Show that this plane wave satisfies the Schrödinger Eqn

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SUMMARY

The discussion centers on demonstrating that the two-dimensional plane wave function, defined as \(\psi(\mathbf{r}) = Ce^{-i\mathbf{k} \cdot \mathbf{r}}\), satisfies the Schrödinger equation \(-\frac{\hbar^{2}}{2m_e}\frac{\mathrm{d}^2 \psi(\mathbf{r})}{\mathrm{d} \mathbf{r}^2} = E\psi(\mathbf{r})\). The key challenge presented is differentiating with respect to the vector \(\mathbf{r}\), which is clarified to mean applying the Laplace operator \(\nabla^2\). Understanding this operator is crucial for correctly applying the differential in the context of quantum mechanics.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the Schrödinger equation.
  • Familiarity with wave functions and their mathematical representations.
  • Knowledge of vector calculus, specifically the Laplace operator \(\nabla^2\).
  • Basic concepts of complex exponentials in physics.
NEXT STEPS
  • Study the application of the Laplace operator \(\nabla^2\) in quantum mechanics.
  • Explore the derivation of the Schrödinger equation for different potential energy scenarios.
  • Learn about plane wave solutions and their physical interpretations in quantum mechanics.
  • Investigate the role of complex numbers in wave functions and their implications in quantum theory.
USEFUL FOR

Students of quantum mechanics, physicists working with wave functions, and anyone interested in the mathematical foundations of the Schrödinger equation.

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Homework Statement


I'm asked to show that the two dimensional plane wave (for constant C)
\psi \left ( \mathbf{r} \right )=Ce^{-i\mathbf{k}\cdot \mathbf{r}}
satisfies the Schrödinger equation:
-\frac{\hbar^{2}}{2m_e}\frac{\mathrm{d}^2 \psi\left ( \mathbf{r} \right )}{\mathrm{d} \mathbf{r}^2}=E\psi\left ( \mathbf{r} \right )
but I'm flummoxed by differentiating wrt a vector, \mathbf{r}. Can someone explain how I should understand this differential, so that I can produce the required equality?

Thanks in advance.
 
Physics news on Phys.org
The intended meaning of ##d^2/d\boldsymbol{r}^2## is the Laplace operator ##\nabla^2##.
 

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