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Physics LibreTexts

3.1: Review - Energy Eigenvalue Problem

  • Page ID
    25703
  • The time-independent wavefunction obeys the time-independent Schrödinger equation:

    \[\boxed{\mathcal{H} \varphi(\vec{x})=E \varphi(\vec{x})} \nonumber\]

    where E is identified as the energy of the system. If the wavefunction is given by just its time-independent part, \( \psi(\vec{x}, t)=\varphi(\vec{x})\), the state is stationary. Thus, the time-independent Schrödinger equation allows us to find stationary states of the system, given a certain Hamiltonian.

    Notice that the time-independent Schrödinger equation is nothing else than the eigenvalue equation for the Hamiltonian operator.

    The energy of a particle has contributions from the kinetic energy as well as the potential energy:

    \[\mathcal{H}=\frac{1}{2 m}\left(\hat{p}_{x}^{2}+\hat{p}_{y}^{2}+\hat{p}_{z}^{2}\right)+V(\hat{x}, \hat{y}, \hat{z}) \nonumber\]

    or more explicitly:

    \[\mathcal{H}=-\frac{\hbar^{2}}{2 m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)+V(x, y, z) \nonumber\]

    which can be written in a compact form as

    \[\boxed{\mathcal{H}=-\frac{\hbar^{2}}{2 m} \nabla^{2}+V(x, y, z)} \nonumber\]

    (Notice that V (x, y, z) is just a multiplicative operator, in the same way as the position is).

    In 1D, for a free particle there is no potential energy, but only kinetic energy that we can rewrite as:

    \[\mathcal{H}=\frac{1}{2 m} p^{2}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}\nonumber\]

    The eigenvalue problem \( \mathcal{H} w_{n}(x)=E_{n} w_{n}(x)\) is then the differential equation

    \[\mathcal{H} w_{n}(x)=E_{n} w_{n}(x) \rightarrow-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} w_{n}(x)}{\partial x^{2}}=E_{n} w_{n}(x) \nonumber\]

    For a free particle there is no restriction on the possible energies, En can be any positive number. The solution to the eigenvalue problem is then the eigenfunction:

    \[w_{n}(x)=A \sin \left(k_{n} x\right)+B \cos \left(k_{n} x\right)=A^{\prime} e^{i k_{n} x}+B^{\prime} e^{-i k_{n} x} \nonumber\]

    which represents two waves traveling in opposite directions.

    We see that there are two independent functions for each eigenvalue En. Also there are two distinct momentum eigenvalues \(\pm k_{n}\) for each energy eigenvalue, which correspond to two different directions of propagation of the wave function \(e^{\pm i k_{n} x}\).

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