While we have focused on the Lagrangian and action for point particles in this chapter, it is possible to formulate the Schödinger equation itself through the principle of least action with a Lagrangian for the wavefunction \(\psi(x, t)\). Consider the Lagrangian

\[\begin{equation*}\mathcal{L}\left(\psi, \psi^{\prime}, \dot{\psi}\right)=i \hbar \psi^{*} \dot{\psi}-\frac{\hbar^{2}}{2 m} \psi^{* \prime} \psi^{\prime}-V(x) \psi^{*} \psi \tag{11.166}\end{equation*}\]

Properly, this is a Lagrangian density, because its dimensions differ from a true Lagrangian by inverse length. Here, we have suppressed temporal and spatial dependence in the wavefunction, \(\psi^{*}\) is the complex conjugate of \(\psi\), and ' and ' represent the temporal and spatial derivative, respectively. \(V(x)\) is the potential, that only depends on spatial position \(x\).

(a) Show that the Schrödinger equation follows from applying the principle of least action to this Lagrangian.

 The wavefunction \(\psi\) and its complex conjugate \(\psi^{*}\) can be thought of as independent. Take the functional derivative of the Lagrangian with respect to \(\psi^{*}\) and set it equal to 0 .

(b) Show that the complex conjugate of the Schrödinger equation follows from applying the principle of least action to this Lagrangian.

(c) The action that follows from this Lagrangian is

\[\begin{equation*}S[\psi]=\int_{0}^{T} d t \int_{-\infty}^{\infty} d x \mathcal{L}\left(\psi, \psi^{\prime}, \dot{\psi}\right) \tag{11.167}\end{equation*}\]

Let's assume that \(\psi\) is an energy eigenstate wavefunction of the (arbitrary) quantum Hamiltonian. What is the action for such an eigenstate wavefunction?