final state with only completely filled shells. This is why if one puts together an atom of Na (which has the n =,n = shells filled, and a single electron in the s level) and one of Cl, which has electronic structure s p 5 meaning n =,n = levels filled, electrons in the s shell and 5 electrons in the p shell, i.e. is one electron short of completely filling the s+p levels the end result is that the Na gives its valence electron to the Cl, so they now both have only completely filled shells. Of course, now the Na will have an effective charge +e and the Cl will have an effective charge e, so there is a strong attraction between them, leading to what is called an ionic bond. If one puts many Na and Cl together, they will crystallize into a solid that we call salt. This is an insulator, because (after the transfer) each electron is in a filled shell and therefore will stay with its own nucleus. This type of result is expected every time one puts together two different kinds of atoms, one with only a few electrons in the last shell, and one missing just a few electrons to fill the last shell. Another possible situation is if one puts together only one species of atoms. There are several possible outcomes. If the atoms are, for example, Si s p or Ge 4s 4p, we see that they have half of the last shell filled (a total of 4 electrons, while 8 is the number of available states is s + p orbitals). Because all atoms are identical, the way to solve the problem now is to share these electrons. In a crystal, the atoms arrange such that each atom is surrounded by 4 other atoms. Each two atoms share two of their valence electrons together, one from each, creating what is called a covalent bond. Because of this, roughly speaking half the time an atom has 8 electrons (taking one electron each from its four neighbor atoms) and half the time is has zero electrons (gives its 4 electrons one each to the neighbors), and everybody is happy. Of course, this is a simplification of what actually happens, which is that orbitals hybridize with one another resulting in new kinds of orbitals which have their own rules as to how many electrons they can accommodate, and thus when are they completely filled. But I think it is a good simplified picture. A crystal of this type would still be an insulator, because now each electron is shared equally by two neighboring atoms, therefore the electrons do not want to move around.

2 To obtain a metal, we must put together atoms which have each only a few valence electrons. Let us use Na as the prototypical example. Each Na has electron in the n= shell, instead of 8, which would make it happy. Because so many are missing, there is no way to create covalent bonds with other Na neighbors, because there are not enough electrons to share around (put it another way, each Na should be surrounded by 7 other Na atoms, and there s no such lattice possible). The way the problem is solved, now, is to let these valence electrons become free. In this case, each Na has a filled shell and effective charge +e, while the extra electrons do not belong to any particular atom, but move freely through the entire crystal. This is why if one applies a small electric field, one gets an electric current in metals all these free electrons start to move along the field lines. By contrast, in an insulator one has to apply a really large electric field before one is able to pull the electrons from their filled shells and make them start moving around. Hopefully you have taken enough chemistry and solid state courses to have a more accurate picture of what is really happening, but this simple picture is enough for our purposes. At this point, we have argued that in a metal, there are a number of electrons (one per each atom in the crystal, in the case of Na) free to move throughout the volume occupied by the metal. They are not free to go outside the metal, because they feel a strong attraction from the positive charges of the nuclei to extract an electron from a metal, one has to give it a certain amount of energy, called the work function, to overcome this attraction. This is generally quite large, so to first order we can think of the electrons as moving freely throughout the metal, and forbidden to go outside. This is very similar to the picture of a gas that we have, and people refer to these electrons as a gas of electrons. So far so good, but we still have to understand why we can pretend that these electrons are not interacting. Of course that in fact the electrons strongly repel one another; as a result of this repulsion, each electron is surrounded by a small volume of empty space where no other electrons come. Except that this is not empty space, but the crystal, which contains the positive charges of the nuclei. Consider now the electric field created by this electron far from its position. If the electron was in vacuum, this would be a field E( r) = eˆ r/(4πǫ r ) pointing radially in the direction ˆ r = r/r. In the crystal, the electron still creates this field, but the positive charges also create their own electric field. In particular, it turns out that the positive charges from inside the bubble surrounding the electron create a field that precisely cancels that created by the electron. In other words, the total positive charge inside the bubble precisely cancels the charge of the electron, so that from far away it looks like an electrically neutral object. Since each electron is surrounded by its own bubble of positive charge, they effectively all behave as if they have no electrical charge, and this is why the approximation that they are non-interacting works extremely well. Comments: (i) what I just said is a very qualitative explanation of what is actually happening. One can do serious and detailed calculations to back up this picture, and to calculate corrections to it, and these have been done. Of course, one can also compare the predictions of this model with those of experiments, and the agreement is good, showing that this is a reasonable starting point. (ii) if you read advanced literature in solid state, you will see that these electrons are not referred to as particles, but as quasi-particles. The reason for this is because the objects for which we are doing statistics are not the bare electrons, but the electrons surrounded by this bubble of crystal empty of other electrons. It s this whole object, electron+bubble, that is electrically neutral, and this is what is meant by a quasi-particle. As the quasi-particle moves through space, its properties will be somewhat different from those of an electron in vacuum, because in the metal the electron must push away the other electrons to protect its bubble. This is why, for example, the mass of the quasi-particle is generally different from the mass of an electron in vacuum. This is true not only

3 in metals, but also in semiconductors the value of the effective mass depends on the details of the material considered. For instance, you may know that electrons doped in GaAs have an effective mass of.7m e. What this means is that these quasi-particles have this effective mass, but people get lazy and refer to them as electrons. We will do the same from now on, for simplicity. (iii) there are materials where this model of non-interacting electrons fails miserably. For example, high-temperature superconductors obtained by doping certain cuprate oxides are in this class, but the list is really long and growing. Understanding how to calculate the properties of such strongly-interacting systems is one of the main research areas in solid state (or condensed matter) physics today. It is a very difficult challenge and so far there is no clear understanding on how to proceed with such calculations but the problem is of great interest due to the remarkable properties of many of these materials, which we would like to be able to understand and use for new technologies. We have good theories for simple systems like simple metals and semiconductors and insulators these were the first solids to be known because they appear naturally on the Earth, around us. Understanding semiconductors and how to use them, for example, is the basis of the extraordinary electronics revolution of the last 6 decades or so, which has changed our lives in fundamental ways. We are now at a point where we are no longer looking at materials surrounding us, but can create new materials in laboratories, using the rules of chemistry and physics that we have learnt from studying the simple materials. For instance, one of the simpler (undoped) cuprate oxides is LaCu O 4, i.e. it has 7 atoms of three different kinds per unit cell, arranged in a very particular way. It is because of this complicated arrangement that this material is unlikely to be found in a pure state in nature there are simpler combinations of these atoms that would occur first. Nowadays people can create much more complicated materials, with 4,5 or more species of atoms arranged in very particular ways. The number of possibilities is enormous and this is why we would like to have the ability to make trustworthy calculations and predict whether certain combinations have properties which are of use for some particular purposes, or not. It is this need and challenge that will keep condensed-matter physicists very busy, for a long time to come. The model Consider N non-interacting electrons (we will call them this, from now on) inside a volume V = L of a metal. Note: in the following we will consider electrons free to move in spatial dimensions. Twodimensional and one-dimensional cases are also possible for instance, electrons in a very thin, long metallic wire will be effectively restricted to one-dimensional motion. I will give you as assignment to repeat some of these calculations for other dimensions. As we ve discussed in the general case, the first step is to find all the possible single-particle levels (what we called e γ ). For this we need to solve for the eigenstates in the case where there is a single electron in the system. Because there are no other interactions, the Hamiltonian for a single electron is: ( ) ĥ = p m = h m x + y + z We know that the eigenfunctions of this Hamiltonian are plane waves: ĥφ k,σ ( r) = ǫ k φ k,σ ( r) () where φ k,σ ( r) = ei k r V χ σ

4 and ǫ k = h k m Here, χ σ is a spinor, which tells us whether the electron has spin-up (if σ = +) or spin-down (if σ = ). In other words, σ is related to the spin projection, m = σ/. Each energy level is spin degenerate, i.e. it can accommodate an electron with spin up and one with spin-down. Note: if there is a magnetic field, then, as we know: ĥ = p m gµ Ŝ z BB z h resulting in the same wavefunctions, but energy levels that depend on the spin as well: ǫ k,σ = h k m gµ σ BB z We will come back to this later on, for the time being we consider zero magnetic field. We must now impose the restriction that electrons can only be inside V. We have already discussed one way to do this, by requiring that the wavefunctions go to zero on the surface. We could do this here as well, by using as wavefunctions combinations of the exponentials (i.e., cos or sin functions) with the allowed k values related to some integers. L Sketch of the space divided in cubes of volume V = L, a few of which are shown. We require that the wavefunction is the same at all points with equivalent locations in all cubes (the black dot shows one such point). At the end, we let the volume expand as L, and obtain a metal without surfaces filling up the entire space. Instead, in solid state we use what are called periodic boundary conditions. The idea is that we want to consider an infinite system, without surfaces (behavior near surfaces can be quite different from that well inside the metal). The way to do this, is to imagine the whole space filled with the metal, which we divide mentally in cubes of volume V. We request that the behavior inside any one cube is identical with that of any other cube, meaning that the wavefunctions must be periodic: φ k,σ ( r) = φ k,σ ( r + L e x ) = φ k,σ ( r + L e y ) = φ k,σ ( r + L e z ) () which means that we have equal probabilities to find electrons in identical positions inside these identical cubes (see figure above). We will then let V and N but such that N/V equals 4

5 the desired finite concentration in this way, the metal really fills up the entire space and it has no surfaces. (If one wants to understand the effects of a surface, one then imposes the other type of boundary conditions, which require the wavefunction to be zero on the surface this changes the allowed energy levels, and one can track the effect of that on the overall behavior. Contrasting the two sets of results, with/without surface, we can deduce the effects of the surface on the system s behavior. We will not consider such effects). Combining Eqs. () and (), we see that the allowed k = (k x,k y,k z ) must satisfy, for example: e i k r = e i k ( r+l e x) where n x =, ±, ±,... is any integer. Similarly, k x = πn x L ; r e ikxl = k x L = πn x k x = πn x L k y = πn y L ; k z = πn z L where n x,n y,n z can be any integers. So we can index any level either by the values (n x,n y,n z,σ) or, more compactly, by ( k,σ), where we understand that the allowed k are the ones given by Eq. (). Now that we know the single-electron energy levels, the problem is basically finished. As discussed, we use the grand-canonical ensemble to study the system, by fixing T,V,µ for the electrons. Each microstate is described by the collection of occupation numbers {n k,σ = or } of electrons in each allowed level. Therefore, the energy and the total number of electrons in the system, in this microstate, are: E {n k,σ } = k,σ ǫ k n k,σ ; N {n k,σ } = k,σ n k,σ (4) () where we use the shorthand notation k,σ to mean sum over all allowed values of k according to Eq. (), and sum over the two values σ = ±. Then, the probability to find the system in this microstate is: p({n k,σ }) = Z e β(e {n k,σ } µn {n k,σ }) (5) where the normalization condition gives: Z = e β(e {n k,σ } µn {n k,σ }) = e β(ǫ k µ)n k,σ k,σ n k,σ = k,σ n k,σ = = k,σ ( + e β(ǫ k µ) ) As a result, ln Z = k,σ ln ( + e β(ǫ k µ) ) In terms of β,α = βµ,v, we have: ln Z = k,σ ln ( + e βǫ k +α ) (6) and we can calculate the average number of particles in the system: N = α ln Z = k,σ e β(ǫ k µ) + (7) 5

6 and the internal energy: U = H = β ln Z = k,σ ǫ k e β(ǫ k µ) + (8) which by comparison with Eq. (4) shows that the average occupation number of a level must be: n k,σ = e β(ǫ k µ) + = f FD(ǫ k ) (9) i.e. the Fermi-Dirac distribution expected for fermions. Of course, we can also calculate the average occupation numbers directly (see assignment 5 of 7). So we just rediscovered the general result (average occupation number for a fermion given by the Fermi-Dirac distribution) we could have started directly from here, in fact, since we knew this is what we should find in any event. In problems in assignments and exams feel free to jump directly here (unless I ask you specifically to derive this result). Equation (7) gives us N for a given chemical potential (and temperature, of course). The problem we re really interested in, is the one where the number of electrons N is fixed (determined by the concentration of electrons in the system, times the volume of the crystal). So the first question we will figure out, is how should we choose µ such that the average number N is precisely the number N that we want? In other words, we re using a grandcanonical ensemble (where we can solve the problem) instead of a canonical one (which describes the system we re really interested in the number of electrons in the metal is really a fixed number). We know that in principle, the grandcanonical system can have any number of electrons in it, but in reality, it is overwhelmingly likely that it has precisely the average number N of electrons in the system, any other possibility has extremely low (but not zero) probability, if N is large enough (remember that the standard deviation N N N N if N is large). Therefore we will choose µ carefully such that the resulting N = N, i.e. we adjust the chemical potential until we re sure that there are precisely as many electrons as we wish to have in the system. To do this, we have to learn one more trick, which is how to evaluate k. Remember that we are interested in a large volume, so L is large and therefore the allowed values of k are close together, see Eq. (). In fact, as L becomes larger and larger the spacing between the values of k become smaller and smaller, and we can think of this sum as being an approximation of an integral. This is similar to saying that if we have a sequence of points x i = iδx, where i =, ±, ±,... and we have to calculate i= f(x i ), in the limit δx we can approximate this as: i= f(x i ) = δx i= f(x i )δx dxf(x) δx We do precisely the same thing for our sum, except that since it is three dimensional, we will have three integrals to deal with. In our case k x = n x π L δk = π L. Then: f( k) = δk δk δkf(k (δk) x,k y,k z ) dk k n x= n y= n z= (δk) x dk y dk z f( k) 6

7 This becomes exact in the limit where δk (L ). The conclusion is that in D, we will always replace: in D: k f( k) = V (π) d kf( k) = V dk (π) x dk y dk z f( k) () Of course, in D and D there is only one, respectively two, integrals, so that: in D: k f(k) = L π dkf(k) () and in D: k f( k) = A dk (π) x dk y f( k) () where L and A are, respectively, the length and the area of the D and D systems considered. Therefore, we have: ln Z = V d (π) k ln ( + e ) β(ǫ k µ) () σ=± which now nicely exhibits the dependence of T,V,µ. Eqs. (7) and (8) become: and N = U = V (π) V (π) σ=± σ=± d k e β(ǫ µ) k + d kǫ k e β(ǫ k µ) + Note that because here the energy levels ǫ k do not depend on σ, the sum over spin projections gives a everywhere. In general, for a spin S, the sum over spin projections would give S +, since that is the number of possible projections. However, if the system is in magnetic field, the energies become functions of σ and the spin-up and spin-down contributions will no longer be equal. The other quantity of interest is the pressure, so let us derive its expression as well. As we know, Φ = k B T ln Z and p = φ, so this comes out easily as: V p = k BT (π) σ=± (4) (5) d k ln ( + e β(ǫ k µ) ) (6) The integrands in all these expressions have the property that they depend on k only through ǫ k, i.e. all integrals are of the form (in D): (π) d kf(ǫ k ) To find an easy way to evaluate these integrals, we use the fact that: F(ǫ k ) = dǫf(ǫ)δ(ǫ ǫ k ) that s just the definition of a delta function. Note the lower limit of the integral is taken to be and not because all ǫ k. In general, the lower limit of this integral is the ground-state energy e GS. 7

8 This is zero in this case (the GS corresponds to n x = n y = n z = k x = k y = k z = ǫ k= =, which is clearly the lowest possible energy. However, if there is a magnetic field applied, then e GS <!, and we would have to use its value as the lower limit in the integral). Then, we rewrite: where (π) d kf(ǫ k ) = (π) g s (ǫ) = (π) d k F(ǫ)dǫδ(ǫ ǫ k ) = dǫf(ǫ)g s (ǫ) d kδ(ǫ ǫ k ) = V k δ(ǫ ǫ k ) (7) if we went back to the discrete sum. This quantity is called the one-electron density of states (or DOS) per spin projection because it counts how many of the levels with a given spin projection have a given energy ǫ (if ǫ k ǫ δ(ǫ ǫ k ) = ). The total density of states, which counts how many levels have a given energy, is just g(ǫ) = V k,σ δ(ǫ ǫ k ) = g s (ǫ) because there is equal number of levels with spin up and spin down at the same energy (in the absence of a magnetic field). Of course, if the fermions had some other spin S, we would have g(ǫ) = (S + )g s (ǫ), i.e. there are (S + ) as many levels of energy ǫ, as levels with a fixed spin projection and energy ǫ. By doing this trick of introducing the density of states, we have reduced the triple integrals over momenta to a single integral over energies. Let us find the DOS in D. First we go to spherical coordinates for the momentum. Because ǫ k depends only on the magnitude of k, the integrals over the angles are trivial and give a 4π, and we are left with: g s (ǫ) = ( (π) 4π dkk δ ǫ h k ) m Here we have a bit of a problem. We know how to deal with things like: dkf(k)δ(k k ) = F(k ) but the delta function that appears in our expression is more complicated that this. However, it can be easily reduced to the simpler form, if we use the identity: δ(f(x)) = x δ(x x ) f (x ) where f(x) is any function, and x are all its simple roots. This makes sense, since the δ function is zero unless its argument is zero, and the argument is zero only when x = x since only F(x ) =. The division by the absolute value of the first derivative is just a needed rescaling. The reason is that in the vicinity of a root, we can approximate f(x) (x x )f (x ), with f (x ) being some specific (finite) number. But: dxf(x)δ(α(x x )) = α F(x ) = dxf(x) δ(x x ) α 8

9 where to calculate the first integral I rescaled its variable αx y. Since this equality holds for any possible function F(x), it follows that we must have δ(α(x x )) = δ(x x )/ α, and putting everything together, we find: δ(f(x)) = x δ(f (x )(x x )) = x δ(x x ) f (x ) For us, f(k) = ǫ h k /m = k = mǫ/ h (remember that both ǫ and k are integrated only over positive values, so the negative root is not going to contribute and we can ignore it). Then, f (k) = h k/m f (k ) = h ǫ/m and thus we can replace: to find: δ ( ǫ h k m in D: g s (ǫ) = π ) = δ ( k ) mǫ h h ǫ m ( dkk δ k ) mǫ h h ǫ m Similar calculations in D and D give (see assignment 5): and = 4π m ǫ (8) h in D: g s (ǫ) = m π h (9) in D: g s (ǫ) = π ( ) m ǫ h () Again, for electrons with spin S = /, the total DOS is just double of the spin-projected DOS. In D, the three basic equations (4),(5) and (6) now become: n = N V = σ=± dǫg s (ǫ) e β(ǫ µ) + = π m h ǫ dǫ e β(ǫ µ) + U = V σ=± dǫg s (ǫ)ǫ e β(ǫ µ) + = V π m h ǫ dǫ e β(ǫ µ) + p =... = k BT m π h dǫ ǫ ln ( + e β(ǫ µ)) = m ǫ π h dǫ e β(ǫ µ) + where the second equality follows after integrating by parts once. Thus, it follows that for a fermionic ideal gas, we always have: pv = U. In principle, we are now done. We use the first equation and adjust the value of µ until its result is the desired density of electrons n. Then, we use this value of µ in the second integral to find U, and then we also have p, for any given n,v,t values. Strictly speaking, these integrals can be evaluated exactly only numerically. However, in the asymptotic limits of very high and of very low 9

10 temperatures, we can actually get good approximations for these integrals (see notes on bosonic and fermionic integrals). As shown in these notes, if we define the fermionic integrals: J ν (+) (β,βµ) = we can show that (a) if βµ (i.e., it is a very negative number), and (b) if βµ : J (+) ν (β,βµ) This is called the Sommerfeld expansion. High-temperature limit ǫ ν dǫ e β(ǫ µ) + Γ(ν + ) β ν+ e βµ +... [ ] J ν (+) (β,βµ) µν+ + π ν(ν + ) +... ν + 6 (βµ) This is the limit where we expect to recover the classical results. If we look at the chemical potential of a classical ideal gas, we get (see write-up on canonical ensembles, or redo the calculation): µ = k B T ln V N ( πmkb T Let me define: λ = πmkb T h This is called the thermal de Broglie wavelength. If we associate it to a a characteristic momentum, h = p p = mπk λ B T p = πk m BT, so this is, roughly speaking, the de Broglie wavelength of a particle with a kinetic energy of the order πk B T, which is close to the average kinetic energy of a particle in equilibrium (from equipartition, that would be /k B T). In the limit of high temperatures, this is a large energy and therefore the de Broglie wavelength is very small. More precisely, we define the classical limit as the limit where λ V N i.e. the electrons look like point masses, since the volume λ over which they spread because of quantum effects is very small compared to the average volume V/N available to each electron, and which defines the average distance between electrons. So then it follows that h ) [ ] V ln µ k Nλ B T in the classical limit. Because of this, we can use approximation (a) for the fermionic integrals, to find immediately: n λ eβµ ; U V λ β eβµ

11 which combine into: U = V n k BT = Nk BT i.e. the expected classical result. Also, since pv = U pv = Nk BT. To be complete, let us consider the value for the chemical potential, which comes out (from the first equation, for n) to be µ = k B T ln(nλ /). Oops! This is not quite right; as I just said, the classical prediction is µ = k B T ln(nλ ), so the argument of the ln differs by a factor of. But there is no real conflict, because the classical calculation was done for spinless particles. To consider the contribution of the spin, we have to include it in the calculation (problem with mixed degrees of freedom, both classical and quantum). If there is no magnetic field, the only difference the spin makes is to bring an extra factor of N to Z (from sum over the spin projections... there are N possible spin arrangements that all have the same energy, if there is no magnetic field, so the Z is just N times bigger than the Z if there was no spin). Then, F decreases by an extra k B T ln N = Nk B T ln, and if you track the effect of this on the chemical potential, you ll see that it precisely brings in the factor of that seemed to be missing. So things are really fine. For those of you who still keep track, this is the final justification that the division by h Nf is the right thing to do in the classical calculations, in that it leads to agreement with the proper results of quantum calculations, i.e. classical results are indeed recovered at high temperatures. Note that unlike for the classical systems, where there was all sorts of fudging (first we pretended that particles are distinguishable and gave them locations r, r,... and then we realized that they are not, so we divided by N! to try to fix the overcounting problem, and we put in the h Nf by hand, arguing that that must be the volume of a microstate, and we used Stirling to simplify lnn!, etc) in the quantum calculation there is no fudging. Once we know the single-particle energies, and whether particles are bosons or fermions, everything can be done exactly. Of course, what really interests us are low temperatures, where quantum behavior should become apparent. The transition to low temperatures should occur roughly when nλ As T decreases, λ increases and this limit will be achieved sooner or later. For such low temperatures we can no longer make the approximations (a), so the results will be quite different from the classical results. Or, in other words, at these temperatures the de-broglie wavelength (the length over which a particle is smeared out by quantum effects) becomes comparable to the average distance between particles, and we really need to use wavefunctions to describe their state, we can t pretend that they are point-like classical particles. Let us estimate what this characteristic temperature should be, in a typical metal. We have: n = λ = mπkb T D kb T h D = h mπ (n) For typical electron densities (say, about e per (5Å), since 5Å is a reasonable distance between atoms in a cubic lattice) and masses, this is an energy scale of the order ev, which corresponds to a temperature T D, K! (the temperature at the surface of our Sun is 6K, just to give you a feeling for how hot this is). And one would have to be well above this temperature to see the electrons behave classically. Needless to say, the metal would melt long before one can heat it that much. This shows that the room temperature T K, which we tend to think of as being rather warm, is in fact extremely cold for electrons, and they are behaving quantum mechanically.

12 In contrast, if we consider a gas of fermionic atoms, the main change we would have to make is to replace m by the mass M of the atom (if the spin S of the atoms is different from /, we would also have to track its effects, but that doesn t change the results too much). Since the mass of a proton or neutron is roughly the electron mass, and there are many protons and neutrons in most atoms, their mass is easily a factor of 4 or more larger, therefore their T D K or less. Of course, the typical densities of atoms in gases at reasonable pressure are also smaller than those of electrons in metal, which further decreases the value of T D. Which is why gases of atoms behave classically at room temperature. Low-temperature limit = the degenerate limit These are temperatures for which: nλ Here we could only use approximation (a) if we kept all higher order terms in the Dirichlet-Riemann expansion, but that would not allow us to do any calculations. What we see from this expansion, though, is that in this limit we must have e βµ βµ. This is possible if µ >, since then as T,βµ.. T = limit Let us look first at this limiting case, and then we will consider finite but small T. As we know, at T =, quantum systems go into their ground states, i.e. the state of lowest possible energy. The way to achieve this, is to completely fill the lowest available levels to their maximum. The energy of the highest filled level, in the ground-state, is called the Fermi energy, E F. Then, the ground state is defined by the occupation numbers: {, if ǫ k E n k,σ = F, otherwise Note that this is what the Fermi-Dirac distribution predicts, if we identify µ = E F (at T = ). In this case, if ǫ k < µ β(ǫ k µ) as β, thus e β(ǫ k µ) and n k,σ. By contrast, for levels above the Fermi energy or chemical potential, ǫ k > µ β(ǫ k µ) as β, thus e β(ǫ k µ) and n k,σ. The distribution is discontinuous for ǫ k = E F, where the occupation number becomes /. The Fermi distribution is pictured as a function of the energy of the level, in the plot on the next page. If temperature is small but non-zero, we expect that results are a bit changed around the Fermi energy, on a distance of order k B T on either sides (continuous line in the plot). Basically, at low T we know that some excited states with energies of up to a few k B T will contribute considerably to the partition function, and these can be obtained by moving a few electrons from under the Fermi energy to just above it this is just the same as the general discussion we had for the Fermi-Dirac distribution at low temperatures. It follows that at T =, the average occupation number of a level of energy E is n(ǫ) = /(e β(ǫ µ) + ) Θ(E F ǫ). Making this replacement in our equations for n and U allows us to evaluate them easily, now: n = π m h ǫ dǫ e β(ǫ µ) + T π m h dǫǫ Θ(EF ǫ) = π m E F h dǫǫ

13 U = V π m h ǫ dǫ e β(ǫ µ) + T V π n(e) m h dǫǫ Θ(EF ǫ) = V π m E F h dǫǫ T= T > E = µ F E Sketch of the average occupation number of a level of energy E at T = (continuous line) and at a small but finite T (dashed line). The integrals are now straightforward, and we find that at T =, n = π m h E F µ = E F = h ( π n ) m and U = V m π h 5 E 5 F = 5 E FnV = 5 E F N i.e., in average each particle contributes /5E F. Note that this is very different from what the classical equipartition theorem would predict, which would be U = at T =. Finally, since p = U/V p = 5 E Fn again, very different from the classical prediction p = nk B T, which would be zero at T =. Of course, classically we find the ground-state (T = state) to have particles not moving (zero kinetic energy) and thus there is no pressure. But for a true quantum system the Pauli principle is obeyed, and because of it electrons occupy states with different k values. The ones with higher energy have a higher momentum p = h k and will contribute to the pressure, which is actually quite large. Note that because the T = average occupation numbers are so simple ( below Fermi energy, above) we do not really need to do all these transformation to integrals over energy etc. We could solve the problem directly in the following (quickest) way. Let k F > be the wavenumber of the particles with energy E F, i.e.: k k F,σ E F = h k F m k F = mef It follows that in the ground state, all levels with momenta k k F are filled, and all other ones are empty. Then: N = = V d (π) kθ(k F k) = V 4πkF (π) h

14 because the D integral is the volume of a sphere of radius k F, so: indeed. Similarly, U = k k F,σ n = N V ǫ k = V (π) = k F π E F = h kf m = h ( π n ) m d kθ(k F k) h k m = V h kf (π) 4π dkk 4 = V h kf 5 m π m 5 after going to spherical coordinates and doing the integrals. If we set aside an h k F/(m) = E F, we are left with V k F/(5π ) = V n/5, and we recover U = 5 N E F. I recommend that for T = calculations you do things this way, since it is the fastest way. However, at finite T the average occupation numbers are no longer so simple (either or ), so we have to go to energy integrals.. Finite, low T limit Here, we know that µβ, since µ will change somewhat from E F (if we want to keep N /V = n fixed), but it will still be a positive quantity of order, K. It follows that we can use the Sommerfeld expansion, which says that if βµ : J ν (+) (β,βµ) = ǫ ν [ dǫ e β(ǫ µ) + = µν+ ν + + π 6 ] ν(ν + ) +... (βµ) The first term just recovers the T = results, since there we integrated only up to µ. The second term will give us the finite T corrections. Let us first find the new value of µ: n = π m h ǫ dǫ e β(ǫ µ) + = π m µ h ( + π 6 (βµ) +... We already know that at T =, n = m E F π h so, since we want to keep the electron density constant, we can rewrite the equation as: E F = µ ( + π 8 ) (βµ) +... µ = EF ( + π 8 ) (βµ) +... which we would like to solve to find µ as a function of E F (which is already known once we choose a concentration n) and T. This is an implicit equation, since µ depends on itself. However, if T is small, then βµ, as discussed, and we must have µ not too different from E F. So, to first order, we can replace µ by E F on the right hand side, which will lead to errors of the same order as the terms already neglected, i.e /(βµ) 4 or higher. After that, we use a Taylor expansion ( + x) x/ +..., since x = π 8 (βe F is small, and we finally find: ) ( ) ( ) µ = E F + π 8 (βe F ) +... = EF π kb T +... E F ) 4

15 This shows that in order to keep n = N /V constant, we must adjust the chemical potential as a function of T. It starts with value µ = E F = h (π n) /(m) at T = (this is the definition of E F ) and then decreases as T increases. This actually makes sense, because remember that we said that in the classical limit (high-t), µ becomes a very large negative number. A sketch of the behavior of µ(t) is shown on the next page. It becomes negative below a temperature T which we can find exactly, because we know that: J (+) (see notes on fermionic integrals), so that: k B T = E F (β, ) = Γ ( ( ) ) ζ ( ) β [ Γ ( ( ζ ) ) ( )] µ Ε F T* T Dependence of the chemical potential on T, if the number of fermions stays constant. At low-t (in the degenerate limit), µ E F, but it decreases and becomes very negative and large as the classical limit is approached. We can also use a comparison with T as a check on when classical limit is achieved, which should happen for T T. Again, since Fermi energies in typical metals are of order ev, we find T, K. Now that we know µ(t,n), we can calculate the internal energy, and then the pressure. Using the Sommerfeld expansion, we have: if βµ, we have: U = V π m h U = V π ǫ dǫ e β(ǫ µ) + = V π m µ 5 h 5 ( + π 6 m h 5 (βµ) +... J (+) ) (β,βµ) We now use the expression for µ, and keep only terms up to order /(βe F ) : U = V 5π m h E 5 F = 5 N E F 5 π π ( kb T E F ( ) kb T +... E F ) π 8 ( + 5π 8 ) (βe F ) +... ( ) kb T +... E F 5

16 and therefore: C V = U = 5 N E F + 5π ( ) U = N T 5 E 5π BT F k E V, N F ( ) kb T +... E F = Nk B π i.e., at low T, the contribution of the electrons to the specific heat increases linearly with temperature. Of course, at very high temperatures where the classical limit becomes correct, we expect, from the equipartition theorem that C V N k B, so the specific heat of the electrons looks as sketched in the next figure. k B T E F C/Nk / Specific heat of the electrons. At low T (compared to,k) it increases linearly with T, while at very high T it reaches the classical limit. A simple qualitative understanding of these last two results can be achieved in the following way. At finite but low temperatures, some electrons with energies within k B T of E F are excited from below to above the Fermi level. We expect that the energy increases by k B T times the typical number of electrons excited. Since only electrons within k B T are likely to be excited, their number should be of order Nk B T/E F, and so we see that we expect an increase of order U = U +N(k B T) /E F, which agrees with the exact derivation up to some factor of π /4. These last two results are important: all metals are expected to have a linear-t dependence in their specific heat, coming from the electrons. From the slope one can find E F, and therefore the effective mass m of the electrons. As we will see later on, the phonons give a T dependence to the specific heat at low T. Finally, since pv = U at all T, we find that in the degenerate limit: p = ( ) 5 E Fn + 5π kb T +... i.e., something very different from the linear dependence expected for a classical gas. Here we considered a very simple structure of the energy levels, like that of free particles. Sometimes things are more complicated, but the formulae we wrote are still valid if one uses the correct density of states g(ǫ) = g s (ǫ). Considering the behavior of D and D cases also proceeds similarly, again using the appropriate g s (ǫ). Other interesting cases, such as undoped and doped semiconductors, can be treated similarly, as we will see next. E F T 6