# 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.

---

## 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.

---

## 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 \(-k^2\).

---

## 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.

---

## 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 \(F=ma\).

---

## 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.