TENSOR PRODUCTS AND PARTIAL TRACES

Transcription

1 Lecture 2 TENSOR PRODUCTS AND PARTIAL TRACES Stéphae ATTAL Abstract This lecture cocers special aspects of Operator Theory which are of much use i Quatum Mechaics, i particular i the theory of Quatum Ope Systems. These are the cocepts of trace-class operators, tesor products of Hilbert spaces ad operators, ad above all of partial traces. Note that a geeral treatmet of partial traces i the ifiite dimesioal case, as we do i this lecture, is ot at all commo i the literature. 2.1 Trace-Class Operators Basic Defiitios Properties Dualities Hilbert-Schmidt Operators Tesor Products Tesor Products of Hilbert Spaces Tesor Products of Operators Coutable Tesor Products Partial Traces Partial Trace with Respect to a Space Partial Trace with Respect to a State Commets Stéphae ATTAL Istitut Camille Jorda, Uiversité Lyo 1, Frace attal@math.uiv-lyo1.fr 1

2 2 Stéphae ATTAL Most of this lecture deals with bouded operators, hece the full theory of ubouded operators, Spectral Theorem, etc is ot ecessary here. We assume the reader is familiar with the usual theory of bouded operators ad i particular with compact operators ad their fudametal represetatio, with the otio of positive (bouded) operators, with cotiuous fuctioal calculus for bouded operators, ad with the polar decompositio. For all these otios, read the correspodig sectios i Lecture 1 if ecessary. 2.1 Trace-Class Operators We first start with the otios of trace, trace-class operators ad their properties. We shall explore the ice duality properties which are attached to these spaces. We ed up with a closely related family of operators: the Hilbert- Schmidt operators Basic Defiitios Defiitio 2.1. Let T be a bouded positive operator o a Hilbert space H. For a fixed orthoormal basis (e ) of H we defie the quatity Tr (T) = N e, Te, which is positive (evetually ifiite). It is called the trace of T. We shall sometimes simply deote it by Tr T. Propositio 2.2. The quatity Tr T is idepedet of the choice of the orthoormal basis (e ). Proof. Let (f ) be aother orthoormal basis of H. As T is a positive operator, it admits a square root T. We the have (the series below beig all made of positive terms, iterchagig the sums is allowed)

3 2 TENSOR PRODUCTS AND PARTIAL TRACES 3 f, T f = 2 T f = 2 T f, e m m = f, 2 T e m m = 2 T e m m = m e m, T e m. This proves the idepedece property. We ow defie the trace-class operators for geeral bouded operators. Defiitio 2.3. A bouded operator T o H is trace-class if Tr T <. Theorem ) Every trace-class operator is compact. 2) A compact operator T, with sigular values (λ ), is trace-class if ad oly if λ <. (2.1) I that case we have Tr T = λ. (2.2) Proof. Assume first that T is positive ad trace-class. Let (e ) be a orthoormal family i H, the Hece 2 T e = e, Te Tr T <. T e coverges to 0 ad T is compact. I particular the operator T = T T is also compact. Now if T is a bouded operator which is trace-class, the T is trace-class by defiitio. Hece T is compact. By the Polar Decompositio the operator T = U T is also compact. We have proved 1). If T is compact the T = U T is also compact. The operator T ca thus be decomposed as T = λ v v,

4 4 Stéphae ATTAL for some positive λ s ad some orthoormal basis (v ). Takig the trace of T with respect to the orthoormal basis (v ) gives Tr T = λ. It is easy to coclude ow. Theorem 2.5. If T is a trace-class operator o H the for every orthoormal basis (e ) of H the series Tr T = e, Te is absolutely coverget. Its sum is idepedet of the choice of the orthoormal basis (e ). We always have Tr T Tr T. (2.3) Proof. By Theorem 2.4 the operator T is compact, hece it admits a decompositio as T = λ u v, =1 for some positive λ s ad some orthoormal bases (u ), (v ) of H. We the have e k, Te k λ e k, u v, e k k k ( ) 1/2 ( ) 1/2 λ e k, u 2 v, e k 2 k k = λ v u = λ which is fiite by Theorem 2.4. We have proved the absolute covergece ad the iequality (2.3). Let us prove the idepedece with respect to the choice of the orthoormal basis. Let (f ) be ay orthoormal basis of H. We the have (usig the absolute covergece above ad Fubii s Theorem)

5 2 TENSOR PRODUCTS AND PARTIAL TRACES 5 f k, Tf k = λ f k, u v, f k k k = λ v, f k f k, u k = λ u, v. This quatity does ot deped o the choice of (f ). By the way, we have proved the followig characterizatio. Corollary 2.6. A bouded operator T o H is trace-class if ad oly if it ca be decomposed as T = λ u v where (u ) ad (v ) are orthoormal families i H ad (λ ) is a sequece of positive umbers such that λ <. Defiitio 2.7. The set of trace-class operators o H is deoted by L 1 (H). We also put T 1 = Tr T Properties Let us start with the very fudametal properties of trace-class operators. Theorem ) The set L 1 (H) is a vector space, the mappig 1 is a orm o L 1 (H). Whe equipped with the orm 1, the space L 1 (H) is a Baach space. This orm always satisfies T T 1. 2) The space L 1 (H) is two-sided ideal of B(H). For every T L 1 (H) ad X B(H) we have 3) The space L 0 (H) is 1 -dese i L 1 (H). X T 1 X T 1. (2.4) 4) If T belogs to L 1 (H) the so does T ad we have Tr (T ) = Tr (T).

6 6 Stéphae ATTAL 5) For ay two bouded operators S ad T such that oe of them is trace-class, we have Tr (S T) = Tr (T S). Proof. 1) I order to prove that L 1 (H) is a vector space ad that 1 is a orm, the oly o-trivial poit to be proved is the followig: for every positive bouded operators S ad T o H we have Let us prove this fact. Let Tr S + T Tr S + Tr T. S = U S, T = V T, S + T = W S + T be the polar decompositios of S, T ad S + T. Let (e ) be a orthoormal basis of H. We have e, S + T e = e, W (S + T) e = e, W U S e + e, W V T e S U W e S e + + T V W e T e ( ) 1/2 ( S U 2 W e 1/2 2) S e + ( + ) 1/2 ( T V 2 W e 1/2 2) T e. But we have S U 2 W e = e, W U S U W e = Tr (W U S U W). Choosig a orthoormal basis (f ) which is adapted to the decompositio H = er W (er W), we see that Tr (W U S U W) Tr (U S U ). Applyig the same trick for U shows that

7 2 TENSOR PRODUCTS AND PARTIAL TRACES 7 Tr (U S U ) Tr ( S ). Altogether, we have proved that e, S + T e Tr ( S ) + Tr ( T ). This proves the vector space structure of L 1 (H) ad also the orm property of 1. By the usual properties of the operator orm we have T 2 = T T = T 2 = T T = T 2. As T is self-adjoit its orm is give by sup x, T x. x =1 The above quatity is clearly smaller tha Tr T. We have proved that T T 1. As a cosequece, if (T ) is Cauchy sequece i B(H) for the orm 1 the it is a Cauchy sequece for the usual operator orm. Hece it coverges i operator orm to a bouded operator T. To each operator T is associated a sequece λ () of sigular values. By hypothesis this family of sequeces is l 1 -coverget. We have proved that it is also l -coverget ad that it coverges to the sequece λ ( ) λ () has to be λ ( ) of sigular values of T. Hece the l 1 -limit of. This meas that the sequece (T ) coverges to T i 1. We have proved 1). 2) We shall eed the followig useful lemma. Lemma 2.9. Every bouded operator is a liear combiatio of four uitary operators Proof. Let B B(H), the we ca write B as a liear combiatio of two self-adjoit operators: B = 1 2 (B + B ) i 2 (B B ). Now, if A is a bouded self-adjoit operator, with A 1 (which we ca assume without loss of geerality), the operators A ± i I A 2 are uitary ad they sum up to 2 A. We are ow back to the proof of Property 2). By Lemma 2.9, it is sufficiet to prove that for all T L 1 (H) ad all uitary operator U we have that U T ad T U belog to L 1 (H). But we have U T = T ad hece U T belog to L 1 (H). We also have T U = U T U ad the same coclusio. We have proved the two-sided ideal property.

8 8 Stéphae ATTAL Let us prove the orm iequality ow. Whe U is uitary, we see from the cosideratios above that U T 1 T 1. O the other had, every bouded B operator is a liear combiatio of four uitary operators with coefficiets i the liear combiatio beig smaller tha B. It is easy to coclude. The property 3) comes immediately from the caoical form of trace-class operators (Corollary 2.6). 4) Cosider the polar decompositios T = U T ad T = V T. We the have T = V T U. By Property 2, the operator T is trace-class, ad so is T. Fially Tr (T ) = e, T e = Te, e = Tr T. 5) Let S ad T be such that S is bouded ad T is trace-class, for example. By Lemma 2.9 we ca reduce the problem to the case where S is uitary. Let (e ) be a orthoormal basis ad f = S e for all. We have Tr (S T) = = = e, S T e S e, T e f, T S f = Tr (T S). We ed up this subsectio with a useful characterizatio. Theorem A bouded operator T o H is trace-class if ad oly if g, T h < for all orthoormal families (g ) ad (h ) i H. Moreover there exists orthoormal families (g ) ad (h ) i H such that g, T h = T 1. Proof. Assume first that T is trace-class. The T ca be decomposed as (Corollary 2.6) T = λ u v, for some orthoormal families (u ) ad (v ) ad a summable sequece (λ ) of positive scalars. Hece we have

9 2 TENSOR PRODUCTS AND PARTIAL TRACES 9 g, T h = λ k g, u k v k, h k ( ) 1/2 ( ) 1/2 λ k g, u k 2 v k, h 2 k k = k λ k u k v k λ k <. We have proved the first part of theorem i oe directio. Note that, i the above computatio, if oe had chose g = u ad h = v for all we would have g, T h = λ k g, u k v k, h = λ k = T 1. k k This proves the secod part of the theorem i that case. Coversely, let T be a bouded operator o H satisfyig g, T h < for all orthoormal families (g ) ad (h ) i H. Let T = U T be its polar decompositio. Choose a orthoormal sequece (h ) i Ra T ad put g = U h for all. Sice U is isometric o Ra T we have U g = h for all. Hece we have g, T h = h, T h = h, T h. By our hypothesis o T this proves that h ), T h is fiite. Complete the family (h ) ito a orthoormal basis ( h of H, by completig with orthoormal vectors i Ra T = er T. The h, T h = h, T h ad hece is fiite. This proves that Tr T < ad that T is trace-class.

10 10 Stéphae ATTAL Dualities Defiitio Let H be ay separable complex Hilbert space. For ay trace-class operator T ad bouded operator X o H we write which is well-defied by Theorem 2.8. Theorem T, X = Tr (T X), 1) The Baach space (L 1 (H), 1 ) is isometrically isomorphic to the dual of (L (H), ) uder the correspodece T T, L (H). 2) The Baach space (B(H), ) is isometrically isomorphic to the dual of (L 1 (H), 1 ) uder the correspodece X, X L1(H). Proof. 1) For all T L 1 (H) the mappig X T, X = Tr (T X) defie a liear form λ T o L (H). By (2.3) ad (2.4) we have Tr (X T) Tr X T X T 1. Hece λ T is a cotiuous liear form, with orm λ T T 1. Takig X = U where U is the partial isometry i the polar decompositio of T, gives the equality of orms: λ T = T 1. Coversely, let λ be a cotiuous liear form o L (H). The mappig B : (u, v) λ( u v ) is a sesquiliear form o H ad it satisfies B(u, v) λ u v. Hece there exists a bouded operator T o H such that λ( u v ) = v, T u = Tr (T u v ). Cosider the polar decompositio T = U T of T ad choose a orthoormal basis (u ) i Ra T. As a cosequece the family {v = U u ; N} forms a orthoormal basis of Ra T. The operator X = is fiite rak ad orm 1. We have This shows that λ(x ) = u i v i i=1 v i, T u i = i=1 u i, T u i. i=1

11 2 TENSOR PRODUCTS AND PARTIAL TRACES 11 u i, T u i λ. j=1 Completig the orthoormal basis (u ) of Ra T ito a orthoormal basis of H (by choosig ay orthoormal basis of (Ra T ) = er T ), we have foud a orthoormal basis (f ) of H such that f, T f λ <. N This shows that T is trace-class ad that T 1 λ. As the fiite rak operators are dese i the compact operators for the operator orm, passig to the limit o the idetity λ(x ) = Tr (T X ) shows that λ(x) = Tr (T X) for all X L (H). Fially, we have λ(x) = Tr (T X) T 1 X. Hece λ T 1. This proves the equality of orms ad the part 1) is completely proved. 2) The proof is very similar to the oe of 1) above. Let B B(H) ad defie the liear form λ B o L 1 (H) by λ B (T) = Tr (B T). Agai, the iequality λ B (T) B T 1 proves that λ B is cotiuous with λ B B. Take T = x y for some x, y H such that x = y = 1, this gives λ B (T) = y, B x. But we have B = sup{ y, B x ; x, y H, x = y = 1} = sup{ λ B (T) ; T = x y, x = y = 1} sup{ λ B (T) ; T L 1 (H), T 1 = 1} = λ B. Hece we have proved the equality λ B = B. Coversely, if λ is a cotiuous liear form o L 1 (H) the (u, v) λ( u v ) defies a bouded sesquiliear form o H. Hece there exists a bouded operator X such that If T L 1 (H) is of the form λ( u v ) = Tr (X u v ).

12 12 Stéphae ATTAL T = λ u v the it is easy to check that λ u v N coverges to T i L 1 (H) whe N teds to +. As λ is cotiuous we get λ(t) = Tr (X T) for every T L 1 (H). It is easy to coclude ow. Defiitio As B(H) is the dual of L 1 (H) it iherits a -weak topology, that is, the topology which makes the mappigs λ T (X) = Tr (X T) beig cotiuous o B(H) for all T L 1 (H). It is easy to see that it coicides with the topology iduced by the semiorms e,f (X) = e, X f where (e ) ad (f ) ru over all sequeces i H such that e 2 < ad f 2 <. This fact ca be easily deduced from the caoical form of trace-class operators (Corollary 2.6). The topology above is ofte called the σ-weak topology o B(H). The followig result is a very classical cosequece of the theory of Baach space duals (see [RS80], Theorem IV.20). Corollary Every σ-weakly cotiuous liear form λ o B(H) is of the form λ(x) = Tr (T X) for some T L 1 (H). This trace-class operator T associated to λ is uique Hilbert-Schmidt Operators We ed up this sectio with the aother special family of operators which are closely coected to the trace-class operators: the Hilbert-Schmidt operators. Defiitio A operator T B(H) is Hilbert-Schmidt if Tr (T T) <.

13 2 TENSOR PRODUCTS AND PARTIAL TRACES 13 The set of Hilbert-Schmidt operators o H is deoted by L 2 (H). The followig properties are obtaied i a similar way as those of L 1 (H). We leave the proofs to the reader. Theorem ) The space L 2 (H) is a two-sided ideal of B(H). 2) A operator T B(H) is Hilbert-Schmidt if ad oly if there exists a orthoormal basis (e ) of H such that T e 2 <. I that case the series coverges to the same sum for all orthoormal basis of H. 3) The mappig S, T = Tr (S T) defies a scalar product o L 2 (H) which gives it a Hilbert space structure. The orm 2 associated to this scalar product satisfies A A 2 A 1. 4) Every Hilbert-Schmidt operator is a compact operator. Coversely, a compact operator is Hilbert-Schmidt if ad oly if the sequece of its sigular values (λ ) satisfies λ 2 <. Every trace-class operator is Hilbert-Schmidt. 2.2 Tesor Products As already explaied i the itroductio of this chapter, we leave the traceclass operators for a while ad start with a completely differet topic: the tesor products of Hilbert spaces ad of operators. The coectios betwee the two otios appear i ext sectio. The otio of tesor product of Hilbert spaces ad of tesor product of operators are key cocepts i Quatum Mechaics, where they appear each time that several quatum systems are ivolved ad iteract with each other.

14 14 Stéphae ATTAL Tesor Products of Hilbert Spaces Defiitio Let H 1, H 2 be two Hilbert spaces. Let ϕ H 1 ad ψ H 2. We defie ϕ ψ to be the bi-atiliear form o H 1 H 2 give by ϕ ψ (ϕ, ψ ) = ϕ, ϕ ψ, ψ. Let E be the set of all fiite liear combiatios of such forms, actig i the obvious way o H 1 H 2. We equip E with a iteral product defied by ϕ ψ, ϕ ψ = ϕ, ϕ ψ, ψ = ϕ ψ (ϕ, ψ), ad its atural extesio to liear combiatios. Propositio The product, o E is well-defied ad positive defiite. Proof. To show that, is well-defied, we eed to show that λ, λ does ot deped o the choice of the liear combiatios represetig λ ad λ. It is thus sufficiet to show that if µ is the ull liear combiatio the λ, µ = 0 for all λ E. Assume λ is of the form i=1 α i ϕ i ψ i, we have λ, µ = α i ϕ i ψ i, µ = i=1 = α i ϕ i ψ i, µ i=1 α i µ (ϕ i, ψ i ) i=1 = 0. This proves that the product is well-defied. Let us prove it is positive defiite. Let λ = i=1 α i ϕ i ψ i agai, ad let (e i ) i=1,...,k ad (f i ) i=1,...,l be orthoormal bases of the spaces geerated by {ϕ 1,..., ϕ } ad {ψ 1,..., ψ } respectively. The k l λ = α ij e i f j, for some coefficiets α ij, ad thus i=1 j=1 λ, λ = k i=1 j=1 l α ij 2. Thus λ, λ is positive ad λ, λ = 0 if ad oly if λ = 0.

15 2 TENSOR PRODUCTS AND PARTIAL TRACES 15 Defiitio As a cosequece (E,,, ) is a pre-hilbert space. We deote by H 1 H 2 its completio. It is called the tesor product of H 1 by H 2. Propositio Let N ad M be sets of idices i N such that (e i ) i N ad (f j ) j M are ay orthoormal basis of H 1 ad H 2 respectively. The (e i f j ) (i,j) N M is a orthoormal basis of H 1 H 2. Proof. The fact that the set {e i f j ; (i, j) N M} is a orthoormal family is clear for the defiitio of the scalar product o H 1 H 2. Let us check it forms a basis. Let S be the vector space geerated by {e i f j ; (i, j) N M}. Let ϕ ψ be a elemet of E, the we have decompositios of the form ϕ = c i e i with c i 2 < i N i N ad ψ = d j f j j M As a cosequece we have with (i,j) N M d j 2 <. j M c i d j 2 < ad the vector i,j c id j e i f j belogs to S, the closure of S. But by direct computatio oe sees that, for all N, M N ψ ϕ i N i N c i d j e i f j j M j M 2 = i N j M c i 2 d j 2 i N i N c i 2 d j 2 j M j M ad hece coverges to 0 whe N ad M ted to +. Thus E is icluded i S. This give the desity of S i H 1 H Tesor Products of Operators Defiitio Let A be a operator o H 1, with a dese domai Dom A, ad B be a operator o H 2, with dese domai Dom B. Let D deote the space Dom A Dom B, that is, the space of fiite liear combiatios of ϕ ψ with ϕ Dom A ad ψ Dom B. Clearly D is dese i H 1 H 2. Oe defies the operator A B o D by (A B)(ϕ ψ) = Aϕ Bψ, ad its obvious liear extesio to all D.

16 16 Stéphae ATTAL Propositio The operator A B is well-defied. If A ad B are closable operators the so is A B. Proof. Beig well-defied i this case meas that if the i=1 m c i ϕ i ψ i = d j ϕ j ψ j (2.5) i=1 j=1 m c i Aϕ i Bψ i = d j Aϕ j Bψ j. (2.6) But, passig through orthorormal bases (e i ) i N ad (f j ) j M for H 1 ad H 2, the idetity (2.5) is equivalet to c i ϕ i, e j ψ i, f k = i=1 j=1 k d i ϕ i, e j ψ i, f k, for all j, k. Applyig A B to i=1 c i ϕ i ψ i but i the basis {e j f k ; j N, k M} gives the idetity (2.6) immediately. i=1 Now, if g Dom A Dom B ad f Dom A Dom B we have obviously (A B) f, g = f, (A B )g. Thus Dom(A B) cotais Dom A Dom B which is dese. This meas that the operator A B is closable. I particular we have also proved that (A B) A B. Defiitio If A ad B are two closable operators we call tesor product of A by B the closure of A B. It is still deoted by A B. Propositio If A ad B are bouded operators o H 1 ad H 2 respectively the A B is a bouded operator o H 1 H 2 ad we have A B = A B. Proof. Assume that f H 1 H 2 ca be decomposed as f = i,j=1 c ij e i f j i some orthoormal basis of H 1 H 2. The we have

17 2 TENSOR PRODUCTS AND PARTIAL TRACES 17 2 (A I)f 2 = (A I) c ij e i f j i,j=1 2 = c ij Ae i f j i,j=1 2 = c ij Ae i j=1 i=1 A 2 c ij 2 j=1 i=1 = A 2 f 2. We have proved that A i is bouded o H 1 H 2, with A I A. Hece its closure is a bouded operator o H 1 H 2, with the same orm estimate. I the same way, we obtai I B is bouded, with I B B. As A B = (A I)(I B) (at least obviously o H 1 H 2 ) we get that A B is bouded ad A B A B. Let ε > 0 be fixed, choose ϕ ad ψ i H 1 ad H 2 respectively, such that Aϕ A ε, Bψ B ε, ϕ = ψ = 1. The (A B)(ϕ ψ) = Aϕ Bψ A B ε A ε B + ε 2. Makig ε go to 0, this proves that (A B) A B. The equality of orms is proved. Tesor products of operators preserve most of the classes of bouded operators. Theorem Let H 1 ad H 2 be (separable) Hilbert spaces. Let A ad B be bouded operators o H 1 ad H 2 respectively. 1) If A ad B are compact operators the A B is a compact operator o H 1 H 2. 2) If A ad B are trace-class operators the A B is trace-class too. I particular we have A B 1 = A 1 B 1 (2.7) ad

18 18 Stéphae ATTAL Tr (A B) = Tr (A) Tr (B). (2.8) 3) If A ad B are Hilbert-Schmidt operators the A B is Hilbert-Schmidt too. I particular we have Proof. We start by oticig that A B 2 = A 2 B 2. (2.9) u 1 v 1 u 2 v 2 = u 1 u 2 v 1 v 2 as ca be easily see by applyig these operators to elemets of the form f g i H 1 H 2, the passig to liear combiatios ad fially passig to the limit to geeral elemets of H 1 H 2. Now the mai characterizatios of compact (resp. trace-class, resp. Hilbert- Schmidt) operators is that they are represeted as λ u v for some orthoormal families (u ) ad (v ) ad a sequece of positive scalars (λ ) which coverges to 0 (resp. is summable, resp. is square summable). It is ow very easy to coclude i all the three cases Coutable Tesor Products All alog this book we shall also make heavy use of the otio of coutable tesor products of Hilbert spaces. Let us detail here how they are costructed. Let (H ) be a sequece of Hilbert spaces. All the fiite tesor products H 0... H are well-defied by the above costructios. But it is ot clear how oe ca rigorously defie the coutable tesor product H. N The idea is the followig: oe defies N H as the iductive limit of the spaces H 0... H, whe teds to +. This ca be achieved oly if oe fids a way to see the space H 0... H as a subspace of H 0... H +1 for all. The way this is obtaied is by choosig a uit vector u H for each, which will costitute a referece vector of H. The elemets f 0... f of H 0... H are idetified to f 0... f u +1 as a elemet of H 0... H +1. This embeddig is isometric, it preserves the scalar

19 2 TENSOR PRODUCTS AND PARTIAL TRACES 19 product, etc... Oe ca easily go the limit + i this costructio. This gives rise to the followig defiitio. Let (H ) be a sequece of Hilbert spaces. Choose a sequece (u ) such that u H ad u = 1, for all. This sequece is called a stabilizig sequece for N H. The space N H is defied as the closure of the pre-hilbert space of vectors of the form N such that f H for all ad f = u for all but a fiite umber of. The scalar product o that space beig obviously defied by g = f, g. N N f, N The above ifiite product is fiite for all but a fiite umber of its terms are equal to u, u = 1. This is all for the defiitio of a coutable tesor product of Hilbert space. Note that the costructio depeds o the choice of the stabilizig sequece (u ) which has bee chose iitially. Hece, whe dealig with such tesor products, oe should be clear about the choice of the stabilizig sequece. f 2.3 Partial Traces The otio of partial trace appears aturally with the tesor products of Hilbert spaces ad of operators. This otio is essetial i the study of ope quatum systems Partial Trace with Respect to a Space Defiitio Let H ad be separable Hilbert spaces. For ay g cosider the operator g : H H f f g. It is clearly a bouded operator from H to H, with orm g. Its adjoit is the operator defied by g : H H u v g, v u,

20 20 Stéphae ATTAL ad its atural extesio by liearity ad cotiuity. Note that its operator orm is also equal to g. I particular, if T is a bouded operator o H the the operator is a bouded operator o H. g T g Lemma If T is a trace-class operator o H the for all f the operator is a trace-class operator o H. f T f Proof. Let f ad assume it is orm 1, without loss of geerality. Let (g ) ad (h ) be ay orthoormal family i H. We have g, f T f h = g f, T h f. We ow use the characterizatio of Theorem 2.10 i both directios. As T is trace-class o H the right had side is fiite for (g f) ad (h f) are particular orthoormal sequeces i H. This meas that the left had side above is fiite for all orthoormal sequeces (g ) ad (h ), which esures that f T f is trace-class. We ca ow prove the mai theorem which defies ad characterizes the partial traces. Theorem Let H ad be two separable Hilbert spaces. Let T be a trace-class operator o H. The, for ay orthoormal basis (g ) of, the series Tr (T) = g T g (2.10) is 1 -coverget. The operator Tr (T) defied this way does ot deped o the choice of the orthoormal basis (g ). The operator Tr (T) is the uique trace-class operator o H such that for all B B(H). Tr (Tr (T) B) = Tr (T (B I)) (2.11) Proof. Let (g ) be a orthoormal basis of. As each g T g is traceclass o H (Lemma 2.27) the, by Theorem 2.10, there exist for all N orthoormal families (e m) m N ad (fm) m N i H such that g T g 1 = m e m, g T g f m.

21 2 TENSOR PRODUCTS AND PARTIAL TRACES 21 Hece we have g T g = 1 = e m, g T g fm m e m g, T fm g. m Note that the families (e m g ),m N ad (fm g ),m N are orthoormal i H, hece, by Theorem 2.10 agai, the above quatity is fiite, for T is trace-class o H. We have proved that the series g T g is 1 -coverget. The operator Tr (T) = g T g is a well-defied, trace-class operator o H. Let us check that this operator is idepedet of the choice of the basis (g ). Let (h ) be aother orthoormal basis of, we have x, Tr (T) y = = = k = k x g, T (y g ) g, h k h l, g x h k, T (y h l ) k l h l, h k x h k, T (y h l ) l x h k, T (y h k ). This proves that Tr (T) is also equal to k h k T h k. We have proved the idepedece property. We prove ow the characterizatio (2.11). Let B be ay bouded operator o H, we have Tr (T (B I)) = e g m, T (B I) e g m m = e g m, T (B e g m ) m = e, g m T g m B e m = Tr (Tr (T)B). We have proved that Tr (T) satisfies the relatio (2.10). We have to prove uiqueess ow, ad the theorem will be completely proved. If S ad S are two trace-class operators o H satisfyig (2.10), the i particular Tr ((S

22 22 Stéphae ATTAL S ) B) = 0 for all B B(H). This is i particular true for all B L (H) hece, by Theorem 2.12, this implies that S S = 0. We ow give a list of the most useful properties of the partial trace. They are all straightforward applicatios of the defiitio ad theorem above, we leave the proofs to the reader. Theorem Let H ad be two separable Hilbert spaces, let T be a trace-class operator o H. 1) If T is of the form A B, with A ad B beig trace-class, the Tr (T) = Tr (B) A. 2) We always have Tr (Tr (T)) = Tr (T). 3) If A ad B are bouded operators o H the Tr ((A I) T (B I)) = A Tr (T) B Partial Trace with Respect to a State I applicatios to Quatum Mechaics oe sometimes also eeds the otio of partial trace with respect to a give trace-class operator, or more precisely with respect to a state. This partial trace is differet, but related to the previous oe. Oce agai this partial trace is better defied through a theorem which characterizes it. Theorem Let H ad be two separable Hilbert spaces. Let T be a fixed trace-class operator o, with caoical form T = λ u v. The, for ay bouded operator X o H, the series λ v X u (2.12) is operator-orm coverget o H. Its limit, deoted by Tr T (X), is a bouded operator o H. The operator Tr T (X) is the oly bouded operator o H satisfyig Tr (Tr T (X) S) = Tr (X (S T)) (2.13)

23 2 TENSOR PRODUCTS AND PARTIAL TRACES 23 for all S L 1 (H). Proof. The series λ v X u is operator-orm coverget for λ v X u λ v X u λ v X u λ X <. Hece it defies a bouded operator, which we deote by Tr T (X). We shall ow check that it satisfies the relatio (2.13). Recal that the family (v ) appearig i the caoical form of T is a orthoormal family o. Hece we ca exted the family (v ) ito a orthoormal basis (ṽ ) of. Recall that T vaishes o the orthogoal complemet of the family (v ). Fially, recall that Tv m = λ m u m for all m. Let S be a trace-class operator o H ad let (e ) be a orthoormal basis of H. We have Tr (X (S T)) =,m e ṽ m, X (S T) e ṽ m =,m e ṽ m, X (S e T ṽ m ) =,m e v m, X (S e T v m ) =,m λ m e v m, X (S e u m ) = λ m e, v m X u m S e,m = e, Tr T (X) S e = Tr (Tr T (X) S). The required relatio is proved. We just have to prove uiqueess ow. If X 1 ad X 2 are two bouded operators o H such that for all S L 1 (H), the we must have Tr (X i S) = Tr (X(S T)) Tr ((X 1 X 2 )S) = 0

24 24 Stéphae ATTAL for all S L 1 (H). By Theorem 2.12, this implies X 1 X 2 = Commets The secod family of partial traces Tr T (X) are called with respect to a state for, i most of the quatum mechaical situatios where they appear, the operator T is a quatum state, a desity matrix (see Lecture 4). The partial traces with respect to a space, Tr (T), are i geeral simply called partial traces (if everyoe is clear about the space which is cocered!). For those with respect to a state T oe geerally makes it more precise: partial trace with respect to T. Note that the two partial traces could be formally related by the followig formulas: Tr T (X) = Tr (X (I T)), (2.14) ad Tr (T) = Tr I (T), (2.15) which ca be obtaied easily from (2.11) ad (2.12), but which are ot quite correct i geeral! Ideed, i full geerality the operator X (I T) is ot traceclass, so its partial trace Tr is ot defied; the operator I is ot trace-class i geeral either, so the trace Tr I is ot defied. Actually these formula are correct oce X (I T) is trace-class, reap. I is trace-class. This is to say, they are meaigful ad true i fiite dimesio typically. Oce ca weake the defiitios so that the idetities (2.14) ad (2.15) are always true, by askig weaker covergeces i the series that defie these partial traces. But it is ot worth developig this here. Notes There is o true refereces which are specially dedicated to trace-class operators or to tesor products. The refereces we have give i Lecture 1 are still valid here. Oly the discussio o partial traces is ot easy to fid i the literature. It is most ofte ot treated, or treated oly i the fiite-dimesioal case. As a cosequece the treatmet we give here o partial trace is origial, above all by the approach we take with partial bras ad kets, which is very close to the way the physicists usually uderstad partial traces.

25 2 TENSOR PRODUCTS AND PARTIAL TRACES 25 Refereces [RS80] Michael Reed ad Barry Simo. Methods of moder mathematical physics. I. Academic Press Ic. [Harcourt Brace Jovaovich Publishers], New York, secod editio, Fuctioal aalysis.