Let   and  be two vector spaces with dimensionality N and M

1. 10. Multiple Particles 10A. Tensor Product of Vector Spaces Definition • Let and  be two vector spaces with dimensionality N and M • Let {|i} and{|j} be orthonormal vector bases for these vector spaces: • Define tensor product space =   of dimension NM with basis • Because the notation gets unwieldy, we abbreviate: • These basis statesare orthonormal • General vector in : • You should think of tijnot as a rectangular matrix, but as a big column vector

2. Some Examples • Three-dimensional vector space is products of one dimensional vector space: • Particle in 3D with spin • Addition of angular momentum • Hydrogen (no spin): • Set of all operators

3. Some Simple Properties • The inner product of two vectors in : • You can combine any vector in  with any vector in  to make a new vector in  • Not all elements of  areof this form • The resulting vector is independent of choice of bases |i and |j • It has easily understood mathematical properties

4. Operators in a Tensor Product Space • A general operator in a tensor product space looks like • Any operator A in vector space  can be combined with operator B in vector space W to make a new operator A B defined by • The product of two such operators is easy: • When one of the operators is the identity, we often abbreviate it: • This is fine as long as it is clear which vector space A or B acts on • We’ve already encountered examples like this:

5. 10B. Multiple Particles The Vector Space • If you have two particles, the first in vectors space 1 and the second in 2, then the two-particle system will lie in vector space • If you have N particles, the system will lie in vector space • Example: two particles with spin have basis states • Then for the two particle system, basis is • If written as a wavefunction, we would have • The square of  tells us the probability that 1st particle is at r1 and second at r2 • Probability particle 1 in region R1 and particle 2 in R2:

6. Operators With Multiple Particles • Operators like Pi and Ri will act on one particle at a time • Typical Hamiltonian will look like

7. Sample Problem Two particles in one dimension with the same mass are in the state , where |n is the n’th harmonic oscillator state. (1) What is N, and write explicitly the wave function (2) What is the probability that one particle has x1 > 0? (3) What is the probability that both particles have x1 , x2 > 0? (4) If the particles repel each other, which state has lower energy? • Normalization: • Wave Function

8. Sample Problem (2) (2) What is the probability that one particle has x1 > 0? x2 • Need to integrate it over the appropriate region • Switch to polarcoordinates: x1

9. Sample Problem (3) (3) What is the probability that both particles have x1 , x2 > 0? (4) If the particles repel each other, which state has lower energy? • This time the region is different • This changes limits on integration, and nothing else • Note that the plus sign increases chance particles are on the same side • Therefore, minus sign almost certainly has lower energy x2 x1

10. 10C. Symmetry and Interchangeable Particles Permutation Operators • Suppose two particles have the same vector space • Define the permutation operator P(12) as the operator that interchanges particle 1 with particle 2 • Effect on basis states: • Interpretation: The object in slot 1 goes to slot 2, and vice versa • Example with three: How to multiply them: • Track what happens to each number, working from right to left Sample Problem Simplify: • P(2  3) leaves 1 alone, then P(12) changes it to 2 • 2 changes to 3, then P(1  2) leaves it alone • 3 changes to 2, then to 1

11. Interchangeable Particles • Interchangeable particles (my term) are particles that can be interchanged without changing the Hamiltonian • Example: Positronium, electron plus anti-electron • If two particles are interchangeable, then the pairswitching operation must commute with the Hamiltonian • It can therefore be simultaneously diagonalized with H • If you pair switch twice, you get back where you started:

12. Multiple Interchangeable Particles • Suppose you have N particles, all of whomcan be interchanged with each other • Can we diagonalize all of themsimultaneously? I.e., can we have • Because the P’s don’t commute,you generally can’t do this • For some states you can, but it only works if ij = kl so they are all  1 • Proof by homework problem Bottom line: There are three cases • States can have eigenvalue +1 under all pair switchings • States can have eigenvalue –1 under all pair switchings • States can have mixed symmetry • Interesting and complicated • But we won’t discuss it

13. Bosons and Fermions • It is pretty obvious that you can buildany permutation out of pair switchings Bosons are particles that have eigenvalue +1 under all pair switchings • Then for all permutations P, it follows that Fermions are particles that have eigenvalue –1 under all pair switchings • Define for any permutation P: • Non-trivial but true: Thisdefinition is unambiguous • Then for all permutations P, it follows that

14. 10D. Identical Particles The Spin-Statistics Theorem • Suppose we have two (or more) identical particles, like two electrons • Then it is inevitable that they are interchangeable in the Hamiltonian • We would expect states to be bosonic, fermionic, or mixed symmetry • But in fact, electrons always are in a fermionic (anti-symmetric) state • Deep result from quantum field theory: • Sometimes ignored in homework problems The Spin Statistics Theorem Particles with integer spin are bosons Particles with half-integer spin are fermions

15. Symmetrization/Antisymmetrization Operators • Define the symmetrization operator: • And the anti-symmetrization operator: • Both are summed over all permutations • These are projection operators that pull out thesymmetric or anti-symmetric parts of a state vector • For fermionic or bosonic states,they leave state vector unchanged • The vector space for N identical bosons is actually: • The vectors space for N identical fermions is actually: • If we have three bosons of type 1, two fermions of type 2, and one boson of type 3, then

16. Basis States for Bosons Suppose we have a set of basis states {|i}for one boson. Find a complete, independent, orthonormal basis set for Nbosons • We know we can write any state as • Problem: these states are not (generally) in • Because they aren’t symmetrized • We know that | = |, so • So these states form a completebasis set • But they aren’t independent • To fix this, insist that the indices are in increasing order i1  i2 etc. • Then normalize them to get acomplete, orthonormal basis • n1 particles in state 1, n2in state 2, …

17. Basis States for Fermions Suppose we have a set of basis states {|i}for one boson. Find a complete, independent, orthonormal basis set for Nbosons • We know we can write any state as • We know that | = |, so • These aren’t independent • If any pair of indices match, this vanishes • To avoid redundancy, i1 < i2 etc. • Complete, orthonormal basis: • Comment: I don’t actually use these formulas as written, I just work out and figure out the normalization factor by hand

18. Dimensionality • If the dimensionality (number of basis vectors) of the space for one particle is infinite, this will also be true for N particles • What if the dimensionality is finite? • For fermions, the dimensionality for N particles is the number of ways you can pick N distinct basis states {|i} out of d choices • For bosons, it’s a little harder

19. Sample Problem A certain particle lives in a three dimensional space, with basis states {|1, |2, |3}. Write explicitly all basis states for three particles if they are (a) bosons (b) fermions First make a list of all possibili-ties ignoring symmetrization,but with i1  i2  i3 Next, symmetrize them • |1,1,1 • |1,1,2 • |1,2,3 Finally, normalize them • |1,1,1: • |1,1,2: • |1,2,3:

20. Sample Problem A certain particle lives in a three dimensional space, with basis states {|1, |2, |3}. Write explicitly all basis states for three particles if they are (a) bosons (b) fermions For fermions: Make a list of all possibilities with i1 < i2 < i3: Next, anti-symmetrize it Finally, normalize it

21. The Pauli Exclusion Principle • Basis states for fermions: • No two particles can occupy the same quantum state • What does this look likeas a wave function? • Ignore spin • Actual state will be somelinear combination of these • With a good choice of basis states, these may approximate the eigenstates of H

22. Composite Particles • Suppose we have elementary particles that satisfy the spin-statistics theorem • How can we tell if composite particles are fermions or bosons? • Let’s try it with 1H: a bound state of a proton and an electron • Let basis states for the electron be |i and for the proton be |j • Then hydrogen in state k might look like • This just says the each hydrogen atomis some combination of an electron and proton • Now consider some state with two hydrogen atoms: • Because the electron is a fermion, the expression must get a minus sign if we swap i  i’. Similarly for the proton if we switch j j’. • Notice that i, i', j, j’, k, and k’ are all dummy variables • We can swap i i', j j’, k  k’

23. Composite Particles (2) • We previously had: • Comparing these, must have • 1H is a boson • This proof generalizes toany composite object: • Does spin-statistics formula apply to composite objects? • Spin of a composite particle is sum ofspins plus internal orbital angular momentum • Since l takes on only integer values, this is an integer (half-integer) if there are an even (odd) number of fermions that comprise it • Spin-statistics theorem still works Anything composed of an even (odd) number of fermions is a boson (fermion)

24. 10E. Non-Interacting Particles States/Energies for Non-Identical Particles • Suppose our Hamiltonian is (exactly or approximately) of the form:where H1is a one-particle Hamiltonian • Assume we can find the eigenstates |i of H1: • Then it is easy to find eigenstates of H: • These states are generally unacceptable for bosons and fermions • For bosons, the eigenstates will be: • Same energies as before • Restricted i1 i2  …  iN • For fermions, the eigenstates will be: • Same energies as before • Restricted i1< i2< … < iN

25. Sample Problem Two non-interacting particles without spin are trapped in a 1D infinite square well of length a. Find the lowest four energy levels, the states, and their degeneracy if the particles are (a) distinguishable (b) bosons (c) fermions • First find the eigenstates and energiesfor the single particle problem • Then for the distinguishable two particles • Make a table: • For bosons, must symmetrize these states

26. Sample Problem (2) Two non-interacting particles without spin are trapped in a 1D infinite square well of length a. Find the lowest four energy levels, the states, and their degeneracy if the particles are (a) distinguishable (b) bosons (c) fermions • For fermions, Pauli exclusion principle means we must skip first and third row • Then anti-symmetrize them • Make a table

27. Ground State for Bosons • For non-interacting bosons, what is the ground state? • Realistic particles always have someinteractions, and they are never exactly at zero temperature • At finite temperature, only some fraction of theparticles end up in the same quantum state • Bose-Einstein condensate • First achieved experimentally in 1995 • Even interacting substances can have similarproperties • Liquid 4He

28. Ground State for Fermions • For non-interacting fermions, what is the ground state? • For finite temperature, it is often that case that there is a sharp boundary, below which almost all states are occupied, above which, almost all states are empty • Let’s study energy for N spin-1/2fermions in a volume V, where N is large • No potential inside the volume, but treat itas cubical infinite square well of size L  L L • The one particle eigenstates for this infinite square well are given by: • The energy for N-particles would be: L L L

29. Counting the States • Include only those states withthe lowest energy, so put acutoffin, the Fermi level: • Helpful to define Fermi energy • We have to have the rightnumber of total fermions: • Sum on spins is 2 • Since N is assumed large, we can estimate the number of lattice points in this region as the volume of this region • Rewrite EFin terms of N and V • Let n = N/V be number density

30. Total Energy To find the total energy: • Sum on spins is 2 • Treat sum on n as if it were an integral • Use spherical coordinates

31. Fermi Pressure • If you adjust the size of the cube, the volume changes • The cube does work and the energy changes: • This lets us calculate the Fermi pressure: • Fermi pressure exists even at zero temperature • It has nothing to do with interactions • It is larger for smaller masses (electrons >> nuclei) • It is responsible for preventing collapse of white dwarfs and neutron stars dL L L L

32. Sample Problem A white dwarf is composed primarily of 12C and 16O nuclei. Provide an order of magnitude estimate relation between the radius R and the mass M of a white dwarf. • Stars are in a balance of pressure vs. gravity • We will ignore all factors of 2, , etc. • Fermi pressure from electrons completely dominates nuclei • Relation between mass M and number of electrons: •  is mass per electron • 12C has 6 electrons and 16O has 8 electrons •  = 2u • Think of the star as having two halves, eachwith mass ½M = M, separated by a distance R • Force of gravity between two halves is: • Pressure is force over area, where area = R2

33. Sample Problem A white dwarf is composed primarily of 12C and 16O nuclei. Provide an order of magnitude estimate relation between the radius R and the mass M of a white dwarf. • Equate pressure to Fermi pressure • Volume of a sphere is R3 • Solve for R • Let’s write M in terms of solar masses and R in terrestrial radii

34. White Dwarfs and Neutron Stars • Final stages of low mass stars are white dwarfs • 12C and 16O stars held up by electron degeneracy • Note that as M increases, size decreases • Density increases as well • At large mass, electrons become semi-relativistic • This creates an instability at about 1.4 solar masses • Catasrophic collapse; electron degeneracy relieved by process • This eliminates protons (and electrons) but neutrons remain • Eventually stopped by neutron degeneracy pressure • Because neutron is 2000 times heavier, neutron star is city-sized

35. White Dwarfs and Neutron Stars: Size

36. 10F. Atoms Hamiltonian and Naïve Approximation • Treat nucleus as fixed at origin • Hamiltonian: • Ignoring spin orbit coupling, etc. Look at the nuclear attraction vs. the electron repulsion • There are Z terms each of size Ze2for the nuclear attraction • There are (Z2 – Z)/2 terms of size e2 for the repulsion • Maybe we can just treat the electronsas non-interacting! Nomenclature: • We will label states by their n and l values • Not by m or ml since this depends on choice of axis • The shell (value of n) will be denoted by giving the integer n • The subshell (value of l) will be denoted by letters llet 0 s 1 p 2 d 3 f 4 g

37. The Ultra-Naïve Picture of Atoms g f d p s Fill them up in order, from bottom to top List all possibly energy states • n determines the energy • For each n, we have l = 0, 1, 2, …, n – 1 subshells • There are 2l + 1 orbitals in each subshell • You can fit two electrons in each orbital How do you decide between 3s, 3p, 3d? Example: Chlorine, Z = 17 n = 5 n = 4 n = 3 n = 2 n = 1

38. Screening and 2s vs. 2p • Let’s rewrite Hamiltonian slightly: • Each electron feels repulsion fromall previously included electrons • Since the 1s electrons arevery close to the origin,the outer electronseffectively see Z reduced,or “screened” • The screened potential looks like this (Z = 3): • At large r, potential Z has been screened a lot (bad) • At very small r, still very negative (good) • There is a big bonus if you can get to small r • Wave functions near origin: • Conclusion: E2p > E2s +Ze -e Unscreened, Z = 3 Unscreened, Z = 1 Screened -2e

39. A Less Naïve Approximation f d p s • Energies in atoms depend on bothn and l • Those with higher l are effectively shifted upwards in energy • Bottom line: lower n + l leads to lower energy, with n as tiebreaker • We can make a table to show in what order we fill things • Make a list with number of states in each subshell 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f n = 7 n = 6 n = 5 n = 4 n = 3 • To figure out which electrons are present, fill up in order n = 2 Sn (Z = 50): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2 n = 1 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d6 7p6

40. The Periodic Table of the Elements p electrons d electrons Noble gases s electrons f electrons • Chemical properties of the elements tend to be governed by the outermost subshell, and how many electrons there are in it • A filled p-subshell is extremely stable – noble gasses • A nearly filled p-subshellwants to borrow anelectron from anotheratom: halogens • The first electron in anew s-subshell is veryeasy to pry away: alkalimetals • d and f subshells arenearly degenerate withother subshells, so theyoften share electrons yieldingmultiple valences

41. The Rules Aren’t Perfect! • The rules I gave you for working out electron configurations are not perfect • Of the first 104 elements in the periodic table: • 83 are exactly right (80%) • 18 are off by one electron (17%) • 3 are off by two electrons (3%) • Overall, 24/5460 electrons are wrong (0.4%) Why aren’t they perfect? • Some levels are extremely close in energy • Subtle interactions can change the electron configuration • Some of these I do not understand, they are just calculated • Let’s try to understand some of these subtleties

42. Some Subtle Points: Filling Orbitals • When I filled up Sn, I went up to 5p2 • When I filled in the orbitals, I did it like this: • Does this mean: • Orbitals one and two have lower energies? • Spin up is lower energy than spin down? No, but there is content in this way of drawing things: • Because the labeling of orbitals depends on choice of axis, it is meaningless to ask which orbitals are filled, they are automatically degenerate • Nonetheless, since electrons repel, they would rather go into different orbitals than the same orbitals • So put one electron in each orbital before putting a second electron in any orbital • After half-filling a subshell, often get more stability • Half filled subshells are especially stable 5p

43. Some Subtle Points: Spins • I also drew the two spins up • I could just as easily have drawn them both down • But I wanted to give the idea that the spins add to make spin 1 • Recall, spin ½ plus spin ½ can yield spin 0 or spin 1 Why does spin make a difference? • Consider these last two electrons (ignore the rest) • They are fermions – wave function must be antisymmetric • State vector can be thought ofas product of space part and spin part • If you exchange the two particles, we must get a minus sign • You choose either: • Space part symmetric, spin part anti-symmetric • Space part anti-symmetric, spin part symmetric 5p

44. Some Subtle Points: Spins (2) 5p • Which configurationhas lower energy? • Hamiltonian doesn’t care about spin, but it does about space • Anti-symmetric space wave function tends to not have electrons near each other • That one has lower energy • Symmetric spin state has lower energy • This corresponds to total s = 1 • As you fill subshells, spins always tend to add until subshell is half filled • Then you have to start cancelling it again

45. Subtle Points: Orbital Angular Momentum 5p • Recall: we wanted space part anti-symmetric • The two valence electrons each have l1 = l2 = 1 • In combining them, the total orbitalangular momentum will be: • Recall: when adding identical angular momentum, maximum value is symmetric, and it alternates as you go down • In this case, the only one that is anti-symmetric is l = 1

46. Subtle Points: Total Angular Momentum 5p • For the Hamiltonian we gave, spin does not directly appear in the Hamiltonian • We can rotate S separately from L • This implies that our states for the entireatom could be labeled like this: • There is also spin orbit coupling terms that look like LS • This means that there is mixing, and we should switch to j and mj • Energy is independent of mj • Generally, lower energy state will have lower j • Lowest j is

47. Spectroscopic Notation for Atoms • For excited states, there are (presumably) many parameters describing which electron is in which configuration • n is complicated • Normally not included in label • If you rotate the atom, mj changes, so also not included • So just need labels for s, l, and j • l is denoted by a letter, this time capitalized • After S, P, D, F, go up alphabetically, skipping J, P, and S • j is denoted by a subscript after the l letter • s is denoted by writing 2s + 1 as a superscript before the letter • So in general would look like: • Hydrogen would be: llet 0 S 1 P 2 D 3 F 4 G 5 H 6 I 7 K 8 L 9 M 10 N 11 O