notes/bad.md

Schrödinger Equation Derivation — Why the Kinetic Term is \hbar^2/(2m)\,\nabla^2

Goal

Start from the classical energy equation

E = \frac{p^2}{2m} + V

and show how this becomes the time-dependent Schrödinger equation

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi

---

1. Start with classical mechanics

For a non-relativistic particle, total energy is

E = \frac{p^2}{2m} + V

where:

• E = total energy
• p = momentum
• m = mass
• V = potential energy

The term

\frac{p^2}{2m}

is just the usual classical kinetic energy.

Since

p = mv

we have

\frac{p^2}{2m} = \frac{m^2 v^2}{2m} = \frac{1}{2}mv^2

So the mysterious divide by (2m) is simply inherited from the ordinary formula for kinetic energy.

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2. Introduce a wave

Quantum mechanics uses a wavefunction \psi(x,t).

Take a simple plane wave in 1D:

\psi(x,t) = e^{i(kx-\omega t)}

This is useful because derivatives acting on exponentials bring down constants.

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3. Differentiate with respect to position

First derivative:

\frac{\partial \psi}{\partial x} = ik\psi

Second derivative:

\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi

So the second derivative returns the same wave multiplied by -k^2.

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4. Relate k to momentum

From de Broglie:

p = \hbar k

so

k = \frac{p}{\hbar}

Substitute this into the second derivative result:

\frac{\partial^2 \psi}{\partial x^2}
=
-\left(\frac{p}{\hbar}\right)^2 \psi
=
-\frac{p^2}{\hbar^2}\psi

Rearrange:

p^2 \psi = -\hbar^2 \frac{\partial^2 \psi}{\partial x^2}

This is the key step:

Momentum squared acting on a wave becomes minus \hbar^2 times the second derivative.

So in operator form,

\hat{p}^2 = -\hbar^2 \frac{\partial^2}{\partial x^2}

and in 3D,

\hat{p}^2 = -\hbar^2 \nabla^2

---

5. Put that into the energy equation

Start from

E = \frac{p^2}{2m} + V

Multiply through by \psi:

E\psi = \frac{p^2}{2m}\psi + V\psi

Now replace p^2\psi by the differential form:

E\psi
=
\frac{1}{2m}\left(-\hbar^2 \frac{\partial^2 \psi}{\partial x^2}\right) + V\psi

So:

E\psi
=
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V\psi

This is exactly where the factor

-\frac{\hbar^2}{2m}

comes from.

Why the divide by 2m stays there

Because the classical kinetic energy was already

\frac{p^2}{2m}

and the quantum substitution only changes what p^2 means:

p^2 \rightarrow -\hbar^2 \frac{\partial^2}{\partial x^2}

It does not change the prefactor 1/(2m).

So:

\frac{p^2}{2m}
\rightarrow
\frac{1}{2m}\left(-\hbar^2 \frac{\partial^2}{\partial x^2}\right)
=
-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}

---

6. Replace energy by a time derivative

For a wave,

\psi(x,t) = e^{i(kx-\omega t)}

differentiate with respect to time:

\frac{\partial \psi}{\partial t} = -i\omega \psi

Multiply by i\hbar:

i\hbar \frac{\partial \psi}{\partial t}
=
i\hbar (-i\omega)\psi
=
\hbar\omega\psi

But Planck says

E = \hbar \omega

so

E\psi = i\hbar \frac{\partial \psi}{\partial t}

Thus the energy operator is

\hat{E} = i\hbar \frac{\partial}{\partial t}

---

7. Substitute into the equation

We had

E\psi
=
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}
+
V\psi

Replace E\psi with i\hbar \partial \psi / \partial t:

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}
+
V\psi

This is the 1D time-dependent Schrödinger equation.

In 3D it becomes

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi

---

8. What the second derivative means physically

The second derivative measures curvature of the wavefunction.

• Gentle curvature -> long wavelength -> small k -> small momentum
• Rapid wiggles -> short wavelength -> large k -> large momentum

Since kinetic energy depends on p^2, it makes sense that it is linked to a second derivative.

So the Laplacian term is really the quantum version of kinetic energy.

---

9. Compact operator summary

The classical energy equation is

E = \frac{p^2}{2m} + V

Quantum substitutions:

E \rightarrow i\hbar \frac{\partial}{\partial t}

p \rightarrow -i\hbar \nabla

Therefore:

p^2 \rightarrow (-i\hbar \nabla)^2 = -\hbar^2 \nabla^2

and so

\frac{p^2}{2m}
\rightarrow
-\frac{\hbar^2}{2m}\nabla^2

Hence:

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi

---

10. Intuition in one sentence

The Schrödinger equation is basically the classical statement

\text{Energy} = \text{kinetic} + \text{potential}

rewritten so that energy and momentum act as differential operators on a wavefunction.

---

11. Tiny memory aid

A good way to remember the kinetic term is:

1. Classical kinetic energy is

   
   \frac{p^2}{2m}
   

2. In quantum mechanics

   
   p \rightarrow -i\hbar \nabla
   

3. Therefore

   
   \frac{p^2}{2m}
   \rightarrow
   \frac{(-i\hbar \nabla)^2}{2m}
   =
   -\frac{\hbar^2}{2m}\nabla^2
   

That is the whole story.

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12. One last link to F=ma

Newton's law tells you how position changes with time.

Schrödinger's equation tells you how the wavefunction changes with time.

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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Possible next page

A natural next step is to derive the time-independent Schrödinger equation

-\frac{\hbar^2}{2m}\nabla^2\phi + V\phi = E\phi

by assuming

\psi(x,t) = \phi(x)e^{-iEt/\hbar}

which turns the problem into an eigenvalue equation.