Ket space as a vector space over the complex numbers

Transcription

1 Ket space as a vector space over the complex numbers kets ϕ> and complex numbers α with two operations Addition of two kets ϕ 1 >+ ϕ 2 > is also a ket ϕ 3 > Multiplication with complex numbers α ϕ 1 > is also a ket ϕ 2 > Satisfying: Commutativity of addition ϕ 1 >+ ϕ 2 >= ϕ 2 >+ ϕ 1 > Associativity of addition ( ϕ 1 >+ ϕ 2 >)+ ϕ 3 >= ϕ 1 >+( ϕ 2 >+ ϕ 3 >) There is a state null> with ϕ>+ null>= ϕ> for all ϕ> Each state ϕ> has an inverse ϕ> with ϕ>+ ϕ>= null> (write + ϕ> as - ϕ>) 1 ϕ>= ϕ> (α+β) ϕ>=α ϕ>+β ϕ> and α( ϕ 1 >+ ϕ 2 >)=α ϕ 1 >+α ϕ 2 > (αβ) ϕ>=α(β ϕ>)

2 The scalar (or inner) product and bra's For ϕ 1 > and ϕ 2 > define <ϕ 1 ϕ 2 >=α with <ϕ ϕ> is real and positive, unless ϕ>= null>, where <null null> =0 <ϕ 1 ϕ 2 >=<ϕ 2 ϕ 1 > * For ϕ 3 >=α ϕ 1 >+β ϕ 2 > we have <ϕ ϕ 3 >=α<ϕ ϕ 1 >+β<ϕ ϕ 2 > <ϕ 3 ϕ>=α * <ϕ 1 ϕ>+β * <ϕ 2 ϕ> Treat <ϕ as an object on its own, called bra. The bras form their own vector space, the dual space dual correspondence ϕ> <ϕ <ϕ 1 ϕ 2 > bra(c)ket Note ϕ 3 >=α ϕ 1 >+β ϕ 2 > <ϕ 3 =α * <ϕ 1 +β * <ϕ 2

3 Bases A basis is a set { a i >} which allows to construct any ϕ> as ϕ>= i c i a i > with unique complex numbers c i An orthonormal (ON) basis satisfies <a i a j >=δ ij For an ON basis we find: Expansion coefficients c i =<a i ϕ> Completeness relation i a i ><a i =1 <ϕ ϕ>= i c i 2

4 Linear Operators Operator  transforms  Φ>= Φ'> Linear operator: Â(α Φ>+β φ>)=αâ Φ>+β φ> Product between operators Ĥ=Ĉ the operator transforming Ĥ Φ>=Ĉ( Φ>) commutator [Ĉ,Â]=ĈÂ-ÂĈ

5 Matrix representation For ON basis { a i >} evaluate State ϕ> coefficients c i =<a i ϕ> Operator  coefficients A ij =<a i  a j > All calculations become matrix-operations Note: Depends on basis used, only for ON basis

6 Important relations for the adjoint operator In matrix representation: A ij =<a i  a j >=<a j  a i >*=A ji * Exercises: If  =Â:  is called hermitian (self-adjoint), A ij =A ji * If  Â=1:  is called unitary

7 Hermitian Operators Matrices with A ij =A ji * can be diagonalized ( A (n) 11 A 12 (n) A 21 A 22 c 2 )(c1 Eigenstates u n = form a new ON basis i (n) n(c1 )=λ (n) c 2 ) with real λ n c (n) i a i with Α un =λ n u n  hermitian Α= n λ n u n u n

8 Common eigenstates of two Hermite operators  and Ĉ [Â,Ĉ]=0 There exists an ON basis { a n >}, where all a n > are simultaneously eigenstates of  and Ĉ (possibly with different eigenvalues) Note: [Â,Ĉ]=0 does not imply that each eigenstate of  is also eigenstate of Ĉ

9 Operators representing observables Measurement: Real measurement values α n associated with physical states a n > providing ON basis Expectation value for state Ψ>: with Hermitian operator Physical observable Hermitian operator

10 Continuous spectrum and spatial representation For continuous measurement values, e.g. spatial position x Eigenstates as before But continuum of states cannot be counted Instead: Any state can be written as Common wavefunction

11 Three spatial dimensions Postulate Thus there is a common set of eigenstates x,y,z> Ψ(r)=<x,y,z Ψ>

12 Function of an operator I Taylor Series Frequently we write f(â) what does it mean? For f (x)=x n we define f (Â)=Ân   For f (x)= n=0 n times f n n! xn we define f (Â)= n=0 f n n! Ân Exercise: [ B, Ân ]=n[ B, Â] An 1 if [[ B, Â], Â]=0 Important implication: [ B, f (Â)]=[ B, Â] f ' (Â) if [[ B, Â], Â]=0

13 Function of an operator II Projection on eigenstates Hermitean operators have a complete set of eigenstates Â= α n a n >< a n n f (Â)= f (α n ) a n ><a n n

14 Change of basis Old basis { a n >}, new basis { b n >} Operator for basis change Û= n Ψ >= n c n a n >= m c' m = b m Ψ = n c ' m b m > b m a n c n = a m U a n b n >< a n unitary! Matrix for basis change

15 Solving the stationary Schrödinger equation by diagonalization Stationary Schrödinger Eq. Search for eigen-energies E n and states φ n > Use matrix representation!

16 One-dimensional harmonic Oscillator Hamiltonian in analogy with classical case Ĥ = p2 2m +1 2 mω2 x 2 P= p m ħ ω, X = x mω satisfying [ P, X ]= 1 ħ i Factorize Define â= 1 2 ( X +i P ) Ĥ= 1 2 ħ ω {( X i P)( X +i P)+i[ P, X ]} Task: Find eigenvalues and -states of N =â â â= m ω 2 ħ x+i 1 Relation to original x,p 2m ω ħ p x= ħ 2 mω (â +â ) p=i mωħ (â â ) 2 Ĥ= 1 2 ħ ω ( P 2 + X 2 ) Ĥ=ħ ω ( â â+ 1 2 ) with [â,â ]=1 Not Hermitian!

17 Eigenvalue problem N φ n >=n φ n > with n R n 0 N =â â [ â,â ]=1 â φ n > is also eigenstate of N with eigenvalue n+1 â φ n > is also eigenstate of N with eigenvalue n 1 Sequence (k φ ) n >=â k φ n > of eigenstates of N with eigenvalues n k (k Stops after k if â φ ) n >= φ death > null> (k requires 0= φ death φ death = φ ) n â (k â φ ) (k ) n =(n k ) φ n (k φ ) n n=k is an integer number! (n=0,1,2,..)

18 Normalized eigenstates N φ n >=n φ n > with n N 0 N =â â [ â,â ]=1 φ n+1 >= 1 n+1 â φ n > and φ n 1 >= 1 n â φ n > Or: φ n >= 1 n! (â ) k φ 0 > where N φ 0 >=0 φ 0 > For each eigenstate, there exists a sequence of states with energy eigenvalues E n =ħω(n+1/2), n=0,1,2,... Ground state φ 0 > with E 0 =ħ ω/2 satisfies â φ 0 >=0

19 Example: calculate m x 2 n x= ħ 2 mω (â +â ) p=i mωħ (â â) 2 â n>= n+1 n+1>, â n>= n n 1>

20 Spatial representation: Ground state Ground state φ 0 > with E 0 =ħ ω/2 satisfies â φ 0 >=0 â= m ω 2ħ x+i 1 2m ωħ p < x â dx ' x ' >< x ' φ 0 >=0 ( mω 2 ħ x+ ħ 2m ω x ) φ 0 ( x)=0 or with ξ= mω ħ x : 1 2 ( ξ+ ξ ) φ 0 (ξ)=0 Normalized solution φ 0 (ξ)=π 1/ 4 e ξ2 /2

21 Spatial representation: Excited states â n>= n+1 n+1>, â n>= n n 1> = â mω 2 ħ x i 1 2m ω ħ p 1 ( 2 ξ ) ξ φ n (ξ)= 1 2 n n! (ξ ξ ) n φ 0 (ξ)= Representation via Hermite- Polynomials 1 π 1/ 4 2 n n! (ξ ξ ) n e ξ2 / 2 ( ξ ξ ) n e ξ2 / 2 =e ξ2 / 2 ( 1) n e ξ2 ( ξ ) n e ξ2 = H n (ξ) Math: These form a basis for the square-integrable functions in 1dimension

22 Eigenstates in spatial coordinates

23 Two dimensions with E=ħ ω (N +1 ) N =0,1,2,.. (N+1) fold degeneracy Spatial representations

24 Three dimensions E=ħ ω ( N +3/2) N =0,1,2,.. (N+1)(N+2)/2 fold degeneracy Alternative solution in spherical coordinates Ansatz spherical harmonics Radial part

25 Commutation relations of (orbital) angular momentum Angular momentum in analogy with classical case L= r p satisfies commutation relations [ L x, L y ]=i ħ L z [ L y, L z ]=i ħ L x [ L z, L x ]=i ħ L y and consequently [ L 2, L i ]=0 for i=x, y, z

26 Generalized angular momentum Common eigenstates Hermitian operators Ĵ x, Ĵ y, Ĵ z satisfying [Ĵ j, Ĵ k ]=i ħ l ϵ jkl Ĵ j Ĵ 2 =Ĵ x 2 +Ĵ y 2 +Ĵ z 2 satisfies [Ĵ 2, Ĵ j ]=0 Eigenvalues of Ĵ 2 are not negative Search set of common eigenstates j, m satisfying Ĵ 2 j, m = j ( j +1)ħ 2 j,m and Ĵ z j,m =m ħ j,m with j 0

27 Properties of Ψ + =Ĵ + j, m Ĵ 2 j, m = j ( j +1)ħ 2 j,m and Ĵ z j, m =m ħ j,m with j 0 Define Ĵ + =Ĵ x +i Ĵ y and Ĵ - =Ĵ x i Ĵ y =Ĵ + Ψ + Ψ + =[ j ( j+1) m(m+1)]ħ 2 (a) m j (b) Ψ + = null m= j or m= j 1 (c) Ĵ 2 Ψ + = j ( j+1)ħ 2 Ψ + and Ĵ z Ψ + =(m+1)ħ Ψ + Identify j, m+1 = 1 ħ j ( j+1) m(m+1) Ĵ + j, m

28 Properties of Ψ - =Ĵ - j, m Ĵ 2 j, m = j ( j +1)ħ 2 j,m and Ĵ z j,m =m ħ j,m with j 0 Define Ĵ + =Ĵ x +i Ĵ y and Ĵ - =Ĵ x i Ĵ y =Ĵ + Ψ - Ψ - =[ j ( j+1) m(m 1)]ħ 2 (a) m j (b) Ψ - = null m= j or m= j+1 (c) Ĵ 2 Ψ - = j ( j +1)ħ 2 Ψ - and Ĵ z Ψ - =(m 1)ħ Ψ - Identify j,m 1 = 1 ħ j ( j +1) m(m 1) Ĵ - j, m

29 Common eigenstates of Ĵ 2 and Ĵ z j-m, j+m N 0 2j N 0 Sequence of states j,m>, m=-j,j+1,...,j 1 j, m+1 = ħ j ( j+1) m(m+1) Ĵ + j,m j, m 1 = 1 ħ j ( j+1) m(m 1) Ĵ - j,m Ĵ 2 j, m = j ( j +1)ħ 2 j,m and Ĵ z j,m =m ħ j,m with j=0, 1 2,1, 3 2, and m= j, j+1, j form multiplets M j with 2j+1 states

30 Eigenstates of Hamiltonian with full rotational symmetry Rotational symmetry: [Ĥ,Ĵ i ]=0 for i=x,y,z and thus [Ĥ,Ĵ 2 ]=0 There exists a common set of eigenstates of Ĥ,Ĵ z, and Ĵ 2 All elements of the multiplet j,m>, m=-j,-j+1,...j have the same energy (2j+1)-fold degeneracy

31 Spin: two-fold degeneracy implies j=1/2 j, m+1 = j, m 1 = 1 ħ j ( j+1) m(m+1) Ĵ + j,m 1 ħ j ( j+1) m(m 1) Ĵ - j,m Notation: Ĵ i Ŝ i

32 Magnetic moment is related to angular momentum of a particle classical homogeneously charged particle μ= q 2m L Define gyromagnetic ratio (Landé-factor, g-factor) atomic physics dominated by electrons μ= g e Ŝ with e>0 2m e Pure electron spin: g e = (quantum electrodynamics) nuclear physics dominated by protons μ=g Proton: g p = Neutron: g n = e 2m p Ŝ

33 Coupling between two spins: The product space Positronium: Bound system of one electron and one positron Arbitrary state Ŝ ħ 2 [( ) e x + ( 0 i i 0 ) e y + ( ) e z] a i Ĥ a j = U 4 ( ) Ground state: E=-3U/4, J=0, para-positronium τ=125ps E=U/4, J=1, ortho-positronium τ=142ns (U=0.8 mev)

34 Spin-Orbit coupling Product space of space and spin spanned by Spinor representation Search for eigenstates classified by angular momentum

35 System 1: Common eigenstates of L 2, L z, Ŝ 2, Ŝ z Possible as [Ŝ i, L j ]=0 Denote states as a,l, m l, s,m s 1 Example

36 The total angular momentum Ĵ= L+Ŝ Ĵ z = L z +Ŝ z Ĵ 2 = L 2 +Ŝ 2 +2 L z Ŝ z + L + Ŝ - + L - Ŝ + Calculation gives System 1 can not be chosen as eigenstates of Ĵ 2

37 System 2: Common eigenstates of Ĵ2, Ĵ z, L 2, Ŝ 2 Ĵ z = L z +Ŝ z Ĵ 2 = L 2 +Ŝ2 +2 L z Ŝ z + L + Ŝ - + L - Ŝ + Ĵ 2, Ĵ z, L 2, Ŝ 2 all commute with each other Denote states as a, j, m j,l, s 2 Change of basis Clebsch-Gordan coefficients

38 Hamiltonian for particle with charge q in electromagnetic fields Lorentz force F =q(e+v B) has no potential V (r) Electrodynamic potentials A(r, t), φ(r, t) with E= φ A/ t B= A Ĥ = ( p q A( r,t )) 2 2m +q φ( r,t) r i and p j satisfy the common commutation relations Two different momenta Canonical momentum p becomes i ħ in spatial representation Kinetic momentum p kin = p q A provides velocity Ψ pkin m Ψ

39 Free particle with charge q in magnetic field Ĥ = p2 2m B=B e z +q φ( r,t) q 2m B L+ q2 B 2 8m r 2 z-direction separates with eigenstates k>~e ikz Ĥ = p 2 z 2m + p 2 2 x+ p y B 2 2m +q2 8m ( x 2 +ŷ 2 ) q 2m B L z = Ĥ osc with ω=qb /2m =ω L z

40 Stationary perturbation theory Hamiltonian Ĥ has typically no analytic solution of Ψ Decompose Ĥ =Ĥ 0 + V known eigenstates a 0 Ĥ 0 a 0 =E a 0 a 0 hopefully small Eigenstate form basis Ψ = a c a a 0

41 Stationary perturbation theory (non-deg.) Eigenvalue problem Expand Choose such that is real. Thus Normalization implies

42 Non-degenerate eigenvalues E⁰ a Admixure of other states