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Equalities with a Large Number of Terms in Mechanics

  1. The quantum-mechanical analogue of the Runge-Lenz vector [Mer, p.268, (12.74)].
    1. Prove that K is a vector operator.
      Proof. [Mer, p.236, (11.19)] (Let A=K) [Lan3, p.124, (36.31)(i)] because n^ = nxx^+nyy^+nzz^.
      By cyclic permutations, it is enough to prove [Lx, Kx] = 0, [Lx, Ky] = iħKz and [Lx, Kz] = -iħKy.
    2. Prove [K, H] = 0.
      Proof. [(pyLz - pzLy) - (Lypz - Lzpy), px2 + py2 + pz2] = 0 [Coh, p.171, (43)].
      [ -(pL - Lp)x , 1/r] = [x/r, px2 + py2 + pz2].
      Remark. The cross product of operators should be defined in the manner of the tensor product as in [Haw, p.208, (14-4)] rather than the manner shown in the table in [Usp, p.193]. This is because the components of the first factor may not commute with those of the second factor. 
    3. Prove [Mer, p.268, (12.77)].
      Proof. By cyclic permutations, it suffices to prove the first of of the three equalities.
      I. Kx = (2me2)-1[(Lypz-Lzpy)-(pyLz-pzLy)]+(x/r) (the determinant is defined in the manner of the tensor product as in [Haw, p.208, (14-4)])
      = (2me2)-1[(Lypz-pyLz)+(pzLy-Lzpy)]+(x/r)
      = (me2)-1(Lypz-pyLz)+(x/r) because Lypz-pyLz =  pzLy-Lzpy.
      Similarly, Ky = (me2)-1(Lzpx-pzLx)+(y/r);
      Kz = (me2)-1(Lxpy-pxLy)+(z/r).
      Using the table in [Usp, p.193] for the definition of the Runge-Lenz vector [Go2, p.103, (3-82)], we discover that the quantum-mechanical analogue of the Runge-Lenz vector [Mer, p.268, (12.74)] actually coincides with the Runge-Lenz vector if we allow the components of p, L and r in [Go2, p.103, (3-82)] to be operators.
      II. Using [Coh, p.168, (11) & (12)], we prove [Lypz-pyLz, Lzpx-pzLx] =  -iħLz(px2+py2+pz2). This encouraging result enables us to see the light at the end of the tunnel.
      III. Using [Coh, p.150, (E-26); p.661, (D-1-a), (D-1-b) & (D-1-c)], we prove
      [Lypz-pyLz, y/r]+[x/r, Lzpx-pzLx] = (2ħ2/r)(x/y-y/x).
      Remark 1. In order to facilitate application, we need a generalized summary [Lan3, p.84, (26.4) & (26.5)] rather than just the proof strategies [Mer, p.233, (11.4)].
      Remark 2. (Debugging) When we prove an equality with a large number of terms, it is quite annoying to discover that an error has resulted in an extra term. The following strategy may help solve the problem: first locate the expression containing the problematic term; then try to find a similar expression and check to see if it contains a term which can cancel the problematic term.
    4. Prove [Mer, p.269, (12.80)].
      Hint. Compare the coefficients of ħ0, ħ2, and ħ4 from both sides of the equality.
      Remark. (Debugging) If an error results  in a problematic term, all we need to do is try to enumerate various ways to produce that term from both sides of the equality. It would save us tremendous time if we ignore other unrelated terms when debugging this problematic term.

  2. A strategy to avoid long calculations: do not expand an expression unless it is necessary.
    Example. Prove the identity given in [Born, p.62, l.10].
    Proof. [(2p1p2)/(p1+p2)2][(2p2p3)/(p2+p3)2] = {1-[(p1-p2)/(p1+p2)]2}{1-[(p2-p3)/(p2+p3)]2}.

  3. Let R be defined as in [Born, p.62, (59)]. Then d2R/db2 contains a large number of terms. However, in order to determine whether the value of R at H = ml0/(4 cos q2) is a local minimum or maximum, only a the few terms given in [Born, p.63, l.-10] need be considered. By ignoring other terms, we avoid a tremendous amount of calculations.

  4. Suppose we want to prove the equality of two mathematical expressions. Each expression may be a product of polynomials. We need not expand these products, but we should put the two expressions in a form so that they could produce as many equal terms as possible when these products are expanded. Then we may ignore their sameness and focus on their differences.
    Example 1. In order to prove the first equality of [Born1, p.44, (36)], we must prove the following identity:
    sin 2 (qi+qt) cos 2 (qi-qt) - sin 2 (qi-qt) cos 2 (qi+qt) = sin 2qi sin 2qt.
    Proof. sin 2 (qi+qt) cos 2 (qi-qt) = (sin qi  cos qt + sin qt cos qi)2 (cos qi  cos qt + sin qi  sin qt)2
    = (sin 2 qi cos 2 qt +sin 2 qt cos 2 qi + 2sin qi cos qi sin qt cos qt)
    (cos 2 qi cos 2 qt +sin 2 qt sin 2 qi + 2sin qi cos qi sin qt cos qt).
    sin 2 (qi-qt) cos 2 (qi+qt) = (sin qi  cos qt - sin qt cos qi)2 (cos qi  cos qt - sin qi  sin qt)2
    = (sin 2 qi cos 2 qt +sin 2 qt cos 2 qi - 2sin qi cos qi sin qt cos qt)
    (cos 2 qi cos 2 qt +sin 2 qt sin 2 qi - 2sin qi cos qi sin qt cos qt).
    All we need to find is the negative contribution from the latter mathematical expression. It is
    -2-1(sin 2qi sin 2qt)(cos 2 qi cos 2 qt + sin 2 qi sin 2 qt )
    -2
  5. -1(sin 2qi sin 2qt)(sin 2 qi cos 2 qt + sin 2 q t cos 2 qi )
    = -2-1(sin 2qi sin 2qt).
    Example 2. In order to prove that the sum of the reflectivity and transmissivity is 1 [Born1, p.66, l.4], we must prove
    r122 + r232 + 2 r12 r23 cos2b = p3 p1-1 t122 t232 = 1+ r122 r232 + 2 r12 r23 cos2b.
    Namely, (p1-p2)2 (p1+p2)-2 + (p2-p3)2 (p2+p3)-2 + 16p1p22p3 (p1+p2)-2 (p2+p3)-2
    = 1 + (p1-p2)2 (p2-p3)2(p1+p2)-2 (p2+p3)-2. It suffices to prove
    (p1-p2)2 (p2+p3)2 + (p12+2p1p2+p22)(p22-2p2p3+p32) + 16p1p22p3
    = (p1+p2)2 (p2+p3)2 +(p12-2p1p2+p22)(p22-2p2p3+p32).
    Seeing the difference between the two sides of the above equality, we only need prove
    -2p1p2 (p2+p3)2 + 2p1p2(p22-2p2p3+p32) + 16p1p22p3 = 2p1p2 (p2+p3)2 - 2p1p2(p22-2p2p3+p32).