Schrödinger equation

Every quantum particle carries a wave function $\psi(x,t)$ , and the Schrödinger equation is the single rule that tells you how it changes in time — the quantum analogue of Newton's second law. Once you separate out the time dependence, the resulting time-independent Schrödinger equation (TISE) becomes the workhorse of the rest of this course: it is the equation you actually solve to find the allowed, quantized energies of a particle in a box, a barrier, an oscillator or an atom. This article builds the TDSE, shows how separation of variables produces the TISE and the idea of a stationary state, states the matching conditions on $\phi(x)$ at a finite potential step, and works through the free-particle solution $\phi(x)=A\sin(kx)$ with real numbers for an electron at $2.0\ \text{eV}$ . Everything here matches the worked numbers of the Pass-1 article — nothing is changed, only unpacked in more detail.

The core insight

The Schrödinger equation $i\hbar\,\partial\psi/\partial t = \hat H\psi$ governs how a quantum wave function evolves, and for a time-independent potential it separates into a time-independent eigenvalue equation $\hat H\phi = E\phi$ whose solutions are stationary states with a fixed, sharply defined energy.

The core idea

1. The time-dependent Schrödinger equation (TDSE). In one dimension, the postulated equation of motion for $\psi(x,t)$ is

i\hbar\,\frac{\partial \psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi(x,t)}{\partial x^2} + V(x)\,\psi(x,t).

where: $\hbar = h/2\pi$ is the reduced Planck constant, $m$ is the particle's mass, $V(x)$ is the potential energy, and $i=\sqrt{-1}$ . The right-hand side comes straight from the classical energy $E = p^2/2m + V(x)$ with momentum turned into the operator $\hat p = -i\hbar\,\partial/\partial x$ , so $\hat p^2 \to -\hbar^2 \partial^2/\partial x^2$ . The whole right side is called the Hamiltonian operator $\hat H$ , and the left side is the energy operator $i\hbar\,\partial/\partial t$ acting on $\psi$ . Because the equation is linear and first-order in time, knowing $\psi(x,0)$ fixes $\psi(x,t)$ at every later instant — this is quantum mechanics' version of determinism, applied to the wave function rather than to a trajectory.

2. Separation of variables and the TISE. When $V(x)$ does not depend on time, try $\psi(x,t) = \phi(x)\,e^{-iEt/\hbar}$ . Substituting into the TDSE, the time-derivative gives a factor $i\hbar\cdot(-iE/\hbar) = E$ on the left, and after cancelling the shared factor $e^{-iEt/\hbar}$ from both sides you are left with an equation purely in $x$ , the time-independent Schrödinger equation:

-\frac{\hbar^2}{2m}\frac{d^2\phi(x)}{dx^2} + V(x)\,\phi(x) = E\,\phi(x).

where: $\phi(x)$ is the spatial part of the wave function and $E$ is the particle's total energy. This is an eigenvalue equation — $\phi(x)$ has to obey boundary conditions (finite everywhere, going to zero far from the particle where required) and, at any point where $V(x)$ stays finite, both $\phi$ and $d\phi/dx$ must be continuous. Only special values of $E$ , the energy eigenvalues, let $\phi(x)$ satisfy all of that at once. This restriction to a discrete list of allowed $E$ is exactly where quantization comes from in every model system ahead: the box, the barrier, the oscillator.

3. Stationary states. A solution built as $\psi(x,t) = \phi(x)e^{-iEt/\hbar}$ is called a stationary state. Its probability density is

|\psi(x,t)|^2 = |\phi(x)|^2\,\big|e^{-iEt/\hbar}\big|^2 = |\phi(x)|^2,

which has no time dependence at all, since the phase factor has magnitude $1$ . So although $\psi$ itself keeps spinning through a complex phase, nothing measurable — probability distributions, expectation values — changes with time. That is precisely why $E$ can be quoted as a fixed, sharply defined energy for such a state.

4. Matching conditions at a step. At a point $x_0$ where $V(x)$ jumps by a finite amount, integrating the TISE across a tiny interval around $x_0$ shows that both $\phi(x)$ and $d\phi/dx$ must be continuous there. The one exception is an infinite jump, as at the wall of an infinite well: then only $\phi$ itself is forced to be continuous (it must go to zero at the wall), while $d\phi/dx$ is allowed to jump.

Figure

A wave function phi of x drawn as a smooth curve continuous across a finite potential step, with the slope also matching on both sides of the step, contrasted with an infinite wall where the curve is forced to zero but its slope changes abruptly

Derivation AVerifying a free-particle wave function solves the TISE

The goal is to check, step by step and with no shortcuts, that the trial spatial wave function $\phi(x) = A\sin(kx)$ actually satisfies the time-independent Schrödinger equation in a field-free region where $V(x)=0$ , and to read off the energy $E$ that this trial solution forces — the same check you will repeat for every model system that follows, so it is worth doing once carefully by hand.

Step 1. Write the TISE for $V=0$ . Dropping the potential term from the general TISE leaves $-\dfrac{\hbar^2}{2m}\dfrac{d^2\phi}{dx^2} = E\phi$ , an equation that says the second derivative of $\phi$ must be proportional to $\phi$ itself — a strong hint that sines, cosines or complex exponentials will work, since they reproduce themselves (up to a constant) under differentiation.

Step 2. Differentiate the trial solution twice. With $\phi(x) = A\sin(kx)$ , the first derivative is $d\phi/dx = Ak\cos(kx)$ , and differentiating again gives $d^2\phi/dx^2 = -Ak^2\sin(kx) = -k^2\phi(x)$ . Notice the second derivative has folded back into a multiple of the original function — exactly the self-reproducing behaviour the equation in Step 1 demands.

Step 3. Substitute back into the TISE. Plugging $d^2\phi/dx^2 = -k^2\phi$ into $-\dfrac{\hbar^2}{2m}\dfrac{d^2\phi}{dx^2} = E\phi$ gives $-\dfrac{\hbar^2}{2m}(-k^2\phi) = E\phi$ , which simplifies to $\dfrac{\hbar^2 k^2}{2m}\phi = E\phi$ . Since this must hold at every value of $x$ , and $\phi(x)$ is not identically zero, the factor $\phi(x)$ cancels from both sides, leaving a plain numerical condition on $E$ rather than a function of $x$ .

Step 4. Read off the energy and interpret it. The condition that survives is $E = \dfrac{\hbar^2 k^2}{2m}$ . This confirms $\phi(x)=A\sin(kx)$ is a genuine solution of the free-particle TISE for any real $k$ , with energy fixed by that $k$ . It is also exactly the classical kinetic energy $p^2/2m$ once you substitute the de Broglie relation $p = \hbar k$ — the Schrödinger equation reproduces the familiar free-particle energy-momentum relation, as it must.

Worked example

An electron ( $m = 9.11\times10^{-31}\ \text{kg}$ ) occupies a stationary state of energy $E = 2.0\ \text{eV}$ in a region where $V(x)=0$ . Find the wavenumber $k$ of its spatial wave function $\phi(x)$ , and the corresponding de Broglie wavelength.

Step 1. Convert the energy to joules, since every other quantity is in SI units. $E = 2.0\ \text{eV} \times 1.6\times10^{-19}\ \text{J/eV} = 3.2\times10^{-19}\ \text{J}$ .

Step 2. Use the result of the derivation above, $E = \hbar^2 k^2/2m$ , solved for $k$ : $k = \sqrt{2mE}/\hbar$ . First compute the product under the square root: $2mE = 2\times(9.11\times10^{-31})\times(3.2\times10^{-19}) = 5.83\times10^{-49}\ \text{kg}\,\cdot\,\text{J}$ , so $\sqrt{2mE} = 7.64\times10^{-25}\ \text{kg}\,\cdot\,\text{m/s}$ .

Step 3. Divide by $\hbar = 1.055\times10^{-34}\ \text{J}\,\cdot\,\text{s}$ to get $k$ . $k = \dfrac{7.64\times10^{-25}}{1.055\times10^{-34}} \approx 7.24\times10^{9}\ \text{m}^{-1}$ .

Step 4. Convert to a de Broglie wavelength using $\lambda = 2\pi/k$ , as a sanity check on the size of $k$ . $\lambda = 2\pi/(7.24\times10^{9}) \approx 0.87\ \text{nm}$ . Answer: $k \approx 7.24\times10^{9}\ \text{m}^{-1}$ and $\lambda \approx 0.87\ \text{nm}$ — a sensible atomic-scale wavelength for a low-energy electron, consistent with electron wave behaviour probed in atomic and crystal-diffraction experiments.

Common traps

Writing the TISE with a plus sign in front of the kinetic term, or dropping the factor of $\tfrac12$ — always rebuild it from $E = p^2/2m + V$ with $\hat p^2 \to -\hbar^2 d^2/dx^2$ rather than trying to recall it directly.
Thinking a stationary state means $\psi(x,t)$ itself is frozen in time — only $|\psi|^2$ (and hence measurable probabilities) is time-independent; the complex phase $e^{-iEt/\hbar}$ keeps rotating.
Forgetting that separation of variables, $\psi = \phi(x)e^{-iEt/\hbar}$ , is only valid when $V(x)$ has no explicit time dependence.
Requiring $d\phi/dx$ to be continuous at an infinite wall — there only $\phi$ itself must vanish, and the slope is allowed to jump.

Ask the tutor: Want to see the separation-of-variables algebra done with every intermediate line, or work through why $d\phi/dx$ is allowed to jump at an infinite wall — ask the AI tutor to walk through it with you.

Why does substituting phi times e to the minus iEt over hbar into the TDSE actually remove the time variable?

Because the time-derivative in the TDSE acts only on the exponential factor and pulls out a constant. Differentiating $e^{-iEt/\hbar}$ with respect to $t$ gives $(-iE/\hbar)e^{-iEt/\hbar}$ , and multiplying by the $i\hbar$ out front on the left side of the TDSE gives $i\hbar\cdot(-iE/\hbar) = E$ , a plain number. The exponential factor then appears on both sides of the equation and cancels, leaving an equation in $x$ alone — the TISE — with $E$ sitting where the time-derivative used to be.

If quantum mechanics is supposed to be probabilistic, why can a stationary state have one exact, sharply defined energy E?

Because $E$ enters the wave function only through the phase factor $e^{-iEt/\hbar}$ , and phase is not a measurable probability — $|\psi|^2$ strips it out completely. The randomness in quantum mechanics shows up in quantities like position, where $|\phi(x)|^2$ gives a genuine probability spread. Energy in a stationary state is different: it is an eigenvalue of the Hamiltonian, and a state built purely from one eigenfunction has that single eigenvalue with certainty, no matter how spread out $\phi(x)$ is in space.

The worked example gives a de Broglie wavelength of about 0.87 nanometres for a 2 electronvolt electron — is that a coincidence, or does it mean something?

It is not a coincidence; it reflects genuinely how the Schrödinger equation reduces to familiar wave mechanics for a free particle. The relation $E = \hbar^2k^2/2m$ derived from the TISE is exactly the classical kinetic energy $p^2/2m$ once you use $p=\hbar k$ , so solving for $k$ and converting to $\lambda = 2\pi/k$ reproduces the ordinary de Broglie wavelength from the previous chapter. The atomic-scale answer, under a nanometre, is why electron waves show up in diffraction from crystal lattices whose spacing is comparable to $\lambda$ .

Schrödinger equation

The core idea

Worked example

Common traps

Before this

Where this leads