Ph125a Homework Assignment #1 with HM s solutions Due 5:00pm on Tuesday, October 3, in the box outside 24 Bridge Annex

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1 Ph5a Homework Assignment # with HM s solutions Due 5:pm on Tuesday October in the box outside 4 Bridge Annex Remember that you must perform all calculations by hand and show your work for full credit (see the course web page for details on the -point grading system) Each student must prepare his/her own writeup based on his/her own understanding of the problems Solution of coupled linear ODE s Consider the set of coupled linear differential equations ẋ Ax where x R A (a) Find the general solution xt in terms of x by matrix exponentiation Answer: First we compute the eigenvalues of A : deta I det Since A is symmetric we know that all of its eigenvalues are real By inspection we can see that and are eigenvalues and so we can solveforthethird: x x x x x Hencewefindx so that eigenvalue is repeated Solving for the eigenvectors using x x x x Ax x x x x x x x x we find for x x x x x x x x x which is satisfied as long as x x x Hence we have two linearly independent solutions

2 v v where with a little foresight we have chosen an orthogonal pair of eigenvectors Likewise for x x x x x x x x x x x x which gives us x x x x x v Hence the basis transformation matrix is 6 P 6 6 Since A is symmetric we should have P P T Just to check P T P Hence A and

3 expat 6 6 e t e t e t 6 6 e t 6 e t 6 e t 6 e t e t 6 et et e t e t et e t et e t et e t et e t et e t et e t et e t et e t et Hence finally e t et e t et e t et xt e t et e t et e t et e t et e t et e t et x (b) Using results from part (a) write the general solution xt by expanding x in eigenvectors of A That is write xt e t c v e t c v e t c v where i v i are eigenvalue-eigenvector pairs for A and the c i are coefficients written in terms of the components of x Answer: Suppose x a b c Then using P we have

4 c c c a b c so xt e t 6 a 6 b 6 c v e t b c v e t a b c Modeling dice: observables and expectation values Suppose we have a pair of six-sided dice If we roll them we get a pair of results a 456 b 456 where a is an observable corresponding to the number of spots on the top face of the first die and b is an observable corresponding to the number of spots on the top face of the second die If the dice are fair then the probabilities for the roll are Pra Pra Pra Pra 4 Pra 5 Pra 6 /6 Prb Prb Prb Prb 4 Prb 5 Prb 6 /6 Thus the expectation values of a and b are 6 a i Pra i i 6 b i Prb i i Let us define two new observables in terms of a and b : s a b p ab 7/ 7/ Note that the possible values of s range from to and the possible values of p range from to 6 Perform an explicit computation of the expectation values of s and p by writing out and s i Prs i i 6 p i Prp i i Do this by explicitly computing all the probabilities Prs i and Prp i You should find that s a b and p a b Why are these results not surprising? Answer: The minimum possible value of s is and its maximum possible 4

5 value is By brute force we then have s Pra Prb Pra Prb Pra Prb 4Pra Prb Pra Prb Pra Prb 5Pra Prb 4 Pra Prb Pra Prb Pra 4Prb Pra Prb 5 Pra Prb 4 Pra Prb Pra 4Prb Pra 5Prb Pra Prb 6 Pra Prb 5 Pra 4Prb Pra Prb 4 Pra 5Prb Pra 6Prb Pra Prb 6 Pra Prb 5 Pra 4Prb 4 Pra 5Prb Pra 6Prb 9Pra Prb 6 Pra 4Prb 5 Pra 5Prb 4 Pra Prb 6 Pra 4Prb 6 Pra 5Prb 5 Pra 6Prb 4 Pra 5Prb 6 Pra 6Prb 5 Pra 6Prb Since a b 7/ we have s a b This has to be the case since expectation is linear Likewise for p the minimum value is and the maximum is 6 In table form and thus p

6 Since a b 49/4 5 we have ab a b This reflects the fact that a and b are statistically independent (uncorrelated) Conditional probabilities for dice You should work Problem before this one You should be able to intuit the correct answers for this problem by straightforward probabilistic reasoning; in class next week we will show how they are consistent with Bayes Rule Pry x Prx Prx y Pry Here Prx y represents "the probability of x given y" where x and y should be propositions or equations (for example Pra s 8 is the probability that a given that s 8) (a) Suppose your friend rolls the pair of dice and without showing you the result tells you that s 8 What is your conditioned probability distribution for a? Answer: If s 8 then the only possible die combinations are Hence the probability distribution for a is /5/5/5/5/5 Via Bayes Rule Prs 8 a ipra i Pra i s 8 Prs 8 The forward conditional probabilities are Prs 8 a Prs 8 a i Prs 8 a i 6 5 Prs 8 a Prb 6 6 Prs 8 a Prb 5 6 Prs 8 a 4 Prb 4 6 Prs 8 a 5 Prb 6 Prs 8 a 6 Prb 6 (b) Suppose your friend rolls the pair of dice and without showing you the result tells you that p What is your conditioned expectation value for s? Answer: If p then the possible combinations ab are 6446 Hence the conditioned probability distribution for s is Prs 7 Prs 8 and the conditioned expectation value is 5/ 4 Matrix observables for classical probability 6

7 Suppose we have a biased coin which has probability p h of landing heads-up and probability p t of landing tails-up Say we flip the biased coin but do not look at the result Just for fun let us represent this preparation procedure by a "classical state vector" x p h p t (a) Define an observable (random variable) r that takes value ifthecoinis heads-up and if the coin is tails-up Find a matrix R such that x T Rx r where r denotes the mean or expectation value of our observable Answer: R x T Rx p h p t p h p t p h p t p h p h p t (b) Now find a matrix F such that the "dynamics" corresponding to turning the coin over (after having flipped it but still without looking at the result) is represented by and x Fx r x T F T RFx Does U F T RF make sense as an observable? If so explain what values it takes for a coin-flip result of heads or tails What about RF and F T R? Answer: p t F Fx p t p h F T RF x T F T RFx p h p t Obviously F T RF is just the observable that takes value for heads and for tails On the other hand 7

8 RF x T RFx p h p t p h p t p h p t p t p h p t p h p t which vanishes for any coin! So it can only be interpreted as a trivial (constant with value zero) observable Likewise F T R p h x T RFx p h p t p h p t p h p t p t p h p t p h p t which also vanishes (c) Let us now define the algebra of flipped-coin observables to be the set V of all matrices of the form p h v ar br ab R Show that this set is closed under matrix multiplication and that it is commutative In other words for any v v V show that v v V v v v v Is U in this set? How should we interpret the observable represented by an arbitrary element v V? Answer: ar br a a b b a b b a By varying a and b we can thus generate any matrix of the form v c d 8

9 where c and d are arbitrary real numbers Likewise all elements of V are of this form Hence v v c d c d c c d d which is still in V and clearly v v v v We get U by setting ab and in general we interpret v as the observable that takes value c for heads and d for tails 9

10 Ph5a HW # with HM s solutions Due 5:pm on Tuesday October in the box outside 4 Bridge Annex Remember that you must perform all calculations by hand and show your work for full credit (see the course web page for details on the -point grading system) Each student must prepare his/her own writeup based on his/her own understanding of the problems Spectral decomposition practice Find the spectral decomposition of the matrix First we find the eigenvalues: A i i deta I det i i so we are looking for roots of the equation so we have and For we have to solve i i a b c which clearly has solution T and then also b ic ib c which implies b ic Hence if we choose a i are two orthonormal eigenvectors Then for we have

11 i i a b c for which we will again choose a and solve b ic which implies b ic Then ib c We have our projection operators i P i i i i P i i i i and spectral decomposition A i i i i which is easily seen to be valid Just for fun we can also verify P P i i i i More on projection operators The basic definition of a projection operator is that it must satisfy P P If P furthermore satisfies P P

12 we say that P is an orthogonal projector As discussed in class the eigenvalues of an orthogonal projector are all equal to either zero or one (a) Show that if P is a projection operator then so is P Simply P P P P P (b) Show that for any orthogonal projector P and any normalized state P Since P is Hermitian we can decompose any state vector a as where Then a c c P P a P a c and if a is normalized then c (c) Show that the singular values of an orthogonal projector are also equal to zero or one The singular values of an arbitrary matrix A are given by the square-roots of the eigenvalues of A A It follows that for every singular value i of a matrix A there exists some unit normalized vector u i such that u i A Au i i Conclude that the action of an orthgonal projection operator never lengthens a vector (increases its norm) We have P P P P so the singular values of P are either zero or one Since u i A Au i i is the square of the norm of the vector obtained by letting A act on u i it follows that the action of an orthogonal projection operator never lengthens a vector For the next two parts we consider the example of a non-orthogonal projection operator mentioned in the class notes N (d) Find the eigenvalues and eigenvectors of N Does the usual spectral decomposition work as a representation of N? First we find its eigenvalues which are the roots of the equation so we find and The corresponding eigenvectors are We note that these are not orthogonal and indeed if we try

13 P we find P i P i i N (e) Find the singular values of N CanyouinterpretthisintermsoftheactionofN on vectors in R? We compute N N The eigenvalues of this matrix satisfy so its eigenvalues are and The singular values of N are thus and meaning that there exists a unit vector that gets lengthened by the action of N Indeed if we consider the (normalized) eigenvector of N N with eigenvalue we find Nu The norm of this vector is now u u N Nu The action of N clearly projects vectors onto the linear span of T However it does not project orthogonally along the T direction Rather it projects along the direction T as can be seen from the fact that vectors parallel to that direction are in the nullspace of N It follows that some vectors will have long shadows when projected this oblique fashion 4

14 The extreme example is the vector u written above which is in fact perpendicular to the direction of projection Hence u the unit vector along T and Nu can be arranged as a right isoceles triangle Operator moments and uncertainty Consider an observable O a for a finite-dimensional quantum system with spectral decomposition O a i P i i (a) Show that the exponential operator E a expo a has spectral decomposition E a e i Pi i Do this by inserting the spectral decomposition for O a into the power series expansion of the exponential We start with the definition expo a O a O a! O a and insert the spectral decomposition O a i P i i This leads to expo a i P i i i P i i P i i i i i n e i Pi i i n n! P i i P i! i i P i! i i P i i P i (b) Prove that for any state a such that ΔO a we automatically have ΔE a In order to have ΔO a it must be the case that P i a a for some eigenspace projector Then E a a e i a E a a e i a ΔE a 4 Uncertainty and dynamics 5

15 Consider the observable and the initial state O x a (a) Compute the uncertainty ΔO x with respect to the initial state a By definition so we start by computing Next O x O x ΔO x O x O x so obviously O x Then finally ΔO x for the initial state a (b) Now let the state evolve according to the Schrödinger Equation with Hamiltonian operator H i i Compute the uncertainty ΔO x as a function of t We have d dt a t i H a t a t exp iht/ a We begin by diagonalizing the Hamiltonian (divided by ): H i i i i H i i i i 6

16 Thus exp iht/ n it n n! H n n it n n! i i n i i i i e it i e it i e it eit ie it ieit ieit ie it e it eit cost sint sint cost so a t cost sint and at time t O x cost sint cost sint cost sint sint cost sintcost sint O x ΔO x sin t cost (c) Repeat part (b) but replace O x with the observable O z That is compute the uncertainty ΔO z as a function of t assuming evolution according to the Schrödinger Equation with the Hamiltonian specified above Noting O z is again the identity we compute at time t 7

17 O z cost sint cost sint cost sint cost sint cos t sin t cost O z ΔO z cos t sint (d) Show that your answers to parts (b) and (c) always respect the Heisenberg Uncertainty Relation ΔO x ΔO z O x O z The right-hand side of this inequality is one-half the absolute magnitude of the expectation value of the commutator of O x and O z Are there any times t at which the Heisenberg Uncertainty Relation is satisfied with equality? We have O x O z so at time t O x O z cost sint cost sint sint cost cost sint Hence with ΔO x ΔO z sint cost we see that the relation is always satisfied The product of the uncertainties actually reaches zero whenever either ΔO x or ΔO z vanishes ie t n 4 n 8

18 Ph5a HW # with HM s solutions Due 5:pm on Tuesday 7 October in the box outside 4 Bridge Annex Remember that you must perform all calculations by hand and show your work for full credit (see the course web page for details on the -point grading system) Each student must prepare his/her own writeup based on his/her own understanding of the problems Density operators (a) For the Stern-Gerlach example discussed in class find a valid density operator for the atomic spin state such that S x S y S z (remember that for a state represented by a density operator O q Tr O q ) Your density operator should be a Hermitian matrix with trace equal to one and eigenvalues Prove that the you find does not correspond to a pure state and therefore cannot be represented by a state vector We recall S x S y i i S z and note that each of these matrices already has trace zero Hence clealry satisfies the requirements to be a valid density operator (trace one eigenvalues both equal to / Hermitian) and achieves Tr S xyz TrS xyz Hence with this state S x S y S z Wefind 4 and Tr / which proves that our is a mixed state (b) Suppose that we perform a measurement of the projection operator P i and obtain a positive result Use the projection postulate for pure states P i i P i to show that in density operator notation maps to

19 i P i P i TrP i Both and here represent the same initial state so the final states are also the same regardless of whether we work in state vector or density matrix representation Thus we can simply take the final state according to the projection postulate for state vectors and form the corresponding density operator: i i i P i P i P i P i P i P i P i P i P i P i There are many ways to show the final step that P i TrP i An easy way is to use the fact that since it represents a pure state: TrP i Tr P i TrP i Tr P i P i Tr P i Tr P i One could also compute the trace in terms of a complete basis k : TrP i k P i k k k k k P i k P i k k P i k k k P i Entanglement and the purity of a reduced density operator Let H A and H B be a pair of two-dimensional Hilbert spaces with given orthonormal bases A A and B B Let AB be the state AB cos A B sin A B For this is an entangled state The purity of A Tr B AB AB

20 Tr A is a good measure of the entangledness of states in H AB For pure states of the above form find the extrema of with respect to ( ) Do entangled states have large or small? We first form the density operator corresponding to AB AB AB cos cossin cossin sin Here stands for A A and so forth Now we take the partial trace over A to find the reduced density operator A B B B B cos A A sin A A and then A cos 4 A A sin 4 A A so Tr A cos 4 sin 4 The extrema correspond to d d 4cos sin 4sin cos cos sin sin cos cos sin For this equation is satisfied for /4 where cos sin / and also / At /4 AB A B A B which we recognize as a highly entangled state Since for we have AB A B which is unentangled and we see clearly that entangled states are associated with smaller purity for the reduced density operator The controlled-not operator Again let H A and H B be a pair of two-dimensional Hilbert spaces with given orthonormal bases A A and B B Consider the controlled-not operator on H AB U AB P A B P A B x Here P A A A P A A A and B x B B B B Write a matrix representation for U AB with respect to the following (ordered) basis for H AB : A B A B A B A B

21 Find the eigenvectors of U AB you should be able to do this by inspection Do any of them correspond to entangled states? We start by writing out U AB P A B P A x B Hence in the basis order given A A B B A A B B A A B B A A B B A B A B A B A B A B A B A B A B U AB We see immediately that A B A B are unentangled eigenstates of U AB Then from prior experience we can also guess A B A B A B B A B A B A B B Neither of these are entangled either 4 Creating entanglement via unitary evolution Working with the same system as in Problems and find a factorizable input state in AB A B 4

22 such that the output state out AB U AB in AB is maximally entangled That is find any factorizable in AB where A Tr B out AB out AB A general factorizable input state has the form in AB such that Tr A a A a A b B b B a b A B a b A B a b A b B a b A B where a a b b If we apply U AB to this state we get out AB U AB in AB a b A B a b A B a b A b B a b A B The simplest ways to get a maximally entangled output state are to set a a and either b or b In the former case we have out AB while in the latter we get out AB A B A B A B A B In either case it is straightforward to verify that Tr A 5

23 Ph5a HW #4 with HM s solutions Due 5:pm on Tuesday 4 October in the box outside 4 Bridge Annex Remember that you must perform all calculations by hand and show your work for full credit (see the course web page for details on the -point grading system) Each student must prepare his/her own writeup based on his/her own understanding of the problems Tensor-product bases Let H A and H B be two-dimensional Hilbert spaces with given bases a a and b b Consider the following entangled state in the joint Hilbert space H AB H A H B AB a b a b where a b is short-hand notation for a b and so on Rewrite this state in terms of a new basis a b a b a b a b where a a cos a sin a sin a cos a and similarly for b b Again a b a b et cetera Is our particular choice of AB special in some way? We first note that so a cos a sin a a sin a cos a a b cos a sin a sin b cos b sin cos a b cos a b sin a b sin cos a b a b sin a cos a cos b sin b and sin cos a b sin a b cos a b sin cos a b AB a b a b a b a b From the calculations we see that it is unusal for a state to have the same

24 expansion coefficients in the old and new bases For example the T coefficients of a b go from to sin cos cos sin sin cos T Matrix representations Let H A and H B be two-dimensional Hilbert spaces with given bases a a and b b Let a b a b et cetera Let the natural tensor product basis kets for the joint state space H ab be represented by column vectors as follows: a b For parts (a-c) let a b a b ab 8 a a b b b b a b 5 8 a a b b b b (a) Find the matrix representation of ab that corresponds to the above vector representation of the basis kets Expanding this out into the joint-state basis ab 8 a a b b b b 5 8 a a b b b b 8 a a b b b b b b b b 5 8 a a b b b b b b b b 6 a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b Alternatively we could have written

25 ab and gotten directly to the final answer (b) Find the matrix representations of the partial projectors A P B and A P B and then use them to compute the matrix representation of A P B ab A P B A P B ab A P B We have so Hence A P B P B A P B ab A P B P B A P B

26 A P B ab A P B and their sum is A P B ab A P B A P B ab A P B (c) Find the matrix representation of a Tr b ab by taking the partial trace using the easy Dirac-notation method discussed in class We start by taking the partial trace in Dirac notation: ab 6 a b a b a b a b a b a b a b a b 5 6 a b a b a b a b a b a b a b a b a 6 a a a a 5 6 a a a a 8 a a 5 8 a a Thus in matrix representation a 8 5 Practice with Dirac notation for joint systems Let H A and H B be two-dimensional Hilbert spaces with given bases a a and b b Let a b a b et cetera Consider the joint state AB a b a b (a) For this particular joint state find the most general form of an observable O A acting only on the A subsystem such that 4

27 where AB O A B AB AB A P B O A B A P B AB Express your answer in Dirac notation We are looking for O A such that P B B B AB O A B AB AB A P B O A B A P B AB AB O A P B AB For the specific state we are given AB O A B AB a b a b O A B a b a b a O A a a O A a AB O A P B AB a b a b O A P B a b a b Hence we simply need a O A a a O A a The most general form of an observable for system A is given in Dirac notation by o a a o a a o a a o a a where o and o are real and o o The constraint we were given requires o so the most general observable has the form O A a a a b ic a a b ic a a abc R (b) Consider the specific operator X A a a a a which satisfies the general form you should have found in part (a) Find the most general form of a joint state vector AB such that AB X A B AB AB A P B X A B A P B AB The most general form of a joint state vector AB c a b c a b c a b c a b so X A B AB c a a a a a b c a a a a a b c a a a a a b c a a a a a b c a b c a b c a b c a b c a b c a b c a b c a b and 5

28 AB X A B AB c c c c c c c c Rec c c c With the projected form however A P B AB c a b c a b and so X A B A P B AB c a a a a a b c a a a a a b c a c a b c a b c a b AB A P B X A B A P B AB c c c c Comparing the two results we only need Rec c Rec c (c) Find an example of a reduced density matrix A for the A subsystem such that no joint state vector AB of the general form of you found in part (b) can satisfy A Tr B AB AB We have AB c a b c a b c a b c a b AB c a b c a b c a b c a b together with the constraint We compute Rec c AB AB c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b c a b and a c a c a c a c a c a c a c a c a c a c c a c a c c a a c c c c a a c c c c a a c c a a Since we require Rec c it follows that c It then becomes clear that for example we cannot achieve 6

29 a a a since this would require c c but by normalization c c c 4 More mixed states Let H A and H B be two-dimensional Hilbert spaces with given bases a a and b b Let a b a b et cetera Suppose that both the A and B system are initially under your control and you prepare the initial joint state AB a b a b (a) Suppose you take the A and B systems prepared in state AB and give them to your friend who then performs the following procedure Your friend flips a biased coin with probability p for heads; if the result of the coin-flip is a head your friend applies a transformation U h a b a b a b a b If the result of the coin-flip is tails your friend does nothing After this procedure what density operator should you use to represent your knowledge of the joint state? We now have a mixed ensemble p : U h AB a b a b p : AB a b a b The corresponding density operator is AB p a b a b a b a b a b a b a b a b p a b a b a b a b a b a b a b a b p a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b a b p a b a b a b a b (b) Suppose you take the A and B systems prepared in state AB and give them to your friend who then performs the following alternate procedure Your friend performs a measurement of the observable O A b b b b but does not tell you the result After this procedure what density operator should you use to represent your knowledge of the joint state? Assume that you can use the projection postulate for state conditioning The given form of O already specifies its spectral decomposition: O A b b A b b 7

30 It is easy to see that the two possible outcomes of the measurement () are equally likely Using the projection postulate we can associate A : b b AB AB a b AB A b b AB : AB so we can simply set A b b AB AB A b b AB AB a b a b a b a b a b 8

31 Ph5a HW #5 with HM s solutions Due 5:pm on Tuesday October in the box outside 4 Bridge Annex Remember that you must perform all calculations by hand and show your work for full credit (see the course web page for details on the -point grading system) Each student must prepare his/her own writeup based on his/her own understanding of the problems This week there are only two problems (since you also have the midterm exam) CHSH/Bell inequality with non-maximally entangled states Following the calculation in the lectures notes from / compute the correlation function g A B A B A B A B for states of the form ab cos a b sin a b Use / in the definitions of A A B and B For what range of values of is our Bell-type inequality violated? With / we have cos and sin so Then Likewise A z a B z b A z a B z b x a x b A B P a P b P a P b P a P b P a P b P a P b P a P b cos sin A B A B x a z b a a a a b b b b a b a b a b a b a b a b a b a b

32 and A B A B z a x b (by symmetry) and finally A B 4 A B x a b z z a b x 4 x a b x 4 4 a a a a b b b b 4 4 a b a b a b a b a b a b a b a b 4 4 a b a b a b a b 4 sincos Putting it all together g 4 sincos sin 4 Hence our Bell-type inequality is violated for such that sin sin This is satisfied by between 7 and 4 radians We note that the corresponding states are entangled while ab becomes factorizable for and / Uncooperative Alice In the quantum state teleportation protocol discussed in the lecture notes of /5 what state should Bob ascribe to his two-level quantum system B if Alice tells him that she has made the Bell-basis measurement but refuses to tell him the result? Give your answer in the form of a density operator From the lecture notes recall the state on H A H B H C just before Alice makes the Bell-basis measurement: a bc ac a b a b ac a b a b ac a b a b ac a b a b Since we know that a a we see that all four outcomes are equally likely It is easy to see what the corresponding post-measurement states are for Bob s system (and they are tabulated in the lecture notes) Hence if Bob knows that this measurement has been performed but doesn t know the outcome he should describe his two-level system by a mixed ensemble:

33 p 4 b a b a b p 4 b a b a b p 4 b a b a b p 4 b a b a b The corresponding density operator is a b b a a b b a a b b a b b b 4 a b b a a b b a a b b a b b a b b a a b b a a b b a b b a b b a a b b a a b b a b b b b b b Note that this is the same as b Tr ac a bc bc a Tr c bc bc c b c b c b c b c c c b c b c b c b c c c b c b c c c b c b c c b b b b

34 Ph5a HW#6 with HM s solutions Due 5:pm on Tuesday 7 November in the box outside 4 Bridge Annex Eigenvalues of H and the range of H Suppose we specify a finite-dimensional Hilbert space H A and Hamiltonian operator H Show that the states H A with minimum and maximum energy H are eigenstates of H Since H is a normal operator we may assume the existence of a complete orthonormal basis j for H A where H j j j Let j min be the index of the smallest eigevalue jmin and let j max be the index of the largest eigenvalue jmax For any state in H A c j j j and Since H jk c j c k j H k c j c k k j k c j j jk j j c j c j it follows that the minimum value of E is obtained when c jmin and the maximum value is obtained when c jmax (with all other coefficients equal to zero) For such assignments of the energy basis expansion coefficients itself is obviously an eigenstate of H The sudden approximation Suppose we specify a three-dimensional Hilbert space H A and a time-dependent Hamiltonian operator Ht t where and t are real-valued parameters (with units of energy) Let t be given by a step function t t t The Schrödinger Equation (SE) can clearly be solved by standard methods in the intervals t and t within each of which H remains constant We can

35 use the so-called sudden approximation to deal with the discontinuity in H at t which simply amounts to assuming that Suppose the system is initially prepared in the ground state of the Hamiltonian at t Use the SE and sudden approximation to compute the subsequent evolution of t and determine the function ft t t ft should be periodic with the largest Bohr frequency of the Hamiltonian We begin by solving for the eigenstates of the Hamiltonian for t Ht Clearly one of the eigenvalues is with corresponding eigenstate T We can also intuit that the other eigenstates are T T also with eigenvalue and with eigenvalue Hence and it is stationary up to time t For positive times we can easily read off the new energy eigenstates and write Then ft t e it e it ft 4 eit e it e it e it 4 e it e it e it e it e it e it cost We see that this is indeed periodic with the largest Bohr frequency which is

36 Rabi frequencies in cavity QED The following problem is loosely based on the paper we looked at in class on / (see lecture notes for reference) You may find it helpful although not necessary to read the paper before working this problem Consider a two-level atom whose pure states can be represented by vectors in a two-dimensional Hilbert space H A Let g and e be a pair of orthonormal basis states for H A representing the ground and excited states of the atom respectively Consider also a microwave cavity whose lowest-energy pure states can be described by vectors in a three-dimensional Hilbert space H C Let be orthonormal basis states representing zero one and two microwave photons in the cavity The experiment is performed by sending a stream of atoms through the microwave cavity The atoms pass through the cavity one-by-one Each atom spends a total time t inside the cavity (which can be varied by adjusting the velocities of the atoms) Immediately upon exiting the cavity each atom hits a detector that measures the atomic projection operator P e e e (Note that in the paper by Brune et al the atoms are initially prepared in the excited state and the detector looks for atoms exiting the cavity in the ground state; we choose the reverse in order to simplify our model) Just before each atom enters the cavity we can assume that the joint state of that atom and the microwave cavity is given by the factorizable pure state g c c c where c c c (a) Suppose the Hamiltonian for the joint atom-cavity system vanishes when the atom is not inside the cavity and when the atom is inside the cavity the Hamiltonian is given by H ac e g g e Show that while the atom is inside the cavity the following joint states are eigenstates of H ac and determine their eigenvalues: E g E E E E g e g e g e g e Then rewrite as a superposition of energy eigenstates It is immediate that H ac E so that is indeed an eigenstate (with zero eigenvalue) Going down the list of other candidates

37 H ac E H ac E e g E e g E H ac E H ac E e g E e g E By inspection we can write g c c c c E c E E c E E (b) Use part (a) to compute the expectation value P e t P e c t P e c t P e c t as a function of the atomic transit time t You should find that your answer is of the form P e Pnsin n t n where Pn is the probability of having n photons in the cavity and n is the n-photon Rabi frequency We can easily propagate the initial state: t c E c e it E e it E c and then compute Using e it E e it E P e c t c e it e it e c e it e it e ic sint e ic sin t e P e t P e c t P e c t P e c t we have P e c sin t c sin t With c P c P and n n we recover the desired expression 4 Rabi frequencies versus Bohr frequencies In the lecture notes of / we saw the general expression 4

38 O m q t O m q t c j c k exp i jk t j O m q k jk where c j are the expansion coefficients for in the energy basis and jk k j / are the system s Bohr frequencies Show that you can use this general expression to reproduce your result from Problem Are Rabi frequencies the same as Bohr frequencies? First we calculate the matrix elements of the partial projector P e c in the energy basis Using and a P e c b P e c a P e c b P e c E P e c E e P e c E e P e c E e P e c E e then in the natural matrix notation P e c Then with c E c E E c E E we would expect P e c c j c k exp i jk t j O m q k jk c eit e it c e it c cost c cos t Using the relevant double-angle formula e it 5

39 P e c c cos t sin t c cos t sin t c sin t c sin t which agrees with our answer from Problem The full set of Bohr frequencies is and in general they are not the same as the Rabi frequencies although some of them happen to coincide 6

40 Ph5a HW#7 with HM s solutions Pauli matrices and the Bloch vector (a) Show that the Pauli operators satisfy x S x y S y z S z Tr i j ij where the indices i and j can take on the values x y or z You will probably want to work with matrix representations of the operators We have so clearly x y i i i z which establishes the i j parts of the equation we are trying to prove For i j we note that Tr i j Tr j i so we only need to compute three matrix products: x y i i i i i z z x i y y z i i i i These are clearly all traceless which completes our proof (b) Show that the Bloch vector for a spin- degree of freedom s S x x S y ŷ S z ẑ i x has length / if and only if the corresponding density operator represents a pure state You may wish to make use of the fact that an abitrary density operator can be parametrized in the following way: x x y y z z where here represents the identity operator We first use the parametrization and the result of part (a) to obtain

41 Tr x y z Since this is equal to the purity we conclude that the state is pure if and only if Since x y z s S x S y S z 4 x y z we conclude that s if and only if the density operator represents a pure state Complete sets of commuting observables Consider a three-dimensional Hilbert space H and the following set of operators: O O O Find all possible complete sets of commuting observables That is determine whether or not each of the sets O O O O O O O O O O O O constitutes a valid CSCO First we check which observables commute O O O O hence O O Likewise

42 O O O O hence O O Finally O O O O and O O We conclude that the possible CSCO s are O O O O O and O O We next check the eigenvalues O : det so we have or the roots of 4 5 Hence we see that all three eigenvalues are distinct and O is okay on its own! This also means that O O is automatically a valid CSCO The matrix forms of O and O makes it clear that each of these observables has eigenvalues and with the latter having two-fold degeneracy Hence neither O nor O is not okay on its own Finally we note that the three obvious basis vectors have distinct pairs of eigenvalues for O and O : : : :

43 So finally the valid CSCO s are O O O and O O Conserved quantum numbers Determine which of the CSCO s from Problem (if any) are conserved by the Schrödinger Equation with Hamiltonian H O O Since the criterion for this is that each of the observables in the CSCO should commute with the Hamiltonian we simply check (using results from above): O H O O O O H O O O O H O O O O O O O O O Hence only O and O O are conserved 4 Stationary perturbation example Suppose we have two spin- degrees of freedom A and B Let the initial Hamiltonian for this joint system be given by H B z S A z B A S B z where A and B are identity operators S A z is the observable for the z-component of spin for system A and S B z is the observable for the z-component of spin for system B Here the notation is meant to emphasize that both spins experience the same magnetic field B z and have the same gyromagnetic ratio (a) Determine the energy eigenvalues and eigenstates for H It is easy to intuit z A z B E L z A z B E z A z B E z A z B E L where L B z and the factors on H A and H B are eigenstates of S z A and S z B (b) Suppose we now add a perturbation term: H tot H W W S A S B S x A S x B S y A S y B S z A S z B 4

44 Compute the first-order corrections to the energy eigenvalues The corrections fo E and E are easy to compute since these are non-degenerate eigenvalues: E E W L 4 E E W L 4 For E and E we need to diagonalize the restriction of W on the corresponding subspace The matrix representation of this restriction is W W W W W and to compute it we first derive the matrix representation of W on the entire joint Hilbert space Using the rules for tensor products of matrices S x A S x B 4 S y A S y B 4 S z A S z B 4 W 4 With the corresponding vector representations of and we have 5

45 W 4 4 and of course W W / Likewise we have W W /4 So in the end The eigenvalue equation is so by 4 W and the corresponding eigenvectors are determined a b a b and so the proper zeroth-order states in the degenerate subspace for H are W W a b a b with corresponding energy corrections W W W 4 W W W 4 Finally the first-order energy corrections are L L L L 4 6

46 Ph5a HW #8 with HM s solutions More perturbation practice Consider two spin- degrees of freedom whose joint pure states can be represented by state vectors in the tensor-product Hilbert space H AB H A H B where H A and H B are each two-dimensional Suppose the initial Hamiltonian for the spins is H a B z S z a b a b B z S z b (a) Compute the eigenstates and eigenenergies of H assuming a b and that both gyromagnetic ratios are non-zero If it is obvious to you what the eigenstates are you can just guess them and compute the corresponding energies The eigenstates are clearly a b a b a b a b with corresponding energies H a b a b B z H a b a b B z H a b a b B z H a b a b B z If a b and a b these eigenenergies are non-degenerate (b) Compute the first-order corrections to the eigenstates under the perturbation W S x a S x b where is a small parameter with appropriate units We use the general formula and note that For the a 4 b state p n S x a S x b a S x a S x b a S x a S x b a S x a S x b a p W n E n E p p b 4 a b b 4 a b b 4 a b b 4 a b a b W a b a b B z a b 4 a b B z a b

47 Similarly for the a b state a b W a b 4 a b a b a b B z 4 a b B z For the a 4 b state a b W a b a b B z a b 4 a b B z a b and for the a b state a b W a b 4 a b a b a b B z 4 a b B z The Rabi Formula Suppose the total Hamiltonian for a spin- particle is H B S z b costs x sints y which includes a static field B in the z direction plus a rotating field in the x y plane Let the state of the particle be written t at z bt z with normalization a b and initial conditions Show that at a b b Δ b sin t Δ b where Δ B This expression is known as the Rabi Formula In the rotating frame t z and (using results from the lecture notes) In matrix representation i d dt t B S z b S x t d dt a b i Δ b b and the eigenvalues of the effective Hamiltonian are as follows: i Δ i Δ 4 b 4 Δ 4 b i Δ b Δ The corresponding eigenvectors are determined by a b

48 so they are (un-normalized) i Δ i b a i b i Δ b i Δ a i b b a ib iδ b a ib iδ i Δ b i Δ b a ib iδ i Δ b i Δ b We see that and so and thus at bt at i Δ b ib i Δ b i Δ b a a e i e i Δ b t a e i Δ b t e i Δ b t a Δ b t ib i Δ b isin Δ b t and finally as advertised at ib Δ b sin Δ b t b Δ b sin t Δ b Aspin- Master Equation Consider the following Master Equation (in the rotating frame) for a resonantly-driven damped two-level system:

49 Here d dt t i Ht Γ t t t and the Hamiltonian (in the rotating frame) is simply H b S x b Here b appears as a driving strength and Γ corresponds to the decay rate of the excited state z Find an exact solution for the steady-state density matrix ss in terms of the parameter b /Γ Steady-state is defined by d dt ss Do not bother to transform back to the stationary frame Show that this is the same state as the steady-state solution of the zero-temperature Bloch equations with driving field b and Δ We have i b Γ i b Hence in steady state Γ Γ i b Γ i b Γ i b Γ i b Γ i b Substituting the latter two equations into the first 4

50 Hence b Γ ss Γ b i b Γ i b Γ i b Γ b Γ b Γ Γ Γ b i b Γ Γ b b ib Γ ib Γ Γ b b Γ b Γ b i i where b /Γ We calculate from this the components of the Bloch vector v x Γ b Tr b ib Γ ib Γ Γ b Γ b Tr ib Γ Γ b b ib Γ v y Γ b Tr i i b ib Γ ib Γ Γ b Γ b Tr b Γ iγ i b i b b Γ b Γ Γ b v z Γ b Tr b ib Γ ib Γ Γ b Γ b Tr b ib Γ ib Γ Γ b The relevant Bloch Equations are Γ Γ b 5

51 v x Γ v x v y b v z Γ v y which at steady state go to v z b v y Γv z Γ Γ v x b v z Γ v y b v y Γv z Γ We immediately see that v x and from the latter two equations as desired v y b Γ v z b b Γ v z Γv z Γ b b v Γ z v z b b Γ Γ b Γ b v y Γ Γ b 4 Solving the undriven Bloch Equations Consider the zero-temperature Bloch Equations without a driving field b : v x Δv y Γ v x v y Δv x Γ v y v z Γv z Γ Γv z Note that if we make a change of variables v z v z v z these can be put in the form of coupled linear ODE s d dt v x v y v z A v x v y v z where the matrix A is given by 6

52 A Γ Δ Δ Γ Γ Use linear algebra methods that you have learned in this course to prove rigorously that all possible initial states evolve to v ss as t Conclude that time evolution according to the (undriven) Bloch Equations is irreversible unlike the unitary evolution corresponding to the Schrödinger Equation for an isolated system We want to compute the matrix exponential of A so we begin by solving for its eigenvectors and eigenvalues Obviously we have a z Γ and in the remaining subspace the eigenvalues are given by Γ Δ : Γ Δ Δ Γ Γ Γ 4 Δ The corresponding eigenvectors are Γ Γ Γ 4Δ Γ iδ Δ i i a b a b i Δ i i a b a b i so in the full vector space a i Γ iδ a i Γ iδ We observe that in general 7

53 v Hence we may write v x v y v z v x a a v y ia a v z a z v t v x e Γ t e iδt a e iδt a v y ie Γ e iδt a e iδt a e Γt v z a z and Hence vt v t v t v ss regardless of the initial state At very long times the system loses all memory of its initial condition and the time evolution is irreversible 8

54 Ph5a HW#9 with HM s solutions Spin rotation pulses Suppose a spin- degree of freedom with gyromagnetic ratio is subjected to a constant magnetic field for a time interval B t b ẑ b (a) Show that if the spin evolves according to the simple Schrödinger Equation the time-development operator from the beginning to the end of the field pulse is where T exp i z b is the pulse area Compute the effect of this operator on an arbitrary Bloch vector During the time interval the (constant) Hamiltonian is H B S b S z b z Hence the time development operator is Then for an abritary initial Bloch vector v we have and sincos sinsin cos cos T exp i b z exp i z cos exp i exp i v z sin exp i z sin exp i sincos sinsin cos z z (b) Consider an ensemble of many spin- particles all prepared in the initial state x Now suppose that the spins experience magnetic field noise modeled as independent magnetic field pulses (of the kind considered in part (a) of this problem) all directed along the ẑ-axis but with randomly distributed areas Let the probability density for the pulse areas be

55 Pr exp Just to be clear the idea here is that each spin experiences a pulse whose area is drawn from Pr; the pulse areas experienced by the different spins in the ensemble are independent random variables Compute the average Bloch vector after the random pulses have acted on the ensemble of spins Your answer should show that the length of the average Bloch vector is more and more attenuated as the variance of the distribution of pulse areas increases Before the pulses act we have average Bloch vector After the pulses act we have v v d cosexp d sinexp The sin integral vanishes because of symmetry and the cos integral can be found in a table or inferred from the lecture notes of /: d cosexp exp 4 Hence the average Bloch vector is indeed more and more attenuated as the variance of the distrubtion of pulse areas increases Bloch sphere geometry Show that rotation pulses do not generally commute by comparing with where XZ X / Z z ZX Z X / z X / exp i 4 x Z exp i z You should calculate XZ and ZX explicitly but also visualize this for yourself on the Bloch sphere to check that your answers make sense What observable could you measure to distinguish XZ from ZX? We can compute an explicit matrix form for X / :

56 x exp i 4 x exp i 4 i 4 expi exp i e i/4 ei/4 ei/4 e i/4 ei/4 e i/4 e i/4 ei/4 cos 4 isin 4 isin 4 cos 4 i i And likewise for Z : exp i x exp i i e i/ e i/ i i Hence for the initial state z we have Z X / z i i i i i i i i y i i X / Z z i i i i i i y Obviously we can distinguish these perfectly by measuring S y On the Bloch sphere if we first apply a / rotation about x then v and a subsequent rotation about ẑ maps this to If we first apply the rotation about ẑ on v then nothing changes while the x rotation then puts us at Since these points are antipodal they are perfectly

57 distinguishable 4