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How to Derive the Schrödinger Equation and Momentum Operator

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.

$$E \to iħ\frac{∂}{∂t}$$

$$p \to -iħ∇$$

Link:Why Do We Need to Solve Schrödinger’s Equation?

Conclusion

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

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