PHYSICS 34 QUANTUM PHYSICS II (25) Assigmet 2 Solutios 1. With respect to a pair of orthoormal vectors ϕ 1 ad ϕ 2 that spa the Hilbert space H of a certai system, the operator  is defied by its actio o these base states as follows:  ϕ 1 = 3 ϕ 1 4i ϕ 2  ϕ 2 = 4i ϕ 1 3 ϕ 2. (a) By use of the defiig coditio for evaluatig the actio of a operator o a bra vector, that is { ψ Â} χ = ψ { χ } for all χ, evaluate ϕ 1  ad ϕ 2 Â. (b) By writig  i matrix form, cofirm the results obtaied i part (a). (a) To evaluate ϕ 1 Â, cosider ϕ 1 { χ } = ϕ 1 { [ ϕ 1 ϕ 1 χ + ϕ 2 ϕ 2 χ ] } = ϕ 1 [  ϕ 1 ϕ 1 χ +  ϕ 2 ϕ 2 χ ] = ϕ 1 [ (3 ϕ 1 4i ϕ 2 ) ϕ 1 χ + ( 4i ϕ 1 3 ϕ 2 ) ϕ 2 χ ] = ϕ 1 [ ϕ 1 (3 ϕ 1 χ 4i ϕ 2 χ ) + ϕ 2 ( 4i ϕ 1 χ 3 ϕ 2 χ ) ] =3 ϕ 1 χ 4i ϕ 2 χ ={3 ϕ 1 4i ϕ 2 } χ. But sice { ϕ 1 Â} χ = ϕ 1 { χ } we have As χ is arbitrary, we coclude that { ϕ 1 Â} χ = {3 ϕ 1 4i ϕ 2 } χ. ϕ 1  = 3 ϕ 1 4i ϕ 2. The other case ca be dealt with i a similar way. ϕ 2 { χ } = ϕ 2 { [ ϕ 1 ϕ 1 χ + ϕ 2 ϕ 2 χ ] } = ϕ 2 [  ϕ 1 ϕ 1 χ +  ϕ 2 ϕ 2 χ ] = ϕ 2 [ (3 ϕ 1 4i ϕ 2 ) ϕ 1 χ + ( 4i ϕ 1 3 ϕ 2 ) ϕ 2 χ ] = ϕ 2 [ ϕ 1 (3 ϕ 1 χ 4i ϕ 2 χ ) + ϕ 2 ( 4i ϕ 1 χ 3 ϕ 2 χ ) ] = 4i ϕ 1 χ 3 ϕ 2 χ ={ 4i ϕ 1 3 ϕ 2 } χ. But sice { ϕ 2 Â} χ = ϕ 2 { χ } we have As χ is arbitrary, we coclude that { ϕ 2 Â} χ = { 4i ϕ 1 3 ϕ 2 } χ. ϕ 2  = 4i ϕ 1 3 ϕ 2. 1
(b) We ca represet the operator  usig the { ϕ 1, ϕ 2 } basis by the followig matrix ( ) ( ) ϕ1  ϕ  1 ϕ 1  ϕ 2 3 4i =. ϕ 2  ϕ 1 ϕ 2  ϕ 2 4i 3 Meawhile, the state ψ is represeted by the colum vector ( ) ϕ1 ψ ψ ϕ 2 ψ so that for the basis state ϕ 1 : ( 1 ϕ 1 ) The bra vector will the be represeted by the row vector ϕ 1 ( 1 ). Thus, i matrix form ϕ 1  ( 1 ) ( ) 3 4i 4i 3 = ( 3 4i ) which is the row vector correspodig to the bra vector 3 ϕ 1 4i ϕ 2. Similarly, we have ϕ 2 ( 1 ) so that ϕ 2  ( 1 ) ( ) 3 4i = ( 4i 3 ) 4i 3 which is the row vector correspodig to the bra vector 4i ϕ 1 3 ϕ 2. 2. Prove the followig: (a) ( ) = Â. (b) ( ˆB) = ˆB  (c) ( ˆB) 1 = ˆB 1  1 (a) Itroducig two arbitrary states φ ad ψ we have φ ( ) ψ = ψ  φ = φ  ψ = φ  ψ. (b) Itroduce two arbitrary states φ ad ψ. The ψ ( ˆB) φ = φ  ˆB ψ = ξ ζ 2
where φ  = ξ ad ˆB ψ = ζ. The so that ξ ζ = ζ ξ = ψ ˆB  φ ψ ( ˆB) φ = ψ ˆB  φ. Sice the states φ ad ψ are arbitrary, we coclude that as required. ( ˆB) = ˆB  (c) The product ( ˆB) 1 is the iverse of  ˆB, i.e. ( ˆB) 1 ( ˆB) = ˆ1. Multiplyig o the right by ˆB 1 ad the  1 gives ad hece so fially ( ˆB) 1 ( ˆB) ˆB 1  1 = ˆB 1  1 ( ˆB) 1  1 = ˆB 1  1 ( ˆB) 1 = ˆB 1  1. 3. If  is ay operator, show that   is Hermitea ad has o-egative eigevalues. Takig the Hermitea cojugate of   gives so that   is Hermitea. ( Â) =  ( ) =   As   is Hermitea it has a complete set of orthoormal eigestates. Call them { ; = 1, 2,... } with   =, ad cosider   = where we have used the fact that the eigestates of   are orthoormal. If we ow put ξ =  the ξ =  ad hece =   = ξ ξ by the properties of the ier product. Thus we have show that as required. 3
4. Show that if two operators  ad ˆB do ot commute, the exp  exp ˆB exp(â+ ˆB). [Hit: expad each expoetial to secod order ad compare the two expressios.] Expadig exp(â) ad exp( ˆB) each to secod order gives, for their product, also to secod order: eâe B = (ˆ1 +  + 1 2 Â2 +... )(ˆ1 + ˆB + 1 2 ˆB 2 +... ) = ˆ1 +  + ˆB + 1 2 Â2 + 1 2 ˆB 2 +  ˆB +.... Similarly, expadig exp(â + ˆB) to secod order, takig care with the order of the o-commutig operators gives: eâ+ ˆB = ˆ1 +  + ˆB + 1 2 ( + ˆB) 2 +... = ˆ1 +  + ˆB + 1 2(Â2 +  ˆB + ˆB + ˆB 2) +.... Subtractig the two results the gives, oce agai to secod order eâe ˆB eâ+ ˆB =  ˆB 1 2( ˆB + ˆB ) +... = 1 2( ˆB ˆB ) = 1 2 [Â, ˆB] +.... Thus, if [Â, ˆB], it must be the case that exp  exp ˆB exp(â + ˆB). 5. Cosider three operators (the Pauli spi operators) ˆσ x, ˆσ y, ad ˆσ z defied to have the properties, for the states ± of a spi half system, give by ˆσ x ± = ˆσ y ± = ±i ˆσ z ± = ± ±. Show that the commutator of ˆσ x ad ˆσ y ca be expressed i terms of ˆσ z. The commutator is [ ˆσx, ˆσ y ] = ˆσx ˆσ y ˆσ y ˆσ x. I order to show if this ca be expressed i terms of ˆσ z, we eed to cosider its actio o each of the basis states ±. Thus we have [ ˆσx ˆσ y ˆσ y ˆσ x ] + = ˆσx ˆσ y + ˆσ y ˆσ x + =i ˆσ x ˆσ y =i + + i + =2i ˆσ z +. Similarly [ ] ˆσx ˆσ y ˆσ y ˆσ x = ˆσx ˆσ y ˆσ y ˆσ x = i ˆσ x + ˆσ y + = i i =2i ˆσ z. 4
I other words we have show that [ ˆσx ˆσ y ˆσ y ˆσ x ] ± = 2i ˆσz ±. Thus, sice the states ± are a set of basis states, we coclude that ˆσ x ˆσ y ˆσ y ˆσ x = 2i ˆσ z. 6. A cavity ca support a electromagetic field of a sigle frequecy. The state of this field ca be specified i terms of the set of basis states { ; =, 1, 2,...}, where is the state i which there are photos i the cavity. Cosider the state (kow as a coheret state) defied by the ifiite series α = e α 2 /2 where α is a complex umber, ad the operator (the photo aihilatio operator) defied such that â = 1. (a) Show that this state is ormalized to uity. (b) Show that α is a eigestate of â with eigevalue α. (c) Show that α α for α α. How does this result compare with that which applies for Hermitea operators? (a) We eed to show that α α = 1. Notig that the bra vector α ca be writte α = e α 2 /2 α ad hece we have α α = e α 2 m= α m m! m. Note the use of differet summatio idices i the two sums. Usig the orthoormality coditio m = δ m we have α α = e α 2 m= α m m! δ m = e α 2 The sum is just the expoetial fuctio, so we get α α = e α 2 e α 2 = 1. α 2. 5
A alterate way of gettig this result without usig a double sum is to write α α = e α 2 /2 α use the fact that α = α, ad that to obtai the same result. α = e α 2 /2 (b) Sice â is a liear operator we ca write â α =e α 2 /2â =e α 2 /2â =e α 2 /2 α â 1. Because of the factor, the term i the sum for = will vaish, so the sum begis at = 1. Further, this factor will cacel the factor appearig i the deomiator i the factorial term, so overall we get: â α = e α 2 /2 =1 ( 1)! 1 = e α 2 /2 +1. These above steps ca be cofirmed (as ca the last below) by simply writig out the first three or so terms of the series ad carryig out the calculatios o each term idividually. Fially we ote that we ca ow extract a commo factor from the sum to give â α = e α 2 /2â +1 ()! = e α 2 /2 α Thus we have show that â α = α α. ()! = α α. (c) To evaluate α α we ca proceed much as was doe i evaluatig α α. Thus we will have α α =e α 2 /2 e α 2 /2 (α ) m δ m m! =e α 2 /2 e α 2 /2 m= (α α) =e α 2 /2 e α 2 /2 e α α which is certaily ot zero. Thus we see that eve though α α, the correspodig eigevectors are ot orthogoal. As this it always the case for Hermitea operators that they have real eigevalues, ad that eigestates with differet eigevalues will always be orthogoal, we coclude that the operator â is ot Hermitea. 6
7. (a) Prove that if a fuctio f (x) ca be expaded i a power series i x, the f (Â) = f (a ) a a where a, = 1, 2, 3,... is the eigestate of  with eigevalue a. (b) Show that if  is Hermitea, the Û = exp(iâ) is uitary. (c) Suppose that U represets a physical process that acts o the system ad that the system is iitially i the ormalized state ψ, so that after the process ceases to act, the ew state of the system is Û ψ. Give that  has oly two eigevalues a 1 ad a 2 (with the correspodig eigestates a 1 ad a 2 ), determie the probability of fidig the system i the state ψ after the process has ceased. (a) The solutio ca be based o the observatio that if a m is a eigestate of  with eigevalue a m the  a m =  1  a m =  1 a m a m = a m  1 a m. Repeatig the procedure a secod time gives ad so o times over to fially give ad hece  a m = a 2 mâ 2 a m a 1 m  a m = a m a m  a m = a m a m. Thus, if f (x) has a power series expasio f (x) = c m x m the f (Â) a = c m  m a m m = c m a m a m = f (a ) a. To complete the proof, we the write this as f (Â) a a = f (a ) a a ad sum over all the eigestates to give f (Â) a a = f (a ) a a. 7
O the left had side of this equatio, we ca use the completeess relatio for the eigestates of Â, that is to write fially yieldig the required result a a = ˆ1 f (Â) a a = f (Â) a a = f (Â) f (Â) = f (a ) a a (b) By expadig Û = exp(iâ) as a power series i Â, we see that Û = (iâ) from which it follows that Û = ( iâ ). But sice  is Hermitea, the  = Â, ad so Û = ( iâ) = e iâ. To show that Û is uitary the requires showig that Û Û = ÛÛ = ˆ1, i.e., that Û = Û 1. We ca do this directly by itroducig the complete set of eigestates of Â, usig part (a), ad writig so that Û = e ia a a Û Û = e ia Û a a. But, from the proof of part (a), we have so that Û a = e iâ a = e ia a Û Û = e ia e ia a a = a a = ˆ1 where the completeess relatio has bee used. Thus we have show that Û Û = ˆ1, ad i the same way, we have = ÛÛ = ˆ1. Thus we coclude that Û = Û 1, i.e. that Û is uitary. 8
(c) As a result of Û actig o the state ψ, the ew state of the system is Û ψ = e iâ[ a 1 a 1 ψ + a 2 a 2 ψ ] = e ia 1 a 1 a 1 ψ + e ia 2 a 2 a 2 ψ. The probability amplitude of fidig the system i the same state ψ is the ψ Û ψ = ψ [ e ia 1 a 1 a 1 ψ + e ia 2 a 2 a 2 ψ ] = e ia 1 ψ a 1 a 1 ψ + e ia 2 ψ a 2 a 2 ψ = e ia 1 a 1 ψ 2 + e ia 2 a 2 ψ 2 The probability of observig the system i the state ψ is the ψ Û ψ 2 = e ia 1 a 1 ψ 2 + e ia 2 a 2 ψ 2 2 = a 1 ψ 4 + a 2 ψ 4 + 2 cos(a 1 a 2 ) a 1 ψ 2 a 2 ψ 2 8. A model for a oe dimesioal metallic crystal cosists of periodically positioed potetial wells a distace a apart correspodig to the sites of the postive ios. A electro i this crystal ca be foud oly at the positios x = a; =, ±1, ±2,.... (a) Costruct a set of basis states for this system. What is the completeess relatio i Dirac otatio for these states? (b) Write dow a expressio i Dirac otatio for the positio operator ˆx for the electro. (c) Derive a expressio for the operator ˆx 2. (a) The electro ca be foud at the positios of the wells, i.e. at x = a, =, ±1, ±2,..., which will the be the eigevalues of the positio eigestates a. The basis states will the be the set of such states, that is { a ; =, ±1, ±2,... } The completeess relatio will the be + = (b) The positio operator will the be ˆx = (c) The operator ˆx 2 will be give by ˆx 2 = ˆx + = a a a = a a = ˆ1. + = + = 9 a a a. a ˆx a a = + = (a) 2 a a
. 9. I the model for a O ( ) 2 io, for which the basis states are the positio eigestates ± a correspodig to the electro beig foud o the oxyge atom at x = ±a, the Hamiltoia takes the form, i the positio represetatio ( ) E A Ĥ. A E Write out this Hamiltoia i Dirac otatio i the { + a, a } basis. Also write out this Hamiltoia i the eergy basis, give that the eergy eigestates of Ĥ are E = ( + a + ( 1) a )/ 2 with eigevalues E = E ( 1) A where = 1, 2, ad show that the two expressios for Ĥ are equal. I geeral, ay operator  ca be writte i the form  = ϕ m ϕ A m m where the states { ϕ, = 1, 2,...} form a set of orthoormal basis states, ad A m = ϕ m  ϕ. I the positio represetatio, the operator Ĥ here will the be Ĥ = +a +a Ĥ +a +a + +a +a Ĥ a a + a a Ĥ +a +a + a a Ĥ a a. What remais is the to read off from the matrix the matrix elemets of Ĥ. This matrix is just ( ) +a Ĥ + a +a Ĥ a a Ĥ + a a Ĥ a which, by compariso with the matrix gives ad hece +a Ĥ + a = E +a Ĥ a = A +a Ĥ a = A a Ĥ a = E Ĥ = E + a +a A + a a A a +a + E a a. I the eergy basis, the Hamiltoia will be, i Dirac otatio, Ĥ = E 1 E 1 E 1 + E 2 E 2 E 2. This ca be rewritte i the positio basis usig the expressios give for the eergy eigestates expressed i terms of the positio eigestates. Thus we have Ĥ = 1 2 (E + A) ( + a a )( +a a ) + 1 2 (E A) ( + a + a )( +a + a ) = 1 2( E + A + E A ) + a +a + 1 2( E A + E A ) + a a + 1 2( E A + E A ) a +a + 1 2( E + A + E A ) a a = E + a +a A + a a A a +a + E a a. 1
1. (a) Show that the fuctio F(x, λ) = 1 2 λe λ x is such that + F(x, λ)dx = 1. (b) Plot this fuctio for λ = 1, 2 ad 4. (c) Calculate the itegral + e x F(x 1, λ)dx ad compare the result obtaied i the limit of λ with the result obtaied from the itegral + e x δ(x 1)dx. (a) I carryig out this itegral, it has to be recogized that the fuctio is discotiuous at x = so the itegral has to be split ito two parts: + + F(x, λ)dx = 1λ e λ x dx 2 = 1 2 λ e λ x dx + 1λ e λ x dx 2 = 1λ e λx dx + 1λ e λx dx 2 2 = 1λ eλx 2 + λ 1λ e λx 2 λ =1. (b) 2. λ = 4 1. λ = 2 λ = 1.5-4 -2 2 4 11
(c) I this itegral, the discotiuity is at x = 1, so the itegral is separated ito two parts at this poit, i.e. + e x F(x 1, λ)dx = 1λ e x e λ x 1 dx 2 1 = 1λ e x e λ(x 1) dx + 1λ e x e λ(x 1) dx 2 2 1 = 1λ e x e λ(x 1) 1 2 + λ 1 1λ e x e λ(x 1) 2 λ 1 = 1 2 λe 1 e λ λ 1 + 1 2 λ e 1 λ + 1. If the limit of λ is ow take of this result we get + lim λ e x F(x 1, λ)dx = e 1 which is the same result that would be foud if we made use of the delta fuctio limit of F(x 1, λ), that is e x δ(x 1)dx = e 1. 1 12