Eigenfunctions of Hermitian operators belonging to different eigenvalues are orthogonal, since: Aψ m = a m ψ m, Aψ n = a n ψ n a n (ψ m, ψ n )=(ψ m, Aψ n )=(Aψ m, ψ n )=a m (ψ m, ψ n ) (a n a m )(ψ m, ψ n )=0 a n = a m (ψ m, ψ n )=0 If several eigenfunctions belong to the same eigenvalue (degeneracy), one can orthogonalize as follows: ψ 1, ψ,... linear independent eigenfunctions C 1 φ 1 = ψ 1 C 1 =(ψ 1, ψ 1 ) 1 C φ = ψ φ 1 (φ 1, ψ ) C = (ψ, ψ ) (φ 1, ψ ) 1 C 3 φ 3 = ψ 3 φ 1 (φ 1, ψ 3 ) φ (φ, ψ 3 )..
Eigenfunctions of a Hermitian operator can always be chosen such that the orthogonality relation (ψ m, ψ n )=δ m,n is satisfied. Moreover, the eigenfunctions of the operators we are considering satisfy the completeness relation ψn(x )ψ n (x) =δ(x x ) ψ n form a complete set of orthogonal eigenfunctions. n general state ψ can be represented as ψ(x) = dx δ(x x )ψ(x )= n dx ψ n(x )ψ n (x)ψ(x ) = n (ψ n, ψ)ψ n (x) = n c n ψ n (x), c n =(ψ n, ψ) normalization n c n = 1.
Expansion in Stationary States Orthogonality and completeness hold in particular for the eigenfunctions of the Hamilton-operators. Hψ n = E n ψ n, ψ n (x, t) =e iet ψn (x) time evolution ψ(x, t) = n c n e ie nt ψn (x), c n =(ψ n, ψ(x, t = 0)) i t ψ(x, t) = n E n c n e ie nt ψn (x) = H n c n e ie nt ψn (x) =Hψ(x, t)
Physical Significance of the Eigenvalues of an Operator operator A, complete orthonormal system of eigenfunctions Ψ n with eigenvalues a m and a wave function ψ(x) = m c m ψ m (x) What is the physical significance of a n and c n? Some Concepts from Probability Theory Let X be a random variable taking values x and w(x)dx the probability that X takes a value in the iintervall [x, x + dx]. m n = χ(τ) = x n w(x)dx = X n e ixτ w(x)dx nth moment of w(x) charakteristic function χ(τ) is the Fourier transform of w(x). Inverse Fourier transform: w(x) = dτ π eixτ χ(τ)
expansion of the exponential function knowledge of moments w(x). χ(τ) = n ( i) n n! τ n m n in generall: F (X ) = example: χ(τ) = e ix τ F (x)w(x)dx
Operators with discrete eigenvalues i) Let the system be in an eigenstate ψ m : A n = (ψ m, A n ψ m )=(a m ) n = χ(τ) = n ( i) n τ n (a m ) n = e iτa m n! w(a) = dτ π eiaτ e iτa m = δ(a a m ) = one measures with certainty the value a m.
ii) Let the system be in state ψ = m c mψ m.then: A n = (ψ, A n ψ)= c m ψ m, A n m m = cmc m (Ψ m, A n ψ m ) m m = cmc m (a m ) n δ m,m m A n = m m c m (a m ) n c m ψ m χ(τ) = n w(a) = ( i) n τ n n! m dτ π eiaτ m c m (a m ) n = m c m e iτa m = m c m e iτa m c m δ(a a m ) the result of the measurement can only be one of the eigenvalues a m. probability to measure a m is given by c m. After the measurement with the value a m,thesystemisinstateψ m. Reduction of the wave function: measurement alters state!. If the wave function after the measurement is known exactly, then this is called an ideal measurement. In this case, the system finds itself in an eigenstate of the operator corresponding to the observable.
Operators with a Continous Spectrum example: momentum operator p = i eigenvalue problem.: i x x ψ p(x) =p ψ p (x) = ψ p (x) =(π) 1/ e ipx/ orthogonality: completeness: dx ψ p(x)ψ p (x) =δ(p p ) dp ψ p(x )ψ p (x) =δ(x x ) expansion in momentum eigenfunctions ψ(x) = dp φ(p ) exp(ip x/) π π comparison with discrete spectra= c m (π) 1 φ(p ), dp m
insert this translation into w(a) = w(p) = dp φ(p ) π δ(p p )= φ(p) π w(p) not just plausible but consequence of the momentum operator Example: position eigenfunctions ψ ξ = δ(x ξ) fulfill eigenvalue eqn. of the position operator: xψ ξ (x) =ξψ ξ (x), EW ξ, continous spectrum orthogonality: (ψ ξ, ψ ξ )=δ(ξ ξ ) completeness: dξψ ξ (x)ψ ξ (x )=δ(x x ) it follows: ψ(x) = dξ ψ(ξ)ψ ξ (x) = expansion coefficients of the position eigenfunctions are just given by the wavefunction = ψ(ξ) probability for the position ξ.
remark: ψ(x) wave function in position representation φ(p) wave function in momentum representation in general: mixed spectrum ψ(x) = n c n ψ n (x)+ da c(a)ψ a (x) =: cn φ n probability density: w(a) = n c n δ(a a n )+ c(a)
Axioms of Quantum Theory i) The state is described by the wave function ψ(x). ii) Observables are represented Hermitian operators A iii) Expectation value of observable represented by the operator A in the state ψ: A =(ψ, Aψ) iv) time evolution of the states i t ψ = Hψ, H = h m + V (x) v) If in a measurement of A the value a n is found, then the wave function changes to the corresponding eigenfunctiion ψ n.
Path integrals W (x) = Φ 1 + Φ complex amplitudes W (x) = i Φ i over all combinations of slits #slits #barriers = sum over all paths
W = x(y) Φ x(y) Feynmans principle idea important: additional information on time dependence paths are characterized by the set of {x(t), y(t)}. W (x) = over all amplitudes corresponding to the paths {x(t), y(t)} If we know the recipe to calculate the amplitudes corresponding to a path = we know the rules of QM Consider a one-dimensional motion of a particle: time at start t a : x(t a )=x a time at end t b : x(t b )=x b
in between we find all paths x(t) Classical Action Prinziple of least action S[x(t)] = tb t a L(x(t), ẋ(t)) dt δs = S[x + δx] S[x] =0
= S[x + δx] = = tb t a tb t a = S[x]+ = δs = δx L t b x + t a L(x + δx,ẋ + δẋ, t) dt L(x,ẋ, t)+ L x tb t a tb t a =0 δx =0beix a, x b δx L x δx L δx + ẋ δẋ L + δẋ dt ẋ L x d dt d dt = L ẋ L ẋ dt Lagrange L x =0
QM: W (b, a) = K(b, a) = probability to start from x a at time t a until x b at time t b K(b, a) = + sum over all paths from a to b Φ[x(t)] = const e (i/)s(x,t) Φ[x(t)] classical limit S 1 eis(x,t) oscillates heavily, averages out only contributions near extrema
Analogy with Riemann integral A i f (x i ) so that limit exists A =lim h 0 h i f (x i ) very similar for summation over all paths: choose subset of all paths time is discretized in : N = t b t a ; = t i+1 t i t 0 = t a ; t N = t b x 0 = x a ; x N = x b
For every timeslice t i (i = 0,..., N) choose x i. Connect points (in space-time) with straight lines K (b, a) Φ[x (t)] dx 1... dx N 1 limiting process: 0, but how? difficult problem, no general solution, but for most practical problmems exists solution, e.g. let L = m ẋ V (x, t) normalization is given by A N with A = (We will prove that in a minute.) πiε m 1/
therefore: K (b, a) = lim A ε 0 1 tb e i S[b,a] dx 1 A dx A dx 3 dx N 1 A A S [b, a] = t a L (x,ẋ, t) dt with piecewise linear trajectories Instead of the piecewise linear trajectories we could have taken the classical trajectories.
There are many more ways to define this integral general notation K(b, a) = b a e i S[b,a] Dx(t) path integral important rules: b c(t c,x c ) S[b, a] =S[b, c] +S[c, a] a K(b, a) = = x c c b a c e i S[b,c]+ i S[c,a] Dx(t) x c K(b, c)k(c, a)dx c
The wave function K(x, t ; x 1, t 1 ) amplitude at coming from 1 ψ(x, t) amplitude at (x, t) Only question of notation! If information (x 1, t) not important, then write ψ(x, t) = wave function. Especially we get: ψ(x, t )= K(x, t ; x 1, t 1 )ψ(x 1, t 1 )dx 1 From this equation we will derive the Schrödinger eqn.: ψ(x, t + ) = = dy A dy A exp exp i t+ t i x + y L dt L( x, x) ψ(y, t), x y ψ(y, t)
now consider L = mẋ ψ(x, t + ) = dy A = substitution y = x + η V (x, t) i m(x y) exp exp strongly oszillating, contributions only from neighborhood of x i V x + y, t ψ(y, t) ϕ(x, t + ) = dη A exp i mη exp i V x + η, t ψ(x + η, t) phase of the exponential function changes to first order for η Taylor expansion: 1. order in,. order in η m V x + η, t ψ(x, t)+ ψ t = V (x, t) + errors of higher order dη A exp i mη 1 i V (x, t) ψ(x, t)+η ψ x + η ψ x
order 0 1 = A = πi m dη da exp 1 i mη = 1 A πi m 1 two integrals: 1 A eimη / η dη =0 1 A eimη / η dη = i m ψ + ψ t = ψ i V ψ im ψ x i ψ t = m ψ + V (x, t)ψ x
One-dimensional Problems Harmonic Oscillator classical Hamiltonian: H = p m + mω x time independent Schrödinger eqn. (p = i ) m usually: solution with Hermite polynomials here: smart algebraic solution Define non-hermitian Operator a d dx + mω x ψ(x) =Eψ(x) a = x = ωmx +ip ωm, a = ωmx ip ωm ωm ωm (a + a ), p = i (a a ) where x 0 := ωm
a = 1 x x 0 + x 0 x, a = 1 x x 0 x 0 x Commutator relations: [a, a ] = 1, since [a, a ] = 1 [ωmx +ip, ωmx ip] ωm = 1 ( iωm)[x, p], ωm [x, p] =i = 1 So we have for the Hamilton operator: H = p m + mω x = ω 4 (a a ) + ω 4 (a + a ) = ω 4 (a + aa + a a + a a + aa + + a a a ) = ω (a a + aa )=ω(a a + 1 )
Problem is reduced to eigenvalue problem of the occupation number operator: ˆn = a a Let ψ ν be an eigenfunction for the eigenvalue ν: ˆnψ ν = νψ ν.thenwehave: ν(ψ ν, ψ ν )=(ψ ν,ˆnψ ν )=(ψ ν, a aψ ν )=(aψ ν, aψ ν ) 0 ν 0 Is there an eigenfunction for the eigenvalue ν =0? aψ 0 =0 (x 0 x + x x 0 )ψ =0 normalized solution: ψ 0 (x) = πx0 1 exp 1 x x 0
Calculation of the remaining eigenfunctions: a a, a = a aa a a a = a a, a = a a a, a = a aa aa a = a, a a = a thus: ˆn, a = a, [ˆn, a] = a consider: ˆn a ψ ν = a ˆn + a ψ ν = a (ˆn + 1) ψ ν = a (ν + 1) ψ ν =(ν + 1) a ψ ν = a ψ ν is eigenfunction with eigenvalue ν +1 normalization: a ψ ν, a ψ ν = ψν, aa ψ ν = ψν, a a +1 ψ ν =(ν + 1) (ψ ν, ψ ν ) > 0
ψ ν+1 = 1 (ν + 1) a ψ ν ψ n = 1 n a ψ n 1 = 1 n! a n ψ0 (nth excited state) Now the other way down: ˆnaψ ν =(aˆn a) ψ ν =(ν 1) aψ ν aψ ν is eigenfunction with eigenvalue ν 1 normalization: (aψ ν, aψ ν )= ψ ν, a aψ ν = ν (ψν, ψ ν ) ν =0: aψ 0 =0 (wealreadyhadthat) ν 1: ψ ν 1 = 1 ν aψ ν
There doesn t exist any more eigenfunctions. To prove this let ν = n + α, n N, 0 <α<1 ˆnψ ν =(n + α)ψ ν ˆn (a n ψ ν )=α (a n ψ ν ) ˆn a n+1 ψ ν =(α 1) <0 operators: a, a ladder operators a creation operator a annihilation operator a n+1 ψ ν Problem energy eigenvalues E = ω(n + 1 ), n = 0, 1,,... ˆn H ω ground state ψ 0 0 1st excited state ψ 1 1 3 ω nd excited state ψ 5 ω. nth excited state ψ n n (n + 1 ) ω...
ψ n = n! 1 πx 0 a n exp 1 x x 0 = n n! 1 πx 0 exp 1 x x H n x 0 x 0 Hermite polynomials H n (x) =e x ( a x 0 =1 = e x x d dx ) n e x n e x = e x e x x d dx n e x e x we have: thus: e x x d dx e x e x e x f = e x (x x)e x f d dx f x d dx x d dx H n (x) =( 1) n e x e x = d dx n e x =( 1) n d n dx n d n dx n e x
H 0 (x) =1 H 1 (x) =x H (x) =4x H 3 (x) =8x 3 1x H 4 (x) = 16x 4 48x + 1 H 5 (x) = 3x 5 160x 3 + 10x Exercises: Orthogonality: dx e x H n (x)h m (x) = π n n! δ mn Generating function: e t +tx 1 = n! tn H n (x) n=0 d Differential eqn.: dx x d dx +n H n (x) =0 Completeness: ψ n (x)ψ n (x )=δ(x x ) n=0
H n and thus ψ n has n nodes (simple real zeros).
Zero-Point energy lowest possible energy: classically 0 QM ω why? calculate uncertainty product x p x =(ψ n, xψ n )= ωm (ψ n,(a + a )ψ n ) = ωm (ψ n, nψ n 1 )+ ωm (ψ n, n +1ψ n+1 )=0 ( x) = x = ωm (ψ n,( a + a a = ωm (n + 1) = x 0 doesn t contribute n + 1 + aa 1+a a ; x 0 = n ωm + a )ψ n ) doesn t contribute quite analog p =0 ; ( p) = p = = x p = n + 1 x 0 n + 1 (minimal for ground state)
now the other way round: start with the uncertainty relation x p (proof at a later time) symmetry : p = x =0 p x 4 for the energy E = H = p m + 1 mω x p m + mω derivative with respect to p gives minimum 4 1 p 1 m mω 8 1 (p min ) =0 p min = mω E mω m + mω 8 mω = ω Zero-point energy is the lowest energy consistent with the uncertainty relation.
Coherent States consider eigenfunction of annihilation operator a: aϕ α = αϕ α (why will become clear later) (ψ n, ϕ α )= 1 n! a +n ψ 0, ϕ α = 1 n! (ψ 0, a n ϕ α )= αn n! (ψ 0, ϕ α ) ϕ α (x) = n=0 (ψ n, ϕ α ) ψ n = C n=0 α n a +n ψ 0 = C n! n! n=0 (αa + ) n n! ψ 0 Normalisation: 1=(ϕ α, ϕ α )=C = C n=0 α n n=0 (n!) = C e α α n n! (n!) ψn, n! ψ n = C = e α /
time evolution follows from time evolution of the stationary states ϕ α (x, t) =e α / = e α / n=0 α n n! ψn e ie nt/ n!, E n = ω αe iωt n ψ n e iωt/ n! n + 1 n=0 define α (t) =αe iωt ϕ α (x, t) =ϕ α(t) (x) e iωt/ position expectation value: x = φ α(t), xφ α(t) = x 0 φα(t),(a + a )φ α(t) = x 0 (α(t)+α (t)) write α = α e iδt x = x 0 α cos(ωt δ) exactly as the classical oszillator probability density (exercises) φ α (x, t) = 1 πx0 exp (x x 0 α cos(ωt δ)) x 0 non spreading Gaussian wave packet
In[8]:= x0 =.; delta =.; omega =.; alpha =.; phiabs@x_, t_d := 1 ê HSqrt@PiD x0l Exp@-Hx - x0 Sqrt@D Abs@alphaD Cos@omega t - deltadl ^ ê x0 ^ D In[33]:= x0 = 1; delta = 0; omega = 1; alpha = 1; In[6]:= Animate@Plot@phiabs@x, td, 8x, -5, 5<D, 8t, 0, 10 Pi<D t 0.5 0.4 Out[6]= 0.3 0. 0.1-4 - 4
Uncertainty relation Schwarz inequality: (ϕ, ψ) (ϕ, ϕ)(ψ, ψ) since i) ϕ =0= ii) ϕ = 0. Decompose ψ into a part parallel to ψ and part perpendicular to ψ ψ = αϕ + χ, (ϕ, χ) =0 = (ϕ, ψ) =α(ϕ, ϕ) (ψ, ψ) = (αϕ + χ, αϕ + χ) =α α (ϕ, ϕ)+(χ, χ) α α (ϕ, ϕ) = (ϕ,ψ) (ϕ,ϕ) = : χ =0 ψ = αϕ
General Uncertainty Relation Hermitian operators H 1, H,arbitrarystateψ define: Ĥ i = H i H i = H i (ψ, H i ψ) ( H i ) =(ψ,(h i H i ) ψ)=(ψ, Ĥi ψ) Schwarz inequality: (Ĥ1ψ, Ĥ1ψ)(Ĥψ, Ĥψ) (Ĥ1ψ, Ĥψ) = (ψ, Ĥ 1 ψ)(ψ, Ĥ ψ) (ψ, Ĥ1Ĥψ) = ( H 1 ) ( H ) (ψ, Ĥ 1 Ĥ ψ)
Ĥ 1 Ĥ = 1 Ĥ1Ĥ + 1 ĤĤ1 + 1 Ĥ1Ĥ 1 ĤĤ1 = 1 {Ĥ1, Ĥ} + 1 [Ĥ1, Ĥ] where {Â, ˆB} = AB + BA. 1 {Ĥ1, Ĥ} = 1 {Ĥ1, Ĥ} Hermitian, real eigenvalues = ( H 1 ) ( H ) 1 (ψ, {Ĥ1, Ĥ}ψ) + 1 (ψ,[ĥ1, Ĥ]ψ) 4 4 0 1 (ψ,[ĥ1, Ĥ]ψ) 4 H 1 H 1 [H 1, H ] general formulation of the Heisenberg uncertainty relation special case: H 1 = x i, H = p j = x i p j δ ij position-momentum uncertainty relation
Under what conditions does the product of the uncertainties attain a minimum? i) The equal sign in the Schwarz inequality holds when Ĥψ = αĥ1ψ ii) anticommutator: {Ĥ 1, Ĥ } =(ψ, Ĥ 1 Ĥ ψ)+(ψ, Ĥ Ĥ 1 ψ)=0 substitution of i) = (ψ, αĥ 1 ψ)+(αĥ1ψ, Ĥ1ψ) =α(ĥ1ψ, Ĥ1ψ)+α (Ĥ1ψ, Ĥ1ψ) =(α + α )(Ĥ 1 ψ, Ĥ 1 ψ)=0 = α = iλ, λ real Ĥ ψ = iλĥ 1 ψ For H 1 = x, H = p we have i x p ψ = iλ(x x)ψ = Gaussian wave packet
Common Eigenfunctions of Commuting Operators Hermitian operators A, B Let ψ n be eigenfunctions of A: Aψ n = a n ψ n arbitrary state: ψ = c n ψ n If [A, B] = 0 = A and B posses common system of eigenfunctions Proof: i) let ψ be a non-degenerate eigenfunction of A: Aψ = aψ A(Bψ) =BAψ = Baψ = a(bψ) now since ψ is non-degenerate, we have Bψ ψ, whichmeansbψ = bψ
ii) Let the eigenvalue a be m-fold degenerate Aψ j = aψ j, j = 1,..., m whereby (ψ j, ψ k )=δ jk it follows A(Bψ j )=BAψ j = Baψ j = a(bψ) so Bψ j eigenfunction of A m = Bψ j = C jk ψ k with C jk =(ψ k, Bψ j )=Ckj k=1 * Matrix C jk is Hermitian = unitary trafo U with U CY = C D U C = C D U diagonal matrix unitary U U = UU =1= CU = UC D
= kth column vector of U: U 1k. C U = U C D is eigenvector of C with eigenvalue C D k U mk multiplying * by Ujr BUjr ψ j = Ujr C jk ψ k = j j,k k C D r U kr ψ k define φ r = k Bφ r = C D r φ r U kr ψ k eigenvalue
now the other way round: Let a complete set of eigenfunctions ψ n, n = 1,,... of operators A and B with eigenvalues a n and b n given. Then [A, B] = 0. Proof: ψ = c n ψ n = [A, B]ψ = c n [A, B]ψ n = c n (ABψ n BAψ n )ψ n = c n (a n b n b n a n )=0 Measurement: state ψ = c n ψ n A: measure with probability c n eigenvalue a n after the measurement, state is now in eigenstate ψ n continue with measurement of observable B which gives b n = A and B can be measured simultanously
Now let [A, B] = 0: after the measurement of A and reduction on the state ψ n, the measurement of B will alter the state to φ k = ψ n = A and B can not be measured simultaneously A complete set of eigenfunctions of the operator A is called a basis of A. The set of Hermitian operators A, B,...,M is called a complete set of operators if these operators all commute with each other and if the common set of eigenfunctions is no longer degenerate. These eigenfunctions can then be characterized by the corresponding eigenvalues a, b,...,m : ψ a,b,...,m
Examples of commuting operators: x 1, x, x 3 ; p 1, p, p 3 ; x 1, p, p 3 but not x 1, p 1 Examples of complete sets of operators are: i) 1D potentials: x or p ii) 3 D potentials: x, y, z or p x, p y, p z iii) 3D spherical symmetric potentials: x, y, z or p x, p y, p z or H, L, L z angular momentum