DIRAC s BRA AND KET NOTATION B. Zwebach October 7, 2013 Cotets 1 From er products to bra-kets 1 2 Operators revsted 5 2.1 Projecto Operators..................................... 6 2.2 Adjot of a lear operator................................. 8 2.3 Hermta ad Utary Operators.............................. 9 3 No-deumerable bass 11 1 From er products to bra-kets Drac veted a useful alteratve otato for er products that leads to the cocepts of bras ad kets. The otato s sometmes more effcet tha the covetoal mathematcal otato we have bee usg. It s also wdely although ot uversally used. It all begs by wrtg the er product dfferetly. The rule s to tur er products to bra-ket pars as follows ( u,v ) (u v). (1.1) Istead of the er product comma we smply put a vertcal bar! We ca traslate our earler dscusso of er products trvally. I order to make you famlar wth the ew look we do t. We ow wrte (u v) = (v u), as well as (v v) 0 for all v, whle (v v) = 0 f ad oly f v = 0. We have learty the secod argumet (u c 1 v 1 + c 2 v 2 ) = c 1 (u v 1 ) +c 2 (u v 2 ), (1.2) for complex costats c 1 ad c 2, but atlearty the frst argumet (c 1 u 1 + c 2 u 2 v) = c 1 (u 1 v) + c 2 (u 2 v). (1.3) Two vectors u ad v for whch (u v) = 0 are orthogoal. For the orm: v 2 = (v v). The Schwarz equalty, for ay par u ad v of vectors reads (u v) u v. For a gve physcal stuato, the er product must be defed ad should satsfy the axoms. Let us cosder two examples: 1
a 1 b 1. Let a = ad b = 1 be two vectors a complex dmesoal vector space of dmeso a 2 b 2 two. We the defe You should cofrm the axoms are satsfed. (a b) a 1 b 1 + a 2 b 2. (1.4) 2. Cosder the complex vector space of complex fucto f (x) C wth x [0,L]. Gve two such fuctos f (x),g(x) we defe L (f g) f (x)g(x)dx. (1.5) 0 The verfcato of the axoms s aga qute straghtforward. A set of bass vectors {e } labelled by the tegers = 1,..., satsfyg (e e j ) = δ j, (1.6) s orthoormal. A arbtrary vector ca be wrtte as a lear superposto of bass states: We the see that the coeffcets are determed by the er product v = α e, (1.7) We ca therefore wrte (e k v) = e k α e = α e k e = α k. (1.8) v = e (e v). (1.9) To obta ow bras ad kets, we reterpret the er product. We wat to splt the er product to two gredets (u v) (u v). (1.10) Here v) s called a ket ad (u s called a bra. We wll vew the ket v) just as aother way to represet the vector v. Ths s a small subtlety wth the otato: we thk of v V as a vector ad also v) V as a vector. It s as f we added some decorato ) aroud the vector v to make t clear by specto that t s a vector, perhaps lke the usual top arrows that are added some cases. The label the ket s a vector ad the ket tself s that vector! Bras are somewhat dfferet objects. We say that bras belog to the space V dual to V. Elemets of V are lear maps from V to C. I covetoal mathematcal otato oe has a v V ad a lear fucto φ V such that φ(v), whch deotes the acto of the fucto of the vector v, s a umber. I the bracket otato we have the replacemets v v), φ (u, (1.11) φ u (v) (u v), 2
where we used the otato (6.6). Our bras are labelled by vectors: the object sde the ( s a vector. But bras are ot vectors. If kets are vewed as colum vectors, the bras are vewed as row vectors. I ths way a bra to the left of a ket makes sese: matrx multplcato of a row vector tmes a colum vector gves a umber. Ideed, for vectors a 1 b 1 we had Now we thk of ths as a 2 b 2 a =, b =.. a b (1.12) (a b) = a 1 b 1 + a 2 b 2 +...a b (1.13) ( ) (a = a 1,a 2...,a, ad matrx multplcato gves us the desred aswer (a b) = ( a 1,a...,a 2 b 1 b 2 b 1 b 2 b) = (1.14). b ) = a 1 b 1 + a 2 b 2 +...a b. (1.15). b Note that the bra labeled by the vector a s obtaed by formg the row vector ad complex cojugatg the etres. More abstractly the bra (u labeled by the vector u s defed by ts acto o arbtrary vectors v) as follows (u : v) (u v). (1.16) As requred by the defto, ay lear map from V to C defes a bra, ad the correspodg uderlyg vector. For example let v be a geerc vector: v 1 v 2 v =, (1.17). v A lear map f (v) that actg o a vector v gves a umber s a expresso of the form f (v) = α 1 v 1 + α 2 v 2 +...α v. (1.18) It s a lear fucto of the compoets of the vector. The lear fucto s specfed by the umbers α, ad for coveece (ad wthout loss of geeralty) we used ther complex cojugates. Note that we eed exactly costats, so they ca be used to assemble a row vector or a bra ( (α = α 1,α 2,...,α ) (1.19) 3
ad the assocated vector or ket Note that, by costructo α) = α 1 α2 (1.20). α f (v) = (α v). (1.21) Ths llustrates the pot that () bras represet dual objects that act o vectors ad () bras are labelled by vectors. Bras ca be added ad ca be multpled by complex umbers ad there s a zero bra defed to gve zero actg o ay vector, so V s also a complex vector space. As a bra, the lear superposto s defed to act o a vector (ket) c) to gve the umber (ω α(a + β(b V, α,β C, (1.22) α(a c) + β(b c). (1.23) For ay vector v) V there s a uque bra (v V. If there would be aother bra (v t would have to act o arbtrary vectors w) just lke (v : (v w) = (v w) (w v) (w v ) = 0 (w v v ) = 0. (1.24) I the frst step we used complex cojugato ad the secod step learty. Now the vector v v must have zero er product wth ay vector w, so v v = 0 ad v = v. We ca ow recosder equato (1.3) ad wrte a extra rght-had sde ( ) (α α α 1 a 1 + α 2 a 2 b) = 1(a 1 b) + α 2(a 2 b) = 1(a 1 + α 2 (a 2 b) (1.25) so that we coclude that the rules to pass from kets to bras clude v) = α α 1 a 1 ) + α 2 a 2 ) (v = 1(a 1 + α 2 (a 2. (1.26) For smplcty of otato we sometmes wrte kets wth labels smpler tha vectors. Let us recosder the bass vectors {e } dscussed (1.6). The ket e ) s smply called ) ad the orthoormal codto reads ( j) = δ j. (1.27) The expaso (1.7) of a vector ow reads v) = )α, (1.28) As (1.8) the expaso coeffcets are α k = (k v) so that v) = )( v). (1.29) 4
2 Operators revsted Let T be a operator a vector space V. Ths meas that actg o vectors o V t gves vectors o V, somethg we wrte as Ω : V V. (2.30) We deote by Ω a) the vector obtaed by actg wth Ω o the vector a): The operator Ω s lear f addtoally we have a) V Ω a) V. (2.31) ( ) Ω a) + b) = Ω a) +Ω b), ad Ω(α a)) = α Ω a). (2.32) Whe kets are labeled by vectors we sometmes wrte Ωa) Ω a), (2.33) It s useful to ote that a lear operator o V s also a lear operator o V Ω : V V, (2.34) We wrte ths as (a (a Ω V. (2.35) The object (a Ω s defed to be the bra that actg o the ket b) gves the umber (a Ω b). We ca wrte operators terms of bras ad kets, wrtte a sutable order. As a example of a operator cosder a bra (a ad a ket b). We clam that the object Ω = a)(b, (2.36) s aturally vewed as a lear operator o V ad o V. Ideed, actg o a vector we let t act as the bra-ket otato suggests: Ω v) a)(b v) a), sce (b v) s a umber. (2.37) Actg o a bra t gves a bra: (w Ω (w a)(b (b, sce (w a) s a umber. (2.38) Let us ow revew the descrpto of operators as matrces. The choce of bass s ours to make. For smplcty, however, we wll usually cosder orthoormal bases. Cosder therefore, two vectors expaded a orthoormal bass { )}: a) = )a, b) = )b. (2.39) 5
Assume b) s obtaed by the acto of Ω o a): Ω a) = b) Ω )a = )b. (2.40) Actg o both sdes of ths vector equato wth the bra (m we fd (m Ω )a = (m )b = b m (2.41) We ow defe the matrx elemets Ω m (m Ω ). (2.42) so that the above equato reads Ω m a = b m, (2.43) whch s the matrx verso of the orgal relato Ω a) = b). The chose bass has allowed us to vew the lear operator Ω as a matrx, also deoted as Ω, wth matrx compoets Ω m : Ω 11 Ω 12...... Ω 1N Ω 21 Ω 22...... Ω 2N Ω, wth Ω j = ( Ω j). (2.44)..... Ω N1 Ω N2...... Ω NN There s oe addtoal clam. The operator tself ca be wrtte terms of the matrx elemets ad bass bras ad kets. We clam that Ω = m)ω m (. (2.45) m, We ca verfy that ths s correct by computg the matrx elemets usg t: (m Ω ) = Ω m (m m)( ) = Ω m δ m m δ = Ω m, (2.46) m, m, as expected from the defto (2.42). 2.1 Projecto Operators Cosder the famlar orthoormal bass { )} of V ad choose oe elemet m) from the bass to form a operator P m defed by P m m)(m. (2.47) Ths operator maps ay vector v) V to a vector alog, ). Ideed, actg o v) t gves P m v) = m)(m v) m). (2.48) 6
Comparg the above expresso for P m wth (2.45) we see that the chose bass, P s represeted by a matrx all of whose elemets are zero, except for the (, ) elemet (P ) whch s oe: 0 0... 0... 0 0 0... 0... 0 P..... 0 0 0... 1... 0. (2.49)..... 0 0 0... 0... 0 A hermta operator P s sad to be a projecto operator f t satsfes the operator equato PP = P. Ths meas that actg twce wth a projecto operator o a vector gves the same as actg oce. The operator P m s a projecto operator sce ( )( ) P m P m = m)(m m)(m = m)(m m)(m = m)(m, (2.50) sce (m m) = 1. The operator P m s sad to be a rak oe projecto operator sce t projects to a oe-dmesoal subspace of V, the subspace geerated by m). Usg the bass vector m) wth m = we ca defe P m, m)(m + )(. (2.51) Actg o ay vector v) V, ths operator gves us a vector the subspace spaed by m) ad ): P m, v) = m)(m v) + )( v). (2.52) Usg the orthogoalty of m) ad ) we quckly fd that P m, P m, = P m, ad therefore P m, s a projector. It s a rak two projector, sce t projects to a two-dmesoal subspace of V, the subspace spaed by m) ad ). Smlarly, we ca costruct a rak three projector by addg a extra term k)(k wth k = m ad k =. If we clude all bass vectors we would have the operator P 1,...,N 1)(1 + 2)(2 +... + N)(N. (2.53) As a matrx P 1,...,N has a oe o every elemet of the dagoal ad a zero everywhere else. Ths s therefore the ut matrx, whch represets the detty operator. Ideed we atcpated ths (1.29), ad we thus wrte 1 = )(. (2.54) Ths s the completeess relato for the chose orthoormal bass. Ths equato s sometmes called the resoluto of the detty. Example. For the sp oe-half system the ut operator ca be wrtte as a sum of two terms sce the vector space s two dmesoal. Usg the orthoormal bass vectors +) ad ) for sps alog the postve ad egatve z drectos, respectvely, we have 1 = +)(+ + )(. (2.55) 7
Example. We ca use the completeess relato to show that our formula (2.42) for matrx elemets s cosstet wth matrx multplcato. Ideed for the product Ω 1 Ω 2 of two operators we wrte (Ω 1 Ω 2 ) m = (m Ω 1 Ω 2 ) = (m Ω 1 1 Ω 2 ) ( N ) N N (2.56) = (m Ω 1 k)(k Ω 2 ) = (m Ω 1 k)(k Ω 2 ) = (Ω 1 ) mk (Ω 2 ) k. k=1 k=1 k=1 Ths s the expected rule for the multplcato of the matrces correspodg to Ω 1 ad Ω 2. 2.2 Adjot of a lear operator A lear operator Ω o V s defed by ts acto o the vectors V. We have oted that Ω ca also be vewed as a lear operator o the dual space V. We defed the lear operator Ω assocated wth Ω. I geeral Ω by (Ω u v) = (u Ωv) (2.57) Flppg the order o the left-had sde we get (v Ω u) = (u Ωv) (2.58) Complex cojugatg, ad wrtg the operators more explctly (v Ω u) = (u Ω v), u,v. (2.59) Flppg the two sdes of (2.57) we also get (v Ω u) = (Ωv u) (2.60) from whch, takg the ket away, we lear that (v Ω (Ωv. (2.61) Aother way to state the acto of the operator Ω s as follows. The lear operator Ω duces a map v) v ) of vectors V ad, fact, s defed by gvg a complete lst of these maps. The operator Ω s defed as the oe that duces the maps (v (v of the correspodg bras. Ideed, v ) = Ωv) = Ω v), (v = (Ωv = (v Ω (2.62) The frst le s just deftos. O the secod le, the frst equalty s obtaed by takg bras of the frst equalty o the frst le. The secod equalty s just (2.61). We say t as The bra assocated wth Ω v) s (v Ω. (2.63) 8
To see what hermtcty meas at the level of matrx elemets, we take u, v to be orthoormal bass vectors (2.59) ( Ω j) = (j Ω ) (Ω ) j = (Ω j ). (2.64) I matrx otato we have Ω = (Ω t ) where the superscrpt t deotes trasposto. Exercse. Show that (Ω 1 Ω 2 ) = Ω Ω by takg matrx elemets. 2 1 Exercse. Gve a operator Ω = a)(b for arbtrary vectors a, b, wrte a bra-ket expresso for Ω. Soluto: Actg wth Ω o v) ad the takg the dual gves Sce ths equato s vald for ay bra (v we read 2.3 Hermta ad Utary Operators Ω v) = a)(b v) (v Ω = (v b)(a, (2.65) A lear operator Ω s sad to be hermta f t s equal to ts adjot: Ω = b)(a. (2.66) Hermta Operator: Ω = Ω. (2.67) I quatum mechacs Hermta operators are assocated wth observables. The egevalues of a Hermta operator are the possble measured values of the observables. As we wll show soo, the egevalues of a Hermta operator are all real. A operator A s sad to be at-hermta f A = A. Exercse: Show that the commutator [Ω 1, Ω 2 ] of two hermta operators Ω 1 ad Ω 2 s at-hermta. There are a couple of equatos that rewrte useful ways the ma property of Hermta operators. Usg Ω = Ω (2.59) we fd If Ω s a Hermta Operator: (v Ω u) = (u Ω v), u,v. (2.68) It follows that the expectato value of a Hermta operator ay state s real (v Ω v) s real for ay hermta Ω. (2.69) Aother eat form of the hermtcty codto s derved as follows: (Ωu v) = (u Ω v) = (u Ω v) = (u Ωv), (2.70) so that all all Hermta Operator: (Ωu v) = (u Ωv). (2.71) 9
I ths expresso we see thatahermta operator moves freelyfrom thebra to theket (ad vceversa). Example: For wavefucto f (x) C we have wrtte (f g) = (f(x)) g(x)dx (2.72) For a Hermta Ω we have (Ωf g) = (f Ωg) or explctly (Ωf(x)) g(x)dx = (f(x)) Ωg(x)dx (2.73) Verfy that the lear operator Ω = d dx s hermta whe we restrct to fuctos that vash at ±. A operator U s sad to be a utary operator f U s a verse for U, that s, U U ad UU are both the detty operator: U s a utary operator: U U = UU = 1 (2.74) I fte dmesoal vector spaces U U = 1 mples UU = 1, but ths s ot always the case for fte dmesoal vector spaces. A key property of utary operators s that they preserve the orm of states. Ideed, assume that ψ ) s obtaed by the acto of U o ψ): ψ ) = U ψ) (2.75) Takg the dual we have ad therefore (ψ = (ψ U, (2.76) (ψ ψ ) = (ψ U U ψ) = (ψ ψ), (2.77) showg that ψ) ad U ψ) are states wth the same orm. More geerally (Ua Ub) = (a U U b) = (a b). (2.78) Aother mportat property of utary operators s that actg o a orthoormal bass they gve aother orthoormal bass. To show ths cosder the orthoormal bass a 1 ), a 2 ),... a N ), (a a j ) = δ j (2.79) Actg wth U we get Ua 1 ), Ua 2 ),... Ua N ), (2.80) To show that ths s a bass we must prove that β Ua ) = 0 (2.81) 10
mples β = 0 for all. Ideed, the above gves β Ua ) = β U a ) = U β a ) = 0. (2.82) Actg wth U from the left we fd that L β a ) = 0 ad, sce the a ) form a bass, we get β = 0 for all, as desred. The ew bass s orthoormal because (Ua Ua j ) = (a U U a j ) = (a a j ) = δ j. (2.83) It follows from the above that the operator U ca be wrtte as sce N U = Ua )(a, (2.84) N =1 U a j ) = Ua )(a a j ) = Ua ). (2.85) =1 I fact for ay utary operator U a vector space V there exst orthoormal bases { a )} ad { b )} such that U ca be wrtte as N U = b )(a. (2.86) Ideed, ths s just a rewrtg of (2.84), wth a ) ay orthoormal bass ad b ) = Ua ). Exercse: Verfy that U (2.86) satsfes U U = UU = 1. Exercse: Prove that (a U a j ) = (b U b j ). 3 No-deumerable bass =1 I ths secto we descrbe the use of bras ad kets for the posto ad mometum states of a partcle movg o the real le x R. Let us beg wth posto. We wll troduce posto states x) where the label x the ket s the value of the posto. Sce x s a cotuous varable ad we posto states x) for all values of x to form a bass, we are dealg wth a fte bass that s ot possble to label as 1), 2),..., t s a o-deumerable bass. So we have Bass states : x), x R. (3.87) Bass states wth dfferet values of x are dfferet vectors the state space (a complex vector space, as always quatum mechacs). Note here that the label o the ket s ot a vector! So ax) = a x), for ay real a = 1. I partcular x) = x) uless x = 0. For quatum mechacs three dmesos, we have posto states x ). Here the label s a vector a three-dmesoal real vector space (our space!) whlethe ket s a vector the ftedmesoal complex vector space of states of thetheory. 11
Aga somethg lke x 1 + x 2 ) has othg to do wth x 1 ) + x 2 ). The ) eclosg the label of the posto egestates plays a crucal role: t helps us see that object lves a fte dmesoal complex vector space. The er product must be defed, so we wll take (x y) = δ(x y). (3.88) It follows that posto states wth dfferet postos are orthogoal to each other. The orm of a posto state s fte: (x x) = δ(0) =, so these are ot allowed states of partcles. We vsualze the state x) as the state of a partcle perfectly localzed at x, but ths s a dealzato. We ca easly costruct ormalzable states usg superpostos of posto states. We also have a completeess relato 1 = dx x)(x. (3.89) Ths s cosstet wth our er product above. Lettg the above equato act o y) we fd a equalty: y) = dx x)(x y) = dx x) δ(x y) = y). (3.90) The posto operator ˆx s defed by ts acto o the posto states. Not surprsgly we let xˆ x) = x x), (3.91) thus declarg that x) are ˆx egestates wth egevalue equal to the posto x. We ca also show that ˆx s a Hermta operator by checkg that ˆx ad ˆx have the same matrx elemets: (x 1 xˆ x 2 ) = (x 2 xˆ x 1 ) = [x 1 δ(x 1 x 2 )] = x 2 δ(x 1 x 2 ) = (x 1 xˆ x 2 ). (3.92) We thus coclude that ˆx = xˆ ad the bra assocated wth (3.91) s (x xˆ = x(x. (3.93) Gve the state ψ) of a partcle, we defe the assocated posto-state wavefucto ψ(x) by ψ(x) (x ψ) C. (3.94) Ths s sesble: (x ψ) s a umber that depeds o the value of x, thus a fucto of x. We ca ow do a umber of basc computatos. Frst we wrte ay state as a superposto of posto egestates, by sertg 1 as the completeess relato ψ) = 1 ψ) = dx x)(x ψ) = dx x) ψ(x). (3.95) As expected, ψ(x) s the compoet of ψ alog the state x). Ovelap of states ca also be wrtte posto space: (φ ψ) = dx (φ x)(x ψ) = dx φ (x)ψ(x). (3.96) 12
Matrx elemets volvg ˆx are also easly evaluated (φ xˆ ψ) = (φ xˆ1 ψ) = dx (φ xˆ x)(x ψ) = dx (φ x) x (x ψ) = dx φ (x)xψ(x). (3.97) We ow troduce mometum states p) that are egestates of the mometum operator ˆp complete aalogy to the posto states Bass states : p), p R. (p p) = δ(p p ), 1 = dp p)(p, (3.98) Just as for coordate space we also have pˆ p) = p p) pˆ = p, ˆ ad (p ˆ p = p(p. (3.99) I order to relate the two bases we eed the value of the overlap (x p). Sce we terpret ths as the wavefucto for a partcle wth mometum p we have from (6.39) of Chapter 1 that e px/ (x p) =. (3.100) 2π The ormalzato was adjusted properly to be compatble wth the completeess relatos. Ideed, for example, cosder the (p p) overlap ad use the completeess x to evaluate t 1 )x/ 1 )u (p p) = dx(p x)(x p) = dxe (p p = du e (p p, (3.101) 2π 2π where we let u = x/ the last step. We clam that the last tegral s precsely the tegral represetato of the delta fucto δ(p p ): 1 du e (p p )u = δ(p p ). (3.102) 2π Ths, the gves the correct value for the overlap (p p ), as we clamed. The tegral (3.102) ca be justfed usg the fact that the fuctos 1 ( 2πx ) f (x) exp, (3.103) L L form a complete orthorormal set of fuctos over the terval x [ L/2, L/2]. Completeess the meas that f (x)f (x ) = δ(x x ). (3.104) We thus have Z 1 ( ) exp 2π (x x ) = δ(x x ). (3.105) L L Z 13
I the lmt as L goes to fty the above sum ca be wrtte as a tegral sce the expoetal s a very slowly varyg fucto of Z. Sce Δ = 1 wth u = 2π/L we have Δu = 2π/L 1 ad the 1 ( ) Δu ( ) 1 ) exp 2π (x x ) = exp u(x x ) due u(x x, (3.106) L L 2π 2π Z u ad back (3.105) we have justfed (3.102). We ca ow ask: What s (p ψ)? We compute 1 dxe px/ (p ψ) = dx(p x)(x ψ) = ψ(x) = ψ(p), 2π (3.107) whch s the Fourer trasform of ψ(x), as defed (6.41) of Chapter 1. Thus the Fourer trasform of ψ(x) s the wavefucto the mometum represetato. It s useful to kow how to evaluate (x pˆ ψ). We do t by sertg a complete set of mometum states: (x pˆ ψ) = dp (x p)(p pˆ ψ) = dp (p(x p))(p ψ) (3.108) Now we otce that d p(x p) = (x p) dx ad thus ( d ) (x pˆ ψ) = dp (x p) (p ψ). dx (3.109) (3.110) The dervatve ca be moved out of the tegral, sce o other part of the tegrad depeds o x: d (x pˆ ψ) = dp (x p)(p ψ) dx (3.111) The completeess sum s ow trval ad ca be dscarded to obta d d (x pˆ ψ) = (x ψ) = ψ(x). dx dx (3.112) Exercse. Show that d (p xˆ ψ) = ψ(p). (3.113) dp 14
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