(a) Using Equation 4.88, work out the first four Laguerre polynomials.

(b) Using Equations 4.86, 4.87, and 4.88, find v(ρ), for the case n=5,I=2.

(c) Find v(ρ)again (for the case role="math" localid="1658315521558" n=5,I=2), but this time get it from the recursion formula (Equation 4.76).

Lq(x)=eqq!(ddx)q(e-x-x9)(4.88)v(ρ)=Ln-2l+1l-1(4.86)Lqp(x)(-1)pddxρLp+q(x)(4.87)cj+1=2(j+l+1-n)(j+1)(j+2l+2)cj(4.76)

Short Answer

Expert verified

(a)TheFirst four Laguerre polynomials are

L0=1L1=1-xL2=2-4x+x2L3=6-18x+9x2-x3

(b)The value ofis v(p) isV(ρ)=21-14ρ+2ρ2

(c)The value of isv(ρ)=c0-23c0ρ+221c0ρ2=c021(21-14ρ+2ρ2)

Step by step solution

01

Definition of the radial wave function

The probability of finding an electron in some finite volume element around a point at a distance of r from the nucleus is given by the radial wave function R(r), which is simply the value of the wave function at some radius r.

02

First four Laguerre polynomials

(a)

The associated Laguerre polynomial is given by equation as

For q=0

L0=exe-xL0=1

Forq=1

L1=exdxd(e-xx)=ex[e-x)-e-xx]=1-x.

for q=2

L2=ex2ddx2exx2=ex2ddx2xe-x-e-x2=ex22e-x-2xe-x+e-xx2-2xe-x=1-2x+12x2

for q=3:

L3=ex6ddx3e-xx3=ex6ddx2-e-xx3+3x2e-x=ex6ddxe-xx3+3x2e-x3x2e-x+6xe-x=ex6-e-xx3+3x2e-x+6x2e-x-12xe-x-6xe-x+6e-x=1-3x+32x2-16x3

Thus,

L0=1L1=1-xL2=2-4x+x2L3=6-18x+9x2-x3

03

Determine v(ρ)

(b)

Need to write vρusing,

v(ρ)=L25(2ρ)L25(x)=(-1)5(dxd)5L7(x)

Need to find L7xas:

L7x=exddx7x7e-x=ex7!ddx67x6e-x-x7e-x=exddx542x5e-x-7x6e-x+x7e-x=exddx4210x4e-x-42x5e-x-84x5e-x+14x6e-x+7x6e-x-x7e-x=exddx22520x2e-x-840+3360x3e-x+840+1260x43-x-252+168x5e-x+228+7x6e-x-x7e-x

=exddx2(2520x2e-x-840+3360x3e-x=840+1260x4e-x-252+168x5e-x+28+7x6e-x-x7e-x)L7x=ex[5040e-x-5040+30240ex-x+15120+37800x2e-xL7x=-12600+8400+8400x3e-x+2100+2100+3150x4e-x-630+252x5e-x+42+7x6e-x-x7e-x]=5040-25280x+52920x2-29400x3+7350x4882x5+49x6-x7

Now, to take the 5th derivative to this to find L25any variable x with power less than 5 will vanish, therefore

localid="1658377935803" L25=-ddx5-882x5+49x6-x7=--8825.4.3.2+49×12x+42x2=60882×2-49×12x+42x2=252042-14x+x2

Thus,

vρ=252042-28ρ+4ρ2=504021-14ρ+2ρ2

04

Calculate v(ρ) using the equation in part (b)

(c)

In this part we will repeat part (b) but using the equations:

v(ρ)=j=0cjpjcj+1=2(j+l+1-n)(j+1)(j+2l+2)cj(4.76)

For n=5andI=2

v52(ρ)=c0+c1ρ+c2ρ2

Where the constants can be determined using the above equation as:

c1=2(3-5)16c0=-23c0.

c2=2(4-5)27c1=-17c1=221c0c3=2(5-5)38

Substitute into, we get

v(ρ)=c0-23c0ρ+221c0ρ2=c021(21-14ρ+2ρ2)

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Most popular questions from this chapter

Suppose two spin -1/2particles are known to be in the singlet configuration (Equation Let Sa(1)be the component of the spin angular momentum of particle number 1 in the direction defined by the unit vectora^ Similarly, letSb(2) be the component of 2’s angular momentum in the directionb^ Show that

Sa(1)Sb(2)=-24cosθ

where θ is the angle between a^ andb^

An electron is at rest in an oscillating magnetic field

B=B0cos(ωt)k^

whereB0 andω are constants.

(a) Construct the Hamiltonian matrix for this system.

(b) The electron starts out (at t=0 ) in the spin-up state with respect to the x-axis (that is:χ(0)=χ+(x)). Determine X(t)at any subsequent time. Beware: This is a time-dependent Hamiltonian, so you cannot get in the usual way from stationary states. Fortunately, in this case you can solve the timedependent Schrödinger equation (Equation 4.162) directly.

(c) Find the probability of getting-h/2 , if you measure Sx. Answer:

sin2(γB02ωsin(ωt))

(d) What is the minimum field(B0) required to force a complete flip inSx ?

(a) Apply S_tolocalid="1656131461017" 10>(Equation4.177), and confirm that you getlocalid="1656131442455" 2h1-1>.

(b) ApplyS+to[00>(Equation4.178), and confirm that you get zero.

(c) Show thatlocalid="1656131424007" 11>andlocalid="1656131406083" 1-1>(Equation4.177) are eigenstates ofS2, with the appropriate eigenvalue

Determine the commutator of S2withSZ(1)(whereSS(1)+S(2)) Generalize your result to show that

[S2,S1]=2Ih(S1×S2)

Comment: Because Sz(1)does not commute with S2, we cannot hope to find states that are simultaneous eigenvectors of both. In order to form eigenstates ofS2weneed linear combinations of eigenstates ofSz(1). This is precisely what the Clebsch-Gordan coefficients (in Equation 4.185) do for us, On the other hand, it follows by obvious inference from Equation 4.187that the sumrole="math" localid="1655980965321" S(1)+S(2)does commute withdata-custom-editor="chemistry" S2, which is a special case of something we already knew (see Equation 4.103).

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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