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Construct the matrixSrrepresenting the component of spin angular momentum along an arbitrary directionr. Use spherical coordinates, for which

r蝉颈苍胃肠辞蝉桅+蝉颈苍胃蝉颈苍桅+肠辞蝉胃k [4.154]

Find the eigenvalues and (normalized) eigen spinors ofSr. Answer:

x+(r)=(cos/2e颈蠒sin/2); x+(r)=(e颈蠒sin(/2)-cos(/2)) [4.155]

Note: You're always free to multiply by an arbitrary phase factor-say,ei-so your answer may not look exactly the same as mine.

Short Answer

Expert verified

The Eigen values and Eigen spinors areabcos2eisin2andab=eisin2cos2

Step by step solution

01

Definition of Eigen values and Eigen spinors

Eigenvalues are a unique set of scalar values associated with a set of linear equations, most commonly found in matrix equations.

Characteristic roots are another name for eigenvectors. It's a non-zero vector that can only be altered by its scalar factor once linear transformations are applied.

In quantum physics, Eigen spinors are considered basis vectors that represent a particle's general spin state.

02

Determination of the Eigen values

Use eigenvalues |Sr-I|=0and find the value the eigenvalues Sr,.

2cos-2e-isin2e-isin2cos-=0-4cos2+2-4sin2=02=4sin2+cos22=-4=2Thus,theeigenvaluesofSris2

03

Determination of the Eigen spinors

Use the eigen-spinor as an exampleab.

S2ab=ab

For =2,

localid="1659013250803" 2cose-isine-isin-cosab=2cose-isine-isin-cosab=ab

Compare the relevant entries on both sides of the matrices.

acos+be-isin=aaeisin-bcos=b

Solve the above two equations.

b=a1-cose-isin=e-i1-cosasin=e-i2sin222sin22cos2a=e-isin2cos2a

For abbecome commonplace,

Apply a2+b2=1.

a2+sin22cos22a2=1

Solve the above expression further.

a2sin22cos22+1=1a2sin2+cos22cos22=1a21cos2=1a=cos2

Substitute the above value in b=eisin2cos2a.

b=eisin2cos2cos2=eisin2

Substitute the values of a, and b in ab.

ab=cos2eisin2

For =-2,

2cose-isine-isin-cosab=-2abacos+be-isin=-aaeisin-bcos=b

Solve the above two equations.

b=-a1+cossinei=-aei2cos222sin2cos2=-aei2cos222sin2cos2Applya2+b2

a2+cos22sin22a2=1a21+cos22sin22=1a=e-isin2

Substitute the above value inb=-aeicos2sin2.

b=-e-isin2e-icos2sin2=-eicos2

Substitute the values of a, and b in ab.

ab=e-isin2-cos2

Thus, the eigen values and the eigen spinors are role="math" localid="1659015532483" ab=cos2e-isin2and

ab=e-isin2-cos2.

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Most popular questions from this chapter

Work out the spin matrices for arbitrary spin , generalizing spin (Equations 4.145 and 4.147), spin 1 (Problem 4.31), and spin (Problem 4.52). Answer:

Sz=(s0000s-10000s-200000-s)Sx=2(0bs0000bs0bs-10000bs-10bs-20000bs-200000000b-s+10000b-s+10)Sy=2(0-ibs0000ibs0-ibs-10000-ibs-10-ibs-20000-ibs-200000000-ibs+10000-ibs+10)

where,bj(s+j)(s+1-j)

Use Equation 4.32 to construct Yll(,)andy32(.) . (You can take P32from Table 4.2, but you'll have to work outPll from Equations 4.27 and 4.28.) Check that they satisfy the angular equation (Equation 4.18), for the appropriate values of l and m .

In classical electrodynamics the force on a particle of charge q

moving with velocity through electric and magnetic fields E and B is given

by the Lorentz force law:F=q(E+vB)

This force cannot be expressed as the gradient of a scalar potential energy

function, and therefore the Schr枚dinger equation in its original form (Equation 1.1)

cannot accommodate it. But in the more sophisticated form iht=H

there is no problem; the classical Hamiltonian isH=12m(p-qA)2+辩蠄where A

is the vector potential(B=A)and is the scalar potential (E=--A/t),

so the Schr枚dinger

equation (making the canonical substitutionp(h/i))becomesiht=[12mhi-qA2+辩蠄]

(a) Show that d<r>dt=1m<(p-qA)>

(b) As always (see Equation ) we identifyd<r>/dtwith<v>. Show that

md<v>dt=q<E>+q2m<(pB-Bp)>-q2m<(AB)>

(c) In particular, if the fields and are uniform over the volume of the wave packet,

show thatmd<v>dt=q(E+<V>B)so the expectation value of (v)moves

according to the Lorentz force law, as we would expect from Ehrenfest's theorem.

A particle of mass m is placed in a finite spherical well:

V(r)={-V0,ra;0,r>a;

Find the ground state, by solving the radial equation withl=0. Show that there is no bound state if V0a2<2k2/8m.

For the most general normalized spinor (Equation 4.139),

compute{Sx},{Sy},{Sz},{Sx2},{Sy2},and{Sx2}.checkthat{Sx2}+{Sy2}+{Sz2}={S2}.

X=(ab)=aX++bX(4.139).

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