A Primer on the Statistics of Longest Increasing Subsequences and Quantum States

Transcription

1 A Primer on the Statistis of Longest Inreasing Subsequenes an Quantum States Ryan O Donnell John Wright Abstrat We give an introution to the statistis of quantum states, with a fous on reent results giving tight bouns for the problems of learning an testing ientity of mixe states. Along the way, we survey the sometimes surprising onnetions between this area an topis as iverse as lassial istribution testing, longest inreasing subsequenes an the RSK algorithm, an representation theory of the symmetri an general linear groups. 1 Spanish ryptograms Suppose you enounter a ryptogram (substitution ipher) written in Spanish. To eipher it, you ll probably want to know the frequeny of letters in Spanish text. So you ownloa Don Quixote [Cer15] an pik out a sample of 500 letters, rawn ranomly with replaement; say, z, v, s, r, ñ,..., q. The resulting histogram of 32 rows might look like this: a á b e é f g h i í ï j l m n ñ o ó p q r s t u ú ü v x y z What an you infer from this sample? It s reasonable for you to estimate that the true frequeny p a of the letter a in Spanish is approximately p a = = 12.8%. Similarly, you might estimate pá = = 0.2%, p b = = 1%, p = = 3.8%, p = = 5.6%, p e = = 13.2%, et.1 Computer Siene Department, Carnegie Mellon University. Supporte by NSF grant CCF oonnell@s.mu.eu Center for Theoretial Physis, Massahusetts Institute of Tehnology. Supporte by NSF grant CCF jswright@mit.eu 1 Gaines s ryptanalysis book [Gai14] reports p a = 12.7%, p b = 1.4%, p = 3.9%, p = 5.6%, p e = 13.2%,

2 Of ourse, the finite sample size n = 500 means there will be some statistial error; for example, with p a p e 13%, the true frequeny of a an e might plausibly be anywhere between 10% an 16%. So on the basis of this sample, you woul be unwise to onfiently elare that e is the most probable letter in Spanish. On the other han, it woul be reasonable for you to onlue that the most frequent letter has frequeny = 13.2%. This question What is the frequeny of the most frequent letter? is an example of a letter-permutation-invariant statisti. That is, it oesn t epen on the names of the letters: it woul be the same if you applie any of the 32! possible permutations to these names (as is one in a ryptogram). Other letter-permutation-invariant statistis inlue: the entropy of the letter frequenies; the total probability of the top-10 most frequent letters; the number of letters with frequeny at least 1%; an so forth. In any long Spanish ryptogram, these statistis woul be approximately the same. Inee, knowing them woul give you a goo way to test whether a new ryptogram is in Spanish or some other language. As in the Don Quixote example, suppose we form a ranom wor w {a,..., z} n by sampling n letters inepenently; say, w 1 = z, w 2 = v, w 3 = s,..., w n = q. On the basis of this, we might wish to estimate some letter-permutation-invariant statisti (e.g., entropy, frequeny of the most frequent letter, et.). It s important to note that there are two symmetries at play. The first symmetry is the position-permutation-invariane of the sample; i.e., the ation of the symmetri group S n. Sine the n raws are inepenent, it oesn t matter that z was the 1st, 107th, an 251st letter, or that v was the 48th, 133r, 338th, an 350th; it only matters that z ourre 3 times, v ourre 4 times, et. This is why we immeiately simplifie to the histogram in our example. The seon symmetry is the letter-permutation-invariane; i.e., the ation of the symmetri group S, where = 32 is the number of letters. This symmetry says that the names of the letter outomes on t matter; in other wors, the statisti only epens on the (multi)set of probabilities {p a, p b,..., p z }. Given this, we an simplify our histogram further by eliminating the letter labels, an then sorting the rows. This proues a sorte histogram like the following: ν 1 = 66 ν 2 = 64 ν 3 = 50 ν 4 = 34 ν 5 = 34 ν 6 = 32 ν 7 = 29 ν 8 = 28 ν 9 = 23 ν 10 = 19 ν 11 = 16 ν 12 = 15 ν 13 = 13 ν 14 = 13 ν 15 = 12 ν 16 = 9 ν 17 = 7 ν 18 = 6 ν 19 = 5 ν 20 = 5 ν 21 = 5 ν 22 = 4 ν 23 = 3 ν 24 = 2 ν 25 = 2 ν 26 = 1 ν 27 = 1 ν 28 = 1 ν 29 = 1 ν 30 = 0 ν 31 = 0 ν 32 = 0 In this sorte histogram ν = SorteHistogram(w), the first row has length ν 1 = 66, iniating that the most frequent letter in the sample ha frequeny 66; the seon row has length ν 2 = 64, iniating that the 2n most frequent letter ha frequeny 64; et. By virtue of the two symmetries in our problem invariane to permuting the n = 500 positions, an invariane to permuting the = 32 letter names the sorte histogram ν enoes all the information we nee to estimate 2

3 any letter-permutation-invariant statisti, suh as entropy, or the probability of the most probable letter. Inee, if we efine p i = ν i /n, it woul be reasonable to estimate these two quantities by p i log(1/ p i ) an p 1, respetively. 2 2 Quantum ontraptions We will now introue the quantum version of the lassial statistis problem esribe in the previous setion. Suppose you waner into a quantum omputing laboratory an fin a ontraption with a button on the sie. Every time you press the button, 5 qubits pop out of the ontraption. If a 5-qubit system is in a pure state, you an represent it as a a a a , where the numbers a i s are omplex amplitues satisfying i a i 2 = 1. In other wors, a 5-qubit pure state an be represente by a unit vetor a C 32. (More generally, a system of q qubits has imension = 2 q, an systems with non-qubit partiles may have imensions that are not powers of 2.) Atually, the ontraption might have some probabilisti omponents insie it; for example, flipping oins, or internal quantum measurement evies. As a onsequene, when you press the button, you may get some some kin of ranomly istribute pure state vetor in other wors, a quantum mixe state. In priniple, the ontraption might proue any probability istribution over any set of unit vetors in C 32. However (see Setion 8) it is a basi fat of quantum mehanis that we may assume, without loss of generality, that the ontraption proues a isrete probability istribution over some basis of 32 orthonormal vetors a, b,,... C 32. Following quantum notation, let s write these unit vetors as 1, 2,..., 32 C 32, an write p 1, p 2,..., p 32 for the assoiate probabilities. In other wors, every time you press the button, the ontraption spits out i with probability p i (i = ). Although we ve numbere them , we may still refer to the vetors as letters. Sine you ve never enountere the ontraption before, both the probabilities p i an the orthonormal vetors i are unknown to you. Not only that, you an t just look at the output vetors to tell what they are; quantum mehanis only allows you to hoose a measurement to perform on them (isusse further in Setion 8), an this measurement itself proues a probabilisti outome. 3 These iffiulties notwithstaning, you may press the button n times, an we ll assume that the resulting outputs are inepenent an unentangle. For example, if you press the button n = 6 times, the ontraption might spit out the sequene 7, 12, 4, 20, 7, 31 ; this woul our with probability p 7 p 12 p 4 p 20 p 7 p 31. At this point, you an perform any measurement you like on the partiles. Quantum tomography refers to the task of using the samples to estimate the mixe state of the ontraption s output. In the general -imensional ase, this (roughly speaking) means estimating the probabilities p 1,..., p an the vetors 1,...,. 2 This strategy of estimating a statisti of p by omputing the statisti for the empirial istribution p is known as the plug-in estimator. Though a goo baseline estimate, it is often suboptimal; see, for example, [WY16, JVHW17] for optimal entropy estimators whih outperform the plug-in estimator. 3 Although, if a little birie tol you the vetors 1,...,, you oul measure in this basis an thereby exatly look at the output vetors. This woul reue you to a lassial senario like that of sampling from unknown Spanish letter frequenies, p 1,..., p 32. 3

4 As in the preeing isussion of Spanish ryptograms, for the moment we ll only onern ourselves with estimating statistis of the (multi)set of probabilities {p 1,..., p }. Most suh statistis have a natural physial meaning; for example, the largest probability gives a measure of how pure the ontraption s output is, an the entropy p i log(1/p i ) is alle the von Neumann entropy of the mixe quantum state. In this ase, we again have two symmetries at play. First, we have the same position-permutation-invariane as before; i.e., the ation of the symmetri group S n. This is beause the n button presses are assume to proue inepenent an unentangle outomes. Seon, sine we only are about statistis epening on {p 1,..., p } an we on t are about the ientity of the orthonormal basis 1,..., of C, we have the symmetry of the unitary group U() ating as rotations/refletions on bases. When estimating properties of the set {p a,..., p z } of Spanish letter frequenies, we fatore out the S n an S symmetries when we reue our sample to its sorte histogram of n boxes an rows. As it turns out (see Setion 11) there is a similar way to fator out the S n an U() symmetries when trying to estimate properties of the probabilities {p 1,..., p } assoiate to the quantum ontraption. In Setion 4, we ll state an Optimal Measurement Theorem, whih esribes a ertain quantum measurement that may be performe without loss of generality when estimating statistis of {p 1,..., p }. Surprisingly, the possible measurement outomes will be sorte histograms of n boxes an rows! The reason for this has to o with the representation theory of the groups S n an U(), whih is intimately onnete with sorte histograms also known as Young iagrams. The later setions of this survey will explain a little representation theory to justify why the Optimal Measurement Theorem is true. Before that, though, we will spen some time analyzing the probability istribution on Young iagrams that arises from the Optimal Measurement Theorem. As we ll see, this istribution is unfortunately not as simple as raw an n-letter wor w from the probability istribution {p 1,..., p } an form its sorte histogram. Rather, it has to o with an interesting ombinatorial property of w: the lengths of its longest inreasing subsequenes. 3 Longest inreasing subsequenes: Robinson, Shenste, Knuth Let w be a length-n wor over the orere alphabet {a, b,, }; for example, suppose n = 10 an w = bbbaab. We efine LIS(w) to be the length of the longest inreasing subsequene of w. (Throughout, inreasing will mean nonereasing ; in other wors, in alphabetial orer.) How an we easily etermine this length? For our example w = bbbaab, a little trial an error will onvine you that the unerline subsequene bbbb is maximal, so LIS(w) = 5. For longer wors, we ll nee to be more systemati. There is a natural ynami program for omputing LIS(w) known as patiene sorting that involves proessing the letters of w one-by-one (see [AD99] for a survey on this topi). As we o this, we maintain a growing array L in whih L[j] = the alphabetially smallest letter that an en a length-j inreasing subsequene. For example, after proessing w = bbbaab, our array will look like L = a a b b. 4

5 This orrespons to the following five inreasing subsequenes: L[1] = a, L[2] = a, L[3] = b, L[4] = b, L[5] =, beause of bbbaab; beause of bbbaab; beause of bbbaab; beause of bbbaab; beause of bbbaab, an it an be heke that there are no subsequenes of length six or greater. The overall longest inreasing subsequene (of the wor proesse so far) is simply the length of the array; an, when a new letter is proesse, it s not har to upate the entries of the array. To test your unerstaning, you might onfirm that if an 11th letter were to arrive at the en of our w, the four possibilities woul be: a a b b + a = a a a b a a b b + b = a a b b b a a b b + = a a a b a a b b + = a a b b The algorithm to upate the iagram (array) an be thought of as follows: Insertion: To proess a new letter, i, fin the rightmost position in whih it an be plae so as to maintain alphabetial orer. If this position is alreay oupie by some letter, then bump that letter out of the iagram. Otherwise, plae i at the en of the iagram, in a new box. This rightmost position, say j, orrespons to the first entry L[j] whih is stritly larger than i. The upate L[j] := i therefore works beause the subsequene ening in L[j 1] an be appene with i to form a subsequene of length j; those letters to the left of L[j] stay the same beause they are alreay less than or equal to i, an those letters to the right stay the same beause there is no inreasing subsequene of length j + 1 or greater ening in i. Consiering the four examples in (1), we see that inserting a new a auses the b in the thir box to be bumpe; inserting a new b auses the in the fifth box to be bumpe; inserting a reates a new box at the en; an inserting a also reates a new box at the en. The value of LIS(w) inreases preisely when a new box is reate. When a letter is bumpe uring the insertion proess, it seems a shame to just throw it in the trash. Following an iea of Robinson [Rob38], Shenste [Sh61], an Knuth [Knu70] ( RSK ), let s instea reursively insert the bumpe letter into a subsequent row of the iagram. When this RSK algorithm is applie to the wor w = bbbaab, we get the following growing sequene of fille Young iagrams: b b b b b b b (1) b b b b b b a a b b b (2) a a a b b b b a a b b b b a a b b b b 5

6 It is not too har to hek that the RSK algorithm, when applie to any wor w of length n, proues what is known as a semistanar Young tableau of size n: a fille n-box Young iagram in whih the rows have inreasing entries an the olumns have stritly inreasing entries. Beause of the seon property, the number of rows will never be more than the number of letters in the alphabet. Given a semistanar Young tableau (SSYT), its shape is the Young iagram (sorte histogram) proue by eleting the entries. We ll write λ = RSKshape(w) for the shape of the SSYT proue by applying the RSK algorithm to wor w; thus, e.g., RSKshape(bbbaab) = (3) As we ve seen, the top row of the iagram graphially enoes the ynami program for etermining the length of the longest inreasing subsequene. Thus if λ = RSKshape(w), then λ 1 = LIS(w). Is there any meaning to the lengths of the subsequent rows of λ? Greene s Theorem [Gre74] implies that there is: Greene s Theorem: If λ = RSKshape(w), then: λ 1 is the length of the longest inreasing subsequene in w; λ 1 + λ 2 is the length of the longest union of 2 inreasing subsequenes in w; λ 1 + λ 2 + λ 3 is the length of the longest union of 3 inreasing subsequenes in w; λ 1 + λ 2 + λ 3 + λ 4 is the length of the longest union of 4 inreasing subsequenes in w; et. For example, in our wor w = bbbaab, Greene s Theorem an (3) tell us that w shoul have 2 isjoint inreasing subsequenes of total length = 8, an inee here they are, unerline/overline: bbbaab. (It s a oiniene that they re both ontiguous.) 4 Symmetri properties of probabilities: lassial vs. quantum Now let s return to quantum ontraptions. Suppose as we were isussing that we have a quantum ontraption that outputs a -imensional mixe state with unknown probabilities p 1,..., p 32 for an unknown orthonormal basis 1,..., of C. (In our example, was 32.) An suppose we want to estimate some statisti only epening on the multiset {p 1,..., p }; for example, the maximum p i (whih we reall is one way of quantifying how pure the ontraption s output is). We press the button n times, obtain n inepenent unentangle outputs, an now must make some kin of quantum measurement. As mentione in Setion 2, it is possible without loss of generality to fator out the S n an U() symmetries, yieling the following (see [CHW07, MW16, OW15, BOW17]): Optimal Measurement Theorem: The optimal 4 quantum measurement when one only ares about {p 1,..., p } has the following property: It reports an n-box, -row Young iagram λ, an the probability istribution of λ (over both the outome of the ontraption an the measurement s ranomness) is exatly the same as that of RSKshape(w) for w p n, meaning that w is a ranom length-n wor in whih eah letter is i {1,..., } inepenently with probability p i. 4 Vis-a-vis either of these two ases: (i) Disriminating between two lasses of multisets, as in Property Testing. (ii) Estimating a statisti with minimal variane (quarati risk). 6

7 This shoul be ompare to the problem of estimating a letter-permutation-invariant statisti of an unknown probability istribution like the frequenies of the = 32 Spanish letters. In that lassial senario, an optimal algorithm also gets an n-box, -row ranom Young iagram ν; however, this ν is simply istribute as the sorte histogram of a ranom wor w. Let s make a loser omparison between the lassial an quantum senarios. In both ases, we want to use n samples to estimate a permutation-invariant property of the probability istribution p = (p 1,..., p ). In both ases, we an imagine that a ranom wor w {1,..., } n is hosen from the prout probability istribution p n. In the lassial ase, we get to see the Young iagram ν = SorteHistogram(w); in the quantum ase, we get to see the LIS information λ = RSKshape(w). For example, if w = bbbaab, then λ = RSKshape(bbbaab) = ν = SorteHistogram(bbbaab) = (4) A first immeiate observation is that the quantum ase is at least as har as the lassial ase. One way to see this is that ν ontains all the information you oul ever want, whereas λ oesn t; another way is via Footnote 3. A seon observation is that the LIS information λ will always be more top-heavy than the sorte histogram ν. More preisely, we will always have that λ majorizes ν, written λ ν, meaning that λ λ k ν ν k for all 1 k, with equality for k =. This follows iretly from Greene s Theorem, sine one an always fin k inreasing subsequenes in w whose union has length at least ν ν k, simply by taking all of the most frequently ourring letter as one subsequene, all of the 2n-most frequently ourring letter as a 2n subsequene,..., all of the kth-most frequently ourring letter as the kth subsequene. A thir observation onerns symmetry with respet to permuting {1,..., }. So far we ve assume we re only intereste in properties of the multiset {p 1,..., p }, suh as the maximum p i, or the entropy of p. This is why we oul reue to the sorte histogram ν in the lassial ase, an why (aoring to the Optimal Measurement Theorem) we an reue to the RSK output λ in the quantum ase. Now it s very lear that the istribution of the sorte histogram ν is invariant to permuting p 1,..., p, but it s far from lear that this is true of the RSK output λ. In fat, it may seem almost efinitely false! The very nature of the RSK algorithm, an the phrase longest inreasing subsequene, are both intimately tie up with the orering on the -letter alphabet. But nevertheless, the following surprising fat is true: The istribution on λ (that is, RSKshape(w) for w p n ) is unhange no matter how the probabilities p 1,..., p are permute. The reason for this will be mentione in Setion 5, but for now you might think about the ase = 2, wherein λ is fully etermine by the length of its first row, LIS(w). Thus the fat says that the length of the longest inreasing subsequene in a ranom wor with 60% 1 s an 40% 2 s has the same istribution as in a ranom wor with 40% 1 s an 60% 2 s... Beause of this symmetry property, we will sometimes assume without loss of generality that p 1 p 2 p. In this ase, we an ombine the previous observations to get an interesting inequality. As mentione, λ always majorizes the sorte histogram ν of w. In turn, the sorte histogram always majorizes the unsorte histogram of ν, all it η. Taking expetations of the statement λ η yiels (E[λ 1 ], E[λ 2 ],..., E[λ ]) (p 1 n, p 2 n,..., p n), (5) a statement we will use several times later. These inequalities help us unerstan lower bouns on the λ i s; we like to get some omparable upper bouns so as to really nail own the istribution 7

8 on λ. This issue will be taken up in Setion 7, but first we igress to esribe a few more properties of the RSK algorithm. 5 The RSK bijetion an Shur symmetri polynomials In fat, we have so far esribe only half of the RSK algorithm. In aition to the semistanar tableau esribe in Setion 3, known as the insertion tableau, the full RSK algorithm applie to a wor w also maintains a seon tableau known as the reoring tableau. This tableau is upate in parallel with the insertion tableau: when the tth new box is ae to the insertion tableau, a new box is ae to the reoring tableau in the same position, fille with timestamp t. In (2) we illustrate how the insertion tableau grows on the example wor w = bbbaab; the full insertion/reoring tableau output of RSK on w = bbbaab woul be: RSK(w) = a a b b b b, (6) For general w, the usual notation is RSK(w) = (P, Q), where P is the insertion tableau an Q is the reoring tableau. Sine the tableaus 5 always grow own-an-to-the-right, Q is always a stanar tableau. This means that both its rows an olumns are stritly inreasing, an that it ontains exatly the numbers 1 through n, where n is the length of wor w. When ombine with the insertion tableau, the reoring tableau gives all the aitional information neee to reverse the steps of the RSK algorithm an thereby invert the RSK mapping. For example, given just the output tableaus in (6), we oul reover w = bbbaab as follows: First, the reoring tableau tells us that the 10th an final box was reate in position 5 of the first row. This an only happen if was the final letter inserte into the first row; hene w 10 = an RSK(w 1 w 2 w 9 ) = a a b b b b, At this step, the reoring tableau tells us that the box in position 3 of the seon row was the final box reate. As a result, must have been inserte into the seon row in the final step, an this oul only have happene if it was previously bumpe own by b in the first row. In onlusion, w 9 = b an RSK(w 1 w 8 ) = a a b b b, Continuing in this manner allows us to reover the entire string w. Remarkably, this argument shows that any pair of tableaus (P, Q) an be inverte into a wor w so long as P is semistanar, Q is stanar, an P an Q have the same shape. Hene, the RSK algorithm gives a bijetion between wors an pairs of tableaus, one stanar an one semistanar, whih we state formally below. Before oing so, we have yet to touh on the most basi appliation of the RSK algorithm, whih is to permutations rather than wors. Given a permutation π S n, if we write it as 5 Sometimes spelle tableaux... 8

9 π = (π(1),..., π(n)), then we an view it as an n-letter wor on the alphabet {1,..., n} whih just happens to have no repetitions. As a result, if RSK(π) = (P, Q), then P has n boxes, ontains eah integer in {1,..., n} exatly one, an is semistanar; this implies that it is in fat stanar, like Q. Conversely, any pair of stanar tableaus (P, Q) of the same shape inverts to a permutation π. As a result, the RSK algorithm also gives a bijetion between permutations an pairs of stanar tableaus. These two bijetions are formalize as follows. Theorem 5.1 (RSK orresponene). Given an integer n an an n-box Young iagram λ, let SYT(λ) be the set of stanar Young tableaus with shape λ. Then the RSK algorithm witnesses the bijetion RSK π S n (P, Q) SYT(λ) SYT(λ). (7) n-box λ Further, for n, let SSYT (λ) be the set of semistanar Young tableaus with shape λ an entries in {1,..., }. Then the RSK algorithm witnesses the bijetion w {1,..., } n RSK (P, Q) n-box λ SSYT (λ) SYT(λ). (8) It s ustomary to write im λ = SYT(λ) for the number of stanar tableaus of shape λ. Taking arinalities of both sies of (7), we see that n! = n-box λ (im λ) 2. (9) (The notation im λ omes from the representation theory of the symmetri group, as we ll see in Setion 11. In this ontext, (9) is also a onsequene of the eomposition of the regular representation of S n into irreuible representations.) A onlusion is that if a permutation π S n is rawn uniformly at ranom, then Pr[RSKshape(π) = λ] = (im λ) 2 /n!. Inientally, there is a famous expliit formula for im λ, the Hook Length formula [FRT54]: n! im λ = where hl( ) = #{boxes in λ ue east an south of, inluing }. λ hl( ), (10) Analogously to (9), suppose we ount both sies of (8) aoring to the prout measure p n on wors forme by a probability istribution p = (p 1,..., p ) on letters. The onlusion is that Pr w p n[rskshape(w) = λ] = s λ(p) im λ, (11) where s λ enotes the Shur polynomial inexe by λ, efine by s λ (x 1,..., x ) = x T ( ), where T ( ) is the entry of tableau T in box. T SSYT (λ) T It is a surprising an non-obvious fat that the Shur polynomials are in fat symmetri in the variables x 1,..., x n. (Hint for the proof: it suffies to show that they are invariant uner interhanging x i an x i+1 ; for this, there s a relatively simple bijetion of tableaus... ) Inee, when ranging over n-box iagrams λ, they form a linear basis for the set of all -variable egree-n symmetri polynomials. (We will enounter another, more familiar, suh basis later in Setion 10: the power sum 9

10 symmetri polynomials.) Finally, we mention an alternative, more ompat formula for the Shur polynomials, whih an be proven using some lassial ombinatoris (see, e.g., [Sta99, Ch. 7]): ( (x λ et j + j) ) i ij s λ (x 1,..., x ) = i<j (x. (12) i x j ) Swapping any two variables x s, x t in the above formula simply reates a negative sign in the numerator an the enominator; thus this formula gives another testament that the Shur polynomials are symmetri. 6 Two majorization theorems for the RSK algorithm In Setion 4 we esribe a majorization result about the RSK algorithm that is an immeiate onsequene of Greene s Theorem: If w is any wor with λ = RSKshape(w) an ν = SorteHistogram(w), then λ ν, meaning that k λ i k ν i for all k. In this setion we mention two aitonal, newly proven [OW16, OW17] majorization results onerning RSK. The first result is highly intuitive. Suppose that p 1 p 2 p is a sorte probability istribution on {1,..., }, an q is another. Further, suppose that q p; roughly speaking, this means that a wor w rawn ranomly from q n tens to have more letters from earlier in the alphabet than if it is rawn from p n. In either ase, the sorteness of p an q ensures that the smaller letters of w ten to ollet up higher in the Young iagram proue by RSK(w), whereas the larger letters, outnumbere by the smaller letters, will be bumpe into the lower rows. As a result, we might expet RSKshape(w) to be more top-heavy for w q n than for w p n. This is exatly what the first majorization theorem says: Coupling Majorization Theorem [OW16]: Let p, q be sorte probability istributions on {1,..., } with q p. Let λ = RSKshape(w) for w p n, an let µ = RSKshape(z) for z q n. Then there is a probabilisti oupling (λ, µ) suh that µ λ always. (As a onsequene, E[µ µ k ] E[λ λ k ] for all k.) Here, a probabilisti oupling (λ, µ) refers to a probability istribution on pairs of Young iagrams suh that the first iagram has marginal λ an the seon iagram has marginal µ. Although this theorem is rather intuitive, a fairly intriate bijetive proof was require. The seon majorization theorem we present is onerne with the lower rows of the Young iagrams proue by RSK. For λ = RSKshape(w), Greene s Theorem tells us an exellent interpretation for the length of the first row, λ 1 : it s equal to LIS(w). The lengths of the lower rows, though, are a little harer to interpret. Let s say we want to unerstan the shape of rows k an below when RSK is applie to wor w. We ll take the example of k = 2 an our favorite wor w = bbbaab, whose growing insertion tableau was shown in (2). We want to fous on the Young iagram forme by rows 2 an below, so we sit next to the entrane of row 2 an wath as letters ome in (after being bumpe from row 1). In the example (2), we see a letter ome in at time 2, a letter at time 6, a letter b at time 7, another letter b at time 8, an a letter at time 9. Let s annotate the original string w with supersripts, iniating these times of being inserte into row 2 : w = 2 b 7 b baab. (To emphasize: the first has supersript 2 beause it was bumpe at time 2 whereas the seon has supersript 9 beause it was bumpe at time 9.) In our example it s a oiniene that all 10

11 letters that mae it to row 2 are from the first half of w. In any ase, let s write w bump for bb, the letters that entere into row 2 in the orer they entere; i.e., sorte aoring to the subsripts above. Then by efinition of the RSK algorithm, the insertion tableau of RSK(w) at rows 2 an below will equal the insertion tableau of RSK(w bump ): b b b b b b b = RSK(w bump ). On the other han, let s write w orig for bb, the letters that entere into row 2 in the orer they originally appear in w. We oul imagine applying RSK to this subsequene of w, although it s sort of ounterfatual: b b b b b b b b b = RSK(w orig ). Comparing the shapes of the Young iagrams proue, we have RSKshape(w orig ) = = RSKshape(w bump ). The seon majorization theorem we present says that this is a general phenomenon: Lower-Row Majorization Theorem [OW17]: Let w {1,..., } n be a wor an let 1 k. When applying RSK to w, some of its letters enter into the kth row. Let w bump enote the sequene of these letters in the orer they enter, an let w orig enote the sequene in the orer they originally appear in w. Then RSKshape(w orig ) RSKshape(w bump ). The proof of this theorem require a rather ompliate analysis of a geometri interpretation of the RSK algorithm known as Viennot s onstrution [Vie81]. Unlike the Coupling Majorization Theorem, we amit to not having great intuition for the Lower-Row Majorization Theorem. (Inee, when realling it, we often forget whether the onlusion shoul have or!) However, it seems to be an invaluable tool for reasoning about the lower rows proue by the RSK algorithm. 7 Probabilisti ombinatoris of longest inreasing subsequenes Let s return to the problem of unerstaning the shape of λ = RSKshape(w) when w p n. 6 Throughout this setion we will assume that p 1 p 2 p, whih is without loss of generality as we have isusse. For onreteness, onsier the following example λ when n = 1000 an p = (0.5, 0.2, 0.2, 0.1): λ 1 = 503 λ 2 = 215 λ 3 = 181 λ 4 = 101 Interestingly, it appears that λ i /n p i for eah 1 i 4, albeit with somewhat large error for p 2 = p 3. If this were typially true in general, then the normalize Young iagram λ/n = 6 Let us mention the istint but relate topi of unerstaning the shape of λ = RSKshape(π) when π S n is a ranom permutation, in whih ase λ is sai to have Planherel istribution. The Planherel istribution is extremely well-stuie; see [IO02] for an exellent overview of this area. 11

12 (λ 1 /n,,..., λ /n) woul provie us with a goo estimate ( p 1,..., p n ) of the sorte probability istribution (p 1,..., p ). In turn, this woul let us estimate any statisti of the multiset {p 1,..., p }. So how might we show that λ i p i n for large n? Let s start with the i = 1 ase. As esribe in Setion 3, λ 1 = LIS(w), so we like to show that the longest inreasing subsequene in w p n has length roughly p 1 n. The lower boun is simple: inee, we alreay etermine (see (5)) that E[λ 1 ] p 1 n. This is beause LIS(w) is always at least the number of 1 s in w, a quantity with mean p 1 n. Let s now heuristially reason about an upper boun for λ 1 = LIS(w). The longest inreasing subsequene in w an always be etermine as follows: First, take some partition of the positions (1,..., n) into ontiguous bloks, B 1,..., B. Next, form an inreasing sequene in w by taking all of the letter-1 s in blok B 1, all of the letter-2 s in blok B 2, an so forth. Finally, maximize this proeure over all partitions into bloks. Now for any partition, the number of letters i that w has in blok B i will be tightly onentrate aroun p i B i. Thus the length of the inreasing subsequene of w forme from the partition shoul be not muh more than p 1 B 1 + p 2 B p B. But p i p 1 for all i, so this is at most p 1 ( B B ) = p 1 n. Inee, one an formalize this argument using the Chernoff boun an get that LIS(w) p 1 n + O( n log n) with high probability. We will later see a notieably tighter upper boun. The fat that inee λ i p i n for all i [] was first shown by Vershik an Kerov [VK81]. Sine then, several works have etermine that in the limit as n, the eviation of the normalize Young iagram λ/n from the probability vetor p is istribute like a ranom vetor arising from the spetrum of ertain ranom matrix ensembles; speifially, it has a partly Gaussian, partly Tray Wiom limiting istribution; this was first shown for the ase of uniform p i s by [Ker03, TW01, Joh01] an generalize to the ase of nonuniform p i s by [ITW01, HX13, M 12]. Unfortunately, these limiting results on t neessarily give us onrete error bouns on the eviations of λ i from p i n: they heavily rely on onsiering p fixe an then taking n. In partiular, the error bouns an have an unontrolle epenene on quantities like an min pi p j (p i p j ) 1. Still, it is very useful to rely on these results for intuition. Most useful has been the following ansatz, whih is suggeste by these limiting results. Ansatz: λ i p i n ± 2 p i i n. Here i is the number of ourrenes of p i in (p 1,..., p ). One of the main goals in [OW16, OW17] is to prove sharp, expliit bouns on the loseness of the normalize Young iagram λ/n to the sorte probability vetor p. For instane, in Setion 12, we sketh a proof of the l 2 -boun E λ [ λ/n p 2 2 ] n, whih is inee onsistent with the ansatz. Going beyon this single global error boun, [OW17] was able to show some per-row error bouns, whih help in analyzing the Hellinger istane an χ 2 - ivergene of λ/n from p. The easiest to state suh boun is the following: p i n 2 τ i n E[λ i ] p i n + 2 τ i n, (13) where τ i = min{1, p i }. We note that this is suggeste by the ansatz, as p i i τ i always. In the remainer of this setion, we will sketh the proof of the upper boun in (13). Our starting point is the fat that muh stronger asymptotis an be obtaine in ase the largest probability p 1 is notieably larger than the seon-largest probability p 2. For example, in a long sequene of ranom English letters, the longest inreasing subsequene will almost surely be essentially the same as the number of e s; thus its istribution will be very lose to having mean p e n 12

13 an stanar eviation p e (1 p e )n. On the other han, in Spanish, where p a p e, the longest inreasing subsequene may involve a mix of a s an e s, an its length has a greater hane of eviating notieably above p a n p e n. To make this observation more formal, let w ( ) = w 1 w 2 w 3 be an infinite ranom wor with eah w i p inepenently, an set w (n) = w 1 w n to be its length-n prefix. Consier the (inefinite) proess of performing RSK on w ( ), an let λ (n) = RSKshape(w (n) ) be the snapshot of the RSK shape at time n. Then Its, Tray, an Wiom [ITW01] showe that E[λ (n) 1 ] p 1n n i>1 p i p 1 p i. (14) The limiting quantity on the right is finite if an only if p 1 > p 2 stritly. Supposing that p 1 p 2 δ, its value is at most p i i=2 δ 1/δ. So at an intuitive level, (14) tells us that in a ranom length-n wor with letter probabilities satisfying p 1 p 2 + δ, the expete length of the longest inreasing subsequene is just an aitive 1/δ larger than the expete length p 1 n of the all-1 s subsequene. Unfortunately, (14) is merely a limiting statement; it oul be true that E[λ (n) 1 ] p 1n only beomes smaller than, say, 2/δ one n 2 2 1/δ or even worse. Inee, the proof of (14) in [ITW01] involves asymptoti haking on the expliit formula (11) (using formulas (10) an (12)) an it heavily relies on an min pi p j (p i p j ) 1 being treate as onstant while n. However, the ombinatorial RSK perspetive allows us a nie trik whih lets us onvert these heavily asymptoti statements to perfetly onrete ones. The trik is to show that E[λ (n) 1 ] p 1n is an inreasing funtion of n; i.e., E[λ (n+1) 1 λ (n) 1 ] p 1. (15) To show this, let δ (n+1) = λ (n+1) 1 λ (n) 1. By efinition, δ(n+1) is the 0/1 iniator ranom variable for the event that, in the infinite RSK proess, inserting letter w n+1 reates a new box in the first row. Thus E[δ (n+1) ] is the probability of this event, an we nee to show the probability is at least p 1. To show this, we reall that the RSK output istribution epens only on the multiset {p 1,..., p }, an not on the orering of the letters; hene, we an reverse the alphabet to 1 > 2 > > without hanging the istribution of λ (t) for any t. But upon oing this, it beomes evient that the probability that the (n + 1)th box is in the first row is at least p 1. This is beause we get a new box in the first row whenever w n+1 = 1 (whih is now the last letter in alphabetial orer ). Thus we have establishe (15). But now we have an inreasing sequene, E[λ (n) 1 ] p 1n, an we know its limiting value thanks to (14). This means that the limiting value must be an upper boun for all n! That is, E[λ (n) 1 ] p 1n i>1 p i p 1 p i, for all n. (16) This alreay gives the upper boun we esire for (13) in the ase when p 1 p n. However it an beome arbitrarily ba when p 2 gets lose to p 1, an it gives nothing at all when p 1 = p 2. To get aroun this, we woul like to slightly shift some probability mass of p onto p 1 so that: (i) the expete LIS is not hange too muh; an, (ii) there is a eent separation between p 1 an p 2. Formally, let δ = 1 n, an onstrut a sorte probability istribution q = (q 1,..., q ) with q 1 = p 1 + δ, q 2 p 2, an q p. (This q an be onstrute by simply moving the bottom δ-mass of p onto p 1. We note q annot be onstrute if p 1 > 1 δ, but in this ase our esire 13

14 boun is trivially true.) We an now apply the Coupling Majorization Theorem from Setion 6 (inee, just the last statement in it, with k = 1.) Using the notation from that theorem, we have E[λ 1 ] E[µ 1 ] q 1 n + q i q 1 n + ( n = p ) n + n = p 1 n + 2 n, (17) q i>1 1 q i n as state in (13). Next, we woul like to generalize this to get a similar upper boun on E[λ k ] for any row 1 k. For this we use the Lower-Row Majorization Theorem. For w p n, it tells us that λ k = RSKshape(w) k = RSKshape(w bump ) 1 RSKshape(w orig ) 1 = LIS(w orig ). Analyzing w orig iretly still seems iffiult, beause it still requires unerstaning whih letters are bumpe to the kth row. However, all the letters bumpe into the kth row are at least k. Hene w orig is a subsequene of w k, the subsequene of w forme by removing all letters less than k. Beause aing letters annot erease the longest inreasing subsequene, we have that LIS(w orig ) LIS(w k ). But w k is simple to analyze: it s istribute exatly as p m k, where 1 m Binomial(n, p k + + p ) an p k is the probability istribution p k + +p (p k,..., p ). So we an onlue that E[λ k ] E[LIS(w k )] E m [(p k ) 1 m + 2 m] p k n + 2 (p k + + p )n, where the seon inequality uses (17) an Jensen s inequality. As p k + + p τ k, we get the laime upper boun in (13). 8 Mehanis of quantum mehanis We have not yet given any justifiation for the Optimal Measurement Theorem, whih onerns a ertain quantum measurement that outputs Young iagrams. Now is the time to elve into the mathematis of quantum states an measurements. In the physial worl, a quantum measurement is a evie that takes in a quantum partile system (of some fixe imension D) an outputs some lassial information. Its output shoul always be onsiere a ranom variable. Even when the input is a eterministi pure state vetor v C D, the output will be ranomly istribute (in a well-efine way, base on the evie itself an the input state v). An on top of this, we will onsier measuring quantum ontraption outputs, whih themselves are ranomize. Speaking of quantum ontraptions, we imagine a senario where, at the push of a button, the ontraption outputs a -imensional state whih is one of the orthonormal vetors 1,..., C with probabilities p 1,..., p. In an effort to learn about these vetors an probabilities, we have onsiere pushing the button n times. Suppose = 32, n = 6, an the output is the sequene v 1, v 2, v 3, v 4, v 5, v 6. Here eah v t is one of 1,..., 32 C 32 although we on t yet know these basis vetors. One thing we might o is buil some leverly hosen measuring evie M that aepts 32-imensional inputs an reas out some lassial information. We oul then apply it to eah of v 1,..., v 6. A more sophistiate thing to o is buil 6 ifferent measuring evies, M 1,..., M 6, eah taking a 32-imensional input, an apply M t to v t, t = An even more sophistiate strategy might involve aaptivity we oul buil an apply ifferent 32-imensional measuring evies base 14

15 on the outomes of previous measurements. However the most sophistiate thing we oul o is buil a single measurement evie M that takes as input all 6 samples simultaneously. If you think of a single v t C 32 as the state of 5 qubits, then olletively v 1,..., v 6 represent the state of 5 6 = 30 qubits. This in turn is efine by some 2 30 = (2 5 ) 6 -imensional vetor. In general, if we have n unentangle -imensional systems with pure states v 1,..., v n C, then their state is efine by a vetor of imension D = n. Speifially, it is the vetor v 1 v 2 v n (C ) n. This situation is least ompliate when eah vetor v t is one of orthonormal possibilities 1,...,, as we have been onsiering. In that ase, (C ) n shoul be thought of as the vetor spae spanne by n vetors that, by fiat, are orthonormal an are name i 1 i 2 i n, i t {1,..., }. For typographial simpliity, we usually write these vetors simply as i 1 i 2 i n, where i 1 i 2 i n ranges over all wors in {1,..., } n. So if, e.g., we have a ontraption with 4-imensional outputs 1, 2, 3, 4 C 4, an we press its button twie, the possible outputs are 4 2 orthonormal vetors in (C 4 ) 2 name 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44. Let s return to the notion of measurement evies for a D-imensional partile system. One of the most general kin of measurement evies works as follows. Let f be an orere orthonormal basis ( frame ) f 1,..., f D for C D. Then we an buil a measurement evie M f that, on input a pure state v C D, proues the following lassial rea-outs: j with probability f j v 2 = f j v i v i f j, j = 1... D. Here we are using the bra-ket notation in whih f j an v enote olumn vetors, an f j enotes the (omplex onjugate-)transpose row vetor of f j. So f j v = f j v is just the usual inner-prout of f j an v, the number f j v 2 = f j v f j v = f j v i v i f j is its square magnitue, an the fat that these quantities sum to 1 is a onsequene of the Pythagorean theorem (an that all the vetors involve have unit length). We have esribe what happens when a pure state v is fe into M f. What happens if we fee in a ranomly hosen pure state? Speifially, say we have a mixe state R, meaning a probability istribution over some pure states v 1,..., v r, in whih outome v i ours with probability q i. Here the v i s are arbitrary unit vetors in C D, an r might be more or less than D. If we make a raw from R, fee the result into the measurement evie M f, an observe the outome, what o we see? We get ( r r r ) j with probability q i f j v i 2 = q i f j v i v i f j = f j q i v i v i f j. (18) Notie that these probabilities only epen on the D D matrix σ = r q i v i v i. (19) This matrix σ is alle the ensity matrix for the mixe state R, an we see that two mixe states R an R with the same ensity matrix proue iential measurement outomes, an thus annot be istinguishe by any measurement evies M f! Aoringly, two suh mixe states are 15

16 onsiere physially iential, an they re mathematially represente by the same objet, the ensity matrix σ. As an example, let D = 2 an efine the unit vetors [ ] [ ] [ ] [ ] 1 0 a =, b =, a 3/5 =, b 4/5 = /5 3/5 Now if we efine the mixe state R = a or b with probability 1 2 eah an the mixe state R = a or b with probability 1 2 eah, they both have the same ensity matrix, namely σ = 1 [ ] [ ] [ ] [ ] = 1 [ ] [ ] 3/5 3/5 4/5 + 1 [ ] [ ] [ ] 4/5 4/5 3/5 1/2 0 = = /5 2 3/5 0 1/2 2 1, (20) where 1 enotes the ientity matrix. In partiular, suppose an engineer esigns an buils a quantum ontraption with 1-qubit (D = 2) output given by R. Then a statistiian waners into the lab, presses the ontraption s button several times, an estimates its output as R. At first it might look like the statistiian estimate the probabilities p 1 = p 2 = 1 2 perfetly but the vetors a, b poorly, sine a, b look quite ifferent. But in fat the statistiian shoul be given full points for a 100% orret estimate! It only makes sense to try to estimate the ensity matrix σ of an unknown mixe state, an the quality of an estimate matrix σ shoul be measure in terms of some matrix-istane between σ an σ. Let s summarize some properties of a D-imensional ensity matrix σ, all of whih follow from (19). First, σ is positive-semiefinite, meaning that it is Hermitian (equal to its omplex onjugate transpose σ ) an that g σ g 0 for all vetors g C D. Seon, σ has trae tr(σ) equal to 1, where the trae is the sum of ρ s iagonal entries. An easy way to see this is to use the linearity of trae, tr(a + B) = tr(a) + tr(b), an the yli property of trae, tr(ab) = tr(ba) = D i,j=1 A ijb ji. Applying these to (19) gives ( r ) r r r tr(σ) = tr q i v i v i = q i tr( v i v i ) = q i tr( v i v i ) = q i tr ([ 1 ]) r = q i = 1. Sine σ is positive-semiefinite, it will always have an orthonormal basis of eigenvetors, all them 1,..., D, with assoiate nonnegative eigenvalues, all them p 1,..., p D 0. Further, the trae of a matrix equals the sum of its eigenvalues. 7 Thus p p D = 1, we an view the p i s as a probability istribution over the eigenvetors i, an σ = D p i i i. In partiular, every positive-semiefinite matrix of trae 1 orrespons to a mixe state over orthonormal pure state outomes, justifying a laim mae in Setion 2. Please note that for a given ensity matrix σ, its spetrum i.e., the multiset of eigenvalues {p 1,..., p D } is uniquely etermine, but it oesn t have an inherent orering. Furthermore, orresponing orthonormal eigenvetors 1,..., D are not uniquely etermine. Taking the example from (20), we see that the 2-imensional ensity matrix σ = has eigenvalues ( 1 2, 1 2 ), but for assoiate eigenvetors we an hoose literally any pair of orthonormal vetors in C 2. The D-imensional analogue of this state, σ = 1 D 1, is alle the maximally mixe state; it is the unique state with spetrum orresponing to the uniform probability istribution ( 1 D,..., 1 D ). Let s make a final observation of relevane for quantum ontraptions. Suppose a quantum ontraption outputs 1,..., with probabilities p 1,..., p, an hene has ensity matrix ρ = 7 This follows beause trae is unitarily invariant: tr(uσu ) = tr(σu U) = tr(σ1) = tr(σ) for any unitary U. Choosing U to be a unitary matrix that moves the orthonormal basis 1,..., D to the stanar basis of C D, we get that UσU is a iagonal matrix with p 1,..., p D on the iagonal, an the laim follows. 16

17 p i i i. If we hit its button n times an view the output olletively, we get w (C ) n with probability Pr p n[w], where w runs over all wors i 1 i 2 i n {1,..., } n. This probability istribution on pure states has ensity matrix ( n )( n ) ( n ) n σ = i t i t = p i i i = ρ = ρ n, (21) i 1,...,i n=1 t=1 p it)( n t=1 t=1 where also enotes the matrix Kroneker prout. Thus quantum tomography problems an be thought of as estimating properties of a ensity matrix ρ given the ability to measure ρ n. 9 Nonommutative probability Before thinking about measurements of the bigger state ρ n, let s first isuss measuring a single ensity matrix ρ C. Measurement an be thought of as a way of generating lassial ranom outomes from a base soure of quantum ranomness, namely a positive matrix ρ with trae 1. In this setion we ll onsistently make an analogy to a similar situation in lassial probability: generating lassial ranom outomes from a base soure of lassial ranomness, namely a probability istribution p R (whih is a vetor of positive numbers aing to 1). Inee, if you restrit attention to iagonal ensity matries ρ, the two situations beome iential. So far we have seen that, given ρ, you an generate lassial ranom outomes with an M f measurement, where f = ( f 1,..., f ) is an orthonormal basis of C. Let s write the resulting outome probabilities from (18) in a slightly ifferent way, using the yli property of trae (an the fat that the trae of a single number is itself): measuring ρ with M f yiels outome j with probability f j ρ f j = tr( f j ρ f j ) = tr(ρ f j f j ). Writing E j = f j f j (the matrix whih projets onto f j ), the above is tr(ρe j ), whih also equals tr(ρ E j ) beause ρ = ρ. Now the trae of a matrix prout X Y is the same as the entrywise ot-prout between matries X an Y : tr(x Y ) = (X ) ij Y ji = i,j=1 t=1 XjiY ji = X, Y, i,j=1 where we use the, notation for matrix ot-prout. Thus we an further write: measuring ρ with M f yiels outome j with probability ρ, E j, E j = f j f j. This an be ompare with the simplest way of generating lassial outomes given a lassial base soure of ranomness p R : namely, simply rawing from p an reporting the outome. If we o this, the probability of outome j is p, e j, where e j = (0,..., 0, 1, 0,..., 0) with the 1 in the jth oorinate. So far we have only use our base soures of ranomness (ρ or p) to generate outomes from the set {1,..., }. In the lassial ase, we oul generate outomes in some other set Ω as follows: First, raw j from p. Next, a some aitional oin flips x. Then form a final outome ω Ω via some eterministi funtion h of j an x. A similar thing is possible in the quantum ase: First, raw v from ρ. Next, a some aitional qubits initialize to, say, 0, thereby inreasing the imension to D. Next, perform a measurement M F using some D-imensional frame F, prouing an outome J {1,..., D}. Lastly, form a final outome ω Ω by applying a eterministi t=1 17