PHYSICS 391 FALL 2017 QUANTUM INFORMATION THEORY I

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1 PHYSICS 391 FALL 2017 QUANTUM INFORMATION THEORY I Jean Well INDEPENDENT STUDY PAPER Professor S.G. Rajeev 10 th December 2017

2 1 Introducton A method s more mportant than a dscovery, snce the rght method wll lead to new and even more mportant dscoveres. Lev Landau ( ) The objectve of ths course was to ntroduce ourselves to quantum nformaton theory, a boomng topc n physcs wth tremendous practcal applcatons. Ths paper s based on a lecture I gave for the Kaptza Socety. In the frst secton we wll learn about the densty operator: a powerful computng tool n quantum mechancs. We wll then have a bref overvew of Lagrange multplers and how to use them to maxmze entropy and retreve the canoncal ensemble s propertes. Fnally, we wll revealed a surprsng relaton between statstcal and quantum mechancs. 2 The Densty Operator 2.1 Quantum Mechancal Ensembles We generally may not have perfect knowledge of a prepared quantum state. Suppose a thrd party, Bob, prepares a state for us and only gves a probablstc descrpton of t,.e., Bob selects ψ x wth probablty p X (x), where p X s the probablty dstrbuton for the random varable X, and x s n some alphabet χ. We can summarze ths nformaton by defnng an ensemble E of quantum states E {p X (x), ψ x } xɛχ (1) For example, E = {( 1, 1 ) ; ( 2, 3 )}. If ths ensemble s span by { 0, 1, 2, 3 }, then the state 3 3 s 0 wth probablty 0, 1 wth probablty 1, 2 wth probablty 0, and 3 wth probablty Of course, the sum of the probabltes adds up to 1. Consder a system wth E = {p, ψ } ɛn (2) wth ψ ψ j = δ j. Suppose = 1, then E = {p 1, ψ 1 } = {1, ψ 1 }. If we measure an observable  n E, then  = ψ 1  ψ 1 = 1 ψ 1  ψ 1 = p 1 ψ 1  ψ 1 Now, what f E = {(p 1, ψ 1 ) ; (p 2, ψ 2 )}? If you thnk as p 1 and p 2 as gvng weght to ther correspondng state s expectaton value, you can guess  = p 1 ψ 1  ψ 1 + p 2 ψ 2  ψ 2 (3) and fnally, f E = {p, ψ } ɛn, then  = p ψ  ψ (4) 1

3 Clam: I = n n n (5) where { n } nɛn s a complete set,. e., δ nm = n m, and I s the dentty operator. Proof: I m = n n n m = n n δ nm = m Thus,  = p ψ  ψ QED = = = = n = n p ψ I ψ p ψ n n  ψ n p n  ψ ψ n n n  p ψ ψ n n ˆρ n (6) where ˆρ = p ψ ψ (7) s the densty operator.  For some matrx B, we have B j = B j, thus very useful and mportant result  = n (ˆρ) nn = T r(âˆρ), whch gves the = T r(âˆρ) (8) Note that the trace s ndependent of whch complete set you use. Suppose { φ } ɛn s a complete set, then usng (5) we get T r(b) = B = BI ( ) = B j ( φ j φ j ) (9) = j φ j B φ j = j φ j B φ j thus, just thnk of the complete bass { δ } ɛn, where for δ the th component s 1 and the other components are 0, to buld your ntuton. 2

4 Why s ths densty operator so mportant to us? We already saw that t makes fndng the expectaton of an observable trval, but that s not t. It also allows you to fnd the probablty of measurng a partcular egenvalue from your ensemble! To see ths, consder an observable Â, so that  a = a a, wth a a = δaa. What s the probablty of measurng a partcular egenvalue a? From the law of total probablty, we get p(a) = (probablty of gettng ψ ) (probablty of gettng a gven that you got ψ ) but the probablty of gettng ψ s just p, and the probablty of gettng a gven that you get ψ s a ψ 2, whch comes from the Bohr nterpretaton. If you re not sure why ths s true, consder the complete set { n } ɛn and some normalzed quantum state ψ, then ψ = n n (10) but ψ ψ = 1, so 1 = n n j n n j = j = n n j δ j j n n = (11) n 2 but we also have p = 1, where n ths case p s the probablty of gettng (measurng) ψ. The dea s to say that n 2 = p, whch s reasonable snce the n terms defntely seem to gve some weght to how bg a role a gven n plays n the constructon of ψ. Fnally, you can see from (10) that n j = n j ψ, thus the probablty of measurng n j gven that your ntal state s ψ would be n j 2 = n j ψ 2. Gong back to our p(a), we now understand why we have from the law of total probablty p(a) = p a ψ 2 = p a ψ ψ a = a p ψ ψ a = a ˆρ a Therefore, f you have ρ, you can fnd some of the most useful quanttes n quantum mechancs trvally. We can make ths a lttle bt more general. Clam: If the ensemble s gven by E = {p X (x), ψ x } xɛχ, we have and ˆρ = xɛχ (12) p X (x) ψ x ψ x (13) p J (j) = T r (Π j ˆρ) (14) 3

5 Proof: (13) s trval to see from (7), but (14) s a bt harder. Recall that Π j = j j, we can show the equvalence between (14) and (12) p J (j) = T r (Π j ˆρ) = = φ Π j ˆρ φ φ j j ˆρ φ = j ˆρ φ φ j = j ˆρ φ φ j = j ˆρI j = j ˆρ j QED Let s see how all of ths theory works n an actual example. Suppose we are workng wth electrons. Bob tells us that an electron has a.5 chance of beng n the state z,, and.5 chance of beng n the state z,. Then E = {( 1, z, ) ; ( 1, z, )}. Thus, from (7), we get 2 2 ˆρ = p ψ ψ = 1 2 z, z, z, z, = (16) What f I ask you to fnd the probablty of gettng x,? Or the expectaton value of Ŝx? Dong ths wthout usng the densty operator s not too dffcult, you should try t before readng the rest. Dd you try yet? I m gong to put my trust n humanknd s rghteousness and contnue. The probablty of gettng x, s gven by ˆρ( x, ) = x, ˆρ x, = = = 1 (17) Hopefully you got the same thng dong t the long way. Now what about Ŝx = h ˆσ 2 x? { ( )} {} 0 ˆσ x = T r(ˆσ x ˆρ) = T r = T r 2 1 = 0 (18) Propertes of the Densty Operator Clam 1: ˆρ s Hermtan Proof: 2 0 (15) ˆρ = p ψ ψ = p ψ ψ = p ψ ψ = ˆρ (19) QED 4

6 Clam 2: T r(ˆρ) = 1 Proof: ( ) T r(ˆρ) = n n ˆρ n = n n ( p ψ ψ ) n = = p ψ n p ψ I ψ n n ψ (20) = p ψ ψ = p = 1 where we used the normalzaton property of the wavefuncton, and (5). Clam 3: ˆρ can be dagonalzed to the followng p p ˆρ =... p χ where χ s the sze of the set. QED (21) Proof: We have that ˆρ = p ψ ψ, thus ˆρ ψ k = p ψ δ k = p k ψ k (22) so ˆρ as egenvectors { ψ k } k wth egenvalues {p k } k. Let Then M = ( ψ 1 ψ 2... ψ χ ) ˆρM = (ˆρ ψ 1 ˆρ ψ 2... ˆρ ψ χ ) = ( p1 ψ 1 p 2 ψ 2... p χ ψ χ ) whch we can rewrte n a more enlghtenng way p p ˆρM = ( 0 p p ) ψ 1 ψ 2... ψ χ.. = M.... p χ p χ f M s nvertble, we get (23) (24) (25) 5

7 p p M 1 ˆρM =... p χ (26) QED 2.3 Pure vs. Mxed Quantum States Pure: We know the system s n a partcular state ψ E = {1, ψ }. Mxed: We don t know the system s n a partcular state ψ,.e., the system could be n several states { ψ } ɛn E = {p, ψ } ɛn. Usng Clam 3, we can defne pure and mxed quantum states usng the densty operator. A mxed state would be gven by (21), and a pure state would be gven by ˆρ =. 1 (27). 0 snce p k = 1 and p = 0 for all k. We wll come back to ths when we wll be dscussng statstcal mechancs and entropy. 2.4 Why s ρ called the Densty Operator  = T r(âˆρ) = m,n n  m m ˆρ n whch becomes for a contnuous bass  = d 3 x d 3 x x  x x ˆρ x (28) Let s focus on x ˆρ x. x ˆρ x = x p ψ ψ x = p x ψ ψ x (29) = p ψ (x )ψ (x) If x = x, then x ˆρ x = p ψ (x) 2. Thus, the dagonal matrx elements are the weghted sums of the probablty denstes of the states n the ensemble. 6

8 3 The Canoncal Ensemble 3.1 Lagrange Multplers Let S be a surface gven by f(x, y, z) = c for some constant c. What f you want to fnd whch pont on ths surface s the closest to the orgn? To do ths we need some background. A curve on ths surface s defned by r(t) = (x(t), y(t), z(t)). Now, the dervatve of ths curve, r (t), s tangent to the curve at any pont P = (x(t), y(t), z(t)). Let s take one of them P 0 = (x(0), y(0), z(0)) = (x 0, y 0, z 0 ). At P 0, r (t) s tangent to the curve and thus tangent to S, whch means t les on the tangent plane of f(x, y, z) at P 0. Ths means that f a vector s perpendcular to r (t) at P 0, then t s perpendcular to the tangent plane of S at P 0. Clam: The gradent of f s perpendcular to r (t) at any pont. If we can prove that f P0 r (t 0 ) = 0 for some arbtrary pont P 0 we would be done. But df dt P 0 = f x dx P 0 dt t 0 + f y dy P 0 dt t 0 + f z dz P 0 dt t 0 = f P0 r (t 0 ) (30) but f(x.y.z) = c where c s constant, thus df dt P 0 = 0. QED FIG. 1. Lagrange Multplers Let s go back to our mnmzaton problem. Try the problem n two dmenson, where S s a curve. Take a compass and start drawng crcles wth a common center at the orgn. Start from a small one and ncrease the radus of the next one. Contnue ncreasng the radus untl a crcle enters n contact wth S. Ths pont on S s the closest one to the orgn. We of course have the exact same procedure n three dmenson wth crcles beng replaced by spheres. Let s call ths last sphere G, defned by g(x, y, z) = x 2 + y 2 + z 2 r 2 = 0. Note that g contans the mnmzaton constrant at the heart of the problem. Now, here s the trck, the tangent plane T, of G and S at the pont of contact s the same one! Therefore, recallng the clam we just proved, ths means that g and f are both perpendcular to ths tangent plane, and so they are parallel to each other. Ths means that f = λ g (31) 7

9 where g s the functon contanng the constrant, and λ s the famous Lagrange multplers. The general case s gven by f = λ g (32) 3.2 Mxed States and Entropy We know that the entropy of a system s gven by S = k p lnp = kt r(ˆρlnˆρ) (33) ˆρ s dagonalzed. What are the {p } n the mnmum entropy state? From (33) t s easy to see that S s a of postve quanttes S 0 S mn = 0. Recall that we need p = 1, whch means that p k = 1 and p = 0 k. Ths mght rng a bell (no pun ntended), we have n ths case a pure state. Thus, n order to have mnmum entropy, an ensemble must be composed of a pure state. Now what about maxmzng entropy? Ths s where Lagrange multplers come handy. In ths case, f = S, and g = j p j 1. Lookng back at (32), we must have ( k ) p lnp = λ p 1 (34) We can smplfy ths qute a bt ( k ) p lnp λ p 1 = but p s arbtrary, thus Usng our constrant p = 1, we get χ =1 = ( ) p k( p )lnp kp λ p p (35) ( k(lnp + 1) λ) p k(lnp + 1) λ = 0 p = e λ k 1 (36) e λ k 1 = 1 e λ k 1 = 1 χ p = 1 χ Ths means that n order to maxmze entropy, we want a unform dstrbuton. Ths gves us that a state s a maxmally mxed state f every egenvalue of the densty operator s equal,.e., ˆρ = χ.. = 1 χ I (38) 1 Note that the entropy for the maxmally mxed state s χ χ 1 S = k p lnp = k χ ln 1 χ =1 =1 (37) = kln χ (39) 8

10 3.3 Statstcal Mechancs To some extent, statstcal mechancs s an assumpton about the densty matrx for a macroscopc system. The assumptons (constrants) are (a) Ĥ p = 1. (b) = E s known. Constrant (b) can be expressed n a more useful way snce from Schroednger equaton Ĥ ψ k = E k ψ k ψ k Ĥ ψ k = E k, whch gves Ĥ = p k ψ k Ĥ ψ k = p k E k = E (40) k k Snce we have two constrants, usng equaton (32) we have that ( k ) p lnp = λ 1 p 1 + λ 2 p E E whch we can smplfy to (41) ( k ) p lnp λ 1 p 1 λ 2 p E E = but p s arbtrary, thus ( k(lnp + 1) λ 1 λ 2 E ) p (42) Usng our constrant p = 1, we get k(lnp + 1) λ 1 λ 2 E = 0 p = e λ 1 k λ 2 E k 1 (43) e λ k 1 = 1 Z (44) wth Z = e λ 2 E k (45) and Lettng λ 2 k = β, we get the well known canoncal ensemble equatons p = e βe Z Z = (46) e βe (47) 9

11 4 Quantum Mechancs and Statstcal Mechancs 4.1 Mxed State, Pure State, and Temperature The nternal energy of a monoatomc gas s gven by E = 3 NkT E T, but lookng back at 2 (47), we must have β 1 snce the exponental must be untless. These two relatons mply that E β 1. Thus, T T β 0 + Z = χ p = 1 χ Ths means that all states are equally probable, whch s what we found for the maxmally mxed state. What about T 0 +? I m sure you can guess what s about to happen Z = =0 e βe = e βe 0 e β(e E 0 ) = e βe 0 + e βe 0 =0 therefore snce E > E 0 > 0, we get =1 e β(e E 0 ) (48) ths means that T 0 + β Z e βe 0 (49) T 0 + β p k = e βe k Z e β(e k E 0 ) (50) thus, p k = 0 k 0 and p 0 = 1. Therefore all the partcles are n the ground state E 0. Ths exactly what we found for a pure state. 4.2 Imagnary Statstcal Mechancs Defnton: (Functon of a Hermtan Operator) Suppose that a Hermtan operator A has a spectral decomposton A = a for some orthonormal bass { }. Then the operator f(a) for some functon f s defned as follows: f(a) f(a ) (51) Recall that Ĥ ψ k = E k ψ k, therefore Ĥ = E ψ ψ. Usng the above defnton, ths means that e βĥ = e βe ψ ψ (52) Note that ths mples ψ k e βĥ ψ k = e βe δ k δ k = e βe k (53) From ths t s trval to see that ˆρ = 1 Z e βĥ (54) and Z = e βe = T r e βĥ (55) 10

12 In quantum mechancs, we can determne how a quantum state evolves wth tme usng the operator U(t). We can get t from lookng at Schroednger equaton. thus, U(t) = e Ĥt h. By makng the change Ĥ ψ(t) = h t ψ(t) = U(t) ψ(0) (56) ψ(t) ψ(t) = e Ĥt h ψ(0) (57) we get t β h (58) whch gves us the startlng result that U(t) U( β h) = e βĥ T r[u( β h)] = T r (59) e βĥ = Z (60) Therefore, statstcal mechancs can be perceved as quantum mechancs n magnary tme. An ncredble result that mght sound famlar f you have studed specal relatvty. 5 Concluson The task s, not so much to see what no one has yet seen; but to thnk what nobody has yet thought, about that whch everybody sees. Erwn Schroednger ( ) Through the use of the densty operator, Lagrange multplers, and entropy, we have dscovered an ncredble relatonshp between statstcal and quantum mechancs. These types of transformatons are known as Wck rotatons and create strange lnks between many areas n physcs. For example, specal relatvty an classcal physcs are lnked by the transformaton t t. To see ths, recall that the dot product nvarance between two nertal frames S and S n specal relatvty gve z 2 t 2 = (z ) 2 (t ) 2 wth S travelng at velocty vẑ wth respect to S. Ths becomes the classcal Eucldean dot product z 2 + T 2 = (z ) 2 + (T ) 2 f you make the followng change t T. It s easy to see from ths last equaton that the two ponts le on the crcumference of a crcle and must therefore be equvalent through a rotaton,.e. ( ) z cos θ sn θ z = (61) sn θ cos θ T thus, T z = z cos θ T sn θ T =T cos θ + z sn θ but we must have z = 0 when z = vt, thus the frst equaton gves (62) tan θ = vt T = v (63) 11

13 but then T = cos θ(z tan θ + T ). We can smplfy ths by usng the fact that sec 2 θ + tan 2 θ = 1 we get sec θ = 1 v 2 = 1. Thus T = 1 cosθ 1 v 2 (t zv). But T = t, so we fnally get t =γ(t vz) z =γ(z vt) Ths approach shows even more clearly that we can see the Lorentz transformatons as a rotaton through an magnary angle θ n the ct z plane, or equvalently, the ct z plane. References [1] Mark M. Wlde, Quantum Informaton Theory, Cambrdge Unv. Press, 2017, Second Edton, pp [2] Segev Benzv Partcle Mxtures, Unversty of Rochester, Department of Physcs & Astronomy, 2017, as part of Prof. Benzv lecture notes for Physcs 407: Quantum Mechancs. (64) 12