4604 solutions set4 fa07

PHY4604 Fall 2007 Problem Set 4 Solutions

Department of Physics of 11

PHY 4604 Problem Set #4 Solutions Problem 1 (25 points):. (a) (2 points) Show that the sum of two hermitian operators is hermitian. Solution: If (Aop)↑ = Aop and (Bop)↑ = Bop then (Aop+Bop)↑ = (Aop)↑+(Bop)↑= (Aop+Bop). (b) (2 points) Suppose that Hop is a hermitian operator, and α is a complex number. Under what condition (on α) is αHop hermitian? Answer: α is real Solution: (αHop) ↑ = α∗(Hop) ↑=α∗Hop. Thus, αHop hermitian provided α = α∗ (i.e. α is real). (c) (2 points) When is the product of two hermitian operators hermitian? Answer: When the two operators commute. Solution: If (Aop)↑ = Aop and (Bop)↑ = Bop then

(AopBop)↑ = (Bop)↑ (Aop)↑= Bop Aop= Aop Bop + [Bop, Aop] and (AopBop)↑ = Aop Bop provided [Bop, Aop] = 0. Note that I used (e).

(d) (2 points) If dxdOop = , what is ↑

opO ?

Answer: dxdOop −=↑

Solution: We know that

[ ] ∫∫∞

∞−

↑∞

∞−

ΨΨ=ΨΨ dxtxOtxdxtxtxO opop ),(),(),(),( 1*21

*2

We see that

∫∫∞

∞−

∞

∞−

∗ ΨΨ−=Ψ

Ψ dxdx

txdtxdxtxdx

txd ),(),(),(),( 1*21

2

where I integrated by parts and dropped the boundary term. Thus,

dxdOop −=↑ .

(e) (2 points) Show that ↑↑↑ = opopopop ABBA )( . Solution: We know that <ψ1|(AopBop)|ψ2>* = <ψ2|(AopBop)↑|ψ1>, but

<ψ1|(AopBop)|ψ2>* = <ψ1|Aop|Bψ2>* = < Bψ2|(Aop)↑| ψ1> = < ψ2|(Bop)↑ (Aop)↑| ψ1>, where I used |Bψ2> = Bop|ψ2> and <Bψ2| = <ψ2|(Bop)↑. (f) (2 points) Prove that [AB,C] = A[B,C] + [A,C]B, where A, B, and C are operators. Solution: We see that

[AB,C] = ABC – CAB A[B,C] + [A,C]B = A(BC-CB) + (AC-CA)B = ABC - CAB

(g) (2 points) Show that dxdfixfp opx h−=)](,)[( , for any function f(x).

Solution: We see that

)(

)()()](,)[(

xdxdfi

dxdfi

dxdfi

dxdfi

dxdfif

dxdixxfp opx

ψψψψ

ψψψ

hhhh

hh

−=+−−=

+−=



(h) (2 points) Show that the anti-hermitian operator, Iop, has at most one real eigenvalue (Note: anti-hermitian means that opop II −=↑ ). Solution: We see that if Iop|ψ> = λ|ψ> then λ = <ψ|Iop|ψ> and

λ∗ = (<ψ|Iop|ψ>)* = <ψ|(Iop) ↑|ψ> = -<ψ|Iop|ψ> = -λ. If we let λ = x+iy then λ∗ = x-iy and λ∗ = -λ implies that x-iy = -x-iy or 2x= 0. The only real eigenvalue is λ = 0. (i) (2 points) If Aop is an hermitian operator, show that 02 >≥< opA . Solution: The norm of any allowed state is positive definite. Namely,

0 ≤ <Aψ|Aψ> = <ψ|(Aop)↑ Aop |ψ> = <ψ|(Aop)2 |ψ> = <(Aop)2>, where I used |Aψ> = Aop|ψ> and <Aψ| = <ψ|(Aop)↑ and (Aop)↑ = Aop. (j) (2 points) The parity operator, Pop, is defined by PopΨ(x,t) = Ψ(-x,t). Prove that the parity operator is hermitian and show that 12 =opP , where 1 is the identity operator. Compute the eigenvalues of the parity operator. Solution: An operator is hermitian if

[ ] ∫∫∞

∞−

∞

∞−


*2

If we let Oop = Pop we get

[ ] ∫∫∫∫∞

∞−

∞

∞−

∞

∞−

∞

∞−

ΨΨ=−ΨΨ=Ψ−Ψ=ΨΨ dxtxPtxdxtxtxdxtxtxdxtxtxP opop ),(),(),(),(),(),(),(),( 1*21

*21

*21

*2

where I changed variables x → -x in the integration. Note that (Pop)2Ψ(x,t) = (Pop) (Pop)Ψ(x,t) = (Pop) Ψ(-x,t) = Ψ(x,t). Thus, 12 =opP .

If Pop|ψ> = λ|ψ> where λ is real since Pop is hermitian. Also, (Pop)2|ψ> = λ2|ψ>, but (Pop)2|ψ> = 1|ψ> which implies that λ2 = 1 and hence λ = ±1 (k) (2 points) Idempotent operators have the property that opop PP =2 . Determine the eigenvalues of the idempotent operator Pop and characterize its eigenvectors. Solution: Here we have Pop|ψ> = λ|ψ> and (Pop)2|ψ> = λ2|ψ> = Pop|ψ> = λ|ψ> thus λ2 = λ or λ(λ -1) = 0 and hence λ = 0 or λ = 1. Let |ψ1> be the eigenket of Pop with eigenvalue 1 and |ψ0> be the eigenket of Pop with eigenvalue 0. Thus, Pop |ψ1> = |ψ1> and Pop|ψ0> = 0. Any (complex) multiple of |ψ1> (i.e. α|ψ1>) is an eigenket of Pop with eigenvalue 1. Any ket orthogonal to |ψ1> is an eigenket of Pop with eigenvalue 0. (l) (3 points) Unitary operators have the property that 1== ↑↑

opopopop UUUU , where 1 is the identity operator. Show that the eigenvalues of a unitary operator have modulus 1. Show that if <Ψ|Ψ> = 1 then <UΨ|UΨ> = 1 (i.e. unitary transformations preserve inner products). Prove that if Aop is hermitian then opiAe is unitary. Solution: We see that 1 = <Ψ|Ψ> = <Ψ|1|Ψ> = <Ψ|(Uop)↑Uop|Ψ> = <UΨ|UΨ>, where I used |Uψ> = Uop|ψ> and <Uψ| = <ψ|(Uop)↑. Now let

∑∞

=

==1 !

)(

n

nopiA

op niA

eU op and opop iAiA

n

nop

op een

iAU −−

∞

=

↑↑ ==

−==

↑

∑1 !

)(,

where I used (Aop)↑ = Aop. Now we see that



1== −↑ opop iAiAopop eeUU and Uop is unitary.

Problem 2 (15 points): Schrödinger’s equation says

ttxitxHop ∂

Ψ∂=Ψ

),(),( h ,

where the Hamiltonian operator is given by )(2/2 xVmpH xop += . (a) (2 points) Prove that Hop is hermitian. Solution: An operator is hermitian if

[ ] ∫∫∞

∞−

∞

∞−


*2

If we let )(21

2

2

xVxm

HO opop +∂∂

== then

[ ]

∫∫

∫∫

∫∫

∫∫

∫∫

∞

∞−

∗∞

∞−

∗

∞

∞−

∗∞

∞−

∗

∞

∞−

∗∞

∞−

∗

∞

∞−

∗∞

∞−

∗

∞

∞−

∞

∞−

ΨΨ=Ψ⎟⎟⎠

⎞⎜⎜⎝

⎛+

∂∂

Ψ=

ΨΨ+∂

Ψ∂Ψ=

ΨΨ+∂

Ψ∂∂

Ψ∂−=

=ΨΨ+Ψ∂

Ψ∂=

Ψ⎥⎦

⎤⎢⎣

⎡Ψ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

∂∂

=ΨΨ

dxtxHtxdxtxxVxm

tx

dxtxxVtxdxx

txtxm

dxtxxVtxdxx

txx

txm

dxtxtxxVdxtxx

txm

dxtxtxxVxm

dxtxtxH

op

op

),(),(),()(21),(

),()(),(),(),(21

),()(),(),(),(21

),(),()(),(),(21

),(),()(21),(),(

1212

2

2

1221

2

2

1212

12122

2

1

*

22

2

1*

2

where I integrated by parts twice and dropped the boundary terms. (b) (5 points) Prove that

>∂

∂<+><>=<

tO

OHiOdtd op

opopop ],[h

.

If the operator Oop does not depend explicitly on time and if it commutes with the Hamiltonian, then <Oop> is constant in time (i.e. it is conserved). Solution: We see that



( )

( )

><+>∂∂

<=

Ψ−Ψ+>∂∂

<

ΨΨ+ΨΨ−+>∂∂

<=

ΨΨ+ΨΨ−+>∂∂

<=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

∂Ψ∂

Ψ+Ψ∂

∂Ψ+Ψ⎟

⎠⎞

⎜⎝⎛

∂Ψ∂

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂Ψ∂

Ψ+Ψ∂

∂Ψ+Ψ

∂Ψ∂

=

ΨΨ=><

∫

∫

∫

∫

∫

∫

∞

∞−

∞

∞−

∞

∞−

∞

∞−

∗

∞

∞−

∞

∞−

],[

)(

)(

),(),(

*

**

**

**

***

*

opop

opopopop

opopopop

opopopop

opop

op

opop

op

op

HOtOi

dxOHHOtOi

dxHOOHtOi

dxHOOHtOi

dxt

iOt

OiO

ti

dxt

Ot

OO

ti

dxtxOtxdtdi

dtOdi

h

h

h

h

hhh

h

hh

where I used

),(),( txHt

txi opΨ=∂

Ψ∂h and ),(),( txH

ttxi op

∗∗

Ψ−=∂

Ψ∂h ,

and since Hop is hermitian we know that

[ ] ∫∫∞

∞−

∞

∞−

ΨΨ=ΨΨ dxtxHtxdxtxtxH opop ),(),(),(),( 1*21

*2 .

Thus,

>∂

∂<+><>=<

tO

OHiOdtd op

opopop ],[h

,

and if Oop does not depend explicitly on time and if [Hop,Oop] = 0 then

0>=< opOdtd which implies that <Oop> is constant (i.e. independent of time).

(c) (5 points) Prove that

dxdVxTpx

dtd

opxop −><>=< 2)( ,

where T is the kinetic energy operator (i.e. H = T + V). Solution: If we let Oop = xop(px)op then from (b)



( )><+><+><+><=

><+><=

>+<>=<>=<

opxopopxopopxopopopxopop

opxopopxopop

opxopopopxopopopxop

pxVpVxpxTpTxi

pxVipxTi

pxVTipxHipxdtd

)](,[])(,[)](,[])(,[

])(,[])(,[

])(,[])(,[)(

h

hh

hh

But 0])(,)[(])(,[ 2

21 == opxopxmopxop pppT

opxopxopopxmopopxopxmopopxmopop pmipxpxppxpxT )(

2)](,)[(],)[()(],)[(],[ 2

1212

21 h−

=+==

dxdVixVppxV opxopx h=−= )](,)[(])(),([

0]),([ =opxxV Thus,

( )

><−>=<⎟⎠⎞

⎜⎝⎛ ><+>

−<=

><+><+><+><>=<

dxdVxT

dxdVixpp

mii

pxVpVxpxTpTxipxdtd

opopxopx

opxopopxopopxopopopxopopopxop

2)()(

)](,[])(,[)](,[])(,[)(

hh

h

h

where I used mpT opxop 2/)( 2= . (d) (3 points) Use (c) to show that in a stationary state

dxdVxT >=<2 .

This is called the virial theorem. Solution: For stationary states h/)(),( iEtextx −=Ψ ψ and hence

∫∫∞

∞−

∞

∞−

=ΨΨ>=< dxxpxxdxtxpxtxpx opxopopxopopxop )()()(),()(),()( ** ψψ

which is independent of time so that for stationary states

dxdVxTpx

dtd

opxop −><=>=< 20)( and dxdVxT >=<2 .

Problem 3 (30 points): Consider an infinite square well defined by V(x) = 0 for 0 < x < L, and V(x) = ∞ otherwise. The stationary state position-space wave functions are given by

h/)(),( tiEnn

nextx −=Ψ ψ with )/sin(2)( LxnL

xn πψ =

and the eigenvalues are 2

222

2mLnEn

hπ= , and n is a positive integer.

(a) (10 points) Find the momentum-space wave function ),( tpxnΦ for the nth stationary state.

Answer: ( )hh

hh/

22

/

)1(1)/()(

),( Lipn

x

tiE

xnx

n

eLpn

neLtp −−

−−−

=Φπ

π



Solution: The momentum-space wave functions are given by

( )

( )

( )

( )

( )hD

hD

hD

hhD

hhD

hhD

hhD

DD

DDD

hh

hh

hh

hh

h

hhh

hhh

hhh

h

h

hh

/22

/

22//

//

///

)//()//(/

0

)//()//(/

0

)//()//(/

0

////

0

///

)1(1)/()(

)/()(21)1(

21

)/)(/()/()/(1)1(

21

)/(1)1(

)/(1)1(2

221

)//(1

)//(12

221

)//()//(2

221

2221

2221

)/sin(221),(

21),(

Lipn

x

tiE

x

LipntiE

xx

xxLipntiE

x

Lipn

x

LipntiE

x

LpLni

x

LpLnitiE

L

x

xpLni

x

xpLnitiE

LxpLnixpLnitiE

LxipLxinLxintiE

LxiptiExip

nxn

xn

xn

xn

xxn

xxn

xxn

xxn

xn

xnx

eLpn

neL

LpnneeL

LpnLpnLpnLpneeL

LLpn

eLLpn

eeL

pLnie

pLniee

Li

pLnie

pLniee

Li

dxeeeLi

dxeeeeLi

dxeLxneL

dxetxtp

−−

−−

−−

−−−

+−−+−

+−−+−

+−−+−

−−+−

−−+∞

∞−

−

−−−

=

−−−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−++

−−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−+

−−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+−

=

−=

−=

=Ψ=Φ

∫

∫

∫∫

ππ

ππ

π

ππππ

π

πππ

πππ

πππ

π

π

πππ

ππ

ππ

ππ

ππ

Note that

( ) ( )⎩⎨⎧

=−−=−− −−+−−

)2/sin()2/cos(

2)1()1(1 )2/()2/()2/()2/(/

h

hhhhhh

LpiLp

eeeeex

xLipLipnLipLipLipn xxxxx

L

L

6,4,25,3,1

==

nn

(b) (5 points) Graph 211 |),(|),( tptp xx Φ=ρ as a function of px for the ground state (i.e. n = 1).

Answer: 222

2

1 ])/([)2/(cos4),(

h

h

h LpLpLtp

x

xx −

=π

πρ

Solution: For n = 1 we get

)2/cos()/(

2),( 22

/)2/(

1 hhh

hh

LpLpeeLtp x

x

tiELip

x

nx

−=Φ

−−

ππ

and

222

22

11 ])/([)2/(cos4|),(|),(

h

h

h LpLpLtptp

x

xxx −

=Φ=π

πρ .

(c) (5 points) Graph 2

22 |),(|),( tptp xx Φ=ρ as a function of px for the 1st excited state (i.e. n = 2).

ρ1(px)

-6.0 0.0 6.0



Answer: 222

2

2 ])/()2[()2/(sin16),(h

h

h LpLpLtpx

xx −

=π

πρ

Solution: For n = 2 we get

)2/sin()/()2(

4),( 22

/)2/(

2 hhh

hh

LpLp

eieLtp xx

tiELip

x

nx

−=Φ

−−

ππ

and

222

22

22 ])/()2[()2/(sin16|),(|),(h

h

h LpLpLtptpx

xxx −

=Φ=π

πρ .

(d) (10 points) Use the momentum-space wave function ),( tpxnΦ to calculate the expectation values of px and px

2 for the nth stationary state and compare your answer with your answer to problem set #2 problem 2(c).

Answer: 0=>< nxp and 2

2222

Lnp nx

hπ=>< same as before

Solution: For even n we get

0])/()[()2/(cos4|),(| 222

222 =

−=Φ=>< ∫∫

+∞

∞−

+∞

∞−x

x

xxxxnxnx dp

LpnLppLndptppph

h

h ππ

and for odd n we get

0])/()[(

)2/(sin4|),(| 222

222 =

−=Φ=>< ∫∫

+∞

∞−

+∞

∞−x

x

xxxxnxnx dp

LpnLppLndptppph

h

h ππ

In both cases the integral vanishes because the integrand is odd under px → -px. For even n we get

∫

∫∫∞+

∞−

+∞

∞−

+∞

∞−

−=

−=Φ=><

dyy

ynyLn

dpLpnLppLndptppp x

x

xxxxnxnx

22

22

2

2

222

222222

)1()2/(cos4

])/()[()2/(cos4|),(|

π

ππ

h

h

h

h

and for odd n we get

∫

∫∫∞+

∞−

+∞

∞−

+∞

∞−

−=

−=Φ=><

dyy

ynyLn

dpLpnLppLndptppp x

x

xxxxnxnx

22

22

2

2

222

222222

)1()2/(sin4

])/()[()2/(sin4|),(|

π

ππ

h

h

h

h

where )/( hπnLpy x= . These integrals can be evaluated by partial fractions as follows

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−

++

+−

=− )1(

1)1(

1)1(

1)1(

141

)1( 2222

2

yyyyyy .

For odd n

ρ2(px)

-1.0 0.0 1.0



∫∫∫∫

∫∞+

∞−

∞+

∞−

∞+

∞−

∞+

∞−

+∞

∞−

+−

−+

++

−

=−

dyy

yndyy

yndyy

yndyy

yn

dyy

yny

)1()2/(cos

41

)1()2/(cos

41

)1()2/(cos

41

)1()2/(cos

41

)1()2/(cos

22

2

2

2

2

22

22

ππππ

π

But

∫∫∫∫+∞

∞−

+∞

∞−

+∞

∞−

+∞

∞−

⎟⎠⎞

⎜⎝⎛===

±du

uundz

zzndz

zzndy

yyn

k

k

kkk)(sin

2)2/(sin)2/)1((cos

)1()2/(cos 2222 ππππ m

Where 1±= yz and 2/znu π= . Now

⎩⎨⎧

=∫+∞

∞− π0)(sin2

duu

uk

21

==

kk

Thus

⎩⎨⎧

=±∫

+∞

∞− 2/0

)1()2/(cos

2

2

ππ

ndy

yyn

k 21

==

kk

and for n odd

2

2222

2

22

44

Lnn

Lnp nx

hh ππ==>< .

For even n we get

∫∫∫∫

∫∞+

∞−

∞+

∞−

∞+

∞−

∞+

∞−

+∞

∞−

+−

−+

++

−

=−

dyy

yndyy

yndyy

yndyy

yn

dyy

yny

)1()2/(sin

41

)1()2/(sin

41

)1()2/(sin

41

)1()2/(sin

41

)1()2/(sin

22

2

2

2

2

22

22

ππππ

π

But

∫∫∫∫+∞

∞−

+∞

∞−

+∞

∞−

+∞

∞−

⎟⎠⎞

⎜⎝⎛===

±du

uundz

zzndz

zzndy

yyn

k

k

kkk)(sin

2)2/(sin)2/)1((sin

)1()2/(sin 2222 ππππ m

where 1±= yz and 2/znu π= . Thus

⎩⎨⎧

=±∫

+∞

∞− 2/0

)1()2/(sin

2

2

ππ

ndy

yyn

k 21

==

kk

and for n even

2

2222

2

22

44

Lnn

Lnp nx

hh ππ==><

Problem 4 (20 points): Consider a three-dimensional vector space spanned by an orthonormal basis |1>, |2>, |3>. “Ket” vectors |α> and |β> are given by

|α> = i|1> -2|2> -i|3> and |β> = i|1> + 2|3>. (a) (5 points) Construct the “bra” vectors <α| and <β| in terms of the dual basis <1|, <2|, <3|. Answer: <α| = -i<1| -2<2| +i<3| and <β| = -i<1| + 2<3| (b) (5 points) Find <α|β> and <β|α> and confirm that <β|α> = <α|β>*.



Answer: <α|β> = 1+2i and <β|α> = 1-2i and <β|α> = <α|β>* Solution:

<α|β> = (-i<1| -2<2| +i<3|)( i|1> + 2|3>) = 1+2i <β|α> = (-i<1| + 2<3|)( i|1> -2|2> -i|3>) = 1-2i

(c) (10 points) Construct all nine matrix elements of the operator Aop = |α><β|, in this basis, and construct the matrix A. Is it hermitian?

Answer: ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−=

ii

iA

201402

201 not hermitian

Solution: A11 = <1|Aop|1> =<1 |α><β|1> = (i)(-i) = +1 A21 = <2|Aop|1> =<2 |α><β|1> = (-2)(-i) = +2i A31 = <3|Aop|1> =<3 |α><β|1> = (-i)(-i) = -1 A12 = <1|Aop|2> =<1 |α><β|2> = (i)(0) = 0 A22 = <2|Aop|2> =<2 |α><β|2> = (-2)(0) = 0 A32 = <3|Aop|2> =<3 |α><β|2> = (-i)(0) = 0 A13 = <1|Aop|3> =<1 |α><β|3> = (i)(2) = +2i A23 = <2|Aop|3> =<2 |α><β|3> = (-2)(2) = -4 A33 = <3|Aop|3> =<3 |α><β|3> = (-i)(2) = -2i

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−=

ii

iA

201402

201 A

ii

iAA T ≠

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−−== ∗↑

242000121

)(

Problem 5 (20 points): The Hamiltonian for a certain three-level system is represented by the matrix

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

abc

baH

000

0

where a, b, and c are real numbers. (a) (5 points) What are the energy eigenvalues of this system (i.e. the allowed energies)? Answer: E1 = c, E2 = a + b, and E3 = a – b. Solution: We determine the eiganvalues by solving the determinate

00

000

=−

−−

λλ

λ

abc

ba

Hence, (a - λ)(c – λ)(a – λ) – b2(c – λ) = 0 and the solution are l = c and (a – λ)2 = b2. Thus a – l = ±b and the three energies are E1 = c, E2 = a + b, and E3 = a – b. (b) (5 points) What are the (normalized) eigenvectors of H?



Answer: ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=

010

| 1E ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=

101

21| 2E

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−>=

101

21| 3E

Solution: For E1 = c we have

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+

+=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

czcycx

zyx

cazbx

cybzax

zyx

abc

ba

000

0

Thus, ax + bz = cx or (a-c)x +bz = 0 and bx + az = cz or bx + (a-c)z = 0 which implies that [(a-c)2 – b2]x = 0 and thus x = 0 (provided that (a-c)2 ≠ b2) and

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=

010

| 1E .

For E2 = a + b we have

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+++

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+

+=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

zbaybaxba

zyx

baazbx

cybzax

zyx

abc

ba

)()()(

)(0

000

Thus, ax + bz = (a+b)x or -bx +bz = 0 and bx+az= (a+b)z or bx - bz = 0 and cy = (a+b)y which implies that y = 0 and x = z.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=

101

21| 2E .

For E3 = a - b we have

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+

+=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

zbaybaxba

zyx

baazbx

cybzax

zyx

abc

ba

)()()(

)(0

000

Thus, ax + bz = (a-b)x or bx +bz = 0 and cy = (a-b)y which implies that y = 0 and x = -z.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−>=

101

21| 3E .

(c) (5 points) If the system starts out at t = 0 in the state

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=

010

)0(| s

what is |s(t)> at later time t.



Answer: ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=>= −

010

)(|)(| /1

hictetEts

Solution: In this case |s(0)> = |E1> and hence

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=>= −−

010

010

)(|)(| //1

1 hh icttiE eetEts .

(d) (5 points) If the system starts out at t = 0 in the state

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=

100

)0(| s

what is |s(t)> at later time t.

Answer: ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−>= −

)/cos(0

)/sin()(| /

h

hh

bt

btiets iat

Solution: In this case

>−>>= 32 |2

1|2

1)0(| EEs .

and hence

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛>=−>>=

−

+−

+−

−

−−+−−−

)/cos(0

)/sin(0

21

101

21

101

21|

21|

21)(|

/

//

//

/

/)(/)(3

/2

/ 32

h

hh

hh

hh

h

hhhh

bt

btie

ee

eee

eeEeEets

iat

ibtibt

ibtibt

iat

tbaitbaitiEtiE

4604 solutions set4 fa07

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