1. Chapter 2 Statistical Description of Systems of Particles

2. We’ve reviewed elementary probability & statistics. Now, we are finally ready to talk about PHYSICS In this chapter (& the rest of the course) we’ll combine statistical ideas with the Laws of Classical or Quantum Mechanics ≡ Statistical Mechanics We can use either a classical or a quantum description of a system. Of course, which is valid obviously depends on the problem!! Discussion of the General Problem

3. Four Essential Ingredients for A Statistical Description of a System with many particles: (~ outline of Ch. 2!) 1. Specify of the State of the System 2. Choose a Statistical Ensemble 3. Formulate a Basic Postulate about à-priori Probabilities. 4. Do Probability Calculations

4. Four Essential Ingredients for A Statistical Description of a Physical System with many particles: (~ an outline of Ch. 2!) 1. Specification of the State of the System: The state referred to here is the “Macrostate of the System”. We’ll thoroughly discuss what we mean by “Macrostate” & we’ll also discuss that this is very different than the “Microstate of the System”! We’ll also need a detailed method for specifying the Macrostate. This is discussed in this chapter.

5. Specification of the System State Microstate  Microscopic System State Quantum Description of the System: This means specifying a (large!) set of quantum numbers. Classical Description of the System: This means specifying a point in a large dimensional phase space.

6. Specification of the System State Macrostate  Macroscopic System State Quantum Description of the System: For an isolated system, this means specifying a subset of the quantum states of the system. The system is described by macroscopic parameters (that can be measured).

7. 2. Statistical Ensemble: We need to decide exactly which ensemble to use. This is also discussed in this chapter. In either Classical Mechanics or Quantum Mechanics: If we had a detailed knowledge of all positions & momenta of all system particles & if we knew all inter-particle forces, we could (in principle) set up & solve the coupled, non-linear differential equations of motion, we could find EXACTLY the behavior of all particles for all time! In practice we don’t have this information. Even if we did, such a problem is Impractical, if not Impossible to solve!

8. Instead, we use Statistical/Probabilistic Methods. Ensemble Now, lets think of doing MANY (≡ N ) similar experiments on the system of particles we are considering. In general, the outcome of each experiment will be different. So, we ask for the PROBABILITY of a particular outcome. This PROBABILITY ≡ the fraction of cases out of N experiments which have that outcome. This is how probability is determined by experiment. A goal of STATISTICAL MECHANICS is to predict this probability theoretically.

9. Next, we need to start somewhere, so we need to assume 3. A Basic Postulate about à-priori Probabilities. “à-priori” ≡ prior (based on our prior knowledge of the system). Our knowledge of a given physical system leads is to expect that there is NOTHING in the laws of mechanics (classical or quantum) which would result in the system preferring to be in any particular one of it’s Accessible States.

10. 3. A Basic Postulate about à-priori Probabilities. There is NOTHING in the laws of mechanics (classical or quantum) which would result in the system preferring to be in any particular one of it’s Accessible States. So, (if we have no contrary experimental evidence) we make the hypothesis that: it is equally probable (equally likely) that the system is in ANY ONE of it’s accessible states.

11. We make the hypothesis that it is equally probable (equally likely) that the system is in ANY ONE of it’s accessible states. This postulate is reasonable & doesn’t contradict any laws of mechanics (classical or quantum). Is it correct? That can only be confirmed by checking theoretical predictions & comparing those to experimental observation! Physics is an experimental science!! Sometimes, this postulate is called the Fundamental Postulate of (Equilibrium) Statistical Mechanics!

12. 4. Probability Calculations Finally, we can do some calculations! Once we have the Fundamental Postulate, we can use Probability Theory to predict the outcome of experiments. Now, we will go through steps 1., 2., 3., 4. again in detail!

13. Statistical Formulation of the Mechanical Problem Section 2.1: Specification of the System State ≡ Microstate Consider any system of particles. We know that the particles will obey the laws of Quantum Mechanics (we’ll discuss the Classical description shortly). We’ll emphasize the Quantum treatment.

14. Section 2.1: Specification of the System State ≡ Microstate Consider any system of particles. Use the Quantum treatment, consider a system with f degrees of freedom can be described by a (many particle!) wavefunction Ψ(q 1,q 2,….q f,t), where q 1,q 2,….q f ≡ set of f generalized coordinates required to characterize the system (needn’t be position coordinates!)

15. For a system with f degrees of freedom, the many particle wavefunction is formally: Ψ(q 1,q 2,….q f,t), q 1,q 2,….q f ≡ a set of f generalized coordinates which are required to characterize the system. A particular quantum state (macrostate) of the system is specified by giving values of some set of f quantum numbers. If we specify Ψ at a given time t, we can (in principle) calculate it at any later time by solving the appropriate Schrödinger Equation

16. Now, lets look at some simple examples, which might also review a little elementary Quantum Mechanics for you.

17. Example 1 The system is a single particle, fixed in position. It has intrinsic spin ½ –Intrinsic angular momentum = ½ ћ. In the Quantum Description of this system, the state of the particle is specified by specifying the projection m of this spin along a fixed axis (which we usually call the z-axis). The quantum number m can thus have 2 values: ½ (“spin up”) or - ½ (“spin down”) So, there are 2 possible states of the system.

18. Example 2 The system is N particles (non-interacting), fixed in position. Each has intrinsic spin ½ so EACH particle’s quantum number m i (i = 1,2,…N) can have one of the 2 values  ½. Suppose that N is HUGE: N ~ 10 24. The state of this system is then specified by specifying the values of EACH of the quantum numbers: m 1,m 2,.. m N.  There are (2) N unique states of the system! With N ~ 10 24, this number is HUGE!!!

19. Example 3 The system is a quantum mechanical, one- dimensional, simple harmonic oscillator, with position coordinate x & classical frequency ω. So the Quantum Energy of this system is: E n = ћω(n + ½ ), (n = 0,1,2,3,….). The quantum states of this oscillator are then specified by specifying the quantum number n. So, there are essentially an  NUMBER of such states!

20. Example 4 The system is N quantum mechanical, one- dimensional, simple harmonic oscillators, at positions x i, with classical frequencies ω i (i = 1,2,.. N). The Quantum Energies of each particle in this system are: E i = ћω i (n i + ½ ), (n i = 0,1,2,3,….). The system’s quantum states are specified by specifying the values of each quantum number n i. Here also, there are essentially an  NUMBER of such states. But, there are also a much larger number of these than in Example 3!

21. Example 5 The system is one particle, of mass m, confined to a rectangular box, but otherwise free. Taking the origin at a corner: 0 ≤ x ≤ L x, 0 ≤ y ≤ L y, 0 ≤ z ≤ L z The particle is described by the QM wavefunction ψ(x,y,z), a solution to the Schrödinger Equation [-ћ 2 /(2m)][(∂ 2 /∂x 2 ) + (∂ 2 /∂y 2 ) + (∂ 2 /∂z 2 )]ψ(x,y,z) = Eψ(x,y,z)

22. A particle of mass m in a rectangular box: 0 ≤ x ≤ L x, 0 ≤ y ≤ L y, 0 ≤ z ≤ L z The QM wavefunction ψ(x,y,z), is a solution to [-ћ 2 /(2m)][(∂ 2 /∂x 2 ) + (∂ 2 /∂y 2 ) + (∂ 2 /∂z 2 )]ψ(x,y,z) = Eψ(x,y,z) Using the boundary condition that ψ = 0 on the box faces, it can be shown that: ψ(x,y,z) = [8/(L x L y L z )] ½ sin(n x π/L x )sin(n y π/L y )sin(n z π/L z ) n x, n y, n z are 3 quantum numbers (positive or negative integers).

23. ψ(x,y,z) = [8/(L x L y L z )] ½ sin(n x π/L x )sin(n y π/L y )sin(n z π/L z ) n x, n y, n z are 3 quantum numbers (positive or negative integers). So, the particle Quantum Energy is: E = [(ћ 2 π 2 )/(2m)][(n x 2 /L x 2 ) + (n y 2 /L y 2 ) + (n z 2 /L z 2 )] The quantum states of this system are found by specifying the values of n x, n y, n z. Again, there are essentially also an  NUMBER of such states.

24. Example 6 N particles, non-interacting, of mass m, confined to a rectangular box. Take the origin at a corner 0 ≤ x ≤ L x, 0 ≤ y ≤ L y, 0 ≤ z ≤ L z. Since they are non-interacting, each particle is described by the QM wavefunction ψ i (x,y,z), which is a solution to the Schrödinger Equation [-ћ 2 /(2m)][(∂ 2 /∂x 2 ) + (∂ 2 /∂y 2 ) + (∂ 2 /∂z 2 )]ψ i (x,y,z) = E i ψ i (x,y,z)

25. N particles of mass m in a rectangular box. 0 ≤ x ≤ L x, 0 ≤ y ≤ L y, 0 ≤ z ≤ L z. The QM wavefunction ψ i (x,y,z), is a solution to [-ћ 2 /(2m)][(∂ 2 /∂x 2 ) + (∂ 2 /∂y 2 ) + (∂ 2 /∂z 2 )]ψ i (x,y,z) = E i ψ i (x,y,z) Using the boundary condition that ψ = 0 on the box faces: ψ i (x,y,z) = [8/(L x L y L z )] ½ sin(n x π/L x )sin(n y π/L y )sin(n z π/L z ) n x, n y, n z are 3 quantum numbers (positive or negative integers).

26. ψ(x,y,z) = [8/(L x L y L z )] ½ sin(n x π/L x )sin(n y π/L y )sin(n z π/L z ) n x, n y, n z are 3 quantum numbers (positive or negative integers). Each particle’s Quantum Energy is: E = [(ћ 2 π 2 )/(2m)][(n x 2 /L x 2 ) + (n y 2 /L y 2 ) + (n z 2 /L z 2 ) The quantum states of this system are found by specifying the values of n x, n y, n z. for each particle. Again, there are essentially also an  NUMBER of such states.

27. What about the Classical Description of the state of a System? Of course, the Quantum Description is always correct! However, it is often useful & convenient to make the Classical Approximation. How do we specify the state of the Classical system then?

28. Lets start with a very simple case: A Single Particle in 1 Dimension: In classical mechanics, it can be completely described in terms of it’s generalized coordinate q & it’s momentum p. The usual case is to consider the Hamiltonian Formulation of classical mechanics, where we talk of generalized coordinates q & generalized momenta p, rather than the Lagrangian Formulation, where we talk of coordinates q & velocities (dq/dt).

29. Of course, the particle obeys Newton’s 2 nd Law under the action of the forces on it. Equivalently, it obeys Hamilton’s Equations of Motion. q & p completely describe the particle classically. Given q, p at any initial time (say, t = 0), they can be determined at any other time t by integrating the equations of motion.

30. q & p completely describe the particle classically. Given q, p at any initial time (say, t = 0), they can be determined at any other time t by integrating the Newton’s 2 nd Law Equations of Motion forward in time.  Knowing q & p at t = 0 in principle allows us to know them for all time t.  q & p completely describe the particle for all time. This situation can be abstractly represented in a geometric way discussed on the next page.

31. Consider the (abstract) 2-dimensional space defined by q, p: ≡ “Classical Phase Space” of the particle. Of course, as q & p change in time, according to the equation of motion, the point representing the particle “State” moves in the plane. At any time t, stating the (q, p) of the particle describes it’s “State” Specification of the “State of the Particle” is done by stating which point in this plane the particle “occupies”.

32. q, p are continuous variables, so an  number of points are in this 2-Dimensional Classical “Phase Space” We’d like to describe the particle “State” classically in a way that the number of states is countable. To do this, it is convenient to subdivide the ranges of q & p into very small rectangles of size:  q   p. Think of this 2-d phase space as divided into cells of equal area:  q  p ≡ h o h o ≡ small constant with units of angular momentum.

33. The 2-d phase space has a large number cells of area:  q  p = h o. The (classical) particle “State” is specified by stating which cell in phase space the q, p of the particle is in. Or, by stating that it’s coordinate lies between q & q +  q & that it’s momentum lies between p & p +  p. “State”  The phase space cell labeled by the (q,p) that the particle “occupies”.

34. This involves the “small” parameter h o, which is arbitrary. However, we can use Quantum Mechanics & the Heisenberg Uncertainty Principle: “It is impossible to SIMULTANEOUSLY specify a particle’s position & momentum to a greater accuracy than  q  p ≥ ½ ћ” So, the minimum value of h o is clearly ½ ћ. As h o ½ ћ, The classical description of the State the quantum description & becomes more & more accurate.

35. Now!! Lets generalize all of this to a MANY PARTICLE SYSTEM 1 particle in 1 dimension means we have to deal with a 2-dimensional phase space. The generalization to N particles is straightforward, but requires thinking in terms of a very abstract Multidimensional phase space. Consider a system with f degrees of freedom:  The system is described classically by f generalized coordinates: q 1,q 2,q 3, …q f f generalized momenta: p 1,p 2,p 3, …p f.

36. A complete description of the classical “State” of the system requires the specification of: f generalized coordinates: q 1,q 2,q 3, …q f. & f generalized momenta: p 1,p 2,p 3, …p f (N particles, 3-dimensions  f = 3N !!)

37. A complete description of the classical “State” of the system requires the specification of: f generalized coordinates: q 1,q 2,q 3, …q f. & f generalized momenta: p 1,p 2,p 3, …p f (N particles, 3-dimensions  f = 3N !!) So, now lets think VERY abstractly in terms of a 2f-dimensional phase space The system’s f generalized coordinates: q 1,q 2,q 3, …q f. & f generalized momenta: p 1,p 2,p 3, …p f are regarded as a point in the 2f-dimensional phase space of the system.

38. 2f-dimensional Phase Space: f q’s & f p’s: Each q & each p label an axis (analogous to the 2-d phase space for 1 particle in 1 dimension). Subdivide this phase space into small “cells” of 2f-dimensional “differential volume”:  q 1  q 2  q 3 …  q f  p 1  p 2  p 3 …  p 1f ≡ (h o ) f The classical “State” of the system is then ≡ the cell in this 2f-dimensional phase space that the system “occupies”.

39. Reif, as all modern texts, takes the viewpoint that the system’s “State” is described by a 2f- dimensional phase space ≡ “The Gibbs Viewpoint” The system “State” ≡ The cell in this phase space that the system “occupies”. Older texts take a different viewpoint ≡ “The Boltzmann Viewpoint”: In this viewpoint, each particle moves in it’s own 6-dimensional phase space In this view, specifying the system “State” requires specifying each cell in this phase space that each particle in the system “occupies”.

40. Summary - Specification of the System State: In Quantum Mechanics: Enumerate & label all possible system quantum states. In Classical Mechanics: Specify which cell in 2f-dimensional phase space Coordinates & momenta of all particles the system occupies. As h o → ½ ћ, the classical & quantum descriptions become the same.

41. Section 2.2: Statistical Ensemble

42. As we know, Statistical Mechanics deals with the behavior of systems of a large number of particles. Because the number of particles is so huge, we give up trying to keep track of individual particles. We can’t solve Schrödinger’s equation in closed form for helium (4 particles), so what hope do we have of solving it for the gas molecules in this room (10 f particles) ?? Statistical Mechanics handles many particles by Calculating the Most Probable Behavior of the System as a Whole rather than by being concerned with the behavior of individual particles.

43. In Statistical Mechanics We assume that the more ways there are to arrange the particles to give a particular distribution of energies, the more probable that distribution is. 6 ways 3 2 1 3 1 2 2 1 3 2 3 1 1 2 3 1 3 2 1 1 4 4 1 1 1 4 1 3 ways Most Probable Distribution Example: 6 energy units, 3 particles to give it to Color Codes for Energy Units ≡ 1 ≡ 2 ≡ 3 ≡ 4

44. Another Example Assuming that All Energy Distributions are Equally Probable If E = 5 and N = 5 then                   All possible configurations have equal probability, but the possible number of ways (weight) is different for each.

45. The Dominant Configuration For a large number of molecules & a large number of energy levels, there is a Dominant Configuration. In the probability distribution, the weight of the dominating configuration is much larger than the weight of the other configurations.  Configurations WiWi {n i } Weight of the Dominant Configuration Weights

46.                   W = 1 = (5!/5!)W = 20 = (5!/3!)W = 5 = (5!/4!) The difference in the W’s becomes larger as N increases! In molecular systems (N~10 23 ) considering the most dominant configuration is certainly enough to calculate averages. The Dominant Configuration

47. The Principle of Equal à-priori Probabilities Statistical thermodynamics is based on the fundamental hypothesis of Equal à-priori Probabilities. That is, all possible configurations of a given system which satisfy the given boundary conditions such as temperature, volume and number of particles, are equally likely to occur. Or the system is equally likely to be found in any one of its accessible states.

48. Example Consider the orientations of three unconstrained & distinguishable spin-1/2 particles. What is the probability that two are spin up & one spin is down at any instant?

49. Example Consider the orientations of three unconstrained & distinguishable spin-1/2 particles. What is the probability that two are spin up & one spin is down at any instant? Solution Of the eight possible spin configurations for the system: ↑↑↑, ↑↑↓, ↑↓↑, ↓↑↑, ↑↓↓, ↓↑↓, ↓↓↑, ↓↓↓ The second, third, & fourth make up the subset "two up and one down". Therefore, The probability of occurrence of this particular configuration is: P = 3/8

50. In principle, the problem of a many particle system is COMPLETELY deterministic: If we specify the many particle wavefunction Ψ (state) of the system (or the classical phase space cell) at time t = 0, we can determine Ψ for all other times t by solving The Time-Dependent Schrödinger Equation & from Ψ(t) we can calculate all observable quantities. Or, classically, if we specify the positions & momenta of all particles at time t = 0, we can predict the future behavior of the system by solving The Coupled Many Particle Newton’s 2 nd Law Equations of Motion.

51. Generally, we usually don’t have such a complete specification of the system available. We need f quantum numbers, but f ≈ 10 24 ! Actually, we usually aren’t interested in such a complete microscopic description anyway. Instead, we’re interested in predictions of MACROSCOPIC properties.  We use Probability & Statistics.  To do this we need the concept of an ENSEMBLE.

52. A Statistical Ensemble is a LARGE number (≡ N ) of identically prepared systems. In general, the systems of this ensemble will be in different states & thus will have different macroscopic properties.  We ask for the probability that a given macroscopic parameter will have a certain value. A Goal or Aim of Statistical Mechanics is to Predict this Probability.

53. Example Consider the spin problem again. But, now, Let The System Have N = 3 Particles, fixed in position, each with spin = ½  Each spin is either “up” (↑, m = ½ ) or “down” (↓, m = - ½ ). Each particle has a vector magnetic moment μ. The projection of μ along a “z-axis” is either: μ z = μ, for spin “up” or μ z = -μ, for spin “down”

54. Put this system into an External Magnetic Field H. Classical E&M tells us that a particle with magnetic moment μ in an external field H has energy: ε = - μ  H Combine this with the Quantum Mechanical result:  This tells us that each particle has 2 possible energies: ε + ≡ - μH for spin “up” ε - ≡ μH for spin “down”  So, for 3 particles, the State of the system is specified by specifying each m =   There are (2) 3 = 8 Possible States!!

55. Possible States of a 3 Spin System Given that we know no other information about this system, all we can say about it is that It has Equal Probability of Being Found in Any One of These 8 States.

56. However, if (as is often the case in real problems) we have a partial knowledge of the system (say, from experiment), then, we know that The system can be only in any one of the states which are COMPATIBLE with our knowledge. (That is, it can only be in one of it’s accessible states) “States Accessible to the System” ≡ those states which are compatible with ALL of the knowledge we have about the system. It is important to use all of the information that we have about the system!

57. Example For our 3 spin system, suppose that we measure the total system energy & we find E ≡ - μH This additional information limits the states which are accessible to the system. From the table, there are clearly Only 3 States which are compatible with this knowledge.  The system must be in one of the 3 states: (+,+,-) (+,-,+) (-,+,+)