Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
3rd Edition
ISBN: 9781107189638
Author: Griffiths, David J., Schroeter, Darrell F.
Publisher: Cambridge University Press
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Chapter 5.1, Problem 5.1P

(a)

To determine

Show that r1=R+(μ/m1)r, r2=R(μ/m2)r, and 1=(μ/m2)R+r, 2=(μ/m1)Rr.

(a)

Expert Solution
Check Mark

Answer to Problem 5.1P

It is proved that r1=R+(μ/m1)r, r2=R(μ/m2)r, and 1=(μ/m2)R+r, 2=(μ/m1)Rr.

Explanation of Solution

Given, the separation between the two particles is

    r=r1r2        (I)

The position of the center of the mass is

    Rm1r1+m2r2m1+m2        (II)

The reduced mass of the system is

    μ=m1m2m1+m2        (III)

Substitute equation (I) in (II) and solve for r1

  R=m1r1+m2(r1r)m1+m2=m1r1+m2r1m2rm1+m2=(m1+m2)r1m2rm1+m2=(m1+m2)r1m1+m2m2rm1+m2

  =r1(m2m1+m2)rR=r1μm1rr1=R+μm1r

Hence it is proved that r1=R+μm1r.

Similarly substitute equation (I) in (II) and solve for r2

  R=m1(r2+r)+m2r2m1+m2=m1r2+m1r+m2r2m1+m2=(m1+m2)r2+m1rm1+m2=(m1+m2)r2m1+m2+m1rm1+m2

  =r2+(m1m1+m2)rR=r2+μm2rr2=Rμm2r

Hence it is prove that r2=Rμm2r.

Let the coordinates of r=(x,y,z) and R=(X,Y,Z). 1 and 2 indicates differentiation with respect to the coordinates of particle 1 and particle 2 respectively.

  (1)x=Xx1X+xx1x        (IV)

Equation (I) and (II) with respect to x

Substitute m1x1+m2x2m1+m2 for X and x1x2 for x in equation (IV) to solve for (1)x

  (1)x=x1(m1x1+m2x2m1+m2)X+x1(x1x2)x=(m1m1+m2)X+x

Similarly equation (I) and (II) with respect to y

  (1)y=(m1m1+m2)Y+y

Similarly equation (I) and (II) with respect to z

  (1)z=(m1m1+m2)Z+z

Adding,

  (1)x+(1)y(1)z=(m1m1+m2)X+x+(m1m1+m2)Y+y+(m1m1+m2)Z+z=(m1m1+m2)(X+Y+Z)+(x+y+z)1=μm2R+r

Hence it is proved that 1=μm2R+r.

Similarly for particle 2

  (2)x=Xx2X+xx2x        (IV)

Equation (I) and (II) with respect to x

Substitute m1x1+m2x2m1+m2 for X and x1x2 for x in equation (IV) to solve for (2)x

  (2)x=x2(m1x1+m2x2m1+m2)X+x2(x1x2)x=(m2m1+m2)Xx

Similarly equation (I) and (II) with respect to y

  (2)y=(m2m1+m2)Yy

Similarly equation (I) and (II) with respect to z

  (2)z=(m2m1+m2)Zz

Adding,

  (2)x+(2)y(2)z=(m2m1+m2)Xx+(m2m1+m2)Yy+(m2m1+m2)Zz=(m2m1+m2)(X+Y+Z)(x+y+z)2=μm1Rr

Hence it is proved that 2=μm1Rr.

Conclusion:

Thus, it is proved that r1=R+(μ/m1)r, r2=R(μ/m2)r, and 1=(μ/m2)R+r, 2=(μ/m1)Rr.

(b)

To determine

Show that the Schrodinger equation becomes 22(m1+m2)R2ψ22μr2ψ+V(r)ψ=Eψ .

(b)

Expert Solution
Check Mark

Answer to Problem 5.1P

It is proved that the Schrodinger equation becomes 22(m1+m2)R2ψ22μr2ψ+V(r)ψ=Eψ .

Explanation of Solution

Write the Schrodinger’s time-independent equation for two-particle system

    22m112ψ22m222ψ+Vψ=Eψ        (I)

From part (a), 1=μm2R+r and 2=μm1Rr. Substituting in the above equation

22m1(μm2R+r)2ψ22m2(μm1Rr)2ψ+Vψ=Eψ22m1(μ2m22R2+r2+2μm2Rr)ψ22m2(μ2m12R2+r22μm1Rr)ψ+Vψ=Eψ22(μ2m1m22R2+r2(1m1)+2μm1m2Rr)ψ22(μ2m12m2R2+r2(1m2)2μm1m2Rr)ψ+Vψ=Eψ22((μ2m1m22+μ2m12m2)R2ψ+r2ψ(1m1+1m2))+Vψ=Eψ

Since, from equation (III), 1μ=1m1+1m2 and μ=m1m2m1+m2.

  22(μ2m1m2(1m1+1m2)R2ψ+r2ψ(1m1+1m2))+Vψ=Eψ22μ2m1m2(1μ)R2ψ22μr2ψ+Vψ=Eψ22μm1m2(m1+m2m1+m2)R2ψ22μr2ψ+Vψ=Eψ22(m1+m2)R2ψ22μr2ψ+Vψ=Eψ

Hence it is proved that 22(m1+m2)R2ψ22μr2ψ+Vψ=Eψ.

Conclusion:

Thus, it is proved that the Schrodinger equation becomes 22(m1+m2)R2ψ22μr2ψ+V(r)ψ=Eψ .

(c)

To determine

The one-particle Schrodinger equation of ψR with the total mass (m1+m2) in place of m and ψr with the reduced mass in place of m. And solve for total energy.

(c)

Expert Solution
Check Mark

Answer to Problem 5.1P

The one-particle Schrodinger equation of ψR with the total mass (m1+m2) in place of m is 22(m1+m2)R2ψR=ERψR and ψr with the reduced mass in place of m is 22μr2ψr+V(r)ψr=Erψr. And the total energy is E=22(m1+m2)(1ψR)R2ψR+22μ(1ψr)r2ψr+V(r).

Explanation of Solution

Write the one-particle Schrodinger equation

    22m2ψ+Vψ=Eψ        (I)

The one-particle Schrodinger equation for ωR with total mass (m1+m2) in place of m, zero potential and energy ER can be written as

  22(m1+m2)R2ψR=ERψR

The one-particle Schrodinger equation for ωr with the reduced mass in place of m, potential V(r) and energy Er can be written as

  22μr2ψr+V(r)ψr=Erψr

The total energy is

  E=Er+ERE=22(m1+m2)(1ψR)R2ψR+22μ(1ψr)r2ψr+V(r)

Hence it is proved.

Conclusion:

The one-particle Schrodinger equation of ψR with the total mass (m1+m2) in place of m is 22(m1+m2)R2ψR=ERψR and ψr with the reduced mass in place of m is 22μr2ψr+V(r)ψr=Erψr. And the total energy is E=22(m1+m2)(1ψR)R2ψR+22μ(1ψr)r2ψr+V(r).

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