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Chapter 1 : SECOND QUANTIZATION Readings and lectures ... You are going to read the chapter at home. I will go over some of the calculations in class, but not all of them. The purpose of the lectures is to understand the results, but not always to derive them in detail. You should read the material before the ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ lecture so that you will know the αβγδεζηθικλμνξοπρςστυφχψω notations. +<=>±×⁄←↑→↓⇒⇔∇∏∑−∓∙√∞ ≤≥≪≫∫∲ ∂≠〈〉ħ 1. THE SCHROEDINGER EQUATION IN FIRST AND SECOND QUANTIZATION Many body theory in first quantization Start with H = ∑Nk=1 T(xk) + ½ ∑’ Nk,l=1 V(xk,xl) ; then find iħ (∂f/∂t) where f is the amplitude for occupation numbers fN ( n 1 n2 … n i … n∞ ) . Notation ∑’ : here the prime means l ≠k The first quantized theory To Solve: iħ ∂Ψ/ ∂t = H Ψ where Ψ = Ψ(x1 x2 x3 … xk … xN) Introduce a complete set of time-independent single-particle wave functions Completeness: We can expand the Nparticle wave function as a product of single-particle wave functions __ So far, this is simple. Now comes the hard part: the N particles are identical particles. 1a. Bosons . Ψ(x1 … xN ; t ) must be symmetric with respect to interchange of any two coordinates; i.e., Ψ( … xk … xl … ; t ) = + Ψ( … xl … xk … ; t ) Then also, C( … Ek … El … ; t ) = + C( … El … Ek … ; t ) for any case of k and l “standard order” (Lecture Jan 14) Summary so far -- the quantum manyparticle problem in first quantized form … ◼ understand that the symmetry of the wave function (for bosons) implies that the basis states depend only on the list of occupation numbers; ◼ expand Ψ( {x} ,t ) in occupation number basis states; ◼ the Schroedinger equation for the occupation number amplitude (eq. 1.25); 1b. The many-particle Hilbert space (a.k. a. Fock space); creation and annihilation operators; SECOND QUANTIZATION bosons H = ∑ij bi+< i |T| j > bj + ½ ∑ij bi+ bj+ < ij |V| kl > bl bk Proof 1/2 Simplify the notation: ψi(x) means ψEi(x); i.e., the single particle wave function with quantum numbers Ei . 1/2 The final result is equivalent to Eq. (1.25). Q.E.D. 1c. Fermions Start again with the first quantized Nbody Hamiltonian , Now add the antisymmetry of the of the wave function Ψ(x1 x2 … xk … xN ; t) , Ψ( … xk … xk … ; t) = − Ψ( … xl … xk … ; t) for any case of k and k , for fermions. **Exchange includes spin indices, suppressed here. Fermions are easier than bosons because the occupation numbers are more limited: for any single-particle state i, ni can only be 0 or 1. That’s the Pauli exclusion principle. It is a consequence of the antisymmetry of the wave function. … ψE (xk) … ψE’ (xl) … must be equal to − … ψE (xl) … ψE’ (xk) … If E’ = E then the N-body wave function could only be 0; not possible. Annihilation and creation operators for fermions. Use notation ci and ci☨ for fermions. FOR FERMIONS, THE DEFINING COMMUTATION RELATIONS ARE ANTI COMMUTATION RELATIONS. { c i , c j } = c i cj + c j ci = 0 { ci☨ , cj☨ } = ci☨ cj☨ + cj☨ ci☨ = 0 { ci , cj☨ } = ci cj☨ + cj☨ ci = δij . Another notation for the anticommutator: {A,B} = [A,B] + The interpretation of fermionic operators (ci and ci ☨ ) is the same as for bosonic operators (bi and bi ☨ ): ci = annihilation operator; annihilates a particle in the state Ei ; ci☨ = creation operator; creates a particle in the state Ei ; ci☨ ci = occupation number operator for the state Ei . Note that this automatically satisfies the Pauli exclusion principle: A state with two particles would be ci ☨cj☨ |0〉; but if i = j then we would have ci ☨ci☨ ; but this is 0 by the second A.C. relation. So two particles cannot occupy the same s.p. state. Antisymmetry of the wave function Theorem. The anti commutation relations of the creation and annihilation operators imply that the N-body wave function will be antisymmetric. ∴ |000...1...1…0> = ½ (ca+cb+ - cb+ca+) |0>