2.3 The time-independent Schrödinger equation

The full Schrödinger equation is

\ii\hbar\frac{\partial\Psi(x,t)}{\partial t} = \hat H\Psi(x,t),\qquad \hat H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\tag{2.3.1}

a partial differential equation. The good news: when $V$ does not depend explicitly on time — which covers almost every textbook problem — separation of variables splits it into two ordinary differential equations, and one of them is completely trivial.

Separating the variables

From the time-dependent equation to the stationary onebasic~5 min

Try a separable solution

\Psi(x,t) = \psi(x)\,\varphi(t)\tag{2.3.2}

and substitute:

\ii\hbar\,\psi(x)\frac{\dd\varphi}{\dd t} = \varphi(t)\left[-\frac{\hbar^2}{2m}\frac{\dd^2\psi}{\dd x^2}+V(x)\psi\right]\tag{2.3.3}

Divide both sides by $\psi\varphi$ :

\underbrace{\ii\hbar\frac{1}{\varphi}\frac{\dd\varphi}{\dd t}}_{\text{depends on }t\text{ only}} =\underbrace{\frac{1}{\psi}\left[-\frac{\hbar^2}{2m}\frac{\dd^2\psi}{\dd x^2}+V\psi\right]}_{\text{depends on }x\text{ only}}\tag{2.3.4}

The left side is a function of $t$ alone, the right side of $x$ alone, and $x$ and $t$ are independent variables. A quantity depending only on $t$ can equal one depending only on $x$ only if both equal the same constant. Call that separation constant $E$ — it will turn out to be the energy.

The time part:

\ii\hbar\frac{\dd\varphi}{\dd t}=E\varphi \;\Longrightarrow\; \varphi(t)=\ee^{-\ii Et/\hbar}\tag{2.3.5}

(the integration constant is absorbed into $\psi$ ). This solution is completely universal and has nothing to do with the shape of the potential. All the physics lives in the spatial part.

The spatial part, which is the time-independent Schrödinger equation:

-\frac{\hbar^2}{2m}\frac{\dd^2\psi}{\dd x^2}+V(x)\psi = E\psi\tag{2.3.6}

or, in operator form, $\hat H\psi = E\psi$ .

So a separable solution always looks like

\Psi_n(x,t)=\psi_n(x)\,\ee^{-\ii E_n t/\hbar}\tag{2.3.7}

This is not an ordinary differential equation

On the surface $\hat H\psi=E\psi$ is a second-order ODE, but it has an extra identity: it is an eigenvalue problem.

The difference is this. An ordinary differential equation has solutions for any value of its parameters; here we additionally demand that $\psi$ satisfy physical boundary conditions (normalisable, continuous, and so on). Those conditions can only be met simultaneously for particular values of $E$ . The allowed values are the energy levels and the corresponding $\psi$ are the energy eigenstates.

Quantisation of energy is not an extra assumption; it is a mathematical consequence of the boundary conditions. This is one of the loveliest things in quantum mechanics: the quantisation conditions that Planck and Bohr had to insert by hand fall out of Schrödinger’s framework on their own.

Why E must be real

Bound-state energy eigenvalues are realadvanced~4 min

Assume $\hat H\psi = E\psi$ with normalisable $\psi$ . Take the complex conjugate (with $V$ real):

-\frac{\hbar^2}{2m}\psi^{*\prime\prime}+V\psi^*=E^*\psi^*\tag{2.3.8}

Multiply the first equation by $\psi^*$ and the second by $\psi$ , then subtract; the $V$ terms cancel:

-\frac{\hbar^2}{2m}\left(\psi^*\psi''-\psi\psi^{*\prime\prime}\right)=(E-E^*)|\psi|^2\tag{2.3.9}

The bracket on the left is again a total derivative, $\frac{\dd}{\dd x}(\psi^*\psi'-\psi\psi^{*\prime})$ . Integrating over all space turns the left side into a boundary term, which vanishes because $\psi$ is normalisable:

0=(E-E^*)\int|\psi|^2\dd x = (E-E^*)\cdot 1\tag{2.3.10}

Hence $E=E^*$ , so $E$ is real.

This is no accident. Its general form is “the eigenvalues of a Hermitian operator are real”, redone in operator language in section 3.5 — in three lines. The long-winded integral proof here shows exactly that: change the language and the same fact becomes much simpler.

Reality of $E$ has an immediate consequence: $|\varphi(t)|^2=|\ee^{-\ii Et/\hbar}|^2=1$ , so the time factor has constant modulus. Therefore $|\Psi_n(x,t)|^2=|\psi_n(x)|^2$ does not change with time — which is where the name “stationary state” comes from. The next section says more.

Reading a wavefunction from its curvature

You can see roughly what $\psi$ looks like without solving anything. Rewrite the stationary equation as

\frac{\dd^2\psi}{\dd x^2}=\frac{2m}{\hbar^2}\left[V(x)-E\right]\psi\tag{2.3.11}

The picture

Classically allowed region ( $E>V$ ): $\psi''$ has the opposite sign to $\psi$ , so the curve bends towards the axis — it oscillates. The larger $E-V$ (the larger the kinetic energy), the sharper the bending and the shorter the wavelength.

Classically forbidden region ( $E<V$ ): $\psi''$ has the same sign as $\psi$ , so the curve bends away from the axis — exponential behaviour. Physically we keep only the decaying branch, or normalisation fails.

Turning point ( $E=V$ ): $\psi''=0$ and the curve changes its concavity there.

The mathematics

\psi''=\frac{2m(V-E)}{\hbar^2}\psi\tag{2.3.12}

$E>V$ : locally $\psi\sim\ee^{\pm\ii kx}$ with $k=\frac{\sqrt{2m(E-V)}}{\hbar}$

$E<V$ : locally $\psi\sim\ee^{\pm\kappa x}$ with $\kappa=\frac{\sqrt{2m(V-E)}}{\hbar}$

These two approximations are the starting point of the WKB method.

This picture alone predicts quantisation

Consider a potential well that traps a particle. Starting from far to the left, $\psi$ must decay exponentially towards zero. Entering the well it starts to oscillate; leaving the well it returns to exponential behaviour.

Here is the catch: the exponential solution on the right generally contains both a growing branch $\ee^{+\kappa x}$ and a decaying branch $\ee^{-\kappa x}$ . For $\psi$ to go to zero far to the right as well, the coefficient of the growing branch must be tuned to exactly zero.

That is only possible for particular values of $E$ , where the oscillation inside the well accumulates just the right phase to match both sides. This is where discrete levels come from. The larger $n$ , the more half-wavelengths fit inside, the more nodes, and the higher the energy.

Key formulas

Time-independent Schrödinger equation
$-\frac{\hbar^2}{2m}\frac{\dd^2\psi}{\dd x^2}+V(x)\psi=E\psi$
An eigenvalue problem: only particular E admit normalisable solutions
Separable solution
$\Psi_n(x,t)=\psi_n(x)\,\ee^{-\ii E_n t/\hbar}$
The time factor is independent of the potential — always this phase
Curvature relation
$\frac{\dd^2\psi}{\dd x^2}=\frac{2m}{\hbar^2}[V(x)-E]\,\psi$
Oscillatory for E>V, exponential for E<V; the whole qualitative picture

Self-check3 questions

What comes next

Separation of variables produces only a family of particular solutions. The genuinely general solution is a superposition of them — and superpositions behave nothing like stationary states. The next section returns to the full time-dependent equation.

The time-independent Schrödinger equation

Recommended first

After this section you should be able to

Separating the variables

Why E must be real

Reading a wavefunction from its curvature

The picture

The mathematics

Key formulas

Self-check3 questions

What comes next

Separating the variables

Why E must be real

Reading a wavefunction from its curvature

◑The picture

∑The mathematics

∎Key formulas

What comes next

The picture

The mathematics

Key formulas