Quantum Mechanics I - Session 4

Transcription

1 Quantum Mechancs I - Sesson 4 Aprl 3, 05 Contents Operators Change of Bass 4 3 Egenvectors and Egenvalues 5 3. Denton Rotaton n D Dagonalzng a matrx 6

2 Operators In order to dene a lnear operator we need to dene how the operator operates on each vector n the space..e for each ket ψ we need to dene A ψ = φ For example the dentty operator s dened by ψ = ψ. If a bass s orthonormal then t has the very useful property that = n n n. () proof: ψ = ( n n n ) ψ = n n n m c m m = n,m n n m c m = n,m c mδ nm n = n c n n = ψ However, t s sucent to dene how a lnear operator acts on all the members of a certan bass nstead of on all the members of the space. That s, f A n = m a m,n m () then we know that A ψ = A n c n n = n,m a m,n c n m. (3) The numbers a m,n are the matrx representaton of the operator A n the bass { n }. Once agan, they are to be found usng the nner-product m A n = l a l,n m l = a m,n, (4) these are called the matrx elements of A. We can use these numbers n order to wrte the operator n a derent way A = a n,m n m (5) n,m Wth ths denton t s easy to see that ψ A φ = ( ψ A) φ = ψ (A ψ ) How does the hermtan conjugate of that operator looks lke? By denton we have m A n == n A m = a n,m. (6) Thnkng of t as a matrx, we have changed the rows and the columns and taken the complexconjugate of each entry.

3 Lnear operators can be multpled, the result beng a lnear operator by tself. B ψ = ϕ and A ϕ = ξ then Suppose AB ψ = A(B ψ ) = A ϕ = ξ. (7) In general of course the operator AB s not equal to the operator BA snce the order of In the matrx representaton of operators ths s convenently done by multplyng the matrces. We'll denote the matrx representaton of the operator AB by the coecents (AB) n,m, so (AB) n,m = n AB m (8) ( ) = n A l l B m = l l n A l l B m = l a n,l b l,m. Queston: Let K be the operator dened by K = φ ψ where φ and ψ are two vectors of the space state.. Under what condton s K Hermtan?. Calculate K. Under what condton s K a projector? 3. Show that K can always be wrtten n the form K = λp P where λ s a constant to be calculated and P and P are projectors. Answer:. The denton of Hermtan s K j = ( j K ) On the one hand, pluggng n K ( j K ) = ( j φ ψ ) = ψ φ j so we see that the requrement s. The denton of a projector s φ ψ = ψ φ K = K K = φ ψ φ ψ so the requrement s ψ φ = 3

4 3. We can always wrte K as K = φ φ ψ ψ / φ ψ Change of Bass Generally there are many derent orthonormal bass we can choose for a gven vector space. Let us revew how we go from one representaton { n } to another { ξ }. We saw that ψ = n c n n, (9) wth c n = n ψ. We can also wrte ψ = ξ e ξ ξ, (0) wth e ξ = ξ ψ. The transformaton s done by usng the dentty operator ψ = c n n = c n ξ ξ n = ( ) c n ξ n ξ () n n ξ n so we conclude that e ξ = n ξ n c n. wth s ξ,n = ξ n we get that ξ e ξ = n s ξ,n c n. () We can thnk of t as a matrx multplyng a vector. A change of bass s lke a rotaton. Denng the operator S = ξ,n s ξ,n ξ n (3) we have that S ψ s the rotated vector. How does the representaton of operators transform from one bass to another? We'll use agan the dentty operator A = a n,m n m = a n,m ξ ξ n m η η = n,m n ξ η = ( ) a n,m ξ n m η ξ η. (4) ξ,η n,m So the new coecents of the matrx representaton α ξ,η are gven by α ξ,η = a n,m ξ n m η. (5) n,m 4

5 Ths can be wrtten agan usng a matrx multplcaton formalsm. Usng the same operator S we get that α ξ,η = n,m s ξ,n a n,m s η,m (6) = n,m s ξ,n a n,m ( s ) m,η or, n matrx form (usng the notaton à to wrte the operator A n the new bass) à = SAS. (7) The operator S s a untary operator, whch means t has an mportant property S S = SS =. (8) 3 Egenvectors and Egenvalues 3. Denton An Egenvector, v, of a matrx, M s dened as M v = λ v (9) where the scalar λ s the Egenvalue of the matrx. The nterpretaton of the egenvector s that of a vector that when s multpled by the matrx remans n the same drecton but only changes ts magntude. The way to nd the egenvalues of a matrx s the followng. Frst you calculate the characterstc polynomal of the matrx dened as M λi = 0 (0) where I s the unt matrx. Solvng for λ gves us the egenvalues of a matrx and then solvng 9 gves us the egenvectors (up to a multplcaton by a scalar). 3. Rotaton n D The rotaton of a coordnate system n D s done by the matrx ( ) cos θ sn θ sn θ cos θ Lets nd the egenvalues and egenvectors of ths matrx. The characterstc polynomal s λ cos θλ + = 0 or n other words λ λ(e θ + e θ ) + = 0 5

6 whose soluton s λ = e θ λ = e θ The egenvectors are gven by ( e θ + e θ e θ e θ ) eθ +e (x ) ( ) θ x = e θ e θ + e θ y y ( + e θ + e θ e θ + e θ whose soluton s ( ) The other egenvector s gven by ) ( ) ( ) x x = y y ( e θ + e θ e θ e θ ) eθ +e (x ) ( ) θ x = e θ e θ + e θ y y and ts soluton s ( ) We see that the egenvalues have a mgantude of whch s expected snce the matrx s untary (preserves the lengthes of vectors). 4 Dagonalzng a matrx Assume that A has n dstnct egenvalues. Then A s dagonalzable. Moreover, f P s the matrx wth the columns C, C,..., and Cn the n egenvectors of A, then the matrx P AP s a dagonal matrx (just lke equaton 7). As an example let us dagonalze the matrx 0 The polynomal s who has the trval soluton λ 3 + 4λ + 4λ = 0 λ = 0 The other two are gven by 6

7 λ = λ 3 = + and the egenvectors are 0 λ 3 λ So we have 0 λ 3 λ P = The nverse of P s gven by 0 P = λ 3 λ λ λ 3 (λ λ λ 3 λ λ λ 3 ) 3 So the dagonal matrx s gven by P P = 0 λ λ The dagonal matrx always has the egenvalues on the dagonal correspondng to the egenvectors column n P. 7