Quantum ideal gases: bosons

Transcription

1 Quantum ideal gases: bosons Any particle with integer spin is a boson. In this notes, we will discuss the main features of the statistics of N non-interacting bosons of spin S (S =,,...). We will only investigate cases without magnetic field. As a result, the energy of the levels will be independent of the spin, and the spin will enter the calculations only through the spin-degeneracy factor S +. Moreover, we will limit ourselves to studying the D case. D and D cases can be pursued similarly, and some questions will show up in assignments. he really interesting new phenomenon, namely the Bose-Einstein condensation, only occurs in D (for non-interacting bosons. With interactions, things are different). he model Consider a gas of N non-interacting, non-relativistic bosons inside a volume V = L. hese could be, for instance, 4 He atoms, or any other bosonic atoms. Other examples of bosons are, for example, the photons (quanta of light) and the phonons (lattice vibrations). We ll discuss these latter examples separately, because they have different Hamiltonians (models) from what we consider here. We only study non-interacting (ideal) systems. As we well know, the first thing we must do is to find out the single-particle eigenenergies. Like for fermions, if there is a single boson in the system, the Hamiltonian is just the kinetic energy: ĥ = p m = h m ( x + y + z It follows that, like for fermions, the single-particle eigenenergies are: ǫ k = h k m he periodic boundary conditions (see discussion for fermions) again select the same allowed levels, given by: 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 again we can index any level by the values (n x,n y,n z,m) or, more compactly, by ( k,m), where we understand that the allowed k are the ones given by Eq. () and the spin-projection m = S, S +,...,S has S + distinct values (because there is no magnetic field, each level is spin-degenerate). So far, everything is similar to what we had for fermions (except for the specific spin degeneracy) this is not surprising, since if there is a single particle in the box, it really makes no difference if it is a fermion or a boson. We need at least two particles to see whether the wavefunction changes sign when we interchange them, or whether they do or do not occupy the same single-particle orbital, before we see clear distinction between fermionic and bosonic behavior. Now that we know the energy levels, we proceed in the usual way. I could actually skip directly to Eqs. (4) and (5), because we know that the Bose-Einstein distribution /[e β(ǫ µ) ] gives the average occupation number of a level with energy ǫ. For completeness, I ll sketch the whole derivation here again. We use the grand-canonical ensemble to study the system, fixing,v,µ. Each microstate is described by the collection of all occupation numbers {n k,m =,,,...} of bosons in each allowed level. herefore, the energy and the total number of bosons in the system, in this microstate, are: E {n k,m } = k,m ǫ k n k,m ; N {n k,m } = k,m n k,m () )

2 where we use the shorthand notation k,m to mean sum over all allowed values of k according to Eq. (), and sum over the g S values of m. hen, the probability to find the system in this microstate is: ρ({n k,m }) = Z e β(e {n k,m } µn {n k,m }) () where the normalization condition gives: Z = e β(e {n k,m } µn {n k,m }) = k,m n k,m = k,m e β(ǫ k µ)n k,m = ( n k,m = k,m ) e β(ǫ k µ) where the last step is possible if and only if k,ǫ k µ µ (here, e GS = when n x = n y = n z =. Remember that in general, we found that for bosons we must have µ e GS ). his is one major difference compared to fermions, where µ can be (and is, at low-) positive! As a result, ln Z = k,m ln ( e β(ǫ k µ) ) and we can calculate the average number of particles in the system: and the internal energy: N = β U = H = β µ ln Z = k,m e β(ǫ µ) k x ln Z(x) = ǫ k x= k,m e β(ǫ µ) k (4) (5) where Since we must have: Z(x) = µstate e βxeµstate+βµnµstate = ( ) e βxǫ k +βµ k,m N = k,m n k,m ; U = k,m ǫ k n k,m we find that the average occupation number of a level must be: n k,m = e β(ǫ k µ) = f BE(ǫ k ) (6) i.e. the Bose-Einstein distribution. Of course, we can also calculate the average occupation numbers directly as an ensemble average (see assignment 5, last problem). Note: in any assignment or exam problem, you can start directly with Eqs. (4) and (5), since the average occupation numbers should be well known. I redid the derivation here just to jog your memory. he next step is to switch to energy integrals, using the same tricks as for fermions. Since the one-particle levels have precisely the same energy as for fermions, we will get precisely the same density of states per spin-projection: g s (ǫ) = 4π m ǫ h

3 (in D; in D and D we have to use the appropriate expressions). herefore, we can make the usual replacements: F(ǫ k ) = V S d (π) kf(ǫ k ) = V (S + ) dǫg s (ǫ)f(ǫ) k,m m= S For simplicity, let me denote the spin degeneracy: g S = S + Using the D density of states per spin projection, we find: and and the internal energy: ln Z = V g S 4π N = V g S 4π U = V g S 4π m h m h m h dǫ ǫ ln ( e β(ǫ µ)) dǫ ǫ e β(ǫ µ) k dǫǫ e β(ǫ k µ) From Φ = k B ln Z and p = φ, we can also find the pressure: V (7) (8) p = k B g S 4π m h dǫ ǫ ln ( e β(ǫ µ)) = U V pv = U if we integrate once by parts (exactly like for fermions). Like for the fermionic case, we see that here all answers depend on bosonic integrals of the general form: J ν ( ) (β,βµ) = ǫ ν dǫ e β(ǫ µ) = Γ(ν + ) β ν+ g ν+ ( e βµ ) (the Riemann-Dirichlet expansion, see notes on fermionic and bosonic integrals). Here, g ν (x) = is the Riemann-Dirichlet function. We re only interested in arguments e βµ, since µ. In this case, this function looks like in the figure below. he essential thing to remember is that g ν () = k= is finite if ν >, and infinite (the series does not converge) if ν. he z = k ν value of this function has its own name, it is called the Riemann function: k= g ν () = ζ(ν) = So, as z, g ν (z) increases from zero to ζ(ν). his largest value possible, ζ(ν), is finite if ν > and infinite if ν. x k k ν k= k ν

4 g (z) ν ν=/ ν= ν= ν= ν=infinity z Fig. Sketch of Riemann-Dirichlet function, g ν(z). he limiting value g ν() = ζ(ν) is finite if ν > and infinite if ν. It is convenient to introduce a short-hand notation: z = e βµ z is called fugacity or activity, depending who you ask. So, in terms of the fugacity and the Riemann- Dirichlet function, we have: n = N V = g S m Γ( ) 4π h g(z) (9) U = V g S 4π m h Γ( 5 ) β 5 β g5(z) () and pv = /U. I remind you that Γ( ) = Γ() = π ; Γ(5) = Γ() = π. 4 As always, the strategy is to use Eq. (9) to adjust the value of µ (or z) so that the boson concentration n has the desired value (whatever we would like it to be). After finding µ (or z) one can use this in Eq. () and other equations to calculate dependence of various quantities on n,v, instead of µ,v,. High-temperature limit his is the limit where we expect to recover the classical results. We already know from the equivalent discussion for fermions that in this limit βµ (the chemical potential is very negative in the classical limit see results for classical grand-canonical ensemble). It follows that z = e βµ, and therefore we can keep only the largest contribution (k = term) to the Riemann-Dirichlet function: his is equivalent with the approximation: J ν ( ) (β,βµ) = dǫ ǫ ν g ν (z) z e β(ǫ µ) dǫǫ ν e β(ǫ µ) = Γ(ν + ) β ν+ e βµ because if µ ǫ µ e β(ǫ µ), so we can ignore the - in the denominator. Note that this is precisely what we had for fermions (there we ignored a + in the denominator, for the same reasons). Because we have the same expressions, we will get the same high- limit as for fermions, in agreement with the classical predictions. 4

5 Let me quickly summarize the results: he density now becomes: n = g S 4π m h Γ( ) β e βµ +... = g S λ eβµ +... () where λ = πmkb h is the thermal de Broglie wavelength. he previous equation looks like the one for fermions, except that there g S =, because we considered only spin /. From this, we find: µ = k B ln g SV πmkb N h which agrees with the classical result for particles with spin S (see similar discussion for fermions). In this high- limit, the energy becomes: and thus: U = V g S 4π m h Γ( 5 ) β 5 e βµ +... = β V n +... = N k B +... pv = U = N k B +... he conclusion is that at high temperatures, the behavior of bosons and fermions agrees with the classical predictions, because in this limit the average occupation numbers e β(ǫ µ) + e β(ǫ µ) e β(ǫ µ) Because the average occupation number of each level is so small, one cannot distinguish between fermions and bosons their behavior is identical and agrees with classical calculations, as expected at high-. he difference between fermions and bosons becomes clear when average occupation numbers are getting close to. For fermions, that is the maximum possible value, while for bosons it is not, so one expects behavior in this limit to be very different. As we know, these differences should be observed in the degenerate (quantum) limit, at low temperatures. Let us see what low temperatures means in this case. he high-temperature approximation holds if z = e βµ nλ g S As for fermions, we can estimate that this approximation will fail, and the crossover to quantum behavior must take place, if becomes low enough that: nλ g S n = g S λ = g S ( mπkb D h ) ( ) kb D = h n mπ g S Again, this looks very similar to the formula we had for electrons, except that g S replaces the spindegeneracy which for electrons is. However, the characteristic numbers are very different. First of all, here m is the mass of a bosonic atom, which is 4 larger than the mass of an electron (nucleons are very heavy compared to electrons). So even if one could manage to get similarly large 5

6 concentrations of bosonic atoms, as one has for electrons in metals, the much larger mass would still lower this crossover temperature by a factor of or 4, i.e. it would bring it to something of the order of a few K. So interesting quantum bosonic behavior can only be expected to be seen at very low temperatures (at least for bosonic atoms. We will study next photons, which are also bosons but with zero rest mass, and we ll see how that changes things). Low-temperature limit = the degenerate limit In this limit, z = e βµ, which is always less than (for bosons µ < ), is not necessarily much less than. So we really need to keep all terms in the Riemann-Dirichlet series, we have no excuse to stop after just a few. In other words, to find the value of µ (or z), we need to solve exactly the equation: n = g S (z) () g(z) = ( ) ( n g S λ g h πmk B ) = a where the value of a depends on the density and various constants and can be calculated. For any value of the temperature, the quantity on the right-hand side is known its value increases as decreases. We can use a graphical solution to find the value of z, as shown in the figure below: g(z) g(z)=a/ / a/ / c ζ(/) =.64 a/ / a/ / z z z Fig. Graphical solution for equation g (z) = a. Because >, g () = ζ ( ) is finite. It is approximately equal to.64. his plot shows that we have a problem at very low. If the temperature is large enough so that a/ g() = ζ( ).64, we have a solution z < µ <, as sketched for two temperatures >. his holds for > C, where the critical temperature satisfies: ( ) ( a n h ) ( = = ζ c = c g S πmk B c ) h πmk B n g S ζ ( ) At this value, the chemical potential becomes µ =, because here the solution is z = e βµ =. What happens if < C? If we take Eq. () as being correct, because it no longer has a solution, it somehow seems to imply that we can not go to lower temperatures?! his is obviously wrong, so something about Eq. () must become wrong when < c. 6

7 A bit of thought shows where the problem is. We know that as, the system should go in its ground state. Because these are bosons, the ground-state here is the state where all bosons are in the k = level, which has the lowest possible energy. It follows that as we lower the temperature towards the ground-state, the occupation number of this level should become larger and larger and therefore its contribution to N is more and more important. But here is the problem: when we switch from a sum over k to an integral, and then an energy integral, the contribution from the ǫ = level is multiplied by the density of states g s (ǫ = ) =. So we actually do not take the contribution of the most important level into account at all! o do things properly, we must consider this contribution separately: N = g S n k= + n k,n = g S e βµ + g V S (π) k,m d k e β(ǫ k µ) n = g S V e βµ + g S (z) () λ g where the first term is the contribution from the lowest energy state, and the second term is what we get after switching to an integral over energies, same as in Eq. (). Remember that this equality only holds in the limit V, because only in that limit we are allowed to switch from sums over momenta to integrals over momenta. Now we can understand what happens: if > C, we have z < e βµ = /z >, so that /(e βµ ) is some positive, finite number. As a result, when we divide by V and take the limit V, the contribution of this term vanishes, and we are left with the same equation we had before, with the same solution. However, for c, we must have µ = (there is no solution with µ <, and we know that µ cannot be positive, so it must stay at ). In this case, the denominator e βµ =, so that the occupation number for the k = state becomes infinite, and therefore the first term is no longer negligible. In fact, the first term: g S V e βµ = n is the concentration of bosons in the lowest energy level available (by definition). he occupation number diverges when µ = because this is an infinite size system V, but the ratio stays finite. So for C, we know that µ = g(z = ) = ζ(/), and therefore we can find what concentration of bosons are in the ground state at any temperature < C : n = n + g S ( πmkb h Since the critical temperature c is defined by: ) ζ ( ) πmkb ( c n = g S ζ h ) we can replace all the constants in the second term, to find. ( ) n = n + n c 7

8 so that: ( n = n c ) So the concentration of bosons in the k = level (the lowest level available to them) is: n = ], if > c, if c [ n c he solutions for µ and n are sketched on the next page. What this shows is that below c, a macroscopic fraction of the atoms occupy the lowest energy level available, with k =. As is lowered towards, more and more atoms occupy this level, so that when = all the atoms occupy it, n = n. his is indeed the expected ground-state for non-interacting bosons. his phenomenon, of having a macroscopic percentage of the bosons condensed in the lowest level available, is called the Bose-Einstein condensation. he temperature c below which this occurs is called the critical temperature, and it is given by the equation at the top of the previous page. For example, for 4 He, for a reasonable concentration of 4 4 mol/m, one finds c K. 4 He has a transition with a measured c K, so the estimate from our simple model is not too bad. he reason it is not better is that in He interactions are not negligible, the transition is into a liquid state and one needs to properly account for that. his liquid phase has extremely interesting properties, because of the Bose-Einstein condensation: for example, the fraction of fluid that is condensed flows without friction, and because of this it is called superfluid. here is an analog phenomenon that happens in metals, called superconductivity, where the resistance of the metal is zero (not very small, but really zero) below a critical temperature. What this means, is that at these low temperatures, electron currents can flow without friction (without resistance) through metal. he reason for this is very interesting and deserves a brief explanation. Remember that when we studied electrons in metals, we argued that to first order, they can be treated as free (non-interacting) particles. his is quite a good approximation and one gets good agreement with experiments at most temperatures. However, the approximation is not perfect, because there are weak interactions between electrons. Interestingly enough, these are attractive why that is so is a long story, and it involves oscillations of the lattice (phonons), see below for a short explanation. So, imagine this very weak attraction between electrons. At high-, the electrons have lots of energy and move very fast past one another, so this weak attraction really has no effects, which is why a model of non-interacting electrons works so well. However, as is lowered, the electrons become colder and less energetic, and if is low enough, even this weak attraction is enough to bind electrons into pairs, called Cooper pairs. Now, even though each electron is a fermion, a pair of two fermions is a boson, because adding up two half-integer spins gives an integer spin. So the Cooper pairs are bosons, and exhibit Bose-Einstein condensation at low-. Below this temperature, Cooper pairs can move through the metal without any friction (analog to the superfluid 4 He), and therefore the resistance is zero. One way people tested this, is to induce a current in a superconducting ring. If the ring was in the normal metal phase, we know that the current would quickly go to zero. In the superconductor, this current flowed for years and years without any measurable decrease - until people got bored and stopped the experiment. 8

9 n n exc µ c c Fig. Left: the concentration n of atoms in the lowest level (blue line) vs. the concentration n exc of atoms in all other (higher energy) levels. As temperature is lowered below c, more and more atoms condense in the ground-state level. Right: the chemical potential. At high-, in the classical limit, it is a very negative number. As is lowered it comes closer and closer to, and for all C, we have µ = (the maximum allowed value in this case). So how can there be attraction between electrons? Well, the rough story is the following: think about an electron traveling through the crystal, at low. Because there is attraction between the electron and the atoms making up the lattice, as the electron goes along, the atoms near its path are attracted to it and move slightly towards it (you should think of the atoms of the lattice as harmonic oscillators, which can move a bit about their equilibrium positions). At high-, the atoms oscillate very strongly and would immediately destroy this correlation. But at low-, they have little energy so they move very slowly, and therefore for awhile, there is a memory of the path taken by the electron all atoms near it are drawn towards this path, and for awhile they stay closer to it. Now, if any other electron happens to be nearby, it will be attracted to this larger concentration of atoms near the path created by the first electron, and will start to follow it in essence, it s as if the electrons are paired, since one will tend to follow the other. o understand things in detail, one needs to learn more condensed matter physics to know how to deal with these interactions properly. However, let me just say, in conclusion, that understanding superconductivity was one of the biggest challenges and successes of the last century. his phenomenon was discovered experimentally by Kamerlingh Onnes in 9, and it was only in 957 that it was fully explained by Bardeen, Cooper and Schrieffer. Part of it is that quantum mechanics and a lot of other physics had to be developed in the meantime; but part of it is that it is very counterintuitive to think that electrons could attract each other, and form pairs which are bosons and Bose-Einstein condense! But it does happen, it is really beautiful quantum physics, and with very important applications as well! Now let s figure out what happens to U and p at these low temperatures, in the condensed phase. First of all, let us see what happens if we properly take into account the contribution of the lowestlevel energy (as we did for the average number N ). Well, not much happens, because: U = k,m ǫ k n k,m Since the energy of the k = level is ǫ k= =, the atoms in this level do not contribute anything to the total internal energy, whether there are just a few of them, or a large number of them. So the general equation: U = V g S 4π m Γ ( ) 5 h β 5 9 g5(z)

10 remains correct at all. Below C we have µ = z =, and we can replace g5(z = ) = ζ( 5).. We can put this in a nicer form using the definition of the critical temperature: to find that for C : n = N V = g S 4π m Γ ( h β c U = ( ) N k ζ ( ) 5 B c ζ ( ) ) ( ζ ) Indeed, if, U because all atoms are in the level of zero energy. he specific heat varies with temperature as: ( ) U C V = V, N which is a rather unusual power law. Of course, we know that at high- we recover the classical prediction of the equipartition theorem, N k B. A sketch of the behavior of C V with is shown in the next figure. Finally, we know that pv = U always, therefore below C we have: ( ) ζ ( ) 5 pv = N k B c ζ ( ) Why does the pressure go to zero as? he reason is that in this limit, all atoms occupy the level with k =, i.e. zero momentum. For pressure to arise, atoms have to collide with the walls and exchange momentum with them. his is possible at higher, where some fraction of the atoms occupy excited states with finite k, but not at =, where they all have zero momentum. C v Nk B / classical limit ~ / c Fig 4. Specific heat. Below c it varies like, while above c is goes towards the classical limit of / N k B. In conclusion, in D, for temperatures below c we see a new state of matter with very interesting properties, coming from the fact that bosons all start to occupy the same level. hings can change if one changes the dimensionality (and therefore, the value of the index ν of the Riemann-Dirichlet function), or if interactions are included.