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Consider two point particles of mass m1 and m2 with position vectors ~r1 and ~r2 and momenta p~1 and p~2 respectively interacting via a central potential. The energy is given by H = p~1 · p~1 p~2 · p~2 + + V (|~r1 − ~r2 |) . 2m1 2m2 We will consider this problem classically and then quantize. ~ ≡ [ m1~r1 + m2~r2 ]/M where M = m1 + m2 We define the center-of-mass coordinate R is the total mass and the relative coordinate is given by ~r = ~r1 − ~r2 . The so-called reduced mass is defined by µ = m1 m2 /M . We will show the key identity p~1 · p~1 p~2 · p~2 P~ · P~ p~ · p~ + = + . 2m1 2m2 2M 2µ The center of mass velocity and the relative velocity are obtained by differentiating the corresponding coordinates: ~˙ ≡ V ~ = m1~v1 + m2~v2 R m1 + m2 and ~r˙ ≡ ~v = ~v1 − ~v2 . The velocities can be solve to obtain ~ + ~v1 = V m2 ~ + µ ~v and ~v2 = V ~ − µ ~v . ~v = V m1 + m2 m1 m2 Therefore, the kinetic energy is given by KE ≡ 1 1 1 m1~v1 · ~v1 + m2~v2 · ~v2 = m1 2 2 2 µ ~ + µ ~v V m1 µ ¶2 + 1 ~ − µ ~v m2 V 2 m2 ¶2 . Simplifying we note that the cross terms vanish1 1 ~ ·V ~ + 1 KE = (m1 + m2 )V 2 2 Ã µ2 µ2 + m1 m2 ! ~v · ~v = M~ ~ µ V · V + ~v · ~v . 2 2 where the last equality follows since µ = m1 m2 1 1 1 ⇒ = + . m1 + m2 µ m1 m2 ~ and is given by P~ = m1~v1 + m2~v2 and the The center of mass momentum is (m1 + m2 )V relative momentum is defined to be p~ = µ (~v1 − ~v2 ) . The expression for the kinetic energy in terms of the momenta is 1 ~ ~ 1 KE = P ·P + p~ · p~ . 2M 2µ 1 m1 µ ~ m2 µ ~ 2 V · ~v − 2 V · ~v = 0 . 2 m1 2 m2 1 For doing quantum mechanics it is useful to write the center-of-mass and relative momenta in terms of the individual momenta: µ µ P~ = p~1 + p~2 and p~ = p~1 − p~2 . m1 m2 How does one go over to quantum mechanics? We know that the components of p~1 , p~2 , ~r1 , and ~r2 are now operators and obey the standard commutation relations. The momenta and coordinates of different particles commute: for example, [x1 , p2x ] = 0. Now we have to convince ourselves that the the components of ~r and p~ obey the correct commutation relations. Calculate [x, px ]: · ¸ x1 − x2 , µ µ µ µ p1x − p2x = [x1 , p1x ] + [x2 , p2x ] = ih̄ µ m1 m2 m1 m2 µ 1 1 + m1 m2 ¶ = ih̄ where we have neglected all commutators that involve different particles since they vanish. Note that center of mass momentum and the relative coordinates commute: we compute [x1 − x2 , Px ]: [x1 − x2 , p1x + p2x ] = [x1 , p1x ] − [x2 , p2x ] = ih̄ − ih̄ = 0 . All the other relations can be similarly verified. Thus we can focus only on the relative coordinates and the Hamiltonian is H = 1 p~ · p~ + V (r) . 2µ Now we can write the kinetic energy as a radial and angular part just as we did before. The m2 only difference is that we use the reduced mass µ = mm11+m . 2 Thus for the hydrogen atom the energy levels are given by (Equation 4.70 in the text) En = − 2 m m µq 4 1 2h̄2 n2 e where q 2 = 4π² and µ = mee+mpp using obvious notation for the electron and proton masses. 0 If we had a muonic hydrogen atom with a µ− in a bound state with a proton we will use the corresponding masses. 2