Department of Physics Temple University

Department of Physics Temple University

Introduction to Quantum Mechanics, Physics 3701 Instructor:Z.-E. Meziani

Solution set for homework # 6 April 16, 2013 Exercise#2, Complement FVI, page 765 Consider an arbitrary physical system whose four-dimensional state space is spanned by a basis of four eigenvectorsjj; mzicommon to ^ J 2 andJz(j= 0 or 1;jmz+j), of eigenvaluesj(j+ 1)h 2 andmzh,

such that:

Jjj; mz>= h q j(j+ 1)mz(mz1jj; mz1>(1) J+jj; j >=Jjj;j >= 0 (2)

a) Express in terms of the ketsjj; mz>, the eigenstates common to

^ J 2 and ^ Jxto be denoted by jj; mx>.We must rst form the matrix of the operator ^ Jxin the basisfjj; mz>g. If we recall the following relation ^ Jx= 1 2 ( ^ J++ ^

J) (3)

then we may use Eqs.??and??to write the matrix of the ^ Jxoperator in the given basis. We rst calculate the individual matrix elements. First, the ^ J+terms ^ J+j1;1>= 0 (4) ^ J+j1;0>= p 2hj1;1>(5) ^ J+j1;1>= p 2hj1;0>(6) ^ J+j0;0>= 0 (7) then theJterms ^ Jj1;1>= p 2hj1;0>(8) ^ Jj1;0>= p 2hj1;1>(9) ^ Jj1;1>= 0 (10) ^ Jj0;0>= 0 (11) Now we may write the operator ^ Jxin matrix form, using Eq.??. in the basisfj1;1>;j1;0>and j1;1>;j0;0>g ^ Jx= h p 2

B B @

0 1 0 0

1 0 1 0

0 1 0 0

0 0 0 0

1 C C A (12) This study source was downloaded by 100000850872992 from CourseHero.com on 04-15-2023 23:56:45 GMT -05:00

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To nd the eigenvalues of this matrix, it is necessary to diagonalize it.

h p 2

• 0

h p 2

h p 2

h p 2

• 0 0

= 0 (13)

The determinant is solved in the usual way, resulting in a characteristic equation given by

h p 2

h p 2

• 0

h p 2

h p 2 h p 2

00

• 0

= 0

h p 2

h p 2

h p 2 h p 2

= 0

2

2

h 2 2

h 2 2 != 0

2

2 h 2

= 0 (14)

This produces four roots, two of which are zero. This= 0 eigenvalue is thus two-fold degenerate.The other two eigenvalues are +handh. Substitution of these eigenvalues into Eq.??allows us to solve for the eigenvectors.^ Jxj >=j >) h p 2

B B @

0 1 0 0

1 0 1 0

0 1 0 0

0 0 0 0

1 C C A

B B @ a b c d 1 C C A =

B B @ a b c d 1 C C A (15) We are therefore able to write the eigenvectors in the new basis,jj; mx>in terms of the old basis, jj; mz>. The following are the eigenvalues that correspond to the eigenvectors in both bases, in the

order:jj; mx>,jj; mz>

0 :j0;0>x=j0;0>(16)

0 :j1;0>x=

1 p 2 (j1;1>j1;1>) (17)

+h:j1;1>x=

1 p 2 (j1;1>+j1;1>+j1;0>) (18)

h:j1;1>x=

1 p 2 (j1;1>+j1;1>j1;0>) (19) These are the eigenstates common to the operators ^ J 2 and ^ Jx

b) Consider a system in the normalized state:

j >=jj= 1; mz= 1>+jj= 1; mz= 0>+jj= 1; mz=1>+jj= 0; mz= 0>(20) Note that this state is normalized. Therefore, we must have the following relation between the

coecients:

jj 2 +jj 2 +jj 2 +jj 2 = 1 (21) This study source was downloaded by 100000850872992 from CourseHero.com on 04-15-2023 23:56:45 GMT -05:00

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i) What is the probability of nding 2h

2 and hif ^ J 2 and ^ Jxare measured simultaneously?In order to nd 2h 2 when acting on a state with ^ J 2 ,j(j+ 1)h 2 must equal 2h 2 ; therefore, j= 1. This restricts the possible states that will return hunder the action of ^ Jx. We must therefore compute the requested probability using Eq. 18. This is done in the usual way. We shall project the eigenstate to state onto the normalized state in Eq. 20 and square the result.P

^ J 2 = 2h 2 ; ^ Jx= h

= =j<1;1j >j 2 =j<1;1j[jj= 1; mz= 1>+jj= 1; mz= 0>+jj= 1; mz=1>+jj= 0; mz= 0>]j 2 =j<1;1jj1;1>j 2 =jj 2 (22) ii) Calculate the mean value of ^ Jzwhen the system is in the statej >, and the probabilities of the various possible results of a measurement bearing only on this observable.The action of ^ Jzon each statejj; mz>returns the same state with the eigenvalue ofmzh.Therefore, the mean value of ^

Jzis given by:

< ^ Jz>=< j ^ Jzj > =< j ^ Jzj[j1;1>+j1;0>+j1;1>+j0;0>]> =< j[(1)hj1;1>+< j(0)hj1;0>+< j(1)hj1;1>+< j(0)hj0;0>] =< jhj1;1>< jhj1;1> = h < j1;1>h < j1;1> = h h jj 2 +jj 2 i (23) Hence, the possible results of a measurement of this observable are as follows P(0) =jj 2 +jj 2 (24) P(+h) =jj 2 (25) P(h) =jj 2 (26) iii) Same questions for the observable ^ J 2 and for ^ Jx.The action of ^ J 2 on a statejj; mz>returns the same state with the eigenvalue ofj(j+ 1)h 2 .Therefore,the mean value of ^ J 2

is given by:

< ^ J 2 >=< j ^ J 2 j > =< jJ 2 j[j1;1>+j1;0>+j1;1>+j0;0>]> =< j h (2h 2 )j1;1>+(2h 2 )j1;0>+(2h 2 )j1;1>+(0)j0;0> i > = 2h 2 < j[j1;1>+j1;0>+j1;1>] = 2h 2 h jj 2 +jj 2 +jj 2 i (27) This study source was downloaded by 100000850872992 from CourseHero.com on 04-15-2023 23:56:45 GMT -05:00

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The action of ^ Jxon a state returns thejj; mx>state with the eigenvalue ofmxh. We must use the transformed eigenstates. Since we have already written the matrix of ^ Jxin thefjj; mz>g representation it is easier to calculate the mean value using matrix form.< ^ Jx>= (

) h p 2

B B @

0 1 0 0

1 0 1 0

0 1 0 0

0 0 0 0

1 C C A

B B @

1 C C A = h p 2 = (

+

• )

B B @

1 C C A =

• (

+

)+

(28) iv) ^ J 2 zis now measured; what are the possible results, their probabilities, and their mean value?The possible results are 0 or h 2 with probabilitiesjj 2 +jj 2 andjj 2 +jj 2 respectively.We have already calculated the mean value of ^ Jz, Eq. 23 . A second action with this operator will only return the same states, but we will pick up an extra h.< ^ J 2 z>=< j ^ J 2 zj > =< j ^ Jzj[Jzjj1;1>+j1;0>+j1;1>+j0;0>]> =< j ^ Jzj[(1)hj1;1>+< j0hj1;0>+< j(1)hj1;1>+< j(0)hj0;0>] = h 2 h jj 2 +jj 2 i (29) Exercise#6, Complement FVI, page 768 Consider a system of angular momentuml= 1. A basis of its state space is formed by the three eigenvectors ofLz:j+ 1>;j0>;j 1>, whose eigenvalues are , respectively +h, 0,hand which satisfy: L= h p 2jm1> L+j+ 1>=Lj 1>= 0 (30) This system posses a quadrupole moment and is placed in an electric eld gradient such that its Hamiltonian

can be written:

H= !0 h

L 2 uL 2 v

(31) whereL 2 uandL 2 vare the components of ~ Lalong the two directionsOuandOvof theXOZplane which form an angle of 45

withOxandOz.a We can rst express the componentsLuandLvin terms of the componentsLxandLy. Looking at gure 1 we see clearly that Lu= 1 p 2 (Lx+Lz) Lv= 1 p 2 (Lx+Lz) (32) This study source was downloaded by 100000850872992 from CourseHero.com on 04-15-2023 23:56:45 GMT -05:00

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