A free electron (mass kg) has momentum-space wave function gathered (p) = (2 ( p)^2…

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · The formalism of quantum mechanics · The formalism of quantum mechanics

The Structure of Quantum Mechanics, Ch Position, momentum and the origin of the canonical commutator

Problem

A free electron (mass m=9.11×10−31m=9.11\times10^{-31} kg) has momentum-space wave function ψ~(p)=(2π(Δp)2)−1/4exp⁡(−p24(Δp)2),Δp=5.0×10−25 kg m s−1,\begin{gathered}\tilde\psi(p) = \left(2\pi(\Delta p)^2\right)^{-1/4}\exp\left(-\dfrac{p^2}{4(\Delta p)^2}\right),\\ \Delta p = 5.0\times10^{-25}\ \text{kg m s}^{-1},\end{gathered} so that ∣ψ~(p)∣2|\tilde\psi(p)|^2 is a Gaussian of standard deviation Δp\Delta p centred at p=0p=0, giving ⟨p^2⟩=(Δp)2\langle\hat p^2\rangle = (\Delta p)^2. ((a)) Using the momentum representation, in which the free-particle Hamiltonian H^=p^2/2m\hat H=\hat p^2/2m acts by multiplication, compute ⟨H^⟩\langle\hat H\rangle in electronvolts. ((b)) State, without carrying out the calculation, what would have to be done to ψ(x)=⟨x∣ψ⟩\psi(x)=\langle x|\psi\rangle to reach the same answer working in the position representation, and say which representation is easier for this operator and why.

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