A Crash Course in Statistical Mechanics, Lecture notes of Quantum Mechanics

Download A Crash Course in Statistical Mechanics and more Lecture notes Quantum Mechanics in PDF only on Docsity! A Crash Course in Statistical Mechanics Noah Miller December 27, 2018 Abstract A friendly introduction to statistical mechanics, geared towards covering the powerful methods physicists have developed for working in the subject. Contents 1 Statistical Mechanics 1 1.1 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Temperature and Equilibrium . . . . . . . . . . . . . . 5 1.3 The Partition Function . . . . . . . . . . . . . . . . . . 8 1.4 Free energy . . . . . . . . . . . . . . . . . . . . . . . . 14 1.5 Phase Transitions . . . . . . . . . . . . . . . . . . . . . 15 1.6 Example: Box of Gas . . . . . . . . . . . . . . . . . . . 17 1.7 Shannon Entropy . . . . . . . . . . . . . . . . . . . . . 18 1.8 Quantum Mechanics, Density Matrices . . . . . . . . . . 19 1.9 Example: Two state system . . . . . . . . . . . . . . . . 21 1.10 Entropy of Mixed States . . . . . . . . . . . . . . . . . 23 1.11 Classicality from environmental entanglement . . . . . . 23 1.12 The Quantum Partition Function . . . . . . . . . . . . . 27 1 Statistical Mechanics 1.1 Entropy Statistical Mechanics is a branch of physics that pervades all other branches. Statistical mechanics is relevant to Newtonian mechanics, relativity, quantum mechanics, and quantum field theory. 1 Figure 1: Statistical mechanics applies to all realms of physics. Its exact incarnation is a little different in each quadrant, but the basic details are identical. The most important quantity in statistical mechanics is called “en- tropy,” which we label by S. People sometimes say that entropy is a measure of the “disorder” of a system, but I don’t think this a good way to think about it. But before we define entropy, we need to discuss two different notions of state: “microstates” and “macrostates.” In physics, we like to describe the real world as mathematical objects. In classical physics, states are points in a “phase space.” Say for example you had N particles moving around in 3 dimensions. It would take 6N real numbers to specify the physical state of this system at a given instant: 3 numbers for each particle’s position and 3 numbers for each particle’s momentum. The phase space for this system would therefore just be R6N . (x1, y1, z1, px1, py1, pz1, . . . xN , yN , zN , pxN , pyN , pzN) ∈ R6N (In quantum mechanics, states are vectors in a Hilbert space H instead of points in a phase space. We’ll return to the quantum case a bit later.) A “microstate” is a state of the above form. It contains absolutely all the physical information that an omniscent observer could know. If you were to know the exact microstate of a system and knew all of the laws of physics, you could in principle deduce what the microstate will be at all future times and what the microstate was at all past times. However, practically speaking, we can never know the true microstate of a system. For example, you could never know the positions and mo- menta of every damn particle in a box of gas. The only things we can actually measure are macroscopic variables such as internal energy, vol- ume, and particle number (U, V,N). A “macrostate” is just a set of 2 1.2 Temperature and Equilibrium Let’s say we label our macrostates by their total internal energy U and some other macroscopic variables like V and N . (Obviously, these other macroscopic variables V and N can be replaced by different quantities in different situations, but let’s just stick with this for now.) Our entropy S depends on all of these variables. S = S(U, V,N) (4) The temperature T of the (U, V,N) macrostate is then be defined to be 1 T ≡ ∂S ∂U ∣∣∣∣ V,N . (5) The partial derivative above means that we just differentiate S(U, V,N) with respect to U while keeping V and N fixed. If your system has a high temperature and you add a bit of energy dU to it, then the entropy S will not change much. If your system has a small temperature and you add a bit of energy, the entropy will increase a lot. Next, say you have two systems A and B which are free to trade energy back and forth. Figure 3: Two systems A and B trading energy. UA + UB is fixed. Say system A could be in one of ΩA possible microstates and system B could be in ΩB possible microstates. Therefore, the total AB system could be in ΩAΩB possible microstates. Therefore, the entropy SAB of both systems combined is just the sum of entropies of both sub-systems. SAB = k log(ΩAΩB) = k log ΩA + k log ΩB = SA + SB (6) 5 The crucial realization of statistical mechanics is that, all else being equal, a system is most likely to find itself in a macrostate corresponding to the largest number of microstates. This is the so-called “Second law of thermodynamics”: for all practical intents and purposes, the entropy of a closed system always increases over time. It is not really a physical “law” in the regular sense, it is more like a profound realization. Therefore, the entropy SAB of our joint AB system will increase as time goes on until it reaches its maximum possible value. In other words, A and B trade energy in a seemingly random fashion that increases SAB on average. When SAB is finally maximized, we say that our systems are in “thermal equilibrium.” Figure 4: SAB is maximized when UA has some particular value. (It should be noted that there will actually be tiny random "thermal" fluctuations around this maximum.) Let’s say that the internal energy of system A is UA and the internal energy of system B is UB. Crucially, note that the total energy of combined system UAB = UA + UB is constant over time! This is because energy of the total system is conserved. Therefore, dUA = −dUB. Now, the combined system will maximize its entropy when UA and UB have some particular values. Knowing the value of UA is enough though, because UB = UAB − UA. Therefore, entropy is maximized when 0 = ∂SAB ∂UA . (7) 6 However, we can rewrite this as 0 = ∂SAB ∂UA = ∂SA ∂UA + ∂SB ∂UA = ∂SA ∂UA − ∂SB ∂UB = 1 TA − 1 TB . Therefore, our two systems are in equilibrium if they have the same temperature! TA = TB (8) If there are other macroscopic variables we are using to define our macrostates, like volume V or particle number N , then there will be other quantities that must be equal in equibrium, assuming our two sys- tems compete for volume or trade particles back and forth. In these cases, we define the quantities P and µ to be P T ≡ ∂S ∂V ∣∣∣∣ U,N µ T ≡ − ∂S ∂N ∣∣∣∣ U,V . (9) P is called “pressure” and µ is called “chemical potential.” In equilib- rium, we would also have PA = PB µA = µB. (10) (You might object that pressure has another definition, namely force di- vided by area. It would be incumbent on us to check that this definition matches that definition in the relevant situation where both definitions have meaning. Thankfully it does.) 7 ΩS(E) is sometimes called the “degeneracy” of E. In any case, we can easily see what the ratio of Prob(E1) and Prob(E2) must be. Prob(E1) Prob(E2) = ΩS(E1)e −E1/kT ΩS(E2)e−E2/kT Furthermore, we can use the fact that all probabilities must sum to 1 in order to calculate the absolute probability. We define Z(T ) ≡ ∑ E ΩS(E)e−E/kT (14) = ∑ s e−Es/kT where ∑ s is a sum over all states of S. Finally, we have Prob(E) = ΩS(E)e−E/kT Z(T ) (15) However, more than being a mere proportionality factor, Z(T ) takes on a life of its own, so it is given the special name of the “partition function.” Interestingly, Z(T ) is a function that depends on T and not E. It is not a function that has anything to do with a particular macrostate. Rather, it is a function that has to with every microstate at some temperature. Oftentimes, we also define β ≡ 1 kT and write Z(β) = ∑ s e−βEs. (16) The partition function Z(β) has many amazing properties. For one, it can be used to write an endless number of clever identities. Here is one. Say you want to compute the expected energy 〈E〉 your system has at temperature T . 〈E〉 = ∑ s EsProb(Es) = ∑ sEse −βEs Z(β) = − 1 Z ∂ ∂β Z = − ∂ ∂β logZ 10 This expresses the expected energy 〈E〉 as a function of temperature. (We could also calculate 〈En〉 for any n if we wanted to.) Where the partition function really shines is in the “thermodynamic limit.” Usually, people define the thermodynamic limit as N →∞ (thermodynamic limit) (17) where N is the number of particles. However, sometimes you might be interested in more abstract systems like a spin chain (the so-called “Ising model”) or something else. There are no “particles” in such a system, however there is still something you would justifiably call the thermodynamic limit. This would be when the number of sites in your spin chain becomes very large. So N should really just be thought of as the number of variables you need to specify a microstate. When someone is “working in the thermodynamic limit,” it just means that they are considering very “big” systems. Of course, in real life N is never infinite. However, I think we can all agree that 1023 is close enough to infinity for all practical purposes. Whenever an equation is true “in the thermodynamic limit,” you can imagine that there are extra terms of order 1 N unwritten in your equation and laugh at them. What is special about the thermodynamic limit is that ΩS becomes, like, really big... ΩS = (something)N Furthermore, the entropy and energy will scale with N SS = NS1 E = NE1 In the above equation, S1 and E1 can be thought of as the average amount of entropy per particle. Therefore, we can rewrite Prob(E) ∝ ΩS(E) exp ( − 1 kTE ) = exp ( 1 kSS − 1 kTE ) = exp ( N ( 1 kS1 − 1 kTE1 )) . The thing to really gawk at in the above equation is that the probability that S has some energy E is given by Prob(E) ∝ eN(...). I want you to appreciate how insanely big eN(...) is in the thermody- namic limit. Furthermore, if there is even a miniscule change in (. . .), 11 Prob(E) will change radically. Therefore, Prob(E) will be extremely concentrated at some particular energy, and deviating slightly from that maximum will cause Prob(E) to plummit. Figure 7: In the thermodynamic limit, the system S will have a well defined energy. We can therefore see that if the energy U maximizes Prob(E), we will essentially have Prob(E) ≈ { 1 if E = U 0 if E 6= U . Let’s now think back to our previously derived equation 〈E〉 = − ∂ ∂β logZ(β). Recall that 〈E〉 is the expected energy of S when it is coupled to a heat bath at some temperature. The beauty is that in the thermodynamic limit where our system S becomes very large, we don’t even have to think about the heat bath anymore! Our system S is basically just in the macrostate where all microstates with energy U are equally likely. Therefore, 〈E〉 = U (thermodynamic limit) and U = − ∂ ∂β logZ(β) (18) is an exact equation in the thermodynamic limit. 12 can let the second law of thermodynamics to do all the hard work, transferring energy into our system at no cost to us! I should warn you that ∆F is actually not equal to the change in internal energy ∆U that occurs during this equilibriation. This is apparent just from its definition. (Although it does turn out that F is equal to the “useful work” you can extract from such a system.) The reason I’m telling you about F is because it is a useful quan- tity for determining what will happen to a system at temperature T . Namely, in the thermodynamic limit, the system will minimize F by equilibriating with the environment. Recall Eq. 19 (reproduced below). Z(β) = Ω(U)e−βU (thermodynamic limit) If our system S is in equilibrium with the heat bath, then Z(β) = exp ( 1 kS − βU ) (at equilibrium in thermodynamic limit) = exp(−βF ). First off, we just derived another amazing identity of the partition func- tion. More importantly, recall that U , as written in Eq. 19, is defined to be the energy that maximizes Ω(U)e−βU , A.K.A. the energy that maximizes the entropy of the world. Because we know that the entropy of the world always wants to be maximized, we can clearly see that F wants to be minimized, as claimed. Therefore, F is a very useful quantity! It always wants to be min- imized at equilibrium. It can therefore be used to detect interesting phenomena, such as phase transitions. 1.5 Phase Transitions Let’s back up a bit and think about a picture we drew, Fig. 7. It’s a very suggestive picture that begs a very interesting question. What if, at some critical temperature Tc, a new peak grows and overtakes our first peak? 15 Figure 9: A phase transition, right below the critical temperature Tc, at Tc, and right above Tc. This can indeed happen, and is in fact what a physicist would call a “first order phase transition.” We can see that will be a discontinuity in the first derivative of Z(T ) at Tc. You might be wondering how this is possible, given the fact that from its definition, Z is clearly an analytic function as it is a sum of analytic functions. The thing to remember is that we are using the thermodynamic limit, and the sum of an infinite number of analytic functions may not be analytic. Because there is a discontinuity in the first derivative of Z(β), there will be a discontinuity in E = − ∂ ∂β logZ. This is just the “latent heat” you learned about in high school. In real life systems, it takes some time for enough energy to be transferred into a system to overcome the latent heat energy barrier. This is why it takes so long for a pot of water to boil or a block of ice to melt. Furthermore, during these lengthy phase transitions, the pot of water or block of ice will actually be at a constant temperature, the “critical temperature” (100◦C and 0◦C respectively). Once the phase transition is complete, the temperature can start changing again. Figure 10: A discontinuity in the first derivative of Z corresponds to a first order phase transition. This means that you must put a fi- nite amount of energy into the system called “latent heat” at the phase transition before the temperature of the system will rise again. 16 1.6 Example: Box of Gas For concreteness, I will compute the partition function for an ideal gas. By ideal, I mean that the particles do not interact with each other. Let N be the number of particles in the box and m be the mass of each particle. Suppose the particles exist in a box of volume V . The positions and momenta of the particles at ~xi and ~pi for i = 1 . . . N . The energy is given by the sum of kinetic energies of all particles. E = N∑ i=1 ~p2i 2m . (21) Therefore, Z(β) = ∑ s e−βEs = 1 N ! 1 h3N ∫ N∏ i=1 d3xid 3pi exp ( −β N∑ i=1 ~p2i 2m ) = 1 N ! V N h3N N∏ i=1 ∫ d3pi exp ( −β ~p 2 i 2m ) = 1 N ! V N h3N (2mπ β )3N/2 If N is large, the thermodynamic limit is satisfied. Therefore, U = − ∂ ∂β logZ = −3 2 N ∂ ∂β log ( N ! −2 3N (V h3 ) 2 3 2mπ β ) = 3 2 N β = 3 2 NkT. You could add interactions between the particles by adding some po- tential energy between V each pair of particles (unrelated to the volume V ). E = N∑ i=1 ~p2i 2m + 1 2 ∑ i,j V (|~xi − ~xj|) (22) The form of V (r) might look something like this. 17 which one it is in. This would be an example of a “classical superposi- tion” of quantum states. Usually, we think of classical superpositions as having a thermodynamical nature, but that doesn’t have to be the case. Anyway, say that your lab mate thinks there’s a 50% chance the system could be in either state. The density matrix corresponding to this classical superposition would be ρ = 1 2 |ψ1〉 〈ψ1|+ 1 2 |ψ2〉 〈ψ2| . More generally, if you have a set of N quantum states |ψi〉 each with a classical probability pi, then the corresponding density matrix would be ρ = N∑ i=1 pi |ψi〉 〈ψi| . (26) This is useful to define because it allows us to extract expectation values of observables Ô in a classical superposition. But before I prove that, I’ll have to explain a very important operation: “tracing.” Say you have quantum state |ψ〉 and you want to calculate the ex- pectation value of Ô. This is just equal to 〈Ô〉 = 〈ψ| Ô |ψ〉 . (27) Now, say we have an orthonormal basis |φs〉 ∈ H. We then have 1 = ∑ s |φs〉 〈φs| . (28) Therefore, inserting the identity, we have 〈Ô〉 = 〈ψ| Ô |ψ〉 = ∑ s 〈ψ| Ô |φs〉 〈φs|ψ〉 = ∑ s 〈φs|ψ〉 〈ψ| Ô |φs〉 . This motivates us to define something called the “trace operation” for any operator H → H. While we are using an orthonormal basis of H to define it, it is actually independent of which basis you choose. Tr(. . .) ≡ ∑ s 〈φs| . . . |φs〉 (29) 20 We can therefore see that for our state |ψ〉, 〈Ô〉 = Tr ( |ψ〉 〈ψ| Ô ) . (30) Returning to our classical superposition and density matrix ρ, we are now ready to see how to compute the expectation values. 〈Ô〉 = ∑ i pi 〈ψi| Ô |ψi〉 = ∑ i pi Tr ( |ψi〉 〈ψi| Ô ) = Tr ( ρÔ ) So I have now proved my claim that we can use density matrices to extract expectation values of observables. Now that I have told you about these density matrices, I should introduce some terminology. A density matrix that is of the form ρ = |ψ〉 〈ψ| for some |ψ〉 is said to represent a “pure state,” because you know with 100% certainty which quantum state your system is in. Note that for a pure state, ρ2 = ρ (for pure state). It turns out that the above condition is a necessary and sufficient con- dition for determining if a density matrix represents a pure state. If a density matrix is instead a combination of different states in a classical superposition, it is said to represent a “mixed state.” This is sort of bad terminology, because a mixed state is not a “state” in the Hilbert space Ĥ, but whatever. 1.9 Example: Two state system Consider the simplest Hilbert space, representing a two state system. H = C2 Let us investigate the difference between a quantum superposition and a classical super position. An orthonormal basis for this Hilbert space is given by |0〉 = ( 0 1 ) |1〉 = ( 1 0 ) 21 Say you have a classical superposition of these two states where you have a 50% probability that your state is in either state. Then ρMixed = 1 2 |0〉 〈0|+ 1 2 |1〉 〈1| = ( 1 2 0 0 1 2 ) . Let’s compare this to the pure state of the quantum super position |ψ〉 = 1√ 2 |0〉+ 1√ 2 |1〉 . The density matrix would be ρPure = ( 1√ 2 |0〉+ 1√ 2 |1〉 )( 1√ 2 〈0|+ 1√ 2 〈1| ) = 1 2 ( |0〉 〈0|+ |1〉 〈1|+ |0〉 〈1|+ |1〉 〈0| ) = ( 1 2 1 2 1 2 1 2 ) The pure state density matrix is different from the mixed state because of the non-zero off diagonal terms. These are sometimes called “inter- ference terms.” The reason is that states in a quantum superposition can “interfere” with each other, while states in a classical superposition can’t. Let’s now look at the expectation value of the following operators for both density matrices. σz = ( 1 0 0 −1 ) σx = ( 0 1 1 0 ) They are given by 〈σz〉Mixed = Tr (( 1 2 0 0 1 2 )( 1 0 0 −1 )) = 0 〈σz〉Pure = Tr (( 1 2 1 2 1 2 1 2 )( 1 0 0 −1 )) = 0 〈σx〉Mixed = Tr (( 1 2 0 0 1 2 )( 0 1 1 0 )) = 0 〈σx〉Pure = Tr (( 1 2 1 2 1 2 1 2 )( 0 1 1 0 )) = 1 22 Figure 12: Air molecules bumping up against a quantum system S will entangle with it. Notice that the experimentalist will not have access to the observ- ables in the environment. Associated with HS is a set of observables ÔS . If you tensor these observables together with the identity, ÔS ⊗ 1E you now have an observable which only measures quantities in the HS subsector of the full Hilbert space. The thing is that entanglement within the environment gets in the way of measuring ÔS ⊗ 1E in the way the experimenter would like. Say, for example, HS = C2 and HE = CN for some very big N . Any state in HS ⊗HE will be of the form c0 |0〉 |ψ0〉+ c1 |1〉 |ψ1〉 (32) for some c0, c1 ∈ C and |ψ0〉 , |ψ1〉 ∈ H. The expectation value for our observable is 〈ÔS ⊗ 1E〉 = ( c∗0 〈0| 〈ψ0|+ c∗1 〈1| 〈ψ1| ) ÔS ⊗ 1E ( c0 |0〉 |ψ0〉+ c1 |1〉 |ψ1〉 ) =|c0|2 〈0| ÔS |0〉+ |c1|2 〈1| ÔS |1〉+ 2 Re ( c∗0c1 〈0| ÔS |1〉 〈ψ0|ψ1〉 ) The thing is that, if the environment E is very big, then any two random given vectors |ψ0〉 , |ψ1〉 ∈ HE will generically have almost no overlap. 〈ψ0|ψ1〉 ≈ e−N (This is just a fact about random vectors in high dimensional vector spaces.) Therefore, the expectation value of this observable will be 〈ÔS ⊗ 1E〉 ≈ |c0|2 〈0| ÔS |0〉+ |c1|2 〈1| ÔS |1〉 . 25 Because there is no cross term between |0〉 and |1〉, we can see that when we measure our observable, our system S seems to be in a classical superposition, A.K.A a mixed state! This can be formalized by what is called a “partial trace.” Say that |φi〉E comprises an orthonormal basis of HE . Say we have some density matrix ρ representating a state in the full Hilbert space. We can “trace over the E degrees of freedom” to recieve a density matrix in the S Hilbert space. ρS ≡ TrE(ρ) ≡ ∑ i E 〈φi| ρ |φi〉E . (33) You be wondering why anyone would want to take this partial trace. Well, I would say that if you can’t perform the E degrees of freedom, why are you describing them? It turns out that the partially traced density matrix gives us the expectation values for any observables in S. Once we compute ρS , by tracing over E , we can then calculate the expectation value of any observable ÔS by just calculating the trace over S of ρSÔS : Tr ( ρÔS ⊗ 1E ) = TrS(ρSÔS). Even though the whole world is in some particular state in HS ⊗ HE , when you only perform measurements on one part of it, that part might as well only be in a mixed state for all you know! Entanglement looks like a mixed state when you only look at one part of a Hilbert space. Furthermore, when the environment is very large, the off diagonal “in- terference terms” in the density matrix are usually very close to zero, meaning the state looks very mixed. This is the idea of “entanglement entropy.” If you have an entangled state, then trace out over the states in one part of the Hilbert space, you will recieve a mixed density matrix. That density matrix will have some Von Neumann entropy, and in this context we would call it “en- tanglement entropy.” The more entanglement entropy your state has, the more entangled it is! And, as we can see, when you can only look at one tiny part of a state when it is heavily entangled, it appears to be in a classical superposition instead of a quantum superposition! The process by which quantum states in real life become entangled with the surrounding environment is called “decoherence.” It is one of the most visciously efficient processes in all of physics, and is the reason why it took the human race so long to discover quantum mechanics. It’s very ironic that entanglement, a quintessentially quantum phenomenon, when taken to dramatic extremes, hides quantum mechanics from view 26 entirely! I would like to point out an important difference between a clas- sical macrostate and a quantum mixed state. In classical mechanics, the subtle perturbing effects of the environment on the system make it difficult to keep track of the exact microstate a system is in. However, in principle you can always just re-measure your system very precisely and figure out what the microstate is all over again. This isn’t the case in quantum mechanics when your system becomes entangled with the environment. The problem is that once your system entangles with the environment, that entanglement is almost certainly never going to undo itself. In fact, it’s just going to spread from the air molecules in your laboratory to the surrounding building, then the whole univeristy, then the state, the country, the planet, the solar system, the galaxy, and then the universe! And unless you “undo” all of that entanglement, the show’s over! You’d just have to start from scratch and prepare your system in a pure state all over again. 1.12 The Quantum Partition Function The quantum analog of the partition function is very straightforward. The partition function is defined to be Z(T ) ≡ Tr exp ( −Ĥ/kT ) (34) = ∑ s e−βEs. Obviously, this is just the same Z(T ) that we saw in classical mechanics! They are really not different at all. However, there is something very interesting in the above expression. The operator exp ( −Ĥ/kT ) looks an awful lot like the time evolution operator exp ( −iĤt/~ ) if we just replace − i ~ t −→ −β. It seems as though β is, in some sense, an “imaginary time.” Rotating the time variable into the imaginary direction is called a “Wick Rotation,” 27