Übungen zur Quantenmechanik (T2)

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1 Arnold Sommerfeld Center LudwigMaximiliansUniversität München Prof Dr Stefan Hofmann Wintersemester 08/9 Übungen zur Quantenmechanik (T) Übungsblatt, Besprechung vom 0 40 Aufgabe Impuls Zeigen Sie für x, a R und eine beliebige Wellenfunktion ψ(x), dass exp ( a x ) ψ(x) = ψ(x a) (ii) (iii) Nutzen Sie diese Eigenschaft und den Satz von Stone, um die Ortsdarstellung des Impulsoperators P zu nden Finden Sie auch die notwendigen Faktoren von, um dem Impuls-Erwartungswert die kanonische Dimension [ P ] = [Wirkung] [Länge] = kg m s zu geben Finden Sie die explizite Ortsraum-Darstellung von e p (x) = x p Betrachten Sie hierfür x P p Bestimmen Sie die Normierungskonstante, indem Sie p q = dx p x x q = δ(p q) fordern (iv) Betrachten Sie nun die drei euklidischen Ortsoperatoren Q a, a =,, 3 und zugehörigen Impulsoperatoren P b, b =,, 3 mit der oben bestimmten Darstellung Zeigen Sie die Kommutatorrelationen [Q a, P b ] = i δ a b, [Q a, (P b ) n ] = in δ a b P n b, [P a, (Q b ) n ] = in δ a b (Q b ) n Was können Sie mithilfe der ersten Relation über die Ortsdarstellung des Impulsoperators folgern, wenn Sie diese nur aus den Kommutatorrelationen bestimmen möchten? Wie sieht die allgemeinst mögliche Form aus und welche Konsequenz hat diese für mögliche Messwerte? (v) Bestimmen Sie die Fourier-Transformation der Funktionen f (x) = exp ( ax + bx + c ), { für a < x < a, f (x) = 0 sonst Solution We expand the expression exp ( a x ) ψ(x) = n=0 where we used Taylor's theorem in the last step n! ( a)n n x ψ(x) = ψ(x a),

2 (ii) Since the norm of ψ is translationally-invariant (cf last sheet), this denes a unitary operator We reparametrise it as ( U (a) = exp ia P ) Taking now the derivative with respect to a and using Stone's theorem, we nd that the operator P dened on the dense subset C 0 (R) L (R), dened as can be extended to a self-adjoint operator P = i x, (iii) We can compute e p (x) := x p with this representation Consider x P p = i e p(x) = pe p (x), so we conclude e p (x) = N p e i px We compute p q = dx Np N q e ipx iqx e = N p N q π δ(p q), so we x N p = N q = π (iv) We rst compute [Q, P] ψ(x) = [x, i x ] ψ(x) = i (xψ (x) ψ(x) xψ (x)) = i ψ(x) This generalises straightforwardly to 3 dimensions, since b x a = δb a, the momenta commute with position operators they are not conjugate to We show the identities by induction The case n = follows from the CCR Suppose now for n the relation [Q a, (P b ) n ] = in δ a b P n b holds Then for n n + we have [ Q a, P n+ ] b = Pb [Q a, P n b ] + [Q a, P b ] P n b = P b in δb a P n b +i δb a P n b = i(n + ) δb a P n b Similarly, we nd for the second relation [ P a, (Q b ) n+] [ = Q b P a, (Q b ) n] [ + P a, Q b] (Q b ) n = Q b ( in δ b a(q b ) n ) i δ b a(q b ) n = i(n + )δ b a(q b ) n We see that the representation satises the canonical commutation relations, but so does any other representation P P +f(x) We are saved by the Stone-von Neumann theorem, which guarantees that all these representations are unitarily equivalent, so amplitudes are not aected In a handwavy way, we can see this for representations given by P P + x f(x), then the unitary transformation relating the dierent wave functions is given by ψ(x) e i f(x) ψ(x) (v) The Fourier transform of the rst function can be computed by completing the square in the exponent We have ax + bx ikx = a(x a (b ik)) + 4a (b ik) Now we can compute ˆf (k) = dx π e ax +bx+c e ikx = e c e 4a (b ik) dx π e a(x a (b ik)) ) = a e k 4a ikb a + b 4a +c

3 The second transformation is computed by ˆf (x) = dx π θ(a x )e ikx = ik = π a e ikx π sin(ka) k a

4 Aufgabe Unschärfe (ii) Die Ortsdarstellung eines Zustands α ist gegeben durch x α = exp ( x (πσ ) /4 σ + ikx Bestätigen Sie für diesen Zustand die Heisenbergsche Unschärferelation für den Orts- und Impulsoperator Was zeichnet das Gauÿ'sche Wellenpaket aus? Zeigen Sie die verallgemeinerte Unbestimmtheitsrelation für zwei Observablen A, B und einen beliebigen Zustand Φ Streu Φ (A )Streu Φ (B) Φ [A, B] Φ ) (iii) Bestimmen Sie die normierten Spin Zustände ψ, für die das Unschärfe-Produkt Streu ψ (S x )Streu ψ (S y ) minimal bzw maximal wird Überprüfen Sie zudem explizit, dass für diese Zustände die Unschärferelation für S x und S y nicht verletzt ist Solution We rst compute the expectation value of Q and Q for the state α We nd x α Q α = dx xe σ = 0 σ π as well as α Q α = dx = dx σ π x e x σ σ π = σ d ( ) da a a= ( σ d da e a x a= ) σ = σ For the expectation values of the momentum, there are two possibilities Either we use the derivative, or we compute the Fourier transformation We can use the result from Ex, that the transformed function is again a Gaussian of the form ( ) ( ) σ 4 p α = exp σ (p k) π Analogous computation to above gives the expectation values σ σ α P α = dp p π e (p k) = k, and α P α σ = dp p σ π e (p k) σ = dp (p + pk + k σ ) π e = σ + k (p k)

5 We compute the uncertainty and nd ( Q) = σ (ii) and and the relation ( P) = σ + k k = σ ( Q) ( P) = σ σ = 4, or, equivalently, Q P =, which is minimal Recall the denitions from PS9, A = A A We will derive an inequality for the variance Consider Φ ( A) Φ = (A A )Φ (A A )Φ = a a, and Φ ( B) Φ = (B B )Φ (B B )Φ = b b Then, by Cauchy-Schwarz, we have Φ ( A) Φ Φ ( B) Φ = a a b b a b Splitting this into real and imaginary part, we nd We now compute ( ) ( ) a b a b + b a a b b a = + i a b = Φ (A A )(B B ) Φ = AB A B With this, we nd the real and imaginary part to be and We nd a b + b a = AB + BA A B = [A, B] + A B, a b = a b b a = AB BA = [A, B] ( ) ( ) [A, B] + A B + [A, B] i Neglecting the rst term and taking the square root, we nd ( A)( B) [A, B] As it stands, this proof only holds for nite-dimensional operators This is due to our ignorance about domains, in particular about the domains of products AB For the innite-dimensional case, this requires more care (iii) First, we parametrise a state α = cos( θ ) + + sin( θ )eiϕ The expectation values are readily computed to be S z = cos(θ), S x = sin(θ) cos(ϕ), S y = sin(θ) sin(ϕ) With this, the variance is given by ( S x ) = ( sin (θ) cos (ϕ) ), 4 and ( S y ) = 4 ( sin (θ) sin (ϕ) ), and the uncertainty is ( S x ) ( S y ) = 4 ( sin (θ) sin 4 (θ) cos (ϕ) sin (ϕ) )

6 We want to nd values of θ, ϕ where this is extremal We compute the derivative with respect to the angles and nd the two equations sin(θ) cos(θ) + 4 sin 3 (θ) cos(θ) cos (ϕ) sin (ϕ) = 0 cos(ϕ) sin 3 (ϕ) cos 3 (ϕ) sin(ϕ) = 0 From the second equation, we nd two solutions The rst one is ϕ = n π for an integer n, and this gives for the rst equation 0 = sin(θ) cos(θ) = sin(θ), which is solved by θ = nπ, ie θ = n π for integer n Half of these solutions, namely θ = nπ, correspond to S z eigenstates, and the phase ϕ is irrelevant The other half, where θ = n+ π, correspond to S x and S y eigenstates, depending on ϕ If ϕ = nπ, they are S x eigenstates, and for ϕ = n+ π, they belong to S y The second solution of the second equation is ϕ = π 4 + n π, however, with this ϕ, there is no solution to the rst equation Fixing now ϕ = n π, we compute the second derivative to check for maxima and minima We obtain d dθ ( S x) ( S y ) = sin (θ) cos (θ) We see that the S z eigenstates have negative second derivative, they are maxima, and the S x and S y eigenstates have positive second derivatives, they are minima To conclude, we nd ϕ = n π, { nπ maximal uncertainty, S z eigenstates, θ = n+ π minimal uncertainty, S x and S y eigenstates Next, we compute explicitly the uncertainty for these states Consider rst the S z eigenstates, given by θ = nπ We nd Thus the bound is S x S y 4, and we nd ( S x ) = ( S y ) = 4, S z = ( ) n ( S x ) ( S y ) = 4 For the eigenstates, we have S z = 0, and either S x = 0 or S y = 0 Thus we have 0 as lower bound and reach it

7 Aufgabe 3 Spin-Algebra Teil II Diese Aufgabe ist der zweite Teil zu Aufgabe 3 auf dem letzten Blatt Stellen Sie sicher, dass Sie die Analyse der Drehimpulsalgebra verstanden haben, denn diese Aufgabe wird eine analoge Analyse für Tensorprodukte durchführen Wir betrachten nun die Zustände s, m deniert wie in Aufgabe 3 auf dem letzten Blatt, und wir denieren sie so, dass sie ein Orthonormalsystem bilden Auf dem letzten Blatt haben wir gezeigt, dass S s, m = N m s, m Bestimmen Sie N m Zeigen Sie, dass S s, m = N m s, m, S + s, m = N m+ s, m +, also dass wir die gleiche Normierungskonstante für beide Operatoren benutzen können Hinweis: Starten Sie mit s, s und nden Sie eine Rekursionsrelation für N m (ii) Welche Dimension hat die Darstellung s =? Welche die Darstellung s =? (iii) Bestimmen Sie die Matrix-Darstellung der Operatoren S a in der Basis der s, m, gegeben durch (S s a) m,m = s, m S a s, m, für s = und s = Überprüfen Sie an einem Beispiel, dass die Algebra erfüllt ist Hinweis: Nutzen Sie die Algebra und bestimmen Sie S x, S y aus S + und S Nun werden wir das Tensorprodukt eines s = Systems mit einem s = System betrachten, und dieses in eine direkte Summe von Darstellungen mit bestimmtem s zerlegen Im Folgenden bezeichnen wir diese Systeme durch bzw, dh ist die Spin- Darstellung von SU() Unser Vorgehen wird analog zu Aufgabe 3 auf dem letzten Blatt sein (iv) Bestimmen Sie die Dimension des Tensorproduktes Ein Zustand s, m ; s, m des Produktsystems transformiert als D (U) s, m ; s, m = D (U) s, m D (U) s, m, wobei D s (U) die Spin-s Darstellung des Elements U SU() ist, dh bezüglich der Matrixdarstellung der Generatoren wie in (iii) (v) Zeigen Sie, dass für innitesimale Transformationen, dh die Drehimpulse addiert werden, dh D s (U) = s + α a S a,s, S a,s s = S a,s s + s S a,s Dies bedeutet, dass für Produktzustände m einfach addiert wird (vi) Überzeugen Sie sich damit, dass der Zustand mit maximalem m in gegeben ist durch,, = 3, 3 (vii) Finden Sie die Produktdarstellung der anderen Zustände 3,, 3, sowie 3, 3 durch Anwenden von S, Spannen diese Zustände auf? Hinweis: Nutzen Sie das Resultat aus (v) um Leiteroperatoren auf dem Tensorprodukt aus den Leiteroperatoren der Faktorenräume zu konstruieren (viii) Überlegen Sie sich aus der Drehimpulsalgebra und durch Betrachten der Dimension, in welcher s- Darstellung die fehlenden Zustände liegen Nutzen Sie dann m = m + m für Produktzustände, um diese Zustände aus Kombinationen von Produktzuständen von zu konstruieren, welche orthogonal zu den 3 -Zuständen sind Dieses Vorgehen funktioniert für beliebige Tensordarstellungen von SU() Wir nden den Zustand mit maximalem m, und zerlegen die Produktdarstellung dann in Darstellungen mit bestimmtem s

8 (ix) Überzeugen Sie sich, dass für beliebige Darstellungen s und s gilt s s = s +s s= s s wobei die direkte Summe ist Im obigen Beispiel entspricht dies = 3 Sie müssen hierfür nicht die explizite Darstellung der Zustände angeben, sondern lediglich das allgemeine Vorgehen für jedes s und die Zustände jeder Darstellung richtig zählen s, Solution We start with the heighest weight state s, s Consider the action s, s S + S s, s = s, s [S +, S ] s, s = s, s S z s, s = s We see that N s = s Note that similarly for S +, we can nd S + s, s = S + N s S s, s = N s [S +, S ] s, s = N s s, s We can now nd a recursive relation for all other N m, with the convention Consider now S s, s = N s s, s, S + s, s = N s s, s N s k = s, s k S + S s, s k, = s, s k [S +, S ] s, s k + s, s k S S + s, s k = (s k) + N s k+ Taking now k = s m and using the recursion relations we nd N m = (s + m)(s m + ) N s = s N s N s = s, N s k N s k+ = s k, (ii) The dimension of the representations is s+ (cf last sheet) So the s = representation has dimension, and s = has dimension 3 (iii) We x the basis, ( =, 0) as well as ( 0, =, ) 0 0, = 0,, 0 =,, = We can obtain S x and S y from the ladder operators We have S x = (S + + S ), S y = i (S + S )

9 To nd the explicit form, we use the algebra, namely S + s, m = N m+ s, m +, S s, m = N m s, m, S z s, m = m s, m The corresponding matrix representations are thus given by ( ) ( ) S + =, S 0 0 =, 0 and S + = , S = With this, we now conclude for s = S z = ( ) 0, S 0 x = ( ) 0, S 0 y = ( ) 0 i i 0 and for s = S z = , S x = 0 0 0, S y = 0 i 0 i 0 i i 0 (iv) The product is of dimension 3 = 6 (v) (vi) (vii) For an innitesimal transformation, we expand both sides and nd + α a S a, = ( + α ) ( ) as a, + α a S a, = + α a S a, + α a S a, + O(α ), so indeed the m-values simply add Since we can add the eigenvalues of S z, the maximal one is given by m = + = 3 The only state available with this m is,, From the consideration of the innitesimal generators above, we can also construct ladder operators as S + = S +, + S +,, S = S, + S, We use the same normalisation as above, and apply the operator S to the state 3 to obtain 3, = 3,, + 3,, 0, 3, = 3,, 0 + 3,,, 3, 3 =,,, 3 =,,, (viii) The above states form only the four-dimensional representation s = 3 We are lacking two dimensions, which corresponds to a s = representation From the angular momentum algebra, we also know that there is a s = representation (in general, for each s there exists a s representation) Hence we need to nd states with m = and s =, which are orthogonal to the four above Good candidates are the states corresponding to m = above with changed sign We nd, = 3,, 3,, 0, = 3,, 0 3,,,

10 (ix) which are indeed related by S ± We found the six states spanning the Hilbert space and see that the tensor representation decomposes as = 3 For a general product, the procedure is analogous to the previous discussion First, we compute the maximal m value by adding the maximal eigenvalues of all factors Next, we construct the corresponding spin-m-representation by acting with S on the maximal state Once this process terminates, we consider the next-lower spin-m -representation and do the same analysis We perform this procedure until we have obtained s s orthogonal vectors The claim, based on our example above, is that this continues until the lowest spin s s We show this inductively The claim is (s + )(s + ) = As base step, consider s = s = We have s +s l= s s = 3 + (l + ) Now suppose s s We rst perform induction in s, ie s s + We have ((s + ) + ) (s + ) = (s + )(s + ) + s + = s +s l=s s (l + ) + s + Note that the sum contains s + terms, so we use the s + to add one to each term, resulting in s +s l=s s (l + ) + s + = s +s l=s s ((l + ) + ) = (s + )+s l=(s + ) s (l + ) Induction in s works analogous, we have to watch out for the case where s + > s, this is where the absolute value comes in We now know that the dimensions match, and we know how to nd the complete set of states in each spin-l-representation, so we conclude that indeed s s = s +s s= s s s

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