763622S ADVANCED QUANTUM MECHANICS Solution Set 1 Spring c n a n. c n 2 = 1.

Transcription

1 7636S ADVANCED QUANTUM MECHANICS Soluton Set 1 Sprng Warm-up Show that the egenvalues of a Hermtan operator  are real and that the egenkets correspondng to dfferent egenvalues are orthogonal (b) Show that f the state ket ψ = n c n a n s normalzed, then the expanson coeffcents c n must satsfy c n = 1 Soluton n A number c s shown to be real f c = c Let us study a Hermtan operator  for whch  =  holds, and  a = a a Based on the evaluaton of the nner product a  a = a a a = a On the other hand, we can use  =  and let that operate to the left: a  a = ( a  ) a = a a a = a a a So we have that a = a and so a s real Let now  b = b b, and examne the nner product a  b : Agan, lettng  act to the left, a  b = b a b b b a  b = a a b a a Takng the dfference of the two above equatons, one gets (b a) a b = 0 By assumpton a and b are dfferent egenvalues, and so we must have a b = 0

2 (b) Straght-forward calculaton ( ) ( ) α α = c a a c a a = c ac a a a = a a a,a a Normalzaton mples that α α = 1, and so c a = 1 a c a Change of bass Prove the theorem regardng the change of bass from the lecture notes: If both the bass { a n } and { b n } are orthonormal and complete, then there exsts a untary operator Û such that b 1 = Û a 1, b = Û a, b 3 = Û a 3, (1) Soluton The proof has three stages: Constructon of the operator Û, proof of property (1), and proof of untarty Frst part s rather easy, the operator should take a state a n and produce b n : ths would be accomplshed by b n a n To make Û work on any gven nput state a n, just take the sum: Û = n b n a n To see f ths gves property (1), let ths Û operate on an arbtrary a : Û a = n b n a n a = b Therefore, Û works as advertsed Next, test whether Û s untary: ÛÛ = j = j = 1, b j a j ( b a ) a b b j a j a δ j b = b b where the last equalty follows from the completeness of the bass { b } 3 Spn operators Consder the spn operators Ŝx, Ŝy, and Ŝz n the { S z ;, S z ; } bass Wrte out the operators Ŝx, Ŝy, and Ŝz n the { S z ;, S z ; } bass (b) Compute the commutators [Ŝx, Ŝy] and [Ŝ, Ŝx] as well as the ant-commutator {Ŝx, Ŝy} (c) Let us defne the ladder operators Ŝ± = Ŝx ± Ŝy Compute Ŝ± S z ; and Ŝ± S z ;

3 Soluton Let us frst summarze the { S z ;, zdn} bass represented wth the egenstates of Ŝx and Ŝy operators wth proper phase choce (see eg J J Sakura, Modern Quantum Mechancs, p 8): S z ; = 1 ( S x ; + S x ; ) S z ; = 1 ( S x ; S x ; ) (a) S z ; = 1 ( S y ; + S y ; ) S z ; = ( S y ; + S y ; ) (b) Representaton of an operator ˆB n the { S z ;, S z ; } bass: ˆB = S z ; j S z ; j ˆB S z ; k S z ; k = B jk S z ; j S z ; k j= k= j= k= and n matrx representaton we use the followng conventon wth the ndces: B = B jk = S z ; j ˆB S z ; k ( ) ( B B Sz ; = ˆB S z ; S z ; ˆB S z ; B B S z ; ˆB S z ; S z ; ˆB S z ; ) Bass representaton for operator Ŝx s after above defntons just the calculaton of the matrx elements B jk : ( ) Sz ; Ŝz S S z = z ; S z ; Ŝz S z ; S z ; Ŝz S z ; S z ; Ŝz S z ; = h ( ) Sz ; S z ; S z ; S z ; = h ( ) 1 0 S z ; S z ; S z ; S z ; 0 1 To do the same for Ŝx,y, we resort to relatons () and fnd out that Ŝ x S z ; = h S z; Ŝ y S z ; = h S z; Ŝ x S z ; = h S z; Ŝ y S z ; = h S z; whch shows that ( ) Sz ; Ŝx S S x = z ; S z ; Ŝx S z ; = h ( ) 0 1 S z ; Ŝx S z ; S z ; Ŝx S z ; 1 0 ( ) Sz ; Ŝy S S y = z ; S z ; Ŝy S z ; = h ( ) 0 S z ; Ŝy S z ; S z ; Ŝy S z ; 0 (b) As we now have the representatons of operators Ŝ n the Ŝz egenstate bass we can use them to calculate the (ant)commutators [( ) ( ) ( ) ( )] [S x, S y ] = S x S y S y S x = h = hs z (The general rule goes [S, S j ] = hɛ jk S k, where ɛ jk s the Lev-Cvta permutaton symbol, and summaton over k s mpled) Then t happens that S = h /4 for all = x, y, z, therefore S = S x + S y + S z = 3 h /4 and t s clear that [S, S x ] = 3 h [I, S x ]/4 = 0 When calculatng [S x, S y ] one notces that S x S y = S y S x whch mples that {S x, S y } = 0

4 (c) Snce matrces are handy objects, let us express the ladder operators Ŝ± also n the famlar { S z ;, S z ; } bass: Ŝ ± = Ŝx ± Ŝy S + = h ( ) h ( ) ( ) = h, S = h ( ) h ( ) ( ) = h, or Ŝ + = h S z ; S z ; Ŝ = h S z ; S z ; and operatons to Ŝz egenstates result: Ŝ + S z ; = 0 Ŝ S z ; = h S z ; Ŝ + S z ; = h S z ; Ŝ S z ; = 0 Now, the physcal meanng of the ladder operators can be read Operator Ŝ+ rases the spn component by h and f the spn component cannot be rased further, we get null state Smlarly, Ŝ lowers the spn component by h Both these operators are non-hermtan 4 Invarance of trace Prove the theorem from lecture note: If T s a untary matrx, then the matrces X and T XT have the same trace and the same egenvalues Soluton Trace of a matrx X s the sum of ts dagonal elements: Tr(X) = X and a remnder the matrx multplcaton expressed n ndex notaton goes (AB) j = k A kb kj Trace of the product then equals Tr AB = (AB) = A k B k = A k B k = Tr BA Wth ths n mnd we are ready to prove the trace nvarance k Tr(T XT ) = Tr(T T X) = Tr X Suppose x s an egenvector of X, wth an egenvalue x : Xx = x x, T T XT T x = x x, T XT [T x ] = x [T x ] k Insert I = T T n front and behnd X Multply both sdes by T from left Fnally, denotng T x = y we have an egenvalue equaton showng that y s an egenvector of T XT wth egenvalue x : Hence, x s also an egenvalue of T XT T XT y = x y 5 Pure ensemble In a pure ensemble, every state s the same, thus statstcal mechancs of such an ensemble reduces to ordnary quantum mechancs (b) Show that the relaton ˆρ = ρ holds for the densty operator of a pure ensemble, and thus Tr ˆρ = 1 Show that the correspondng matrx representaton s gven by a matrx where one of the dagonal elements s one and the rest are zero Is ths always true?

5 Soluton We defne the densty operator ˆρ as follows ˆρ = ω α () α () As stated n the text, ths operator contans all the physcally sgnfcant nformaton we can possbly obtan about the ensemble n queston, and the expectaton value of an observable  s gven by A pure ensemble s specfed by ω =  = Tr(ˆρÂ) { 1, for some = k 0, for k The correspondng densty operator can be wrtten as ρ = α () α () Clearly, the densty operator for a pure ensemble s dempotent, that s, ˆρ = α () α () α () α () = α () α () = ˆρ Takng the trace on both sdes obvously gves Tr ˆρ = Tr ˆρ = 1 Note that the calculaton goes backwards as well! Suppose we do not know whether or not ˆρ s a pure state densty operator, but we do know that Tr ˆρ = 1 Snce Tr ˆρ = 1 always, we have that ω = 1 (3) The square of ˆρ n the bass where t s dagonal s of course ˆρ = ω α () α (), and ts trace then becomes ω = 1 (4) Now make a counter clam: Suppose ˆρ s not pure, that s, ω < 1 for all But then we have that ω < ω for all, and so the sum of squares of ω must be less than the sum of ω Ths s n contradcton wth Eq (4), and so the counter assumpton s false Clearly then, at least one ω must be one, and snce ther sum s one, there can be only sngle for whch ths s true, say = k So, and ˆρ s pure ˆρ = α (k) α (k)

6 (b) The elements of the densty matrx take n ths case the form α () ˆρ α (j) = α () α (k) α (k) α (j) = δ k δ kj e the dagonal element correspondng to the sngle populated state (ndex k) s one whle all the other elements are zero But ths need not be always the case Although the densty operator has a representaton as a matrx wth a sngle one on the dagonal n the bass {α () }, n some other bass the representaton s not dagonal As an example, suppose n bass {β () }, we have so that ˆρ = α (k) α (k) α (k) = c 1 β (1) + c β (), = (c 1 β (1) + c β () )(c 1 β (1) + c β () ) = c 1 β (1) β (1) + c 1c β () β (1) + c 1 c β (1) β () + c β () β () In the new bass the densty matrx reads c 1 c 1 c 0 0 c 1c c 0 0 ρ = In general, the matrx ρ can be fully flled wth numbers, but stll have that ρ = ρ, Tr ρ = 1, and represent a pure ensemble