Introduction to Quantum Mechanics · Problem 3.13

Formalism — David J. Griffiths

Physics — Quantum Mechanics · momentum_space

Problem

Show that

\langle x \rangle = \int \Phi^* \left( i\hbar \frac{\partial}{\partial p} \right) \Phi \, dp.

Hint: Notice that

x \exp(ipx/\hbar) = -i\hbar (\partial/\partial p) \exp(ipx/\hbar)

, and use Equation 2.147. In momentum space, then, the position operator is

i\hbar \partial/\partial p

. More generally,

\langle Q(x, p, t) \rangle = \begin{cases} \int \Psi^* \hat{Q}\left(x, -i\hbar\frac{\partial}{\partial x}, t\right) \Psi \, dx, & \text{in position space;} \\ \int \Phi^* \hat{Q}\left(i\hbar\frac{\partial}{\partial p}, p, t\right) \Phi \, dp, & \text{in momentum space.} \end{cases}

In principle you can do all calculations in momentum space just as well (though not always as easily) as in position space.

Intuitive explanation

The key insight is that position and momentum play dual roles in position and momentum space: just as $p$ becomes $-i\hbar \partial/\partial x$ in position space, $x$ becomes $i\hbar \partial/\partial p$ in momentum space. To prove this, use the hint's differential relation to rewrite the position operator acting on the momentum-space wavefunction $\Phi(p)$ , then apply the definition of expectation value in momentum space. This demonstrates the deep symmetry between the two representations.

Intuitive explanation

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