* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download PART 1 Identical particles, fermions and bosons. Pauli exclusion
History of quantum field theory wikipedia , lookup
Hidden variable theory wikipedia , lookup
Probability amplitude wikipedia , lookup
Renormalization wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Schrödinger equation wikipedia , lookup
Quantum electrodynamics wikipedia , lookup
Quantum state wikipedia , lookup
Renormalization group wikipedia , lookup
Copenhagen interpretation wikipedia , lookup
Coupled cluster wikipedia , lookup
Canonical quantization wikipedia , lookup
Hartree–Fock method wikipedia , lookup
Bohr–Einstein debates wikipedia , lookup
Bell's theorem wikipedia , lookup
Dirac equation wikipedia , lookup
EPR paradox wikipedia , lookup
Particle in a box wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Spin (physics) wikipedia , lookup
Double-slit experiment wikipedia , lookup
Molecular Hamiltonian wikipedia , lookup
Identical particles wikipedia , lookup
Atomic orbital wikipedia , lookup
Hydrogen atom wikipedia , lookup
Elementary particle wikipedia , lookup
Electron scattering wikipedia , lookup
Electron configuration wikipedia , lookup
Matter wave wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Tight binding wikipedia , lookup
Wave–particle duality wikipedia , lookup
Wave function wikipedia , lookup
Atomic theory wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
O. P. Sushkov School of Physics, The University of New South Wales, Sydney, NSW 2052, Australia (Dated: March 1, 2016) PART 1 Identical particles, fermions and bosons. Pauli exclusion principle. Slater determinant. Variational method. He atom. Multielectron atoms, effective potential. Exchange interaction. 2 Identical particles and quantum statistics. Consider two identical particles. 2 electrons 2 protons 2 12 C nuclei 2 protons . . . . The wave function of the pair reads ψ = ψ(~r1 , ~s1 , ~t1 ...; ~r2 , ~s2 , ~t2 ...) where r1 , s1 , t1 , ... are variables of the 1st particle. r2 , s2 , t2 , ... are variables of the 2nd particle. r - spatial coordinate s - spin t - isospin . . . - other internal quantum numbers, if any. Omit isospin and other internal quantum numbers for now. The particles are identical hence the quantum state is not changed under the permutation : ψ (r2 , s2 ; r1 , s1 ) = Rψ(r1 , s1 ; r2 , s2 ) , where R is a coefficient. Double permutation returns the wave function back, R2 = 1, hence R = ±1. The spin-statistics theorem claims: * Particles with integer spin have R = 1, they are called bosons (Bose - Einstein statistics). * Particles with half-integer spins have R = −1, they are called fermions (Fermi - Dirac statistics). 3 The spin-statistics theorem can be proven in relativistic quantum mechanics. Technically the theorem is based on the fact that due to the structure of the Lorentz transformation the wave equation for a particle with spin 1/2 is of the first order in time derivative (see discussion of the Dirac equation later in the course). At the same time the wave equation for a particle with integer spin is of the second order in time derivative. An example: The vector potential in electrodynamics is to some extent equivalent to the wave equation of photon. The photon has spin S = 1. Maxwell’s equation for the vector ~ reads potential A h 1 ∂2 c2 ∂t2 i 2 ~ ~ − ∇ A = 4π~j. here ~j is electric current. The equation contains the second time derivative. Comment: Do not mix permutation with parity, these are different operations. permutation : ψ (r1 , s1 ; r2 , s2 ) → ψ(r2 , s2 ; r1 , s1 ) space reflection: ψ (r1 , s1 ; r2 , s2 ) → ψ(−r1 , s1 ; −r2 , s2 ) 4 Example: Statistics influence rotational spectra of diatomic molecules. Consider rotational spectrum of Carbon2 molecule that consists of two A 12 12 C isotops. C nucleus has spin S = 0, hence it is a boson. 12 C r1 1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 12 C 1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 r2 FIG. 1: Rotation of Carbon2 molecule The wave function of the molecule reads Ψ(1, 2) = U [(~r1 + ~r2 )/2] V ~r1 − ~r] ϕ1 , ϕ2 Here ϕ1 and ϕ2 are spin wave functions of the first and the second nucleus respectively. U is the wave function of the center of mass motion. V is the wave function of the relative motion. Spin of the nucleus is zero, S = 0. Hence ϕ1 = ϕ2 = 1. V (~r1 − ~r2 ) = χ(| r~1 − r~2 |)Ylm (~ r1 − r~2 ) where Ylm is spherical harmonic. Let us perform permutation of the particles V (2, 1) = χ (|~r2 − ~r1 |) Ylm (~r2 − ~r1 ) = (−1)l χ(| ~r1 − ~r2 |)Ylm (~r1 − ~r2 ) = (−1)l V (1, 2) . Here I have used the exact mathematical relation (see 3d year quantum mechanics) Ylm (−~r) = (−1)l Ylm (~r) . Requirement of Bose statistics: ψ (2, 1) = ψ (1, 2) Hence only even values of l are allowed in the rotational spectrum of C2 consisting of two 12 C isotops. 5 In a molecule consisting of two identical 12 C isotops only even values of l are allowed. In a molecule consisting of two different isotops, say 12 C and 13 C or 12 C and 14 C all values of l are allowed. 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 l=4 l=4 l=3 12 l=2 l=2 l=0 l=1 l=0 12 C C 12 14 C C FIG. 2: Rotational spectra of C2 molecule. Left: two identical nuclei each with spin S=0. Right: two distinguishable nuclei each with spin S=0. 6 Two noninteracting fermions in an external potential. Consider any external potential: Coulomb field of a nucleus, a potential well, etc. Let ϕa and ϕb be single particle states in the potential. For an infinite potential well these are simple standing wave as it is illustrated in the Fig. below ϕb ϕa FIG. 3: Two lowest single particle orbital quantum states in an infinite square well potential. Let us put two electrons with parallel spins in the potential. Requirement of Fermi statistics: ψ (1, 2) = −ψ (2, 1). Therefore, the many-body (in this case “many” = 2) wave function is 1 Ψ (1, 2) = √ [ϕa (r1 ) ϕb (r2 ) − ϕa (r2 ) ϕb (r1 )] | ↑i1 | ↑i2 2 (1) If ϕa = ϕb then Ψ ≡ 0. Thus, one cannot put two fermions in the same single-particle quantum state. This is Pauli exclusion principle. If spins of the electrons are opposite then the single particle states are different, ϕa↑ 6= ϕa↓ , even if the coordinate states are identical. So, such two-electron state is possible. 1 Ψ (1, 2) = ϕa (r1 ) ϕa (r2 ) √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ] 2 (2) 7 Slater determinant Let us introduce spin in the definition of the single particle orbitals, and let us enumerate these orbitals by index i : In these notations orbitals of the previous example are ϕ1 (r, s) = ϕa↑ ≡ ϕa (r) | ↑i ϕ2 (r, s) = ϕa↓ ≡ ϕa (r) | ↓i ϕ3 (r, s) = ϕb↑ ≡ ϕb (r) | ↑i ϕ4 (r, s) = ϕb↓ ≡ ϕb (r) | ↓i Hence the wave function (1), page 6 can be written as determinant 1 1 ϕ1 (1) ϕ1 (2) Ψ (1, 2) = √ [ϕ1 (1) ϕ3 (2) − ϕ1 (2) ϕ3 (1)] ≡ √ 2 2 ϕ3 (1) ϕ3 (2) Similarly, the wave function (2), page 6 also can be written as determinant. 1 1 ϕ1 (1) ϕ1 (2) Ψ (1, 2) = √ [ϕ1 (1) ϕ2 (2) − ϕ1 (2) ϕ2 (1)] ≡ √ 2 2 ϕ2 (1) ϕ2 (2) I repeat the meaning of the notation ϕi (x)). Here the index i enumerates single particle orbitals and x shows coordinates (spatial, spin, etc.) of a particle. Using these notations we can write the many-body wave function for arbitrary number of noninteracting fermions ϕ (1) ϕ (2) .... ϕ (N) 1 1 1 ϕ (1) ϕ (2) .... ϕ (N) 2 2 2 1 Ψ (1, 2, ...N) = √ .... .... .... .... N! .... .... .... .... ϕN (1) ϕN (2) .... ϕN (N) This form was suggested by Slater and it is called Slater determinant. 8 Permutation of two particles is equivalent to permutation of two columns. For Example ϕ (2) ϕ (1) .... ϕ (N) 1 1 1 ϕ (2) ϕ (1) .... ϕ (N) 2 2 2 1 Ψ (2, 1, ...N) = √ .... .... .... .... = −Ψ (1, 2, ...N) . N! .... .... .... .... ϕN (2) ϕN (1) .... ϕN (N) So, the Fermi statistics requirement is automatically satisfied. If two orbitals in the determinant coincide: i = j; then the determinant vanishes because there are two identical lines. This describes the Pauli exclusion principle. Slater determinant is a very convenient form for the wave function. Unfortunately this form is exact only for noninteracting fermions. 9 Interacting fermions. Variational solution for He atom ground state Hamiltonian of He atom e2 p~2 2 Ze2 Ze2 p~1 2 − + + − 2m 2m r1 r2 |r~1 − r~2 | Here we neglect all relativistic/magnetic effects. The relative magnitude of these effects is Ĥ = ∼ v 2 /c2 ∼ (Zα)2 = (Z/137)2 ≪ 1. Schrodinger equation Hψ (1, 2) = Eψ (1, 2) can be solved exactly numerically but: 1) The solution is very involved technically. 2) There is no exact solution for three (Li atom) and more electrons. So, we need an approximate but relatively simple method which can be propagated to multi-electron systems. Variational method Energy of the system reads D E Z Ψ|Ĥ|Ψ = Ψ∗ ĤΨd3 r1 d3 r2 d3 r3 ... The wave function is normalized hΨ|Ψi = Z Ψ∗ Ψd3 r1 d3 r2 d3 r3 ... = 1 Let us find minimum of energy with respect to variation of Ψ∗ . To account for the normalization constraint let us use the Lagrange multiplier method with λ being the Lagrange multiplier δ δΨ∗ (x) [hΨ∗ H|Ψi − λ hΨ∗ |Ψi] = 0 ⇒ Ĥψ − λψ = 0 ⇒ Ĥψ = λψ ⇒ λ = E. So, we end up with usual Schrodinger equation. In this case the number of variational parameters = ∞, Ψ∗ (x) at each point x. The variational method does not bring anything new. 10 For practical applications we choose a finite number of parameters and hence the variational method gives an approximate answer. Example He-like ion ground state Hydrogen-like ion: single electron in Coulomb field of a nucleus with charge Z (see 3d -year quantum mechanics.) The Hamiltonian: Z 11 00 00 11 00 11 00 11 FIG. 4: Hydrogen-like ion H= p2 Ze2 − 2m r The electron ground state energy: ǫ=− Z 2 me4 2h̄4 The electron wave function: ϕ(r) = Ae−Zr/aB s Z3 A= πa3B aB is Bohr radius, and A is the normalization constant. Atomic units: Eatomic = E me4 /h̄4 me4 = 27.2eV h̄4 ratomic = r/aB h̄2 aB = ≈ 0.53Å = 0.53 10−8 cm me2 11 Below I omit the subscript “atomic”. The electron Hamiltonian, energy, and wave function in atomic units are p2 Z ∆ Z − =− − 2 r 2 r Z2 ǫ=− 2 ϕ(r) = Ae−Zr r Z3 A= π H= Now consider He-like ion. He-like ion Hamiltonian in atomic units reads Ĥ = − ∆1 ∆2 Z 1 Z − − − + 2 2 r1 r2 |r1 − r2 | Let us consider the two electron variational wave function of the following form Ψ = ϕ (1) ϕ (2) Φs 1 Φs = √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ] 2 where Φs is the spin wave function corresponding to total spin zero. For electron orbital ϕ we use the hydrogen-like ansatz, ϕ (r) = s 3 Zef f −Zef f r e , π where Zef f is a variational parameter. According to the variational method we have to calculate energy of the system and then minimize it with respect to variation of Zef f . Z 1 1 Z Ψ E = hΨ|H|Ψi = − hΨ|∆1 + ∆2 |Ψi − Ψ + Ψ + Ψ 2 r1 r2 |~r1 − ~r2 Z Φ†s ϕ∗ (2) ϕ∗ (1) ∆1 ϕ (1) ϕ (2) Φs d3 r1 d3 r2 Z Z † ∗ 3 ∗ 3 ϕ (1) ∆1 ϕ (1) d r1 ϕ (2) ϕ (2) d r2 = Φs |Φs Z = ϕ∗ (1) ∆1 ϕ (1) d3 r1 hΨ|∆1 |Ψi = 12 Assignment problem: Z 2 ϕ∗ (1) ∆1 ϕ (1) d3 r1 = Zef f Remember that ∆= 1 d 2d r r 2 dr dr Z Z Z Ψ Ψ = Φ†s ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ∗ (2) Φs d3 r1 d3 r2 r1 r1 Z Z † Z ∗ 3 ∗ 3 = Φs |Φs ϕ (2) ϕ (2) d r2 ϕ (1) ϕ (1) d r1 r1 Z Z = ϕ∗ (1) ϕ (1) d3 r1 r1 Assignment problem: Z Ψ ϕ∗ (1) Z ϕ (1) d3 r1 = ZZef f r1 Z 1 1 Ψ = Φ† ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ (2) Φs d3 r1 d3 r2 |r~1 − r~2 | |r1 − r2 | Z 1 ∗ ϕ (1) ϕ (2) d3 r1 d3 r2 = hΦs |Φs i ϕ∗ (2) ϕ∗ (1) |r~1 − r~2 | Z 1 = ϕ∗ (2) ϕ∗ (1) ϕ (1) ϕ (2) d3 r1 d3 r2 |r~1 − r~2 | Assignment problem: Z 1 ϕ (1) ϕ (2) d3 r1 d3 r2 ϕ∗ (2) ϕ∗ (1) |r~1 − r~2 | Z 6 Zef 1 f = 2 e−2Zef f r1 e−2Zef f r2 r12 dr1 dΩ1 r22 dr2 dΩ2 π |r1 − r2 | 5 = Zef f 8 13 Altogether the energy is 5 2 E = hΨ|H|Ψi = Zef f − 2ZZef f + Zef f 8 It is minimum at Zef f = Z − 5 . 16 The physical energy which is the minimum energy is 2 5 E = hΨ|H|Ψi = − Z − 16 Z=2 27 →− 16 2 In electron volts 27 E=− 16 2 27.2 = −77.38eV . Experiment: E = −78.9eV , so the simple variational solution works remarkably well. 14 Multi-electron atom and effective self-consistent potential A particular electron Averaged electron cloud Z 11 00 00 11 00 11 00 11 FIG. 5: A cartoon of multielectron atom. A particular electron which we consider is shown by the black line. Other electrons which together with nucleus produce the effective potential for the “black” electron are shown by red. The effective potential method reduces (approximately) the many-body problem to a single particle problem. H many-body → Hsp , “sp” stands for single-particle Hsp = p2 + Vef f (r) 2m (3) For a neutral atom r << aB : V = − Ze2 ef f Z r r >> aB : Vef f = − e2 r Veff r −e2 r −Ze2 r FIG. 6: A sketch of the effective potential of a neutral atom. (4) 15 The effective potential is different from the simple Coulomb potential of a point-like nucleus. Therefore the hydrogen degeneracy of states with the same principal quantum numbers is lifted. Energy levels in a Coulomb potential 3s 3p 2s 2p Energy levels in the effective atomic potential 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 3d 1s 3s 3p 2s 2p 1s FIG. 7: In the periodic table of elements the levels are filled from down to up. 3d 16 Exchange interaction Consider two lowest excitations of He atom. The experimental spectrum is as follows Spectroscopic Energy notation 1s2s 1 S0 ——————— 20.62eV 1s2s 3 S1 ——————— 19.82eV ground state 1s2 1 S0 ——————— The spectroscopic notation is 0.eV 2S+1 ˆ ~ˆ ~ˆ LJ , J~ = L +S For example 1 S0 means that Letter S ⇒ L = 0 Lef t superscript = 1 ⇒ 2S + 1 = 1 ⇒ S = 0 spin Right subscript = 0 ⇒ J = L + S = 0 + 0 = 0 The states |1s2s, 1 S0 > and |1s2s, 3 S4 > differ by total spin only, 0 S= 1 Question: The interaction is spin independent (Coulomb). Why the states have different energies? 17 Single particle orbitals ϕ1 (r) ≡ ϕ1s (r) = q Z 3 ef f −Zef f r e π ϕ2 (r) = ϕ2s (r) − some f unction with one node We do not need an explicit form of ϕ1 (r) and ϕ1 (r) for our discussion. The two electron wave functions (Slater determinants) are 1 1 |1s2s, 1 S 0 i = √ [ϕ1 (r1 )ϕ2 (r2 ) + ϕ1 (r2 )ϕ2 (r1 )] × √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ] 2 2 1 |1s2s,3 S1 i = √ [ϕ1 (r1 )ϕ2 (r2 ) − ϕ1 (r2 )ϕ2 (r1 )] × | ↑i1 | ↑i2 2 Reminder: Spin wave functions of two electrons are 1 |S = 0, Sz = 0i = √ [| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ] 2 |S = 1, Sz = 1i = | ↑i1 | ↑i2 1 |S = 1, Sz = 0i = √ [| ↑i1 | ↓i2 + | ↓i1 | ↑i2 ] 2 |S = 1, Sz = −1i = | ↓i1 | ↓i2 Let us calculate contributions to energies that come from the Coulomb interaction between electrons E1 S0 1 1 S0 = S0 |r~1 − r~2 Z 1 1 = [ϕ∗1 (1)ϕ∗2 (2) + ϕ∗1 (2)ϕ∗2 (1)] [ϕ1 (1)ϕ2 (2) + ϕ1 (2)ϕ2 (1)] 2 |~r1 − ~r2 | Z Z ∗ |ϕ1 (r1 )|2 |ϕ2 (r2 )|2 3 3 ϕ1 (1)ϕ1 (2)ϕ∗2 (2)ϕ2 (1) 3 3 = d r1 d r2 + d r1 d r2 |~r1 − ~r2 | |~r1 − ~r2 | 1 ϕ1 r1 ϕ1 ϕ1 r1 ϕ2 ϕ2 r2 ϕ2 ϕ2 r2 ϕ1 FIG. 8: Feynman-like diagrams for the direct and the exchange interaction 18 E3 S1 1 3 S1 = S1 ~r1 − ~r2 Z 1 1 [ϕ∗1 (1)ϕ∗2 (2) − ϕ∗1 (2)ϕ∗2 (1)] [ϕ1 (1)ϕ2 (2) − ϕ1 (2)ϕ2 (1)] = 2 |~r1 − ~r2 | Z ∗ Z ϕ1 (1)ϕ1 (2)ϕ∗2 (2)ϕ2 (1) 3 3 |ϕ1 (1)|2 |ϕ2 (2)|2 3 3 d r1 d r2 − d r1 d r2 = |r~1 − r~2 | |~r1 − ~r2 | 3 ϕ1 r1 ϕ1 ϕ1 r1 ϕ2 ϕ2 r2 ϕ2 ϕ2 r2 ϕ1 FIG. 9: Feynman-like diagrams for the direct and the exchange interaction Thus, the energy is ϕ1 r1 ϕ1 ϕ1 r1 ϕ2 ϕ2 r2 ϕ2 ϕ2 r2 ϕ1 E exchange contribution direct contribution Due to statistics the exchange term depends on total spin in spite of the fact that the interaction is spin independent. Comparing with experimental data on He we find: ϕ1 r1 ϕ2 2 0.76 eV ϕ2 r2 ϕ1 Magnetism in solids is due to exchange interaction.