 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s3/3/1/solution

- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} ψ}{d x ^{2}} + V (x) ψ (x) = E ψ (x) .

The differential term is the kinetic energy operator acting on the wave function,

V (x)

supplies the multiplicative potential energy, and their sum is the Hamiltonian operator. The eigenvalue

E

is the stationary state's total energy. This form assumes a nonrelativistic particle, a time-independent scalar potential, and no additional internal degrees of freedom in the stated model.

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