CogitaVerse

Menu

Derivation of Schrödinger Wave Equation | Quantum Mechanics Postulates

Derivation of the Schrödinger Wave Equation

From Quantum Mechanics Postulates to the Time‑Independent Wave Equation

The Schrödinger wave equation is the fundamental equation of quantum mechanics. It can be derived from the first three postulates of quantum mechanics, which state that:

  • Postulate 1: The state of a quantum system is completely described by a wavefunction \(\psi\) (or state vector).
  • Postulate 2: Every observable physical quantity corresponds to a linear Hermitian operator. The operator for momentum is \(\hat{p} = \frac{h}{2\pi i}\nabla\).
  • Postulate 3: The only possible results of a measurement of an observable are the eigenvalues of its operator. For energy, the eigenvalue equation is \(\hat{H}\psi = E\psi\).

We derive the time‑independent Schrödinger equation by constructing the Hamiltonian operator \(\hat{H}\) (energy operator) from the classical expression for total energy and then applying the eigenvalue equation.

Step 1: Classical Energy Expression

Consider a single particle of mass \(m\) moving in three dimensions under a potential \(V(x,y,z)\). The total energy (Hamiltonian) is the sum of kinetic and potential energies:

\[ E = T + V \tag{2} \]

Kinetic energy in terms of linear momentum \(p\) is:

\[ T = \frac{1}{2}mv^2 = \frac{p^2}{2m} \tag{3} \]

In three dimensions, the total momentum squared is the sum of squares of components:

\[ p^2 = p_x^2 + p_y^2 + p_z^2 \tag{4} \]

Thus, the classical energy becomes:

\[ E = \frac{p_x^2 + p_y^2 + p_z^2}{2m} + V \tag{6} \]

Step 2: Quantum Mechanical Operators

According to quantum mechanics (Postulate 2), the momentum operators are:

\[ \hat{p}_x = \frac{h}{2\pi i} \frac{\partial}{\partial x} \tag{7} \]
\[ \hat{p}_y = \frac{h}{2\pi i} \frac{\partial}{\partial y} \tag{8} \]
\[ \hat{p}_z = \frac{h}{2\pi i} \frac{\partial}{\partial z} \tag{9} \]

The potential energy operator is simply multiplication by \(V(x,y,z)\). Substituting these operators into the classical energy expression gives the Hamiltonian operator \(\hat{H}\):

\[ \hat{H} = \frac{1}{2m} \left[ \left( \frac{h}{2\pi i} \frac{\partial}{\partial x} \right)^2 + \left( \frac{h}{2\pi i} \frac{\partial}{\partial y} \right)^2 + \left( \frac{h}{2\pi i} \frac{\partial}{\partial z} \right)^2 \right] + V \tag{10} \]

When the squares are expanded, we use the fact that \(i^2 = -1\). The square of each momentum operator becomes:

\[ \hat{p}_x^2 = \left( \frac{h}{2\pi i} \right)^2 \frac{\partial^2}{\partial x^2} = \frac{h^2}{4\pi^2 i^2} \frac{\partial^2}{\partial x^2} = -\frac{h^2}{4\pi^2} \frac{\partial^2}{\partial x^2} \]

Therefore, the Hamiltonian simplifies to:

\[ \hat{H} = \frac{-h^2}{8\pi^2 m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) + V \tag{11} \]

Using the Laplacian operator \(\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\), we write compactly:

\[ \hat{H} = -\frac{h^2}{8\pi^2 m} \nabla^2 + V \tag{12} \]

Step 3: Eigenvalue Equation for Energy (Postulate 3)

According to Postulate 3, the allowed energy values \(E\) and the corresponding wavefunctions \(\psi\) satisfy the eigenvalue equation:

\[ \hat{H} \psi = E \psi \tag{1} \]

Substitute the Hamiltonian from equation (11) into (1):

\[ \left[ -\frac{h^2}{8\pi^2 m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) + V \right] \psi = E \psi \]

Bring all terms to one side:

\[ -\frac{h^2}{8\pi^2 m} \left( \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} \right) + V\psi – E\psi = 0 \]

Multiply both sides by \(-\frac{8\pi^2 m}{h^2}\) to obtain the standard form:

\[ \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} – \frac{8\pi^2 m}{h^2} V\psi + \frac{8\pi^2 m}{h^2} E\psi = 0 \tag{16} \]

Grouping the terms involving \((E – V)\psi\):

\[ \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} + \frac{8\pi^2 m}{h^2} (E – V)\psi = 0 \tag{17} \]

Equation (17) is the time‑independent Schrödinger wave equation for a particle in three dimensions.

Final Compact Form

Using the Laplacian \(\nabla^2\), the equation becomes:

\[ \nabla^2 \psi + \frac{8\pi^2 m}{h^2} (E – V)\psi = 0 \]

For one‑dimensional problems (motion along \(x\) only), the equation reduces to:

\[ \frac{d^2 \psi}{dx^2} + \frac{8\pi^2 m}{h^2} (E – V)\psi = 0 \]

Note: The derivation above assumes that the wavefunction \(\psi\) is time‑independent. The full time‑dependent Schrödinger equation includes the time derivative and is written as \(\hat{H}\psi = i\hbar \frac{\partial \psi}{\partial t}\). The time‑independent form is sufficient for stationary states where energy is constant.

Summary of Key Equations in the Derivation

\(E = T + V\)
\(\hat{p}_x = \frac{h}{2\pi i}\frac{\partial}{\partial x}\)
\(\hat{H} = \frac{1}{2m}(\hat{p}_x^2 + \hat{p}_y^2 + \hat{p}_z^2) + V\)
\(\hat{H} = -\frac{h^2}{8\pi^2 m}\nabla^2 + V\)
\(\hat{H}\psi = E\psi\)
\(\nabla^2 \psi + \frac{8\pi^2 m}{h^2}(E – V)\psi = 0\)

© 2025 — Complete step‑by‑step derivation of the Schrödinger wave equation from quantum mechanical postulates. All equations included as per the original derivation. Optimized for clarity and mathematical accuracy.

Related Posts

Scroll to Top