Internatonal Mathematcal Forum, Vol 3, 08, no, 507-57 HIKARI Ltd, wwwm-hkarcom https://doorg/0988/mf088055 The Order Relaton and Trace Inequaltes for Hermtan Operators Y Huang School of Informaton Scence and Engneerng Chengdu Unversty, Chengdu 6006, Chna & Key Laboratory of Pattern Recognton and Intellgent Informaton Processng of Schuan Provnce, Chengdu Unversty, Chengdu 6006, Chna & The Research Insttute of Bg Data Chengdu Unversty, Chengdu 6006, Chna Copyrght 08 Y Huang Ths artcle s dstrbuted under the Creatve Commons Attrbuton Lcense, whch permts unrestrcted use, dstrbuton, and reproducton n any medum, provded the orgnal work s properly cted Abstract In ths paper, we gve the order relaton and trace nequaltes of Hermtan operators by comparson wth ther counterparts of real numbers Keywords: Hermtan operators, order relaton, trace nequaltes Introducton Hermtan operators play an mportant role n mathematcs and some scentfc areas In quantum nformaton theory, they are frequently encountered [] Because Hermtan operators have real egenvalues, they can be seen as the counterpart of real numbers For example, there exsts comparson relaton between any two real numbers so that we can know whch number s larger than the other Lkewse, we can defne the order relaton for Hermtan operators In ths paper, we wll present the order relaton and some trace nequaltes of Hermtan operators by comparson wth ther counterparts of real numbers
508 Y Huang Prelmnares In ths paper, we assume that all operators are n one Hlbert space The symbol stands for Hermtan adjont operaton and the symbol stands for complex conjugate operaton Suppose A s an operator on Hlbert space H We have the followng defntons of three types of specal operators [] Defnton A s a Hermtan operator f A A Defnton A s a postve operator f for any vector n H, A 0 Defnton 3 A s a projecton operator f t has the form k A where s an orthonormal set n H Lemma Suppose A s an operator on Hlbert space H, then for any vector n H, A 0 f and only f A 0 The converse s obvous For the forward statement, suppose A 0 for any vector n H Because any operator A on one Hlbert space can be wrtten as A B C where B ( A A ) and C ( A A ) are both Hermtan operators Thus A 0 ( B C) 0 B C 0 For a Hermtan operator H and an arbtrary vector, the complex conjugate of the complex number H s H * H H, thus H s real So both B and C are all real numbers Therefore B C 0 B = C =0
Order relaton and trace nequaltes for Hermtan operators 509 Because s arbtrary, we can choose one egenvector v correspondng to any egenvalue of B Hence v B v v 0 0 Thus all egenvalues of B are zeros By the spectral decomposton of B, we mmedately get B 0 Smlarly, C 0 Therefore A B C 0 Note that we can conclude all egenvalues of A are zeros drectly from the condton A 0 for all n H by the smlar way above (replace by one egenvector of any egenvalue of A) wthout wrtng t as A B C But can we get A 0 from ths fact? The answer s negatve The reason arses from the fact that all egenvalues of one operator are zeros s not suffcent to ensure ths operator s defntely a zero operator For example, the operator 0 on a qubt, whch has matrx representaton 0 0 0 wth respect to the bass 0,, has only zero egenvalues But 0 0 Hence, although we can conclude that all egenvalues of A are zeros, we cannot get A 0 unless A has a spectral decomposton lke the Hermtan operators B and C n the proof of ths lemma Ths s why we wrte A as A B C wth two Hermtan operators B and C Theorem Suppose A s an operator on Hlbert space H, then for any vector n H, A s a real number f and only f A s a Hermtan operator The converse has been shown n the proof of Lemma For the forward statement, f A s a real number for any vector n H, equvalently we have A * A A A A A ( A A) 0
50 Y Huang By Lemma, we get A A0 A A A s a Hermtan operator Corollary A postve operator s necessarly a Hermtan operator Suppose A s a postve operator satsfyng, by Defnton, A s real and non-negatve for any vector n one Hlbert space So by Theorem, A s a Hermtan operator By Defnton 3 and Corollary, t s easy to verfy that a projecton operator s necessarly a postve operator and hence a Hermtan operator We can fnd from Lemma, Defnton and Theorem that the equvalent condtons for the zero operator, postve operators and Hermtan operators all focus on the same number A, where s arbtrary, and dffer from each other only on takng a specal value of real number and a real number, respectvely A, whch s 0, a non-negatve 3 Order relaton and trace nequaltes for Hermtan operators In ths secton, we restrct all operators to be Hermtan operators 3 Order relaton for Hermtan operators Hermtan operators have real egenvalues Hence, they can be seen as the counterpart of real numbers We have the concept of the lager and smaller comparson between any two real numbers as well as nequaltes for real numbers Naturally, we expect that ths knd of noton of comparson relaton can be extended from real numbers to Hermtan operators Consequently, there are nequaltes for Hermtan operators In fact, we can establsh ths noton as follows Defnton 3 Suppose both A and B are Hermtan operators We wrte A 0 f A s a postve operator In addton, we wrte A B f A B 0 Equvalently, we may wrte the nverse relaton B A f A B One can readly verfy the operaton propertes of Hermtan operators and postve operators as follows:
Order relaton and trace nequaltes for Hermtan operators 5 If both A and B are Hermtan operators and c s a real number, then () both A B and A B are Hermtan operators; () ca s a Hermtan operator If both A and B are postve operators, and c s a non-negatve real number, e f A 0, B 0, and c 0, then () AB 0; () ca 0 Example 3 A projecton operator P, by Defnton 3, satsfes where I s the dentty operator 0 P I Example 3 The three Paul operators x 0 0, y 0 0 and z 0 0 whch are three Hermtan operators satsfy I I, I I x, I I y z To see ths pont and for smplcty, we denote any of these three Paul operators as Snce has egenvalues, t has spectral decomposton where and are orthonormal egenvectors correspondng to egenvalues and, respectvely On the other hand, Thus I ( I) 0 and I 0 Therefore I I Property 3 (reflexve property) The order relaton for Hermtan operators s reflexve, e for any Hermtan operator A, A A Snce for any Hermtan operator A, A A 0 where the zero operator 0 s a postve operator Thus by Defnton 3, A A Property 3 (transtve property) The order relaton for Hermtan operators s transtve, e If A A and A A3, then A A3
5 Y Huang If A A and A A3, then A 3 A ( A3 A ) ( A A ) 0 Hence A A3 Property 33 (antsymmetrc property) By defnton, A B and B A mply that A B A B ( B A) 0 for any and B A ( B A) 0 for any Thus ( B A) 0 for any By Lemma, we get A B These three propertes of order relaton for Hermtan operators are completely smlar to those of order relaton for real numbers However, unlke real numbers, not every par of Hermtan operators s comparable For example, there exsts no order relaton between Hermtan operators A 0 0 and B, because nether A B 0 0 nor B A 0 0 s a postve operator (note that 0 ( AB) 0 0, ( AB) 0, 0 ( B A) 0 0, and ( B A) 0 ) so that nether A B nor B A s true In ths sense, we can draw the concluson: Theorem 3 The order relaton for Hermtan operators s a partal order rather than a total order 3 The trace nequaltes for Hermtan operators Although Hermtan operators can be seen as the counterpart of real numbers, when t comes to the multplcaton operaton, they are qute dstnct from each other The product of any two real numbers s defntely a real number However, the product of two Hermtan operators may not be a Hermtan operator agan
Order relaton and trace nequaltes for Hermtan operators 53 One usual counterexample for ths case s that the product of any two dfferent Paul operators n x, y, z s not a Hermtan operator but actually a skew-hermtan operator (an operator A s a skew-hermtan operator f A A ) For example, x y z, where z has non-real but purely magnary egenvalues, whch mples that the product x y s not a Hermtan operator Nevertheless, what s the result f we consder takng the trace functon (the trace of an operator s the sum of all ts egenvalues) of the product of two Hermtan operators nstead of the product tself? For nstance, we check the last counterexample We have tr( ) tr( ) 0 (note that every Paul x operator n x, y, z has egenvalues ) The result 0 s a real number Is that a fact rather than chance? Next we wll show that t s actually a realty Moreover, by usng the spectral decomposton, we wll prove and establsh some trace nequaltes for Hermtan operators and manly for the product of two Hermtan operators whch are smlar to correspondng nequaltes for real numbers Theorem 3 If both A and B are Hermtan operators, then tr (AB) s a real number Snce y B s a Hermtan operator, t has the spectral decomposton B wth real egenvalues and correspondng orthonormal egenvectors Thus tr (AB) = tr( A ) tr( A ) A z Snce A s a Hermtan operator, A s a real number for any by Theorem Therefore tr (AB) s a real number Ths result ndcates that the product of two Hermtan operators may not be a Hermtan operator, whle the trace of ther product, however, s always a real number But ths fact does not hold for the product of three or more Hermtan
54 Y Huang operators For nstance, tr( ) x y z Based on ths theorem, we wll prove the followng nequaltes Theorem 3 (The trace nequaltes for Hermtan operators) If all operators n the followng statements are Hermtan operators, we have nequaltes below: () If A B, then tr( A) tr( B) ; () If A 0 and B 0, then tr ( AB) 0 ; (3) If B C and A 0, then tr( BA) tr( CA) and tr( AB) tr( AC) ; (4) If 0 A A and 0 B B, then 0 tr( AB) tr( AB ) () A B B A 0 tr ( B A) 0 tr( B) tr( A) 0 tr( A) tr( B) () Snce B s a postve operator, t has the spectral decomposton B wth non-negatve egenvalues and correspondng orthonormal egenvectors Thus tr (AB) = tr( A ) tr( A ) A Snce A s a postve operator, A s non-negatve for any Therefore tr ( AB) 0 (3) B C C B 0 By (), tr(( C B) A) 0 tr( CA) tr( BA) 0 tr( BA) tr( CA) Smlarly (or by tr( AB) tr( BA) and tr( AC) tr( CA) ), we can obtan tr( AB) tr( AC) (4) The frst nequalty 0 tr( AB) can be obtaned by () We can get the second nequalty by (3) as follows: tr(ab) tr ( A B) and tr( A B) tr( A B ) tr(ab) tr ( A B )
Order relaton and trace nequaltes for Hermtan operators 55 Now let s compare these trace nequaltes ()-(4) to correspondng basc nequaltes for real numbers as follows: Inequalty (): ths nequalty s the counterpart of the nequalty for real numbers: f a 0 and b 0, then ab 0 Note that for Hermtan operators A and B satsfyng A 0 and B 0, we cannot defntely get AB 0 Ths s because we even cannot ensure AB s a Hermtan operator, let alone t s a postve operator For nstance, set operator A 0 0 0 and operator B 0 where ( 0 ) But the product AB 0 0 0 s not a Hermtan operator snce ts adjont s not tself However we can see, when we take trace functon to the product AB, the correspondng nequalty, e nequalty (), holds: tr ( AB) tr( 0 0 ) tr( 0 ) 0 0 Inequalty (3): ths nequalty s the counterpart of the nequalty for real numbers: f a b and c 0, then ac bc Note that for Hermtan operators A, B, and C satsfyng B C and A 0, we cannot defntely get BA CA In fact, lke the above dscusson on Inequalty (), we even cannot ensure that both BA and CA are Hermtan operators n ths case For nstance, set B C 0 0, and A 0 Nether the product BA ( ) nor the product CA 0 0 0 s a Hermtan operator snce nether of ther adjonts s tself However we can see, when we take trace functon to these two products BA and CA, the correspondng nequalty, e nequalty (3), holds: tr ( BA) tr(( ) ) tr( ) tr( CA) tr( 0 0 ) tr( 0 ) Inequalty (4): ths nequalty s the counterpart of the nequalty for real numbers:
56 Y Huang f 0 a b and 0 a b, then 0 aa bb Note that agan, for Hermtan operators A, B, A and B satsfyng 0 A A and 0 B B, we cannot defntely get For nstance, 0 A 0 0 A I and 0 AB A B 0 B B I But the product AB 0 0 0 s not a Hermtan operator However we can see, when we take trace functon to these two products AB and A B, the correspondng nequalty, e nequalty (4), holds: 0 tr ( AB) tr( 0 0 ) tr( A B ) tr( I) Fnally, we gve two examples to show the applcaton of these trace nequaltes for Hermtan operators Suppose both Q and S are postve operators, and P s a projecton operator Thus Q S Q and 0 P I By trace nequalty (3) n Theorem 3, we get tr( P( Q S)) tr( PQ) tr( Q) Another example s that we can get tr( ) where s a densty operator and x, y, z s one of the three Paul operators We can obtan ths nequalty by trace nequalty (3) n Theorem 3 and the nequalty I I whch we have mentoned n Example 3 4 Concluson In ths paper, we have presented and proved the order relaton and some trace nequaltes of Hermtan operators by comparson wth ther counterparts of real numbers We have shown that the order relaton for Hermtan operators has reflexvty, transtvty and antsymmetry, but t has no comparablty, whch means that not every par of Hermtan operators s comparable Hence unlke real numbers, Hermtan operators are not well ordered and have a partal order rather than a total order Addtonally, the set of real numbers s closed under the multplcaton operaton, whle the set of Hermtan operators s not In other words, the product of any two real numbers s always a real number, but the product of two Hermtan operators may not be a Hermtan operator Nevertheless, consderng takng the trace functon for the product of two Hermtan operators
Order relaton and trace nequaltes for Hermtan operators 57 nstead of the product tself, real numbers and Hermtan operators exhbt some smlar aspects once agan, whch are the trace nequaltes for Hermtan operators n ths artcle Acknowledgements Ths work s supported n part by the Scentfc Research Fund of the Educaton Department of Schuan Provnce of Chna (No 8ZB035) and the Research Fund of Key Laboratory of Pattern Recognton and Intellgent Informaton Processng of Schuan Provnce n Chengdu Unversty (No MSSB-05-0) References [] M A Nelsen and I L Chuang, Quantum Computaton and Quantum Informaton, Cambrdge Unversty Press, Cambrdge, 000 Receved: October 8, 08; Publshed: November 5, 08