HydrogenAtom

HydrogenAtom - The Hydrogen Atom Michael Fowler 11/22/06...

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The Hydrogen Atom Michael Fowler 11/22/06 Factoring Out the Center of Mass Motion The hydrogen atom consists of two particles, the proton and the electron, interacting via the Coulomb potential , where as usual () 2 12 / Vr r e r −= GG rrr = G G . Writing the masses of the two particles as Schrödinger’s equation for the atom is: 1 , mm 2 22 2 1 2 ,, e rr E rr r ψψ ⎛⎞ −∇ = ⎜⎟ ⎝⎠ == . G G But are not the most natural position variables for describing this system: since the potential depends only on the relative position, a better choice is , , rR G G defined by: 11 2 2 , mr mr rrr R + =− = + G G G GGG so R G is the center of mass of the system. It is convenient at the same time to denote the total mass by , M =+ and the reduced mass by . m = + Transforming in straightforward fashion to the variables , G G Schrödinger’s equation becomes 2 Rr e . R rE R r Mm r = G G G G Writing the wave function ( ) ( ) () , R r ψ G G G G we can split the equation into two: () () () () () 2 2 2 2 2 2 RR RE R M Vr r E r m Ψ= Ψ + = G = G = G
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2 and the total system energy is Note that the motion of the center of mass is (of course) just that of a free particle, having a trivial plane wave solution. From now on, we shall only be concerned with the relative motion of the particles. Since the proton is far heavier than the electron, we will almost always ignore the difference between the electron mass and the reduced mass, but it should be noted that the difference is easily detectable spectroscopically: for example, the lines shift if the proton is replaced by a deuteron (heavy hydrogen). . Rr EE E =+ We’re ready to write Schrödinger’s equation for the hydrogen atom, dropping the r suffixes in the second equation above, and writing out 2 G explicitly in spherical coordinates: 2 22 2 2 2 2 11 1 sin . 2s i n s i n e rE mr r r r r r ψψ ψ θ θθ θϕ ⎛⎞ ∂∂ −+ + ⎜⎟ ⎝⎠ = = Factoring Out the Angular Dependence: the Radial Equation Since the potential is spherically symmetric, the Hamiltonian H commutes with the angular momentum operators 2 , z L L so we can construct a common set of eigenkets of the three operators 2 ,, z . H LL The angular dependence of these eigenkets is therefore that of the ’s, so the solutions must be of the form m l Y ( ) ( ) ( ) , . m Elm Elm l rR r Y θφ = Now, notice that in the Schrödinger equation above, the angular part of 2 G is exactly the differential operator 2 /2 2 L mr , so operating on ( ) ( ) ( ) , m Elm Elm l r Y = it will give . The spherical harmonic can then be cancelled from the two sides of the equation leaving: () 2 1/2 ll m r + = 2 m l Y ( ) () () 2 1 1 2 El El El dd e r R r E m r dr dr r r + −− = = R r it now being apparent that R ( r ) cannot depend on m . The radial derivatives simplify if one factors out 1/ r from the function R , writing ( ) El ur r = and temporarily suppressing the E and l to reduce clutter.
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HydrogenAtom - The Hydrogen Atom Michael Fowler 11/22/06...

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