# Schrödinger Equation Derivation 

 Why the Kinetic Term is $-\hbar^2/(2m)\,\nabla^2$

## Goal

Start from the **classical energy equation**, E=total energy, p=momentum, m=mass, V=potential energy. So the Hamiltonian (not the Larangian)...

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

Note this is $E=\frac{1}{2} m v^2$. P is momentum so $p^2 = {m^2v^2}$, so dividing by 2m converts the kinetic energy a momentum equation

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. Slowly 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 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 **divide by $2m$** simply comes from the classical kinetic energy formula.

---

## 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 a plane wave solution. In quantum mechanics, position is not definite — the probability of finding the particle is given by $|\psi|^2$. This is similar to the position in classical physics.

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

[Quantum Phase Corkscrew](Quantum%20Phase%20Corkscrew.md)



---

## 3. Differentiate with respect to position

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

Note this differentiates by time. It's looking at how fast the particle waveform oscillates which is directly related to the energy via the plank constant. As for a photon this means the frequency is $\propto$ E.


Second derivative w.r.t. distance:

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



 So the second derivative returns the same wave multiplied by $-k^2$ because $i^2$ is -1. The first derivative gives momentum, and the second derivative gives momentum squared, which corresponds to kinetic energy.
 
 ---
## 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:
$$
p^2 \psi = -\hbar^2 \frac{\partial^2 \psi}{\partial x^2}
$$
Key idea:

> **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 $\psi$:
$$
E\psi = \frac{p^2}{2m}\psi + V\psi
$$
Replace $p^2\psi$:
$$
E\psi
$$

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

So
$$
E\psi =
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V\psi
$$
This is where the factor
$$
-\frac{\hbar^2}{2m}
$$

comes from.

---

## 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's relation says

$$
E = \hbar \omega
$$
So
$$
E\psi = i\hbar \frac{\partial \psi}{\partial t}
$$
Thus
$$
\hat{E} = i\hbar \frac{\partial}{\partial t}
$$

---

## 7. Substitute into the equation

We had

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

Replace $E\psi$:

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

This is the **1D time‑dependent Schrödinger equation**.

In 3D:

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

---

## 8. Physical meaning of the second derivative

The second derivative measures **curvature** of the wavefunction.

- Gentle curvature → long wavelength → small $k$ → small momentum  
- Rapid oscillation → short wavelength → large $k$ → large momentum  

Since kinetic energy depends on $p^2$, it naturally connects to a **second derivative**.

---

## 9. Operator summary

Classical equation:
$$
E = \frac{p^2}{2m} + V
$$
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
$$
\frac{p^2}{2m}
\rightarrow
-\frac{\hbar^2}{2m}\nabla^2
$$
So we obtain

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

---

## 10. Intuition in one sentence

The Schrödinger equation is simply

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

rewritten so **energy and momentum become differential operators acting on a wavefunction**.

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