Max Diem

Quantum Mechanical Foundations of Molecular Spectroscopy


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psi 3"/>; and so forth that is, of course, obtained by carrying out the matrix multiplication indicated in Eq. (2.21).

      2.3.2 Solution of the Particle‐in‐a‐Box Schrödinger Equation

      Rearranging Eqs. (2.19) and (2.20) yields

      (2.22)left-brace minus StartFraction normal h with stroke squared Over 2 m EndFraction StartFraction d squared Over normal d x squared EndFraction right-brace normal psi left-parenthesis x right-parenthesis equals upper E normal psi left-parenthesis x right-parenthesis

      which is a simple differential equation that can be used to obtain the eigenfunctions ψ(x):

      could be solution of the differential Eq. (2.23),

      since

      (2.25)StartFraction normal d squared y Over normal d x squared EndFraction equals minus upper A b squared cosine italic b x

      Here, the term b2 would correspond to 2mE / ħ2, and A is a yet undefined amplitude factor. Similarly,

      and

      (2.28)StartFraction d squared Over normal d x squared EndFraction normal psi left-parenthesis x right-parenthesis equals minus upper A StartFraction 2 italic m upper E Over normal h with stroke squared EndFraction sine left-parenthesis StartFraction 2 italic m upper E Over normal h with stroke squared EndFraction right-parenthesis Superscript one half Baseline x equals minus upper A StartFraction 2 italic m upper E Over normal h with stroke squared EndFraction normal psi left-parenthesis x right-parenthesis

      (2.29)normal psi left-parenthesis x right-parenthesis equals 0 a t x equals 0 and a t x equals upper L

      Because of these conditions, the cosine function proposed as possible solutions (Eq. [2.24]) of Eq. (2.23) was rejected, since the cosine function is nonzero at x = 0. Because of the required continuity at x = L, the value of the function

normal psi left-parenthesis x right-parenthesis equals upper A sine left-parenthesis StartFraction 2 italic m upper E Over normal h with stroke squared EndFraction right-parenthesis Superscript one half Baseline x

      must be zero at x = L as well. This can happen in two ways: The first possibility occurs if the amplitude A is zero. This case is of no further interest, since a zero amplitude of the wavefunction would imply that the particle is not inside the box. The second possibility for the wavefunction to be zero at x = L occurs if

      (2.30)sine left-parenthesis StartFraction 2 italic m upper E Over normal h with stroke squared EndFraction right-parenthesis Superscript one half Baseline upper L equals 0

      Since the sine function is zero at multiples of π radians, it follows that