Oppgave 1. Norges teknisk-naturvitenskapelige universitet NTNU Institutt for fysikk EKSAMEN I: MNFFY 245 INNFØRING I KVANTEMEKANIKK

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1 Norges teknisk-naturvitenskapelige universitet NTNU Institutt for fysikk EKSAMEN I: MNFFY 45 INNFØRING I KVANTEMEKANIKK DATO: Fredag 4 desember TID: 9 5 Antall vekttall: 4 Antall sider: 5 Tillatte hjelpemidler: Kalkulator, matematisk formelsamling Faglig kontakt under eksamen: Ingjald Øverbø, telefon Sensurdato: 4 januar English version on pages 3 5 Oppgave a Skriv ned Hamilton-operatoren H z for en partikkel med masse m som beveger seg i det éndimensjonale potensialet V z z = mω z Anta at systemet ved t = prepareres i begynnelsestilstanden Ψ z z, = C e mωz / Bestem C slik at Ψ z z, er normert, og finn helst uten regning forventningsverdien z ved t = b Vis at Ψ z z, er en egenfunksjon til H z og finn egenverdien Hva blir da bølgefunksjonen Ψ z z, t for t >? c Betrakt et nytt system, hvor partikkelen beveger seg i det éndimensjonale potensialet V x x = mω x, og anta at begynnelsestilstanden er Ψ x x, = C e mωx x / Forklar hvorfor vi kan bruke samme normeringskonstant som ovenfor, og argumentér for at x = x Vis så enkelt som du kan at forventningsverdien for impulsen p x ved t = er p x = Argumentér for at tilstanden Ψ x x, t er en ikke-stasjonær tilstand når x d Vis at usikkerhetene x og p x er uavhengige av x, og vis at x p x = Hva er da z og p z for tilstanden Ψ z z, i a? e Betrakt enda et nytt system, hvor partikkelen beveger seg i det éndimensjonale potensialet V y y = mω y, og anta at begynnelsestilstanden er Ψ y y, = C e mωy / e iymωx /

2 Forklar hvorfor vi kan bruke den samme konstanten C her også Argumentér for at y er lik null og for at y har samme verdi som x ovenfor Finn p y f Anta nå at partikkelen beveger seg i det tredimensjonale potensialet V x, y, z = V x x + V y y + V z z = mω r, og anta at dette systemet prepareres i begynnelsestilstanden Ψx, y, z, = Ψ x x, Ψ y y, Ψ z z,, hvor Ψ x x, osv er begynnelsestilstandene som ble introdusert ovenfor Det er mulig å finne bølgefunksjonen Ψx, y, z, t for t > for dette systemet, og en kan også vise at usikkerhetene i posisjoner og impulser er tidsuavhengige, men dette skal du ikke gjøre Bruk i stedet Ehrenfests teorem til å finne forventningsverdiene x t, y t og z t for t >, og vis at r t = ê x x t + ê y y t + ê z z t følger en sirkelbane Oppgitt: I n β x n e βx dx = / β n I β ; I β = π/β / ; d dt r = p m ; d p = V dt Oppgave a La F betegne en observabel, {f n } dens egenverdispektrum som antas å være diskret og ikke-degenerert, og n dens normerte egenvektorer Betrakt et fysisk system som er i en vilkårlig tilstand ψ Ved å bruke fullstendighetsrelasjonen kan vi utvikle tilstandsvektoren ψ i egenvektorsettet n til F : ψ = n n n ψ = n n ψ n Anta nå at det foretas en måling av observabelen F på dette systemet i Hva er de mulige målte verdiene for F? ii Hva er sannsynlighetsamplituden, og hva er sannsynligheten, for å måle verdien f n? iii Dersom måleresultatet er f 7, hva er da resultatet ved en ny måling av F umiddelbart etter den første målingen? I resten av denne oppgaven betrakter vi et spinn- -system, hvor vi bruker egentilstandene ẑ og ẑ som basis Dette leder til en matriserepresentasjon, hvor basis-tilstandene representeres av Pauli-spinorene χẑ = og χ ẑ = b Vis eksplisitt at χẑ og χ ẑ er egentilstander til S z = σ z, og bestem egenverdiene Finn lengde og retning av vektoren S = S x ê x + S y ê y + S z ê z for tilstanden χẑ

3 c Hva er generelt de mulige resultatene ved en måling av S x på spinn- -systemet? Hva er sannsynlighetene for å finne hvert av disse resultatene dersom systemet er i tilstanden χẑ før målingen? [Hint: Bruk resultatet for S x fra b] Hva kan du si om tilstanden til systemet umiddelbart etter målingen av S x? d Finn en normert løsning χ ˆx av egenverdiligningen S xχ ˆx = χ ˆx Hva er den fysiske tolkningen av skalar-produktet χ ˆx χ ẑ? Bruk dette til å verifisere resultatet for en av sannsynlighetene funnet i b e Hva er den generelle definisjonen av spinn-retningen for spinn- -systemet? [Hint: Sjekk hvordan definisjonen virker for tilfellet under b] Finn spinn-retningen for tilstanden χ = / + i/ Oppgitt: σ x =, σ y = i i, σ z = English version: Problem a Write down the Hamiltonian operator H z for a particle with mass m which moves in the one-dimensional potential V z z = mω z Assume that the system is at t = prepared in the initial state Ψ z z, = C e mωz / Determine C such that Ψ z z, is normalized, and find preferably without calculation the expectation value z at t = b Show that Ψ z z, is an eigenfunction of H z and find the eigenvalue What is then the wavefunction Ψ z z, t for t >? 3

4 c Consider a new system, where the particle moves in the one-dimensional potential V x x = mω x, and assume that the initial state is Ψ x x, = C e mωx x / Explain why we can use the same normalization constant as above, and argue that x = x Show in the simplest way you can find that the expectation value of the momentum p x at t = is p x = Argue that the state Ψ x x, t is a non-stationary state when x d Show that the uncertainties x and p x are independent of x, and show that x p x = What are then z and p z for the state Ψ z z, in a? e Consider yet another system, where the particle moves in the one-dimensional potential V y y = mω y, and assume that the initial state is Ψ y y, = C e mωy / e iymωx / Explain why we can use the same constant C here too Argue that y is equal to zero and that y has the same value as x above Find p y f Assume now that the particle moves in the three-dimensional potential V x, y, z = V x x + V y y + V z z = mω r, and assume that this system is prepared in the initial state Ψx, y, z, = Ψ x x, Ψ y y, Ψ z z,, where Ψ x x, etc are the initial states introduced above It is possible to find the wavefunction Ψx, y, z, t for t > for this system, and one can also show that the uncertainties in positions and momenta are time independent, but you are not supposed to do that Use, instead, Ehrenfest s theorem to find the expectation values x t, y t and z t for t >, and show that r t = ê x x t + ê y y t + ê z z t follows a circular orbit Given: I n β x n e βx dx = / β n I β ; I β = π/β / ; d dt r = p m ; d p = V dt 4

5 Problem a Let F denote an observable, {f n } its eigenvalue spectrum which is assumed to be discrete and non-degenerate, and n its normalized eigenvectors Consider a physical system which is in an arbitrary state ψ Using the completeness relation, we may expand the state vector ψ in terms of the eigenvectors n of F : ψ = n n n ψ = n n ψ n Suppose now that a measurement of the observable F is carried out on this system i What are the possible measured values of F? ii What is the probability amplitude, and what is the probability, of measuring the value f n? iii If the measured value is f 7, what is then the result of a new measurement of F immediately after the first measurement? In the remaining part of this problem, we consider a spin- system, where we use the eigenstates ẑ and ẑ as a basis This leads to a matrix representation, where the basis states are represented by the Pauli spinors χẑ = and χ ẑ = b Show explicitly that χẑ and χ ẑ are eigenstates of S z = σ z, and determine the eigenvalues Find the length and direction of the vector S = S x ê x + S y ê y + S z ê z for the state χẑ c What are in general the possible results of a measurement of S x on the spin- system? What are the probabilities of getting each of these results if the system is in the state χẑ before the measurement? [Hint: Use the result for S x from b] What can you say about the state of the system immediately after the measurement of S x? d Find a normalized solution χ ˆx of the eigenvalue equation S xχ ˆx = χ ˆx What is the physical interpretation of the scalar product χ ˆx χ ẑ? Use this to confirm the result for one of the probabilities found in b e What is the general definition of the spin direction for the spin- how your definition works for the case in b] Find the spin direction for the state / χ = + i/ system? [Hint: Check Given: σ x =, σ y = i i, σ z = 5

6 NTNU INSTITUTT FOR FYSIKK EKSAMEN I : MNFFY 45 Innføring i kvantemekanikk 4 des LØSNINGSFORSLAG Problem a The Hamiltonian operator is H z = p z m + V zz = m z + mω z The normalization integral is = Ψ z z, dz = C e mωz / dz = C π / mω/ / Choosing C real and positive, we find C = mω /4 π Since the probability density Ψ z z, = Ce mωz / is symmetric with respect to z =, the expectation value is z = z Ψ z z, dz = [Technically, this follows because the integrand is antisymmetric with respect to z = ] b Calculating Ψ z z, z we find that = Ψ z z, mω z and Ψ z z, = Ψ z z z, m ω z mω, H z Ψ z z, = [ ω m m z mω ] + mω z Ψ z z, = ω Ψ zz, Thus, Ψ z z, is an eigenfunction of H z with the eigenvalue ω, which is the ground-state energy For t >, this oscillator will continue to be in the ground state, Ψz, t = Ψ z z, e ie t/ = C e mωz / e iωt/ c The probability density Ψ x x, = Ce mωx x / has the same form as in a, apart from being shifted such that it is symmetric with respect to x = x It follows that we can use the same normalization constant as in a, and that The expectation value of p x is p x = Ψ x, i x = x x Ψ xx, dx = i Ψ x x, x Ψ xx, dx =

7 [The integral has to be equal to zero because the expectation value is real Technically, it follows from the facts that Ψ x x, is symmetric, while Ψ x x, / x is antisymmetric, with respect to x = x, such that the integrand is antisymmetric] Since the eigenfunctions of H x are all either symmetric or antisymmetric, whereas the initial state Ψ x x, is asymmetric, it follows that the wavefunction Ψx, t for the present system will be non-stationary except when x = d Changing integration variable, from x to u = x x, we have Thus, Furthermore, with Ψ x x, x we find x = x x = Ψ x x, x x Ψ x x, dx = C [ = Ψ x x, mω ] x x u e mωu / du = mω π / π / mω 3/ = x = mω and mω [ Ψ x x, m ω = Ψ x x x, x x mω ], p x = p x p x = p x [ m = Ψ ω x, x x mω ] dx = m ω x + mω = mω Thus, both x and p x = mω are independent of x For the uncertainty product we find x p x =, which is the minimal value allowed by the uncertainty relation Since the z-dependence of Ψ z z, = C e mωz / is the same as the x-dependence of Ψ x x, for x =, it follows from the results above that z = x = mω and p z = p x = mω e We note that the y-dependence of Ψ y y, = Ce mωy / is the same as the x- dependence of Ψ x x, for x = Hence, Ψ y y, is normalized, like Ψ x x, It also follows that y = and y = x = mω

8 For p y we find p y = = = Ψ y, i y Ψ yy, dy Ψ y, [ i Ψ yy, mω ] y + imωx / dy Ψ y y, imωy + mωx dy = imω y + mωx = mωx f From the expression x = Ψ x, y, z, x Ψx, y, z, dxdydz = Ψ x x, x Ψ x x, dx Ψ y y, dy Ψ z z, dz, where the last two integrals are normalization integrals, we understand that the expectation values x etc are equal to those found above Thus, the expectation values for t = are From Ehrenfest s theorem it follows that with the general solution x = x, p x =, y =, p y = mωx, z =, p z = d x t dt = d p x t dt m = V m x t = ω x t, x t = A cos ωt + B sin ωt, p x t = m d dt x t = mωa sin ωt + mωb cos ωt, and similarly for y and z Inserting the initial values, we then find that x = A and = mωb = x t = x cos ωt Similarly, replacing x by y and at last by z, we find and = A and mωx = mωb = y t = x sin ωt, = A and = mωb = z t = Thus, r t follows a circular orbit, with radius x Problem a i According to one of the basic postulates of quantum mechanics, the possible measured values of the observable F are given by the spectrum of eigenvalues: {f n } ii The probability amplitude of measuring the eigenvalue f n or to find the system in the state n is given by the n -component of ψ : A n = n ψ, 3

9 and the probability to get f n is P n = A n = n ψ iii According to another postulate, a measurement giving F = f 7 will in general change the state of the system, leaving it in the state 7 A new measurement immediately after this measurement will then return the value f 7 b We have S z χẑ = = χ ẑ and S z χ ẑ = = χ ẑ Thus, χẑ and χ ẑ are eigenstates of S z with eigenvalues + and, respectively For the state χẑ we find the expectation values S z = χ ẑ S zχẑ =, S x = χ ẑ χẑ =, S y = χ i ẑ χẑ = i Thus, the length of the vector S is, and its direction is along the z-axis c The possible results of a measurement of S x are + and, as for the measurement of any component of S Since the expectation value of S x for the state χẑ is zero, the probabilities of getting + and must be equal, and therefore equal to After the measurement of S x, the system will be left in the state χ ˆx if we measure S x =, and in the state χ ˆx if we measure S x = d With S x = σ x, the eigenvalue equation becomes a = σ x χ ˆx = = b Thus, the solution is χ ˆx = a a + b a b Choosing a = /, we get a normalized solution: χ ˆx = Physical interpretation: If the system is in the state χẑ and we measure S x, the the probability amplitude to get + and to leave the system in the state χ ˆx is χ ˆx χ ẑ, and the probability is χ ˆx χ ẑ Explicitly, we find for this probability: confirming the result in b χ ˆx χ ẑ = /, 4 =,

10 e The general definition of the spin direction for the spin- system is that it is given by the vector σ = S We note that σ = ê z for the state χẑ in a, as it should For the state χ, we find σ x = χ σ x χ = /, i/ σ y = χ σ y χ = = /, σ z = χ σ z χ = = / + i/ = /, Thus, the spin direction is σ = ê x + ê y We note that it lies in the xy-plane, at an angle of 45 degrees with the x-axis 5