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A.11 Checking units and scaling

In a general expression, multiply top and bottom by powers of c until, as far as possible, ℏ only occurs as ℏc, m as mc2 and ω as ℏω or ω∕c. Then use the table

ℏcmc2ℏω ω∕c ℏ
[Energy][Length] [Length] [Energy] [Length−1] [Energy][Time]

If the electric charge enters without external fields, write in terms of α. Use the combinations eE and μB B (both with dimensions of energy) when external fields enter. (See section A.12 for more on units in EM.)

In calculations use eV or MeV as much as possible (eg instead of using me in kg, use me c2 = 0.511 MeV). Remember ℏc = 1973 eV Å = 197.3 MeV fm; also useful is ℏ = 6.582 × 10−22 eV s−1.

Often we need to cast the Schrödinger equation in dimensionless units to recognise the solution in terms of special functions. Suppose we have a potential of the form V (x) = βxn for some integer n. Then the dimensions of β are [Energy][Length−n]. The other scales in the problem are ℏ2∕2m which has dimensions [Energy][Length2] and the particle energy. We proceed by forming a length scale x0 = (ℏ2 ∕(2mβ))1∕(n+2) and an energy scale ℰ = (ℏ2β2∕n∕2m)n∕(n+2). Writing x = x0z and E = ℰμ, the Schrödinger equation for y(z) ≡ ψ(x) reads

f″ − znf − μf = 0

or its 3-d equivalent. For the harmonic oscillator, n = 2 and β = 1 2mω2, so x0 = (ℏ∕mω)1∕2 and ℰ = 1 2ℏω as expected. For the hydrogen atom, n = −1 and β = ℏcα, so r0 = 1 2a0 and ℰ = 2mc2 α2 = 4ERy, which illustrates the fact that we might have to play with numerical factors in the length scale to obtain the standard form.

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