The Schrödinger Equation

$$iħ\frac{∂}{∂t}Ψ({\bf r},t)=-\frac{ħ^2}{2m}∇^2Ψ({\bf r},t)$$

Using Hamiltonian,

$$iħ\frac{∂}{∂t}Ψ({\bf r},t)=\hat{H}Ψ({\bf r},t)$$

In this post, after deriving the Schrödinger Equation equation, we obtain the momentum operator.

## Preparation of Derivation of Schrodinger Equation

### Rewrite the wave function

The expression of the wave function generally has the following form.

$$ψ(x,t)=Cexp[i(kx-ωt)]$$

It is known that the following relation holds in quantum mechanics.

$$E=hν=ħω.$$

So,

$$ω=\frac{E}{ħ}.$$

In addition, the de Broglie wavelength $λ$ is expressed as follows.

$$λ=\frac{h}{p}$$

Therefore, introducing wave number $k$ yields the following equation.

$$k=\frac{2π}{λ}=\frac{2πp}{h}=\frac{p}{ħ}$$

By substituting these into the expression of wave function, the equation of wave using momentum $p$ and energy $E$ is obtained.

$$ψ_p(x,t)=Cexp\left[\frac{i}{ħ}(px-Et)\right]$$

## One-dimensional Schrodinger equation

As shown above, free particles with momentum $p$ and energy $E$ can be represented by wave function $ψ_p$ using the constant $C$ as follows.

$$ψ_p(x,t)=Cexp\left[\frac{i}{ħ}(px-Et)\right]$$

From the superposition principle, the general wave function $ψ$ is the sum of $C_a ψ_a, C_b ψ_b, C_c ψ_c, …$.

$$ψ(x,t)=C_aψ_a+C_bψ_b+C_cψ_c+…$$
$$\left( ψ_{a}(x,t)=exp\left[\frac{i}{ħ}(p_ax-E_at)\right] \right)$$

This wave function \ (ψ (x, t) \) can be expressed using integral with respect to momentum $p$ as follows. However, note that coefficients such as $C_a,C_b$ do not depend on position $x$ and time $t$ but may depend on momentum $p$. Therefore, $C$ in the following expression may also depend on momentum $p$.

$$ψ(x,t)=\int_{-∞}^∞ C(p)exp\left[\frac{i}{ħ}(px-Et)\right] dp \cdots (1)$$

Differentiate both sides of equation (1) with respect to t.

$$iħ\frac{∂ψ(x,t)}{∂t}=\int_{-∞}^∞ EC(p)exp\left[\frac{i}{ħ}(px-Et)\right] dp \cdots (2)$$

Differentiate both sides of equation (1) with respect to x twice.

$$\frac{∂^2ψ(x,t)}{∂x^2}=\int_{-∞}^∞ \left( -\frac{p^2}{ħ^2} \right) × C(p)exp\left[\frac{i}{ħ}(px-Et)\right] dp$$

Multiply the both sides of this expression by $-\frac{ħ ^ 2}{2m}$.

$$-\frac{ħ^2}{2m}\frac{∂^2ψ(x,t)}{∂x^2}=\int_{-∞}^∞ \left( \frac{p^2}{2m} \right) × C(p)exp\left[\frac{i}{ħ}(px-Et)\right] dp \cdots (3)$$

According to classical mechanics, $\frac{p^2}{2m}$ on the right side of equation (3) represents free particle energy $E$. Then, the right side of the equation (2) is equal to the right side of the equation (3). Therefore, we obtain the following equation.

$$iħ\frac{∂ψ(x,t)}{∂t}=-\frac{ħ^2}{2m}\frac{∂^2ψ(x,t)}{∂x^2}$$

Thus, the one-dimensional Schrodinger equation is obtained.

## Three-dimensional Schrodinger equation

The three-dimensional Schrodinger equation can be obtained in the same way as in the one-dimensional case.

Free particles with momentum ${\ bf p}$ and energy $E$ are represented by the following wave function $ψ_{\bf p}$ with a constant $C$.

$$ψ_{\bf p}({\bf r},t)=Cexp\left[\frac{i}{ħ}({\bf p}･{\bf r}-Et)\right]$$

Because of the superposition principle,

$$ψ({\bf r},t)=C_aψ_a+C_bψ_b+C_cψ_c+…$$
$$\left( ψ_{a}({\bf r},t)=C_aexp\left[\frac{i}{ħ}({\bf p}_a･{\bf r}-E_at)\right] \right)$$

Express the wave function $ψ({\bf r},t)$ using integral with respect to momentum.

$$ψ({\bf r},t)=\int_{-∞}^∞ C({\bf p})exp\left[\frac{i}{ħ}({\bf p}･{\bf r}-Et)\right] d^3p \cdots (4)$$

Differentiate both sides of the equation (4) with respect to t.

$$iħ\frac{∂ψ(x,t)}{∂t}=\int_{-∞}^∞ EC({\bf p})exp\left[\frac{i}{ħ}({\bf p}･{\bf r}-Et)\right] d^3p \cdots (5)$$

Use  Laplace Operator $\nabla ^2$

$$-\frac{ħ^2}{2m}∇^2ψ({\bf r},t)=\int_{-∞}^∞ \left( \frac{|{\bf p}|^2}{2m} \right) × C(p)exp\left[\frac{i}{ħ}({\bf p}･{\bf r}-Et)\right] dp・・・(6)$$

The right side of the equation (5) is equal to that of the equation (6).

$$iħ\frac{∂ψ({\bf r},t)}{∂t}=-\frac{ħ^2}{2m}∇^2ψ({\bf r},t)$$

## Momentum operator

### One-dimensional momentum operator

$$ψ_p(x,t)=Cexp\left[\frac{i}{ħ}(px-Et)\right]$$

By differentiating both sides of the above equation by t or x, we obtain the following equations.

$$iħ\frac{∂}{∂t}ψ_p(x,t)=Eψ_p(x,t)$$

$$-iħ\frac{∂}{∂x}ψ_p(x,t)=pψ_p(x,t)$$

By comparing the two formulas, the following correspondence relationships can be found.

$$E \to iħ\frac{∂}{∂t}$$

$$p \to -iħ\frac{∂}{∂x}$$

These are called energy operator and momentum operator, respectively.

### Three-dimensional momentum operator

Three-dimensional energy operator and momentum operator are obtained in the same way.

$$E \to iħ\frac{∂}{∂t}$$

$$p \to -iħ∇$$

## Conclusion

I showed how to derive the Schrödinger Equation and the momentum operator.