403 lines
5.8 KiB
Markdown
403 lines
5.8 KiB
Markdown
# Schrödinger Equation Derivation — Why the Kinetic Term is $\hbar^2/(2m)\,\nabla^2$
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## Goal
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Start from the **classical energy equation**
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$$
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E = \frac{p^2}{2m} + V
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$$
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and show how this becomes the **time-dependent Schrödinger equation**
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$$
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i\hbar \frac{\partial \psi}{\partial t}
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=
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-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi
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$$
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---
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## 1. Start with classical mechanics
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For a non-relativistic particle, total energy is
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$$
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E = \frac{p^2}{2m} + V
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$$
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where:
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- \(E\) = total energy
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- \(p\) = momentum
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- \(m\) = mass
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- \(V\) = potential energy
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The term
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$$
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\frac{p^2}{2m}
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$$
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is just the usual classical kinetic energy.
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Since
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$$
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p = mv
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$$
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we have
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$$
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\frac{p^2}{2m} = \frac{m^2 v^2}{2m} = \frac{1}{2}mv^2
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$$
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So the mysterious **divide by \(2m\)** is simply inherited from the ordinary formula for kinetic energy.
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---
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## 2. Introduce a wave
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Quantum mechanics uses a **wavefunction** \(\psi(x,t)\).
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Take a simple plane wave in 1D:
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$$
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\psi(x,t) = e^{i(kx-\omega t)}
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$$
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This is useful because derivatives acting on exponentials bring down constants.
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---
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## 3. Differentiate with respect to position
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First derivative:
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$$
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\frac{\partial \psi}{\partial x} = ik\psi
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$$
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Second derivative:
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$$
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\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi
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$$
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So the second derivative returns the same wave multiplied by \(-k^2\).
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---
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## 4. Relate \(k\) to momentum
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From de Broglie:
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$$
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p = \hbar k
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$$
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so
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$$
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k = \frac{p}{\hbar}
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$$
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Substitute this into the second derivative result:
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$$
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\frac{\partial^2 \psi}{\partial x^2}
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=
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-\left(\frac{p}{\hbar}\right)^2 \psi
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=
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-\frac{p^2}{\hbar^2}\psi
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$$
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Rearrange:
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$$
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p^2 \psi = -\hbar^2 \frac{\partial^2 \psi}{\partial x^2}
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$$
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This is the key step:
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> **Momentum squared acting on a wave becomes minus \(\hbar^2\) times the second derivative.**
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So in operator form,
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$$
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\hat{p}^2 = -\hbar^2 \frac{\partial^2}{\partial x^2}
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$$
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and in 3D,
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$$
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\hat{p}^2 = -\hbar^2 \nabla^2
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$$
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---
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## 5. Put that into the energy equation
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Start from
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$$
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E = \frac{p^2}{2m} + V
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$$
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Multiply through by \(\psi\):
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$$
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E\psi = \frac{p^2}{2m}\psi + V\psi
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$$
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Now replace \(p^2\psi\) by the differential form:
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$$
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E\psi
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=
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\frac{1}{2m}\left(-\hbar^2 \frac{\partial^2 \psi}{\partial x^2}\right) + V\psi
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$$
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So:
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$$
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E\psi
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=
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-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V\psi
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$$
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This is exactly where the factor
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$$
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-\frac{\hbar^2}{2m}
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$$
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comes from.
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### Why the divide by \(2m\) stays there
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Because the classical kinetic energy was already
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$$
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\frac{p^2}{2m}
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$$
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and the quantum substitution only changes **what \(p^2\) means**:
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$$
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p^2 \rightarrow -\hbar^2 \frac{\partial^2}{\partial x^2}
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$$
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It does **not** change the prefactor \(1/(2m)\).
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So:
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$$
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\frac{p^2}{2m}
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\rightarrow
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\frac{1}{2m}\left(-\hbar^2 \frac{\partial^2}{\partial x^2}\right)
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=
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-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}
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$$
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---
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## 6. Replace energy by a time derivative
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For a wave,
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$$
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\psi(x,t) = e^{i(kx-\omega t)}
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$$
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differentiate with respect to time:
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$$
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\frac{\partial \psi}{\partial t} = -i\omega \psi
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$$
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Multiply by \(i\hbar\):
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$$
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i\hbar \frac{\partial \psi}{\partial t}
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=
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i\hbar (-i\omega)\psi
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=
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\hbar\omega\psi
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$$
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But Planck says
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$$
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E = \hbar \omega
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$$
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so
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$$
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E\psi = i\hbar \frac{\partial \psi}{\partial t}
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$$
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Thus the energy operator is
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$$
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\hat{E} = i\hbar \frac{\partial}{\partial t}
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$$
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---
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## 7. Substitute into the equation
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We had
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$$
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E\psi
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=
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-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}
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+
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V\psi
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$$
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Replace $E\psi$ with $i\hbar \partial \psi / \partial t$:
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$$
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i\hbar \frac{\partial \psi}{\partial t}
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=
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-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}
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+
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V\psi
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$$
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This is the **1D time-dependent Schrödinger equation**.
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In 3D it becomes
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$$
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i\hbar \frac{\partial \psi}{\partial t}
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=
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-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi
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$$
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---
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## 8. What the second derivative means physically
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The second derivative measures **curvature** of the wavefunction.
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- Gentle curvature -> long wavelength -> small \(k\) -> small momentum
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- Rapid wiggles -> short wavelength -> large \(k\) -> large momentum
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Since kinetic energy depends on \(p^2\), it makes sense that it is linked to a **second** derivative.
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So the Laplacian term is really the quantum version of kinetic energy.
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---
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## 9. Compact operator summary
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The classical energy equation is
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$$
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E = \frac{p^2}{2m} + V
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$$
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Quantum substitutions:
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$$
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E \rightarrow i\hbar \frac{\partial}{\partial t}
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$$
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$$
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p \rightarrow -i\hbar \nabla
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$$
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Therefore:
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$$
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p^2 \rightarrow (-i\hbar \nabla)^2 = -\hbar^2 \nabla^2
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$$
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and so
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$$
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\frac{p^2}{2m}
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\rightarrow
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-\frac{\hbar^2}{2m}\nabla^2
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$$
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Hence:
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$$
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i\hbar \frac{\partial \psi}{\partial t}
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=
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-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi
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$$
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---
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## 10. Intuition in one sentence
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The Schrödinger equation is basically the classical statement
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$$
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\text{Energy} = \text{kinetic} + \text{potential}
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$$
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rewritten so that **energy and momentum act as differential operators on a wavefunction**.
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---
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## 11. Tiny memory aid
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A good way to remember the kinetic term is:
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1. Classical kinetic energy is
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$$
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\frac{p^2}{2m}
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$$
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2. In quantum mechanics
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$$
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p \rightarrow -i\hbar \nabla
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$$
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3. Therefore
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$$
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\frac{p^2}{2m}
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\rightarrow
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\frac{(-i\hbar \nabla)^2}{2m}
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=
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-\frac{\hbar^2}{2m}\nabla^2
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$$
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That is the whole story.
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---
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## 12. One last link to \(F=ma\)
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Newton's law tells you how **position** changes with time.
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Schrödinger's equation tells you how the **wavefunction** changes with time.
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So it is fair to think of it as the quantum analogue of a fundamental equation of motion — even though mathematically it comes more directly from the **energy equation** than from \(F=ma\).
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---
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## Possible next page
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A natural next step is to derive the **time-independent Schrödinger equation**
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$$
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-\frac{\hbar^2}{2m}\nabla^2\phi + V\phi = E\phi
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$$
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by assuming
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$$
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\psi(x,t) = \phi(x)e^{-iEt/\hbar}
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$$
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which turns the problem into an eigenvalue equation.
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