1 Vectors over the complex numbers

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1 Vectors for quantum mechancs 1 D. E. Soper 2 Unversty of Oregon 5 October 2011 I offer here some background for Chapter 1 of J. J. Sakura, Modern Quantum Mechancs. 1 Vectors over the complex numbers What s a vector? One can take two approaches, one very concrete, one abstract. Ultmately, the abstract approach s most useful and the concrete approach s even a bt msleadng. However, both are useful. We consder vectors ψ n an N-dmensonal vector space over the complex numbers. In the concrete approach, such a vector s a lst of N complex numbers: ψ { ψ 1, ψ 2,..., ψ N }. (1) There s a specal vector, the zero vector, defned by 0 {0, 0, 0, 0}. (2) If c s a complex number, we can defne multplcaton by c as c ψ {c ψ 1, c ψ 2,..., c ψ N }. (3) If ψ and { 1, 2,..., N } are two vectors, we can defne addton of these vectors as ψ + { ψ 1 + 1, ψ 2 + 2,..., ψ N + N }. (4) In the abstract approach, we say that an N-dmensonal vector space V over the complex numbers s a set of vectors ψ V together wth two operatons: multplcaton by a complex number and addton of vectors. Thus we postulate that f ψ V and c s a complex number, then cψ V. Furthermore, f ψ V and V, then ψ + V. We also postulate that there s a vector 0 V wth 0 ψ 0. Fnally, 1 Copyrght, 2011, D. E. Soper 2 soper@uoregon.edu (c 1 c 2 )ψ c 1 (c 2 ψ), (5) 1

2 and (c 1 + c 2 )ψ c 1 ψ + c 2 ψ (6) c(ψ + ) cψ + c. (7) These propertes say what a vector s, but do not say what N-dmensonal means. To defne ths, we frst defne lnearly ndependent and lnearly dependent. A set of vectors 3 ψ 1, ψ 2,..., ψ n s lnearly ndependent f, for any complex coeffcents c 1, c 2,..., c n, only f c 1 ψ 1 + c 2 ψ c n ψ n 0 (8) c 1 c 2 c n 0. (9) If a set of vectors s not lnearly ndependent, then t s lnearly dependent. Thus ψ 1, ψ 2,..., ψ n s lnearly dependent f there are some coeffcents c 1, c 2,..., c n that are not all zero such that c 1 ψ 1 + c 2 ψ c n ψ n 0. (10) In ths case, at least one of the coeffcents, call t c a, s non-zero. Then ψ a a (c /c a ) ψ. (11) That s, f ψ 1, ψ 2,..., ψ n are lnearly dependent, then one of them can be expressed as a lnear combnaton of the others. Ths gves us the needed defnton: V s N-dmensonal f there s at least one set of N vectors ψ 1, ψ 2,..., ψ N that s lnearly ndependent and f any set of N + 1 vectors s lnearly dependent. A set of N lnearly ndependent vectors ψ j n a vector space of dmenson N s called a bass for V. Once we have a bass, any other vector can be expressed as a lnear combnaton of the bass vectors, N c ψ. (12) 1 3 I apologze that here ψ 1, ψ 2, etc. are vectors, whereas n the concrete example I used a very smlar notaton ψ 1, ψ2, etc. for complex numbers that are the components of a sngle vector. Fortunately, the bra-ket notaton that we wll use generally does not suffer from ths notatonal dffculty. 2

3 A bass for our concrete example s ψ 1 {1, 0, 0,..., 0}, ψ 2 {0, 1, 0,..., 0}, ψ N {0, 0, 0,..., 1}. (13) In ths course, we wll (usually) use a notaton for vectors n whch a vector s represented as, called a ket-vector or just a ket. Here the ψ can be any name that dstngushes the vector from other vectors. Thus, for nstance, f we have vectors ψ 1, ψ 2,..., ψ N, we can denote them by kets 1, 2,..., N. 2 Inner product For quantum mechancs, we use a vector space V over the complex numbers that has an nner product. For any two vectors ψ and, the nner product (, ψ) s a complex number. The nner product s lnear n the frst (rght hand) factor and conjugate-lnear n the second factor: Furthermore (, c 1 ψ 1 + c 2 ψ 2 ) c 1 (, ψ 1 ) + c 2 (, ψ 2 ), (c c 2 2, ψ) c 1( 1, ψ) + c 2( 2, ψ). Evdently, (ψ, ψ) s real. We postulate that and f ψ 0, (14) (, ψ) (ψ, ). (15) (ψ, ψ) 0 (16) (ψ, ψ) > 0. (17) In our concrete example, the nner product could be (, ψ) N ψ. (18) j1 In ths course, we wll usually use the notaton (, ψ) ψ 3 (19)

4 for the nner product. We call a bra-vector, or just a bra. I usually thnk of bras as kets just wrtten backwards so as to provde a convenent notaton for nner products. Sakura nterprets the bras as belongng to a dfferent space from the kets. Then there s a mappng that he calls DC that maps a ket nto ts correspondng bra. If you thnk of t that way, then the mappng s conjugate-lnear: f c1 1 + c2 2, (20) then c c 2 2. (21) Wth the bra-ket notaton ψ ψ. (22) Wth the bra-ket notaton, the property that the nner product s lnear n the ket vectors s stated as (c c2 2 ) c c2 2 (23) The property (c 1 1 +c 2 2, ψ) c 1( 1, ψ)+c 2( 2, ψ) s a lttle more awkward to state. If c1 1 + c2 2, then, usng Eq. (21), we can wrte ( c c 2 2 ) ψ c c 2 2. (24) Thus the bra-ket notaton makes t easy to get thngs rght as long as we remember Eq. (21), although t s a lttle trcky to express what the postulates are. Wth the bra-ket notaton, the zero vector s usually represented as just 0 wthout a ket symbol. Thus we mght wrte + 0. We could wrte + zero, but people usually don t do that. One often sees a notaton 0, but that can mean ground state of a harmonc oscllator or vacuum state or somethng else other than zero. 3 Interpretaton of the nner product Suppose that s a state vector normalzed to 1. Suppose further that we know how to measure whether our physcal system s n state. 4

5 For nstance, we can determne f an atom s n a state S z, + by sendng t through a Stern-Gerlach apparatus wth ts magnetc feld n the z-drecton and seeng whether ts path s bent n the z-drecton. Suppose that ψ s a state vector normalzed to ψ 1 and that we know how to prepare a system so that we know t s n state ψ. For nstance, we can prepare an atom n a state Sz, + by sendng some atoms through a Stern-Gerlach apparatus wth ts magnetc feld n the z-drecton and selectng one whose path s bent n the z-drecton. Suppose n addton that s a state vector normalzed to 1 and that we know how to measure whether the system s n state. For nstance, we can measure whether an atom n a state Sx, + by sendng the atom through a Stern-Gerlach apparatus wth ts magnetc feld n the x-drecton and seeng f the path of the atom s bent n the x-drecton. Then f we prepare a system n state ψ, the probablty that we wll fnd upon the approprate measurement that t s n state s 2. Note that to make ths nterpretaton work, we always represent a system that s n a defnte quantum state by a vector ψ wth the normalzaton ψ 1. 4 Orthonormal bass If 1, 2,..., N form a bass for our space, there s, n general, no partcular behavor of the nner product j. However, a bass can be orthonormal, whch means that j { 1 j. (25) 0 j Ths connects to the physcal nterpretaton. Suppose that we know how to prepare states represented by vectors wth the normalzaton 1 for each. Suppose further that experment shows that f the system s prepared n state and we measure whether t s n state j for j, then the probablty that the system wll be found to be n state j s zero. That means that j 0. That s, the vectors form an orthonormal bass for the space of quantum states. An example of ths s the two states Sz, + and Sz, descrbng the spn of an electron. 5

6 In Sakura, the term base kets refers to an orthonormal bass. I wll usually also be a lttle sloppy n language and refer to bass kets wth 1,..., N when I mean an orthonormal bass. There s a useful relaton for an orthonormal bass. Snce t s a bass, any vector can be expressed as a lnear combnaton of the bass elements, N c. (26) From ths, we can fnd the expanson coeffcents. We have Thus the expanson can be wrtten 1 N j c j cj. (27) 1 N. (28) 1 We usually wrte ths n the form N. (29) 1 Notce how ths works wth the probablstc nterpretaton of nner products. Let us assume that, normalzed to 1, represents a physcal state. Suppose that the bass states correspond to possble results of a measurement. For nstance, the bass states could be Sx, + and Sx,. Recall that the condton j 0 when j means that f measurement shows that our system s n state, then we are sure that t s not n state j. For example, f we send an atom through a Stern-Gerlach apparatus, then f t went up, t dd not go down and f t went down, t dd not go up. Now the normalzaton condton 1 gves 1 ( N ) 1 N N 2. (30) 1 1 Ths says that the sum of the probabltes to get the varous results s 1. 6

7 Exercse 1.1 Let us postulate that the state space for a spn 1/2 system s spanned by an orthonormal bass consstng of two bass vectors, Sz ; + and Sz ;. Defne Sx ; Sz ; Sz ;, S x ; 1 2 S z ; S z ;. (31) Show that the vectors Sx ; + and Sx ; form an orthonormal bass. Exercse 1.2 Snce Sx ; + and Sx ; form an orthonormal bass, we can expand Sz ; + n ths bass, Sz ; + c ++ Sx ; + + c + Sx ;. (32) Smlarly, we can expand S z ; n ths bass, Sz ; c + Sx ; + + c Sx ;. (33) Use Eq. (29) to fnd the coeffcents c j. Exercse 1.3 Another bass for the state space for a spn 1/2 partcle conssts of the states Sy ; + 1 Sz ; + + Sz ;, 2 2 Sy ; 1 Sz ; + Sz ;. 2 2 (34) These vectors can be wrtten n terms of the bass vectors Sx ; ±, Sy ; + d ++ Sx ; + + d + Sx ;, Sy ; d + Sx ; + + d Sx ;. (35) Use Eq. (29) to fnd the coeffcents d j. 7

8 I should pont out that the sgn conventons used n these exercses are from Sakura (1.4.17). Ths agrees wth Sakura s tentatve choce n (1.1.14) but not wth the tentatve choce n (1.1.9). The states Sy, ± can be prepared and measured wth a Stern-Gerlach apparatus wth magnetc feld orented along the y drecton. Note that we need factors of n the coeffcent relatng these vectors to Sx, ±. Ths llustrates why we choose a vector space over the complex numbers rather than over the real numbers for representng a quantum system. 5 Operators A lnear operator A on the space V maps a vector to another vector ψ A ψ. The mappng s lnear, ( ) A c ψ1 1 + c2 ψ2 c 1 A ψ1 + c2 A ψ2. (36) Evdently, any lnear combnaton of lnear operators defnes another lnear operator: (αa + βb) ψ αa + βb ;. (37) There are two specal operators, the null operator 0, and the unt operator 1, 0 ψ 0, (38) 1 ψ. (39) One can defne the product of two lnear operators n a straghtforward fashon: the product C AB s defned by C A(B ). Note that n general AB and BA are not equal to each other. The bra-ket notaton provdes a slck way to represent certan operators. We can nterpret the symbol ψ as an operator. If A 2 1, the acton of A on a state ψ s A ψ ( 2 1 ) ψ 2 ( 1 ) 2 1. (40) For nstance, f we have an orthonormal bass, then Eq. (29) gves us a way to wrte the unt operator 1. (41) 8

9 We wll use ths often. A good example s to express the acton of an arbtrary operator A n terms of an (orthonormal) bass wth bass vectors. Let A ψ. Usng the bass vectors, the j-component of s ψ j j ψ (42) and the -component of s. (43) These related by A ψ A 1 ψ A j j ψ. (44) Then ths has the form of a matrx multplcaton, j j A j ψj. (45) where the matrx elements are A j A j. (46) We can also use the unt operator nserton method to fnd the matrx elements of a product of two operators AB j A 1 B j A k k B j. (47) k That s, the matrx that represents AB s the matrx product of the two matrces that represent A and B. Exercse 1.4 The operators that represent the three components of spn for a spn 1/2 system are S x, S y and S z. The matrx elements of these operators n the Sz, ± bass are Sz, Sx Sz, j 1 2 (σ x) j, Sz, Sy Sz, j 1 2 (σ y) j, (48) Sz, Sz Sz, j 1 2 (σ z) j, 9

10 where (dsplayng the + elements frst and the elements second) ( ) 0 1 σ x, 1 0 ( ) 0 σ y, (49) 0 ( ) 1 0 σ z. 0 1 Defne the comutator of S x wth S y by [S x, S y ] S x S y S y S x. Show that [S x, S y ] S z. What s [S y, S z ]? What s [S z, S x ]? 6 The adjont of an operator Suppose that A s a lnear operator. Then we can defne another operator A, called the adjont of A. We say that A appled to a state gves another state A. To say what s, we gve ts nner product ψ wth any state ψ : ψ A. (50) If we use our alternatve notaton for vectors and nner products, the defnton s (A, ψ) (, Aψ). (51) Unfortunately, the bra-ket notaton does not allow us to wrte the defnton of the adjont of an operator n ths smple way. The best that we can do s to use ψ A A. (52) Note that takng the adjont twce s the same as dong nothng: ψ (A ) A ψ ( ψ A ) ψ A, (53) for every par of vectors ψ and, so (A ) A. (54) If we have two operators A and B and defne C AB, then what s C? We have (C, ψ) (, Cψ) (, ABψ) (A, Bψ) (B A, ψ), (55) 10

11 so C B A. (56) There s a smple relaton between the matrx elements of A and those of A. Usng Eq. (52), we have A j j A. (57) Thus the matrx elements of A are the complex conjugates of the matrx elements of A wth the ndces reversed. When we use the bra-ket notaton, we can say that an operator A can act backward on a bra, A. (58) The new bra vector s defned by Eq. (50). Then the correspondng ket vector s A. Thus also f A (59) then A. (60) An operator A s self-adjont (or hermtan ) f An operator U s untary f U U 1, that s f A A. (61) U U 1. (62) We wll explore self-adjont operators and untary operators n subsequent sectons. 7 Untary operators Suppose that ψ and are two vectors and that U s a untary operator. Defne ψ U ψ and U. Then ψ U U ψ 1. (63) That s, a untary operator preserves nner products. 11

12 For ths reason, we use untary operators to represent symmetry operatons. For nstance, there wll be a untary operator to represent the effect of rotatng a system about the y-axs through an angle π/3 (say). If we rotate ψ and, we don t get the same states. We get dfferent states ψ and. However, ψ has a physcal meanng: ts square s the probablty for to look lke ψ f we perform the approprate measurement. It should not matter whether we perform the measurement on state ψ or we rotate to get ψ and perform the rotated measurement, correspondng to. Thus the rotaton should be represented by a untary operator. We wll nvestgate rotaton operators n some detal later n ths course. We can also use a untary operator to represent any change of bass. Suppose that the vectors A,, 1,..., N form an orthonormal bass, the A bass, for the space of states of a quantum system. Suppose that the vectors B,, 1,..., N form another bass. Defne a lnear operator U by U A, B,. (64) Wth ths defnton, we do not know at the moment whether U s untary. Let us nvestgate ths queston. If c A,, (65) we have U ψ c U A, c B,. (66) Smlarly, for any other vector wth d A,, (67) we have U d B,. (68) The nner product between U and U s U U d c j B, B, j,j,j,j d c j δ,j d c j A, A, j (69) ψ. 12

13 Thus f we defne Ψ U U, we have Ψ for every vector. Thus ( Ψ ) 0 for every vector. Takng ( Ψ ), we have ( Ψ ψ )( Ψ ) 0. (70) However, the nner product of a vector wth tself can be zero only f the vector s zero. Thus Ψ 0. That s U U ψ (71) for every vector ψ. Ths means that U U 1. That s, the operator U that takes one orthonormal bass nto another s untary. 8 Self-adjont operators Physcal quanttes lke angular momentum, energy, momentum, and poston are represented n quantum mechancs by self-adjont operators. Thus we should know some propertes of self-adjont or hermtan operators. The spectral theorem. Let A be a lnear operator on the space of quantum states of a system and let A be self-adjont: A A. Then there s an orthonormal bass such that the vectors are egenvectors of A, A a. (72) Here a s a real number, the egenvalue of A correspondng to the egenvector. Ths theorem has an mmedate consequence, A A 1 A a. (73) Ths gves us the spectral representaton of A: A a. (74) Ths theorem s pretty easy to prove, but we wll not do t n these notes. Instead, let us look at a couple of the ngredents. Suppose that we look for egenvectors of A and ther correspondng egenvalues a. We can choose 13

14 to normalze all egenvectors that we fnd to 1. We see mmedately that any egenvalue must be real: a a ( A ) A ( ) A A a (75) a. Furthermore, egenvectors correspondng to dfferent egenvalues must be orthogonal: (a a j ) j ( A ) j ( A j ) A A j 0. (76) If (a a j ) 0 then we must have j 0. Furthermore, f there are several egenvectors that correspond to the same egenvalue, then any lnear combnaton of these egenvectors wll also be an egenvector wth ths same egenvalue. Thus the egenvectors wth ths egenvalue form a vector space that s a subspace of the complete space. We can choose any orthonormal bass that we lke for ths subspace. That s, we can smply choose egenvectors wth a gven egenvalue to be orthogonal to each other. These remarks prove part of the theorem. It remans to be proved that there are enough egenvectors so that they form a bass for the complete space. The spectral theorem s mportant for the nterpretaton of quantum mechancs. We often assocate a measurement wth a self-adjont operator. To see how ths assocaton works, consder a self-adjont operator A that s assocated wth a certan measurement. (For example, we mght measure the y-component of the spn of a partcle.) If we have a state ψ and we make a measurement that corresponds to the operator A a, (77) then we do not know to start wth what result we wll get. The possble results are the egenvalues a. If the state happens to be, then the result of measurng A s certanly a. Conversely, f the result of the measurement s a, then the state after the measurement 4 s. In general, we have ψ ψ. (78) 4 Ths s the standard prncple and apples to our Stern-Gerlach examples. However, one needs to be a lttle careful. We should be sure that nothng happened after the measurement that could change the state. 14

15 The probablty that we wll fnd n the state and thus get a measurement a s ψ 2. As we have earler seen wth a smple calculaton, these probabltes sum to 1. Exercse 1.5 Let us postulate that the state space for a spn 1/2 system s spanned by an orthonormal bass consstng of two bass vectors, Sz ; + and Sz ; and that the phases of these vectors are defned usng the standard conventon such that the matrx elements of S x and S y n ths bass are gven by 1/2 tmes the standard Paul spn matrces, Eq. (49). Show that the vectors Sx, ± gven by Eq. (31) are egenvectors of S x : S x Sx, ± ± 1 2 Sx, ±. (79) Show that the vectors Sy, ± gven by Eq. (34) are egenvectors of S y : S y Sy, ± ± 1 2 Sy, ±. (80) I recommend usng a matrx notaton n whch the Sz ; ± vectors are represented by ( ) ( ) 1 0 and. (81) 0 1 Exercse 1.6 Let us agan postulate that the state space for a spn 1/2 system s spanned by an orthonormal bass consstng of two bass vectors, Sz ; + and Sz ; and that the phases of these vectors are defned usng the standard conventon such that the matrx elements of S x and S y n ths bass are gven by 1/2 tmes the standard Paul spn matrces. Consder the operator S n cos θ S z + sn θ S x. (82) Ths s the operator that measures spn n the drecton of a vector n that les n the z-x plane and makes an angle θ wth the z-axs. 15

16 Fnd the egenvalues of S n and the correspondng egenvectors, expressed as lnear combnatons of Sz ; + and Sz ; : S n; c(, j) Sz ; j. (83) j I recommend usng a matrx notaton n whch the Sz ; ± vectors are represented as n the prevous problem. Then you need to solve a 2 2 matrx egenvalue equaton, whch I presume that you know how to do. Exercse 1.7 Suppose that we send a spn-1/2 atom through a Stern-Gerlach apparatus that measures spn n the drecton n, as n Exercse 1.6. Suppose that we select an atom that goes n the + n drecton. That s, the value of S n for ths atom s +1/2. Now we send ths atom through a Stern-Gerlach apparatus that measures spn n the ˆx drecton. What s the probablty that when we measure S ˆx for ths atom, the value S ˆx wll be 1/2? One often speaks of the expectaton value A of A n the state ψ. The defnton s A a Probablty a ψ 2. (84) We can evaluate ths to obtan a very smple expresson, A a ψ ( ψ a ) ψ ψ A. (85) 9 More about self-adjont operators We can look at ths more closely by ntroducng another concept. Consder the operator P. (86) If ψ α, then P. If s orthogonal to,.e. 0, then P 0. If Ψ α +, then P Ψ α. That s P pcks out 16

17 the part of Ψ that s proportonal to. We call P the projecton operator onto the space spanned by the vector. More generally, suppose M s a subspace of the complete vector space V that we are usng to descrbe our quantum states. That s, M s a subset of V and s tself a vector space n the sense that f ψ1 and ψ2 are n M, then so s α ψ1 1 + α2 ψ2. Then there s a complementary vector space M that conssts of vectors such that 0 for every vector n M. Every vector Ψ can be wrtten as the sum of a vector n M and a vector n M. To see ths, choose an orthonormal bass of vectors ψ n M. Defne ψ ψ ψ Ψ. (87) Further defne Ψ. (88) Then ψj ψj Ψ ψj ψ ψ Ψ ψj Ψ ψj Ψ 0. (89) Therefore s n M. Thus Ψ +, (90) where ψ s n M and s n M, as clamed. Choose an orthonormal bass of vectors ψ n M and an orthonormal bass of vectors n M. Then t follows from the constructon above that the decomposton of a general vector Ψ nto a part n M and a part n M s Ψ ψ ψ Ψ + j j Ψ. (91) Ths leads us to defne the operator that projects onto M by P (M) ψ ψ (92) j and the operator that projects onto M by P (M ) j j. (93) 17

18 We have P (M) Ψ, P (M ) Ψ. (94) Ths may all seem a lttle abstract, but once you get used to t, I thnk that you wll fnd t very helpful to thnk about subspaces usng these deas. You may want to draw lttle pctures of the subspaces, even though you are pretty much lmted to two or three real dmensons and you mght want many complex dmensons nstead. Wth ths notaton, we can consder agan our self-adjont operator A agan. There could be more than one egenvector wth a gven egenvalue a. In ths case, we say that the egenvalue s degenerate. We can defne a projecton operator P (a) Θ(a a). (95) Here Θ(a a) 1 f has egenvalue a equal to a and Θ(a a) 0 f has egenvalue a not equal to a. Thus only egenvectors whose egenvalues equal a are ncluded n the sum. The operator P (a) projects onto the subspace consstng of vectors that are egenvectors of A wth egenvalue a. We can wrte the spectral theorem as A a a P (a). (96) Here we sum over the dstnct egenvalues a of A. Ths form of the spectral theorem has the advantage that the operator A unquely determnes the egenvalues a and the subspaces projected by P (a). In each subspace, we can fnd an orthonormal bass of egenvectors wth egenvalue a, but t s our choce whch bass to pck. If the egenvalue a s degenerate, someone else mght pck a dfferent bass. Here s an mportant applcaton. Suppose that we have two self-adjont operators A and B. Suppose further that AB BA, that s [A, B] 0. (97) Then we can dagonalze A and B at the same tme. That s, there s an orthornormal bass such that A a, B b. (98) 18

19 Thus A B a, b. (99) The proof of ths uses what we have learned n ths secton. Snce A s self-adjont, we can fnd subspaces M(a) consstng of egenvectors of A wth egenvalues a: A a P (a). (100) a Let ψ be a vector n M(a). Consder the vector B. We have AB ψ BA ab. (101) Thus B ψ s n M(a). We can thus consder B to be an operator just on the subspace M(a). Snce t s self-adjont, there s an orthonormal bass of vectors n M(a) that are egenvectors of B. Of course, these vectors are also egenvectors of A. We construct egenvectors of B n each subspace M(a). In ths way we construct a bass for the whole space V such that the bass vectors are egenvectors of both A and B. We wll fnd n ths course that t happens pretty frequently that we want to fnd egenvectors of some operator A and that s another operator B that commutes wth A, so that we can fnd egenvectors of A and B together. Ths helps n the problem of fndng the egenvectors of A. In fact, there may be more than two operators, all of whch commute wth each other. Then, wth an evdent extenson of the results of ths secton, there s a bass of egenvectors of all of the operators. Exercse 1.8 Suppose a self-adjont lnear operator s represented by the matrx A (102) Suppose another self-adjont lnear operator actng on the same space s represented by the matrx B (103)

20 Show that these operators commute. Snce they commute, you should be able to fnd an orthonormal bass consstng of three egenvectors ψ(), 1, 2, 3, whch mght convenently be wrtten as column vectors ψ 1 () ψ() ψ 2 (), (104) ψ 3 () such that these vectors are egenvectors of both A and B: Aψ() a ψ(), Bψ() b ψ(). (105) Fnd the egenvectors ψ() and the correspondng egenvalues a and b. 10 Trace and determnant In calculatons, one often needs the trace and the determnant of an operator A. The trace of A s defned by usng the matrx representaton of A n an orthonormal bass, tr[a] A. (106) The trace has the mportant property tr[ab] tr[ba]. (107) To see ths, wrte AB,j A j j B tr[ab],j (108) j B A j j BA j tr[ba]. j What f we use a dfferent bass, new,? Then tr new [A] new, A new,. (109) 20

21 Let new, U. As we have seen, U s untary. We have tr new [A] U 1 AU tr[u 1 AU]. (110) Usng Eq. (107), ths s tr new [A] tr[u 1 AU] tr[uu 1 A] tr[a]. (111) Thus we get the same result no matter whch bass we use. Note that f A s self-adjont, then tr[a] s the sum of the egenvalues of A. We can also defne the determnant of an operator A by usng the matrx elements of A n some bass: det[a] {} ɛ( 1, 2,..., N ) 1 A 1 2 A 2 N A N. (112) Here N s the dmenson of our vector space and ɛ( 1, 2,..., N ) s defned by two propertes: ɛ(1, 2,..., N) 1 and ɛ( 1, 2,..., N ) s antsymmetrc under exchange of any two of ts ndces. Thus ɛ( 1, 2,..., N ) 0 f any two of the ndces are the same. The only way that ɛ( 1, 2,..., N ) can be nonzero s f { 1, 2,..., N } s one of the N! permutatons of {1, 2,..., N}. Then ɛ( 1, 2,..., N ) s 1 f { 1, 2,..., N } s an even permutaton {1, 2,..., N} and ɛ( 1, 2,..., N ) s 1 f { 1, 2,..., N } s an odd permutaton {1, 2,..., N}. Notce that det[1] 1. (113) We can also wrte ɛ( 1,..., N ) j 1 A 1 jn A N ɛ(j1,..., j N ) det[a] (114) {} and ɛ(j 1,..., j N ) j 1 A 1 jn A N ɛ(1,..., N ) det[a] (115) {j} and also ɛ( 1,..., N )ɛ(j 1,..., j N ) j 1 A 1 jn A N N! det[a]. {} {j} (116) 21

22 To prove these, we can frst note that Eq. (114) follows n a straghtforward fashon from the defnton, Eq. (112). Then Eq. (116) follows by multplyng Eq. (114) by ɛ(j 1,..., j N ) and summng. Fnally, to derve Eq. (115), we can note that the left hand sde of Eq. (115) s completely antsymmetrc n the ndces n, so the left hand sde has to be ɛ( 1,..., N ) tmes somethng, call t X. To fnd out what X s, we can multply by ɛ( 1,..., N ) and sum over the ndces n. The left hand sde of the resultng equaton s now the left hand sde of Eq. (116), whle the rght hand sde s N! X. We thus fnd that X det[a]. The determnant has the mportant property det[ab] det[a] det[b]. (117) To see ths, wrte det[ab] 1 ɛ( 1,..., N )ɛ(j 1,..., j N ) j 1 AB 1 jn AB N N! {} {j} 1 ɛ( 1,..., N )ɛ(j 1,..., j N ) N! {} {j} j1 A k1 k1 B 1 jn A k1 kn B N {k} 1 N! {k} ɛ(j 1,..., j N ) j 1 A k1 jn A k1 {} ɛ( 1,..., N ) k 1 B 1 kn B N {j} 1 ɛ(k 1,..., k N )det[a]ɛ(k 1,..., k N )det[b] N! {k} det[a] det[b]. (118) What f we use a dfferent bass, new, wth new, U. Then det new [A] det[u 1 AU] det[a]det[u]det[u 1 ] det[a]det[uu 1 ] det[a]det[1] det[a]. (119) 22

23 Thus we get the same result no matter whch bass we use. Note that f A s self-adjont, then det[a] s the product of the egenvalues of A. One smple applcaton of ths s that f we call the egenvalues a and the egenvectors, we can wrte, for any number λ, A λ1 (a λ). (120) Thus det[a λ1] (a λ). (121) Ths s an Nth order polynomal functon of λ that vanshes when λ equals any egenvalue of A. Thus to fnd the egenvalues, we smply need to calculate det[a λ1] and fnd where t vanshes. 23

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