The Schrödinger Equation and the Wave Function: From the Born Rule to the Evolution of Expectation Values

0. Key points

• In quantum mechanics a state is represented by a wave function $\psi(x,t)$, whose evolution in time is governed by the time-dependent Schrödinger equation $i\hbar\,\partial_t \psi = \hat{H}\psi$. The equation is of first order in time, so $\psi$ at a single instant determines the entire future and the entire past.
• The quantity $|\psi(x,t)|^2$ is the probability density for position (the Born rule). This interpretation does not collapse because, when the potential is real-valued, the total probability $\int |\psi|^2 dx$ stays constant in time, a fact that follows from a local conservation law for probability (a continuity equation); see Theorem 4.2 and Corollary 4.3.
• The momentum operator $\hat{p} = -i\hbar\,\partial_x$ is not a rule imposed from above. Computing $m \, d\langle x\rangle/dt$ directly from the Schrödinger equation forces this form upon us (Proposition 5.1).
• The time derivative of an expectation value can be written with a commutator (Lemma 5.4). Applying this to position and momentum yields Ehrenfest’s theorem, according to which expectation values satisfy something closely resembling Newton’s equation of motion (Theorem 5.5). In general, however, $\langle V'(x)\rangle \neq V'(\langle x\rangle)$, so classical mechanics genuinely reappears only when the wave packet is sufficiently narrow.
• When the Hamiltonian does not depend on time, the variables separate and the problem reduces to the eigenvalue problem $\hat{H}\varphi = E\varphi$, the time-independent Schrödinger equation. Its solutions (stationary states) have a time-independent probability density, and the general solution is a superposition $\sum_n c_n \varphi_n(x)\, e^{-iE_n t/\hbar}$ of them.

1. Motivation: giving the wave an equation of motion

As we saw in The birth of quantum mechanics, the “old quantum theory” of the years 1900 to 1925 explained the experimental facts of black-body radiation, the photoelectric effect and atomic spectra by pasting quantum conditions as side rules onto classical mechanics. Bohr’s model of the hydrogen atom is the standard example: impose the condition that the angular momentum of a circular orbit be an integer multiple of $\hbar$, and the wavelengths of the Balmer series come out exactly right.

It is hard to call this a theory. Nothing explains why angular momentum should be quantised, and systems more complicated than hydrogen (even the helium atom) are entirely out of reach. Above all, a quantum condition is not an equation of motion. In classical mechanics Newton’s equation, or equivalently the canonical equations（Theorem 4.2）［ハミルトン形式の力学］ of the Hamiltonian formalism,

$$\dot{x} = \frac{\partial H}{\partial p}, \qquad \dot{p} = -\frac{\partial H}{\partial x}$$

determine the state at any time from the state $(x, p)$ at one time. The old quantum theory had nothing to play this role.

The turning point was de Broglie’s proposal of 1924. If light is a wave and simultaneously a particle, then a particle such as an electron ought to have a wave aspect too, and its wavelength ought to be tied to its momentum by

$$\lambda = \frac{h}{p}, \qquad \text{that is,} \qquad p = \hbar k \quad (k = 2\pi/\lambda) .$$

This can be checked experimentally (the Davisson–Germer electron diffraction experiment). And once the proposal is granted, one question becomes unavoidable: what equation does this wave obey?

Schrödinger, confronted with exactly this question in Zurich, published his answer in 1926. He first tried a relativistic form (what we now call the Klein–Gordon equation), but the fine structure of hydrogen it predicted disagreed with experiment, so he retreated to a non-relativistic version and published that instead. The irony is that electron spin, unknown at the time, was what governed the fine structure, which made the non-relativistic version look “correct”.

One problem as serious as the equation itself remained: what is $\psi$? Schrödinger himself initially took $|\psi|^2$ to be a charge density, that is, an entity genuinely spread out in space. That reading is incompatible with the fact that the wave packet of a free particle spreads without limit as time goes on, whereas an electron is always found at a single point when it is observed. In 1926, in a paper on scattering, Born proposed reading $|\psi|^2$ as a probability density, and this has been the standard ever since. The goal of this article is to build up these two things carefully: the equation of motion and the probability interpretation.

flowchart TB
subgraph CL["Classical mechanics"]
  C1["State: the pair of position x and momentum p"] --> C2["Equation of motion: Hamilton's canonical equations"]
  C2 --> C3["Measurement: the values of x, p themselves are obtained"]
end
subgraph QM["Quantum mechanics"]
  Q1["State: the wave function ψ"] --> Q2["Equation of motion: the Schrödinger equation"]
  Q2 --> Q3["Measurement: the Born rule fixes probabilities only"]
end
CL -.->|"Correspondence principle: narrow wave packets"| QM

The structural correspondence between classical and quantum mechanics. The three layers — state, equation of motion, measurement — survive intact, but their contents are all replaced

2. Preliminaries: where wave functions live, and Dirac notation

Throughout we treat a single particle of mass $m$ moving in one dimension. Generalising to three dimensions only requires replacing $\partial_x^2$ by the Laplacian $\nabla^2$ and $dx$ by $d^3\boldsymbol{r}$.

Definition 2.1（States and wave functions）

The state of the system at time $t$ is represented by a complex-valued function $\psi(\cdot, t) : \mathbb{R} \to \mathbb{C}$ that is square integrable, that is, satisfies

$$\int_{-\infty}^{\infty} |\psi(x,t)|^2 \, dx < \infty .$$

This $\psi$ is called the wave function. We write $L^2(\mathbb{R})$ for the complex vector space of all square-integrable functions.

The space $L^2(\mathbb{R})$ carries an inner product. Following the convention of physics, we take it to be antilinear in the first argument (that is the slot carrying the complex conjugate):

$$\langle \varphi \mid \psi \rangle := \int_{-\infty}^{\infty} \overline{\varphi(x)}\, \psi(x) \, dx, \qquad \|\psi\| := \sqrt{\langle \psi \mid \psi\rangle}.$$

That this satisfies the inner-product axioms (positive definiteness, linearity in the second argument, and $\langle \varphi \mid \psi\rangle = \overline{\langle \psi \mid \varphi\rangle}$) is immediate from the definition. For instance the last property reads $\overline{\langle \psi \mid \varphi\rangle} = \overline{\int \overline{\psi}\varphi\,dx} = \int \psi \overline{\varphi}\, dx = \langle \varphi\mid\psi\rangle$. For the general theory of inner-product spaces see Inner product spaces and Gram–Schmidt orthogonalisation (the list of axioms is Definition 3.1［内積空間とグラム・シュミット直交化］).

In Dirac’s notation one writes the state itself as a ket $|\psi\rangle$ and reads the function $\psi(x)$ as its “component in the position representation”,

$$\psi(x) = \langle x \mid \psi \rangle .$$

This is the same idea as choosing a basis $\{\boldsymbol{e}_i\}$ in finite dimensions and writing components $v_i = \langle \boldsymbol{e}_i, \boldsymbol{v}\rangle$; the only difference is that the “basis” is indexed by the continuous label $x$ (the care needed to treat this basis rigorously is collected in the Appendix). A bra $\langle \varphi |$ is the linear functional $|\psi\rangle \mapsto \langle \varphi \mid \psi\rangle$ acting on kets to return a number.

Remark 2.2

A wave function $\psi$ and the function $e^{i\theta}\psi$, with a constant $\theta \in \mathbb{R}$, represent the same physical state. Indeed $|e^{i\theta}\psi|^2 = |\psi|^2$, and the expectation values defined below are unchanged as well: $\langle e^{i\theta}\psi \mid \hat{A} \mid e^{i\theta}\psi\rangle = \overline{e^{i\theta}}e^{i\theta}\langle \psi \mid \hat{A}\mid \psi\rangle = \langle \psi \mid \hat{A}\mid\psi\rangle$. This factor $e^{i\theta}$ is called a global phase. By contrast, the relative phase $\theta$ in a superposition $\psi_1 + e^{i\theta}\psi_2$ is observable, because it shifts the position of the interference fringes. “Phase cannot be measured” is wrong; the correct statement is that only the global phase cannot be measured.

3. The time-dependent Schrödinger equation

3.1. Reading operators off a plane wave

Granting the de Broglie relation（Axiom 6.1）［The Birth of Quantum Mechanics］ $p = \hbar k$ and the Planck–Einstein relation（Axiom 5.1）［The Birth of Quantum Mechanics］ $E = \hbar\omega$, let us differentiate the simplest wave corresponding to a free particle, the plane wave

$$\psi(x,t) = A\, e^{i(kx - \omega t)} .$$

We find

$$\begin{aligned} i\hbar \frac{\partial \psi}{\partial t} &= i\hbar \cdot (-i\omega)\psi = \hbar\omega\, \psi = E\,\psi, \\ -i\hbar \frac{\partial \psi}{\partial x} &= -i\hbar \cdot (ik)\psi = \hbar k\, \psi = p\,\psi, \\ -\hbar^2 \frac{\partial^2 \psi}{\partial x^2} &= -\hbar^2 \cdot (ik)^2 \psi = \hbar^2 k^2 \psi = p^2 \psi . \end{aligned}$$

In other words, on plane waves multiplication by the energy is replaced by $i\hbar\,\partial_t$, and multiplication by the momentum by $-i\hbar\,\partial_x$. Using the non-relativistic relation $E = p^2/(2m)$ between energy and momentum for a free particle, we obtain

$$i\hbar \frac{\partial \psi}{\partial t} = E\psi = \frac{p^2}{2m}\psi = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}$$

identically for plane waves. For a particle moving in a potential $V(x)$ the classical relation is $E = p^2/(2m) + V(x)$, so adding $V(x)\psi$ on the right-hand side is the natural guess.

3.2. The equation and the Hamiltonian

Definition 3.1（The time-dependent Schrödinger equation）

Let a real-valued function $V : \mathbb{R} \to \mathbb{R}$ (the potential) be given. The wave function $\psi(x,t)$ of a single particle of mass $m$ obeys

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

This partial differential equation is the time-dependent Schrödinger equation, and the operator $\hat{H}$ is the Hamiltonian operator. In Dirac notation it reads

$$i\hbar \frac{d}{dt}|\psi(t)\rangle = \hat{H}\,|\psi(t)\rangle .$$

The form of $\hat{H}$ is obtained from the classical Hamiltonian $H(x,p) = p^2/(2m) + V(x)$ by the substitutions $x \to \hat{x}$ (the operator of multiplication by $x$) and $p \to \hat{p} = -i\hbar\,\partial_x$. This substitution is called canonical quantisation. The fact that in classical mechanics the Hamiltonian is the generator of time evolution (Hamiltonian mechanics) survives here untouched.

Remark 3.2

The argument of §3.1 is not a derivation. We started from a single family of solutions, the plane waves, and guessed a linear partial differential equation containing them. At the stage of adding $V$ there is no ground beyond “we would like it to be so”. The Schrödinger equation is, like Newton’s equation of motion, a fundamental law (an axiom), and the case for it rests not on a derivation but on the agreement of its predictions with experiment. In fact the equation has succeeded without exception in the non-relativistic domain, from the hydrogen spectrum through chemical bonding and the band structure of solids to superconductivity.

Proposition 3.3（The superposition principle）

Fix $V$. If $\psi_1$ and $\psi_2$ are both solutions of Definition 3.1 and $c_1, c_2 \in \mathbb{C}$ are constants (independent of both time and position), then $c_1\psi_1 + c_2\psi_2$ is also a solution of the same equation.

Proof（Proposition 3.3）

The operator $\partial_t$ is linear. So is $\hat{H}$, since $\partial_x^2$ is linear and multiplication by $V(x)$ is linear. Hence

$$\begin{aligned} i\hbar \frac{\partial}{\partial t}\left(c_1\psi_1 + c_2\psi_2\right) &= c_1 \left(i\hbar \frac{\partial \psi_1}{\partial t}\right) + c_2\left(i\hbar\frac{\partial \psi_2}{\partial t}\right) \\ &= c_1 \hat{H}\psi_1 + c_2 \hat{H}\psi_2 \qquad (\text{using that $\psi_1, \psi_2$ are solutions}) \\ &= \hat{H}\left(c_1\psi_1 + c_2\psi_2\right) . \end{aligned}$$∎

This simple proposition is the source of almost every “mystery” of quantum mechanics. Two-slit interference, and the oscillation produced by superposing stationary states that we shall meet below, both issue from the single fact that the Schrödinger equation is linear. Note that $c_1$ and $c_2$ must be constants: if $c_1(t)$ depended on time, an extra term $i\hbar\dot{c}_1\psi_1$ would appear and the computation would fail.

Remark 3.4

Two remarks on the shape of the equation.

It is of first order in time. The classical wave equation $\partial_t^2 u = c^2\partial_x^2 u$ is of second order in time, so it needs both $u$ and $\partial_t u$ as initial data. The Schrödinger equation is of first order, so the single function $\psi(x, t_0)$ determines the future and the past completely. This is determinism in exactly the same sense in which a single pair $(x,p)$ of initial data determines a classical trajectory, and it shows that what is “essentially indeterministic” in quantum mechanics is not the time evolution but the measurement.

The factor $i$ is essential. If $i\hbar\,\partial_t$ were replaced by $-\hbar\,\partial_t$, the equation would become $\partial_t \psi = (\hbar/2m)\partial_x^2\psi - (V/\hbar)\psi$, of diffusion (heat-conduction) type. A plane wave would then simply decay monotonically as $e^{ikx}e^{-(\hbar k^2/2m)t}$, with no oscillation and no interference, and $\int|\psi|^2dx$ would not be conserved. The imaginary unit is what makes the wave function something that oscillates rather than something that diffuses.

4. The Born interpretation and normalisation

Definition 4.1（The Born rule (probability density for position)）

A wave function $\psi$ is said to be normalised when

$$\langle \psi \mid \psi \rangle = \int_{-\infty}^{\infty} |\psi(x,t)|^2\, dx = 1 .$$

In that case the probability of finding the particle in the interval $[a,b]$ when its position is measured at time $t$ is

$$P(a \le x \le b; t) = \int_a^b |\psi(x,t)|^2\, dx .$$

The function $\rho(x,t) := |\psi(x,t)|^2$ is called the probability density.

The Born rule. The area under the probability density ρ over the interval [a, b] is the probability of finding the particle in that interval

For this interpretation to make sense, $\int|\psi|^2 dx = 1$ must hold at every time. If we normalised at $t = 0$ and the total came to $0.8$ one second later, the probability interpretation would collapse. This is not a postulate: it is a fact provable from the Schrödinger equation.

Theorem 4.2（The continuity equation for probability）

Let $V$ be real-valued and let $\psi$ be a solution of Definition 3.1 that is twice continuously differentiable in $x$ and once continuously differentiable in $t$. Put

$$\rho(x,t) := |\psi(x,t)|^2, \qquad j(x,t) := \frac{\hbar}{2mi}\left(\overline{\psi}\,\frac{\partial \psi}{\partial x} - \frac{\partial\overline{\psi}}{\partial x}\,\psi\right) = \frac{\hbar}{m}\,\mathrm{Im}\!\left(\overline{\psi}\,\frac{\partial \psi}{\partial x}\right) .$$

Then

$$\frac{\partial \rho}{\partial t} + \frac{\partial j}{\partial x} = 0$$

holds for all $x$ and $t$. The quantity $j$ is called the probability current density.

Proof（Theorem 4.2）

First we check that the two expressions for $j$ agree. Setting $z := \overline{\psi}\,\partial_x\psi$ we have $\overline{z} = \psi\,\partial_x\overline{\psi}$, so from $z - \overline{z} = 2i\,\mathrm{Im}(z)$,

$$\frac{\hbar}{2mi}(z - \overline{z}) = \frac{\hbar}{2mi}\cdot 2i\,\mathrm{Im}(z) = \frac{\hbar}{m}\mathrm{Im}(z) .$$

Now for the main claim. Dividing both sides of Definition 3.1 by $i\hbar$,

$$\frac{\partial \psi}{\partial t} = \frac{1}{i\hbar}\hat{H}\psi = \frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2} - \frac{i}{\hbar}V\psi .$$

Take the complex conjugate of this. Here we use the hypothesis that $V$ is real-valued, so that $\overline{V} = V$:

$$\frac{\partial \overline{\psi}}{\partial t} = -\frac{i\hbar}{2m}\frac{\partial^2\overline{\psi}}{\partial x^2} + \frac{i}{\hbar}V\overline{\psi}.$$

Differentiating $\rho = \overline{\psi}\psi$ in time and substituting both expressions,

$$\begin{aligned} \frac{\partial \rho}{\partial t} &= \frac{\partial \overline{\psi}}{\partial t}\psi + \overline{\psi}\frac{\partial \psi}{\partial t} \\ &= \left(-\frac{i\hbar}{2m}\overline{\psi}_{xx} + \frac{i}{\hbar}V\overline{\psi}\right)\psi + \overline{\psi}\left(\frac{i\hbar}{2m}\psi_{xx} - \frac{i}{\hbar}V\psi\right) \\ &= \frac{i\hbar}{2m}\left(\overline{\psi}\,\psi_{xx} - \overline{\psi}_{xx}\,\psi\right) + \frac{i}{\hbar}V\overline{\psi}\psi - \frac{i}{\hbar}V\overline{\psi}\psi \\ &= \frac{i\hbar}{2m}\left(\overline{\psi}\,\psi_{xx} - \overline{\psi}_{xx}\,\psi\right). \end{aligned}$$

Note that the potential terms cancelled exactly; this is a direct consequence of $V$ being real-valued. Finally, by the product rule,

$$\frac{\partial}{\partial x}\left(\overline{\psi}\psi_x - \overline{\psi}_x\psi\right) = \overline{\psi}_x\psi_x + \overline{\psi}\psi_{xx} - \overline{\psi}_{xx}\psi - \overline{\psi}_x\psi_x = \overline{\psi}\psi_{xx} - \overline{\psi}_{xx}\psi ,$$

so that

$$\frac{\partial \rho}{\partial t} = \frac{i\hbar}{2m}\frac{\partial}{\partial x}\left(\overline{\psi}\psi_x - \overline{\psi}_x\psi\right) = -\frac{\partial}{\partial x}\left[\frac{\hbar}{2mi}\left(\overline{\psi}\psi_x - \overline{\psi}_x\psi\right)\right] = -\frac{\partial j}{\partial x} ,$$

where along the way we used $i\hbar/(2m) = -\hbar/(2mi)$ (because $1/i = -i$).

∎

This is exactly the form of the continuity equation in fluid mechanics and of charge conservation in electromagnetism. It says that probability is neither created nor destroyed, but only transported as a flow.

Corollary 4.3（Conservation of normalisation）

In addition to the hypotheses of Theorem 4.2, suppose that at each time $\psi(x,t) \to 0$ and $\partial_x \psi(x,t) \to 0$ as $|x| \to \infty$, and that the time derivative and the integral in $\int |\psi|^2 dx$ may be interchanged. Then

$$\frac{d}{dt}\int_{-\infty}^{\infty} |\psi(x,t)|^2 dx = 0 .$$

In particular, if $\psi$ is normalised at $t = 0$ it is normalised at every time.

Proof（Corollary 4.3）

Granting the interchange of derivative and integral and using Theorem 4.2,

$$\frac{d}{dt}\int_{-\infty}^{\infty}\rho\, dx = \int_{-\infty}^{\infty}\frac{\partial \rho}{\partial t}\, dx = -\int_{-\infty}^{\infty}\frac{\partial j}{\partial x}\, dx = -\Bigl[\, j(x,t) \,\Bigr]_{x=-\infty}^{x=+\infty}.$$

Since $j = (\hbar/m)\,\mathrm{Im}(\overline{\psi}\,\partial_x\psi)$ is a product of $\psi$ and $\partial_x\psi$, both of which tend to $0$ by hypothesis, we have $j \to 0$. Hence the right-hand side vanishes.

∎

So normalisation need only be carried out once, at the beginning. This is far from obvious, and it rests on $V$ being real-valued.

Remark 4.4

Seeing what happens when $V$ has an imaginary part makes the mechanism clear. Take $V = V_R - i\Gamma/2$ with a real constant $\Gamma > 0$. Then the potential terms in the proof above no longer cancel and we get $\partial_t \rho = -\partial_x j - (\Gamma/\hbar)\rho$. Integrating, $\int\rho\,dx$ decays like $e^{-\Gamma t/\hbar}$: particles are disappearing, and the probability interpretation fails. Conversely, one sometimes introduces an imaginary part deliberately, as an “optical potential” giving a phenomenological description of the decay of unstable nuclei or of absorbers.

Example 4.5（Normalising a Gaussian wave packet and computing a probability）

Let $a > 0$ be a constant and normalise $\psi(x) = A\,e^{-x^2/(2a^2)}$ with $A > 0$. We have

$$\int_{-\infty}^{\infty}|\psi|^2 dx = A^2\int_{-\infty}^{\infty} e^{-x^2/a^2}\,dx = A^2 \cdot a\sqrt{\pi}$$

(substituting $\alpha = 1/a^2$ into the Gaussian integral $\int_{-\infty}^{\infty}e^{-\alpha x^2}dx = \sqrt{\pi/\alpha}$). Setting this equal to $1$,

$$A = \frac{1}{(\pi a^2)^{1/4}}, \qquad \psi(x) = \frac{1}{(\pi a^2)^{1/4}}\,e^{-x^2/(2a^2)} .$$

Let us find the probability of locating the particle in $|x| \le a$ in this state. Substituting $u = x/a$,

$$P(|x| \le a) = \frac{1}{a\sqrt{\pi}}\int_{-a}^{a} e^{-x^2/a^2}dx = \frac{1}{\sqrt{\pi}}\int_{-1}^{1}e^{-u^2}du = \frac{2}{\sqrt{\pi}}\int_0^1 e^{-u^2}du = \operatorname{erf}(1) \approx 0.8427 ,$$

where we used that the integrand is even. Here $\operatorname{erf}$ is the error function $\operatorname{erf}(z) = (2/\sqrt{\pi})\int_0^z e^{-u^2}du$. So the probability is about 84%.

5. Expectation values, the momentum operator, and Ehrenfest’s theorem

5.1. Defining expectation values

Once the probability density is known, the mean position can be written down exactly as probability theory prescribes:

$$\langle x \rangle = \int_{-\infty}^{\infty} x\,|\psi(x,t)|^2\,dx = \int_{-\infty}^{\infty}\overline{\psi}\,x\,\psi\,dx = \langle \psi \mid \hat{x} \mid \psi\rangle .$$

Momentum is the problem. Momentum is not a function of position, so an expression such as $\int p\,|\psi|^2 dx$ is meaningless. How, then, should it be defined? The clue is the correspondence with classical mechanics. Classically $p = m\dot{x}$, so in quantum mechanics we would like $\langle p \rangle$ to be $m\,d\langle x\rangle/dt$. In fact, carrying this out as a computation rather than as a definition produces the form of the momentum operator.

Proposition 5.1（Deriving the momentum operator）

Under the same hypotheses as in Corollary 4.3 ($V$ real-valued, $\psi$ a normalised solution, $\psi \to 0$ and $\partial_x\psi \to 0$ as $|x|\to\infty$, and in addition $x\,j(x,t) \to 0$),

$$m\frac{d\langle x\rangle}{dt} = \int_{-\infty}^{\infty} \overline{\psi}\left(-i\hbar\frac{\partial}{\partial x}\right)\psi\, dx .$$

Proof（Proposition 5.1）

Interchange derivative and integral and use Theorem 4.2:

$$\frac{d\langle x\rangle}{dt} = \int_{-\infty}^{\infty} x\,\frac{\partial \rho}{\partial t}\,dx = -\int_{-\infty}^{\infty} x\,\frac{\partial j}{\partial x}\,dx .$$

Integrating by parts,

$$-\int_{-\infty}^{\infty} x\,\frac{\partial j}{\partial x}\,dx = -\Bigl[\,x\,j\,\Bigr]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} j\,dx = \int_{-\infty}^{\infty} j\,dx$$

(the first term vanishes by the hypothesis $x\,j \to 0$). Next substitute the definition of $j$ and integrate the second term by parts:

$$\int_{-\infty}^{\infty} \frac{\partial \overline{\psi}}{\partial x}\,\psi\, dx = \Bigl[\,\overline{\psi}\psi\,\Bigr]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}\overline{\psi}\,\frac{\partial \psi}{\partial x}\,dx = -\int_{-\infty}^{\infty}\overline{\psi}\,\frac{\partial\psi}{\partial x}\,dx$$

(the boundary term vanishes because $\psi \to 0$). Hence

$$\int_{-\infty}^{\infty} j\,dx = \frac{\hbar}{2mi}\int \left(\overline{\psi}\psi_x - \overline{\psi}_x\psi\right)dx = \frac{\hbar}{2mi}\cdot 2\int \overline{\psi}\,\psi_x\,dx = \frac{1}{m}\int \overline{\psi}\left(\frac{\hbar}{i}\frac{\partial}{\partial x}\right)\psi\,dx .$$

Since $\hbar/i = -i\hbar$, multiplying both sides by $m$ gives the claim.

∎

This result tells us to introduce the operator $\hat{p} := -i\hbar\,\partial_x$ and to define $\langle p\rangle := \langle \psi \mid \hat{p}\mid\psi\rangle$. It agrees precisely with the form read off from plane waves in §3.1. In general we set the following.

Definition 5.2（The expectation value of an observable）

For a normalised wave function $\psi$ and a linear operator $\hat{A}$ acting on $\psi$, the expectation value of $\hat{A}$ is

$$\langle \hat{A}\rangle_\psi := \langle \psi \mid \hat{A} \mid \psi\rangle = \int_{-\infty}^{\infty}\overline{\psi(x)}\,\bigl(\hat{A}\psi\bigr)(x)\,dx .$$

In particular, taking $\hat{x}$ (multiplication by $x$), $\hat{p} = -i\hbar\,\partial_x$ and $\hat{H} = \hat{p}^2/(2m) + V(\hat{x})$ determines the expectation values of position, momentum and energy.

5.2. Why Hermitian operators?

The mean of measured values must be real. This requirement severely restricts the operators that may correspond to observables.

Proposition 5.3（Expectation values of Hermitian operators are real）

Suppose a linear operator $\hat{A}$ satisfies $\langle \varphi \mid \hat{A}\psi\rangle = \langle \hat{A}\varphi \mid \psi\rangle$ for all $\varphi, \psi$ in the family of wave functions under consideration (in which case $\hat{A}$ is called Hermitian). Then $\langle \hat{A}\rangle_\psi \in \mathbb{R}$ for every normalised $\psi$. Moreover $\hat{x}$ and $\hat{p} = -i\hbar\partial_x$ are Hermitian on the family of smooth functions tending to $0$ at infinity.

Proof（Proposition 5.3）

First the initial claim. By the property $\langle \varphi\mid\psi\rangle = \overline{\langle \psi\mid\varphi\rangle}$ of the inner product (§2) together with Hermiticity,

$$\langle \hat{A}\rangle_\psi = \langle \psi \mid \hat{A}\psi \rangle = \langle \hat{A}\psi \mid \psi\rangle = \overline{\langle \psi \mid \hat{A}\psi\rangle} = \overline{\langle \hat{A}\rangle_\psi} ,$$

and a complex number equal to its own conjugate is real.

Next $\hat{x}$. Since $x$ is real,

$$\langle \varphi \mid \hat{x}\psi\rangle = \int \overline{\varphi}\,x\psi\,dx = \int \overline{x\varphi}\,\psi\,dx = \langle \hat{x}\varphi\mid\psi\rangle .$$

Finally $\hat{p}$. Integration by parts gives

$$\begin{aligned} \langle \varphi \mid \hat{p}\psi\rangle &= \int_{-\infty}^{\infty} \overline{\varphi}\,\left(-i\hbar\,\psi_x\right) dx \\ &= -i\hbar\Bigl[\,\overline{\varphi}\psi\,\Bigr]_{-\infty}^{\infty} + i\hbar\int_{-\infty}^{\infty}\overline{\varphi}_x\,\psi\,dx \\ &= \int_{-\infty}^{\infty}\overline{\left(-i\hbar\,\varphi_x\right)}\,\psi\,dx = \langle \hat{p}\varphi\mid\psi\rangle . \end{aligned}$$

The boundary term vanished because $\varphi, \psi \to 0$, and in the penultimate equality we used $\overline{-i\hbar\varphi_x} = +i\hbar\,\overline{\varphi}_x$. Note how the $-i$ in the definition of $\hat{p}$ does the work here: had we defined $\hat{p}$ as $\partial_x$ (without the $i$), the integration by parts would flip the sign, giving $\langle \varphi\mid\partial_x\psi\rangle = -\langle \partial_x\varphi\mid\psi\rangle$, and the operator would not be Hermitian.

∎

Hermiticity of operators, the reality of eigenvalues and the orthogonality of eigenfunctions are treated in earnest in Operators and observables (see Theorem 3.5［Operators and Observables］). The finite-dimensional version of the mathematical background is the spectral theorem.

5.3. Time evolution of expectation values

Lemma 5.4（The time derivative of an expectation value）

Let $\psi$ be a normalised solution of Definition 3.1, let $\hat{A}$ be a Hermitian operator (possibly depending explicitly on time), and suppose $\hat{H}$ is Hermitian as well. Assuming the necessary interchanges of derivative and integral, and the vanishing of the boundary terms in the integrations by parts,

$$\frac{d}{dt}\langle \hat{A}\rangle_\psi = \frac{i}{\hbar}\bigl\langle\, [\hat{H}, \hat{A}\,]\,\bigr\rangle_\psi + \Bigl\langle \frac{\partial \hat{A}}{\partial t}\Bigr\rangle_\psi , \qquad [\hat{H},\hat{A}] := \hat{H}\hat{A} - \hat{A}\hat{H} .$$

Proof（Lemma 5.4）

By the product rule,

$$\frac{d}{dt}\langle \psi \mid \hat{A}\mid \psi\rangle = \Bigl\langle \frac{\partial \psi}{\partial t}\;\Bigm|\; \hat{A}\,\psi\Bigr\rangle + \Bigl\langle \psi \Bigm| \frac{\partial \hat{A}}{\partial t}\,\psi\Bigr\rangle + \Bigl\langle \psi \Bigm| \hat{A}\,\frac{\partial \psi}{\partial t}\Bigr\rangle .$$

By Definition 3.1, $\partial_t\psi = -(i/\hbar)\hat{H}\psi$. Substituting this straight into the third term,

$$\Bigl\langle \psi \Bigm| \hat{A}\left(-\frac{i}{\hbar}\hat{H}\psi\right)\Bigr\rangle = -\frac{i}{\hbar}\langle \psi \mid \hat{A}\hat{H}\mid\psi\rangle .$$

For the first term, recall that the inner product is antilinear in its first argument. The coefficient $-i/\hbar$ is conjugated to $+i/\hbar$, so

$$\Bigl\langle -\frac{i}{\hbar}\hat{H}\psi \;\Bigm|\; \hat{A}\psi\Bigr\rangle = +\frac{i}{\hbar}\bigl\langle \hat{H}\psi \mid \hat{A}\psi\bigr\rangle = +\frac{i}{\hbar}\bigl\langle \psi \mid \hat{H}\hat{A}\psi\bigr\rangle ,$$

where the last equality uses the Hermiticity of $\hat{H}$ (in the sense of Proposition 5.3). Adding these up,

$$\frac{d}{dt}\langle \hat{A}\rangle_\psi = \frac{i}{\hbar}\langle \psi\mid \hat{H}\hat{A} - \hat{A}\hat{H}\mid\psi\rangle + \Bigl\langle \frac{\partial\hat{A}}{\partial t}\Bigr\rangle_\psi .$$∎

This lemma contains the general principle that an observable is conserved exactly when it commutes with the Hamiltonian: if $\hat{A}$ has no explicit time dependence and $[\hat{H},\hat{A}] = 0$, then $\langle \hat{A}\rangle$ is constant in time. Compare this with the same role played in classical mechanics by the Poisson bracket $\{H, A\}$ (see Theorem 6.1［正準変換とポアソン括弧］ in Canonical transformations and Poisson brackets, and Symmetries and conservation laws). The commutator is the quantum version of the Poisson bracket.

Theorem 5.5（Ehrenfest's theorem）

Let $V$ be real-valued and differentiable, let $\psi$ be a normalised solution of Definition 3.1, and assume the technical hypotheses of Lemma 5.4. Then

$$\frac{d\langle \hat{x}\rangle}{dt} = \frac{\langle \hat{p}\rangle}{m}, \qquad \frac{d\langle \hat{p}\rangle}{dt} = -\bigl\langle V'(\hat{x})\bigr\rangle ,$$

where $V'(\hat{x})$ denotes the operator of multiplication by the function $V'(x) = dV/dx$.

Proof（Theorem 5.5）

As preparation we compute a commutator. For any smooth $\psi$,

$$[\hat{x},\hat{p}]\psi = x(-i\hbar\psi_x) - (-i\hbar)\frac{\partial}{\partial x}(x\psi) = -i\hbar x\psi_x + i\hbar(\psi + x\psi_x) = i\hbar\,\psi ,$$

which gives the canonical commutation relation（Theorem 4.3）［Operators and Observables］ $[\hat{x},\hat{p}] = i\hbar$ ($i\hbar$ times the identity operator). Consequently $[\hat{p},\hat{x}] = -i\hbar$.

First identity. Since $\hat{A} = \hat{x}$ has no explicit time dependence, Lemma 5.4 gives $d\langle\hat x\rangle/dt = (i/\hbar)\langle[\hat{H},\hat{x}]\rangle$. As $V(\hat x)$ commutes with $\hat x$ (both are multiplication by a function of $x$, so the order is immaterial),

$$[\hat{H},\hat{x}] = \frac{1}{2m}[\hat{p}^2, \hat{x}] .$$

Applying the commutator identity $[\hat{B}\hat{C}, \hat{D}] = \hat{B}[\hat{C},\hat{D}] + [\hat{B},\hat{D}]\hat{C}$ (verified by expanding both sides) with $\hat B = \hat C = \hat p$ and $\hat D = \hat x$,

$$[\hat{p}^2,\hat{x}] = \hat{p}[\hat{p},\hat{x}] + [\hat{p},\hat{x}]\hat{p} = -i\hbar\hat{p} - i\hbar\hat{p} = -2i\hbar\,\hat{p} .$$

Hence

$$\frac{d\langle \hat x\rangle}{dt} = \frac{i}{\hbar}\cdot\frac{1}{2m}\cdot(-2i\hbar)\langle \hat{p}\rangle = \frac{-2i^2}{2m}\langle \hat p\rangle = \frac{\langle \hat{p}\rangle}{m}.$$

Second identity. Since $\hat{A} = \hat{p}$ likewise has no explicit time dependence, $d\langle \hat p\rangle/dt = (i/\hbar)\langle [\hat{H},\hat{p}]\rangle$. As $\hat{p}^2$ commutes with $\hat p$, we have $[\hat{H},\hat{p}] = [V(\hat x),\hat p]$, and for any smooth $\psi$,

$$[V,\hat{p}]\psi = V\cdot(-i\hbar\psi_x) - (-i\hbar)\frac{\partial}{\partial x}(V\psi) = -i\hbar V\psi_x + i\hbar\left(V'\psi + V\psi_x\right) = i\hbar V'(x)\,\psi$$

(using the product rule). Therefore

$$\frac{d\langle\hat p\rangle}{dt} = \frac{i}{\hbar}\cdot i\hbar\,\langle V'(\hat x)\rangle = -\langle V'(\hat x)\rangle .$$∎

Remark 5.6

Combining the two identities of Theorem 5.5 gives

$$m\frac{d^2\langle \hat x\rangle}{dt^2} = -\langle V'(\hat x)\rangle ,$$

which looks just like Newton’s equation of motion $m\ddot{x} = -V'(x)$. This is not, however, a recovery of classical mechanics. The right-hand side is $-\langle V'(\hat x)\rangle$, the expectation value of the force, and not $-V'(\langle \hat x\rangle)$, the force at the expectation value. The two agree only when $V'$ is affine, that is, when $V$ is at most quadratic (free particle, uniform force, harmonic oscillator).

In general, expanding $V'$ in a Taylor series about $\langle \hat x\rangle$ and setting $\xi := x - \langle \hat x\rangle$, so that $\langle \xi\rangle = 0$,

$$\langle V'(\hat x)\rangle = V'(\langle \hat x\rangle) + \tfrac{1}{2}V'''(\langle \hat x\rangle)\,\langle \xi^2\rangle + \cdots = V'(\langle \hat x\rangle) + \tfrac{1}{2}V'''(\langle \hat x\rangle)\,(\Delta x)^2 + \cdots$$

where $\Delta x := \sqrt{\langle \hat x^2\rangle - \langle \hat x\rangle^2}$ is the standard deviation of position. Hence the condition under which the classical approximation is justified is

$$\left|V'''(\langle \hat x\rangle)\right|(\Delta x)^2 \ll \left|V'(\langle \hat x\rangle)\right|,$$

that is, the gradient of the potential must be nearly constant across the spread of the wave packet. Tunnelling and interference are the typical situations in which this condition fails, and there tracking the motion of expectation values does not capture the phenomenon.

Example 5.7（Variances and uncertainty for a Gaussian wave packet）

For $\psi(x) = (\pi a^2)^{-1/4}e^{-x^2/(2a^2)}$ from Example 4.5, let us compute the variances of position and momentum all the way. Write $N^2 = 1/(a\sqrt{\pi})$.

Position. Since $x|\psi|^2$ is odd, $\langle \hat x\rangle = 0$. Next, putting $\alpha = 1/a^2$ into the Gaussian integral $\int_{-\infty}^{\infty}x^2 e^{-\alpha x^2}dx = \tfrac12\sqrt{\pi/\alpha^3}$ gives $\int x^2 e^{-x^2/a^2}dx = \tfrac12 a^3\sqrt{\pi}$, so

$$\langle \hat x^2\rangle = \frac{1}{a\sqrt{\pi}}\cdot\frac{a^3\sqrt{\pi}}{2} = \frac{a^2}{2}, \qquad \Delta x = \frac{a}{\sqrt{2}} .$$

Momentum. Since $\psi$ is real-valued,

$$\langle \hat p\rangle = \int \psi\,(-i\hbar\,\psi_x)\,dx = -i\hbar\int \psi\psi_x\,dx = -i\hbar\left[\frac{\psi^2}{2}\right]_{-\infty}^{\infty} = 0 .$$

For $\langle \hat p^2\rangle$, integrate by parts:

$$\langle \hat p^2\rangle = \int \psi\left(-\hbar^2\psi_{xx}\right)dx = -\hbar^2\left(\Bigl[\psi\psi_x\Bigr]_{-\infty}^{\infty} - \int \psi_x^2\,dx\right) = \hbar^2\int_{-\infty}^{\infty}\psi_x^2\,dx .$$

Since $\psi_x = -(x/a^2)\psi$, we get $\int\psi_x^2 dx = a^{-4}\int x^2\psi^2 dx = a^{-4}\langle \hat x^2\rangle = a^{-4}\cdot a^2/2 = 1/(2a^2)$. Therefore

$$\langle \hat p^2\rangle = \frac{\hbar^2}{2a^2}, \qquad \Delta p = \frac{\hbar}{\sqrt{2}\,a} .$$

The product.

$$\Delta x\,\Delta p = \frac{a}{\sqrt 2}\cdot\frac{\hbar}{\sqrt2 a} = \frac{\hbar}{2} .$$

The parameter $a$ has cancelled. Narrowing the wave packet makes $\Delta x$ smaller, but $\Delta p$ grows correspondingly and the product stays fixed at $\hbar/2$. This is a state realising equality in the uncertainty relation $\Delta x\,\Delta p \ge \hbar/2$, which is why the Gaussian packet is called a “minimum-uncertainty state”. The uncertainty relation itself is proved in Operators and observables (see Corollary 5.4［Operators and Observables］; that the Gaussian packet attains equality is Example 5.5［Operators and Observables］).

6. The time-independent Schrödinger equation and stationary states

6.1. Separation of variables

When the potential $V$ does not depend on time, neither does $\hat{H}$. In that case the standard technique for partial differential equations, separation of variables, is available. Assuming $\psi(x,t) = \varphi(x)T(t)$ (with neither factor identically $0$) and substituting into Definition 3.1,

$$i\hbar\,\varphi(x)\,T'(t) = T(t)\,\bigl(\hat{H}\varphi\bigr)(x) .$$

Dividing both sides by $\varphi(x)T(t)$ at points where $\varphi(x)T(t) \ne 0$,

$$\underbrace{i\hbar\frac{T'(t)}{T(t)}}_{\text{a function of } t \text{ alone}} = \underbrace{\frac{(\hat{H}\varphi)(x)}{\varphi(x)}}_{\text{a function of } x \text{ alone}} .$$

The left-hand side does not depend on $x$ and the right-hand side does not depend on $t$. Since they are equal, both must be constant. The constant has the dimensions of energy, so we call it $E$. The equation then splits into two:

$$i\hbar\,T'(t) = E\,T(t) \quad\Longrightarrow\quad T(t) = T(0)\,e^{-iEt/\hbar},$$ $$\hat{H}\varphi = E\varphi, \qquad\text{that is,}\qquad -\frac{\hbar^2}{2m}\varphi''(x) + V(x)\varphi(x) = E\,\varphi(x) .$$

Definition 6.1（The time-independent Schrödinger equation and stationary states）

For a time-independent potential $V$, the eigenvalue problem

$$\hat{H}\varphi = E\varphi, \qquad \hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$$

is called the time-independent Schrödinger equation. When $\varphi \in L^2(\mathbb{R})$ with $\varphi \neq 0$ satisfies it, $\varphi$ is called an eigenfunction and $E$ an eigenvalue (an energy level), and the corresponding

$$\psi(x,t) = \varphi(x)\,e^{-iEt/\hbar}$$

is called a stationary state.

The following theorem explains the name “stationary”.

Theorem 6.2（Properties of stationary states）

Let $\varphi$ be a normalised eigenfunction of Definition 6.1 with eigenvalue $E$, and put $\psi(x,t) = \varphi(x)e^{-iEt/\hbar}$. Then:

1. $\psi$ is normalised at every time.
2. The probability density $\rho(x,t) = |\psi(x,t)|^2 = |\varphi(x)|^2$ does not depend on time.
3. For every operator $\hat{A}$ without explicit time dependence, the expectation value $\langle \hat{A}\rangle_{\psi(t)}$ does not depend on time.
4. The energy has no spread: $\langle \hat{H}\rangle = E$ and $\langle \hat{H}^2\rangle = E^2$, hence $\Delta E = 0$.

Proof（Theorem 6.2）

The eigenvalue $E$ is real: since $\hat H$ is Hermitian, Proposition 5.3 shows that $\langle \hat H\rangle_\varphi = \langle \varphi\mid E\varphi\rangle = E\|\varphi\|^2 = E$ is real. Hence $|e^{-iEt/\hbar}| = 1$.

(2) $|\psi(x,t)|^2 = |\varphi(x)|^2\,|e^{-iEt/\hbar}|^2 = |\varphi(x)|^2$, which contains no $t$.

(1) By (2), $\int|\psi(x,t)|^2 dx = \int|\varphi|^2dx = 1$ for every $t$.

(3) The phase factor is conjugated on the bra side, so

$$\langle \psi(t)\mid \hat{A}\mid\psi(t)\rangle = \overline{e^{-iEt/\hbar}}\,e^{-iEt/\hbar}\,\langle \varphi\mid\hat{A}\mid\varphi\rangle = \langle \varphi\mid \hat A\mid\varphi\rangle ,$$

which contains no $t$ (the same computation as in Remark 2.2).

(4) From $\hat{H}\varphi = E\varphi$ we get $\langle \hat H\rangle = \langle \varphi\mid E\varphi\rangle = E$. Also $\hat{H}^2\varphi = \hat{H}(E\varphi) = E\hat{H}\varphi = E^2\varphi$, so $\langle \hat H^2\rangle = E^2$. Hence $(\Delta E)^2 = E^2 - E^2 = 0$.

∎

Caution

A stationary state is not a state at rest. The function $\psi$ itself oscillates furiously as $e^{-iEt/\hbar}$ (for $E = 1\,\mathrm{eV}$ the period is about $4\times 10^{-15}$ seconds). What is time-independent is only $|\psi|^2$ and the expectation values. And, as we see next, the moment two stationary states are superposed this oscillation becomes observable.

6.2. The general solution is a superposition of stationary states

Theorem 6.3（The general solution by eigenfunction expansion）

Suppose $\hat H$ does not depend on time and that there is a family of eigenfunctions $\{\varphi_n\}_{n\in\mathbb{N}}$ forming an orthonormal system, with $\hat{H}\varphi_n = E_n\varphi_n$ and $\langle \varphi_m\mid\varphi_n\rangle = \delta_{mn}$, which is complete in the space of wave functions under consideration (every $\psi \in L^2$ is a limit of linear combinations of the $\varphi_n$). Then the solution of Definition 3.1 with initial condition $\psi(x,0) = \psi_0(x)$ is

$$\psi(x,t) = \sum_{n} c_n\,\varphi_n(x)\,e^{-iE_n t/\hbar}, \qquad c_n = \langle \varphi_n\mid\psi_0\rangle .$$

Moreover, if $\psi_0$ is normalised then $\sum_n |c_n|^2 = 1$.

Proof（Theorem 6.3）

It is a solution. Each term $\varphi_n(x)e^{-iE_nt/\hbar}$ is a solution of Definition 3.1 by the derivation of Definition 6.1. Indeed

$$i\hbar\frac{\partial}{\partial t}\left(\varphi_n e^{-iE_nt/\hbar}\right) = i\hbar\left(-\frac{iE_n}{\hbar}\right)\varphi_n e^{-iE_nt/\hbar} = E_n\varphi_n e^{-iE_nt/\hbar} = \hat{H}\left(\varphi_n e^{-iE_nt/\hbar}\right) .$$

By Proposition 3.3 (and the assumption that term-by-term differentiation is permitted), any linear combination is a solution too.

It satisfies the initial condition. At $t = 0$ we have $\psi(x,0) = \sum_n c_n\varphi_n(x)$. By completeness we may expand $\psi_0 = \sum_n a_n\varphi_n$, and applying $\langle \varphi_m\mid\cdot\rangle$ to both sides gives, by orthonormality,

$$\langle \varphi_m\mid\psi_0\rangle = \sum_n a_n\langle\varphi_m\mid\varphi_n\rangle = \sum_n a_n\delta_{mn} = a_m ,$$

so $a_m = c_m$.

The sum of the squared coefficients. Using orthonormality,

$$1 = \langle \psi_0\mid\psi_0\rangle = \Bigl\langle \sum_m c_m\varphi_m \Bigm| \sum_n c_n\varphi_n\Bigr\rangle = \sum_{m,n}\overline{c_m}c_n\,\delta_{mn} = \sum_n |c_n|^2$$

(the conjugate falls on $c_m$ because the inner product is antilinear in the first argument).

∎

The completeness hypothesis is not a light one. In finite dimensions the spectral theorem (Theorem 4.2［スペクトル定理］) guarantees the existence of an orthonormal eigenbasis for a Hermitian matrix; in infinite dimensions one needs the spectral theorem for self-adjoint operators, and when continuous spectrum is present the sum is moreover replaced by an integral (the free particle is such a case; see the Appendix).

The number $|c_n|^2$ is interpreted as “the probability of obtaining the value $E_n$ when the energy is measured”. That $\sum_n|c_n|^2 = 1$ corresponds to these probabilities summing to $1$, and suggests that the Born rule for position holds in the same form for energy. This generalisation is formulated in Operators and observables.

6.3. Example: the infinite square well

Example 6.4（Energy levels and probabilities in an infinite square well）

Let $L > 0$ and take

$$V(x) = \begin{cases} 0 & (0 < x < L) \\ +\infty & (\text{otherwise}).\end{cases}$$

In the region where the potential is infinite we must have $\varphi = 0$ (otherwise $\hat H\varphi$ would diverge), and continuity of $\varphi$ imposes the boundary conditions $\varphi(0) = \varphi(L) = 0$. Inside the well $V = 0$, so

$$-\frac{\hbar^2}{2m}\varphi'' = E\varphi \quad\Longleftrightarrow\quad \varphi'' = -k^2\varphi, \qquad k := \frac{\sqrt{2mE}}{\hbar} .$$

For $E > 0$ the general solution is $\varphi(x) = A\sin kx + B\cos kx$. From $\varphi(0) = 0$ we get $B = 0$. From $\varphi(L) = A\sin kL = 0$ with $A \ne 0$ (otherwise $\varphi \equiv 0$, which is not a state) we get $\sin kL = 0$, that is,

$$k L = n\pi \quad (n = 1, 2, 3, \ldots) .$$

We exclude $n = 0$ because it gives $\varphi\equiv 0$, and negative $n$ merely changes the overall sign, giving the same state (Remark 2.2). The energies are

$$E_n = \frac{\hbar^2 k^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2} .$$

Normalisation gives $\int_0^L A^2\sin^2(n\pi x/L)dx = A^2 L/2 = 1$, so $A = \sqrt{2/L}$ and

$$\varphi_n(x) = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L} .$$

Numbers. Confining an electron ($m = 9.109\times10^{-31}\,\mathrm{kg}$) to $L = 0.5\,\mathrm{nm}$ gives

$$E_1 = \frac{\pi^2 (1.055\times10^{-34})^2}{2(9.109\times10^{-31})(0.5\times10^{-9})^2} \approx 2.41\times10^{-19}\,\mathrm{J} \approx 1.50\,\mathrm{eV} .$$

Then $E_2 - E_1 = 3E_1 \approx 4.51\,\mathrm{eV}$, and the photon wavelength corresponding to this gap is $\lambda = hc/\Delta E \approx (1240\,\mathrm{eV\cdot nm})/(4.51\,\mathrm{eV}) \approx 275\,\mathrm{nm}$, in the ultraviolet. Here is the most basic numerical intuition of quantum mechanics: confinement on atomic and molecular scales is tied to visible and ultraviolet light.

A probability. In the ground state $n = 1$, let us compute the probability of finding the particle in the middle third, $[L/3, 2L/3]$. Using $\sin^2\theta = (1-\cos 2\theta)/2$,

$$\begin{aligned} P &= \int_{L/3}^{2L/3}\frac{2}{L}\sin^2\frac{\pi x}{L}dx = \frac{1}{L}\int_{L/3}^{2L/3}\left(1 - \cos\frac{2\pi x}{L}\right)dx \\ &= \frac{1}{L}\left[x - \frac{L}{2\pi}\sin\frac{2\pi x}{L}\right]_{L/3}^{2L/3} = \frac{1}{L}\left[\frac{L}{3} - \frac{L}{2\pi}\left(\sin\frac{4\pi}{3} - \sin\frac{2\pi}{3}\right)\right]. \end{aligned}$$

Since $\sin(4\pi/3) = -\sqrt3/2$ and $\sin(2\pi/3) = +\sqrt3/2$, the bracket equals $-\sqrt3$. Hence

$$P = \frac{1}{3} + \frac{\sqrt3}{2\pi} \approx 0.333 + 0.276 = 0.609 .$$

A classical particle bouncing back and forth in the well at constant speed would be in the middle third with probability $1/3 \approx 0.333$. The quantum ground state gives about $0.609$: it is strongly concentrated towards the centre.

Example 6.5（Superposing two stationary states and the oscillation of expectation values）

In the same well, take the initial state

$$\psi(x,0) = \frac{1}{\sqrt2}\bigl(\varphi_1(x) + \varphi_2(x)\bigr) .$$

Since $\varphi_1$ and $\varphi_2$ are orthonormal, $\|\psi(0)\|^2 = (1/2)(1+1) = 1$ and the state is normalised. By Theorem 6.3,

$$\psi(x,t) = \frac{1}{\sqrt2}\left(\varphi_1(x)e^{-iE_1t/\hbar} + \varphi_2(x)e^{-iE_2t/\hbar}\right).$$

Let us compute the expectation value of position. Writing $\omega := (E_2 - E_1)/\hbar$,

$$\langle \hat x\rangle(t) = \frac{1}{2}\Bigl(\langle\varphi_1|\hat x|\varphi_1\rangle + \langle\varphi_2|\hat x|\varphi_2\rangle + e^{-i\omega t}\langle\varphi_1|\hat x|\varphi_2\rangle + e^{i\omega t}\langle\varphi_2|\hat x|\varphi_1\rangle\Bigr).$$

We need the relevant integrals. First, integrating $\int_0^L x\cos(k\pi x/L)dx$ by parts and using $\sin k\pi = 0$, $\cos k\pi = (-1)^k$ for integer $k$,

$$\int_0^L x\cos\frac{k\pi x}{L}dx = \left[\frac{Lx}{k\pi}\sin\frac{k\pi x}{L}\right]_0^L + \left[\frac{L^2}{k^2\pi^2}\cos\frac{k\pi x}{L}\right]_0^L = \frac{\bigl((-1)^k - 1\bigr)L^2}{k^2\pi^2} .$$

The diagonal elements are $\langle\varphi_n|\hat x|\varphi_n\rangle = (1/L)\int_0^L x(1 - \cos(2n\pi x/L))dx = (1/L)(L^2/2 - 0) = L/2$ (the integral above vanishes because $k = 2n$ is even). For the off-diagonal element, the product formula $\sin A\sin B = \tfrac12[\cos(A-B) - \cos(A+B)]$ gives

$$\begin{aligned} \langle \varphi_1|\hat x|\varphi_2\rangle &= \frac{2}{L}\int_0^L x\sin\frac{\pi x}{L}\sin\frac{2\pi x}{L}dx = \frac{1}{L}\int_0^L x\left(\cos\frac{\pi x}{L} - \cos\frac{3\pi x}{L}\right)dx \\ &= \frac{1}{L}\left(\frac{-2L^2}{\pi^2} - \frac{-2L^2}{9\pi^2}\right) = -\frac{16L}{9\pi^2} \approx -0.180\,L . \end{aligned}$$

This is real, and since $\hat x$ is Hermitian and the $\varphi_n$ are real-valued, $\langle\varphi_2|\hat x|\varphi_1\rangle$ has the same value. Using $e^{-i\omega t} + e^{i\omega t} = 2\cos\omega t$,

$$\langle \hat x\rangle(t) = \frac{L}{2} - \frac{16L}{9\pi^2}\cos\omega t, \qquad \omega = \frac{E_2 - E_1}{\hbar} = \frac{3\pi^2\hbar}{2mL^2} .$$

The expectation value oscillates back and forth in the well with amplitude $0.180L$. By Theorem 6.2 a single stationary state has motionless expectation values, but a superposition of them moves. The frequency is fixed by the energy difference alone, which is exactly Bohr’s frequency condition $\Delta E = \hbar\omega$. With the numbers used above (an electron, $L = 0.5\,\mathrm{nm}$) we have $\Delta E \approx 4.51\,\mathrm{eV}$, so the period is $T = 2\pi/\omega = h/\Delta E \approx 9.2\times10^{-16}\,\mathrm{s}$, about $0.92$ femtoseconds.

Remark 6.6

If in Example 6.5 we superpose $\varphi_3$ instead of $\varphi_2$, the expectation value does not move. Looking at the symmetry about the centre $x = L/2$ of the well, we have $\varphi_n(L-x) = (-1)^{n+1}\varphi_n(x)$, so $\varphi_1$ and $\varphi_3$ are both symmetric about the centre while $x - L/2$ is antisymmetric. Hence $\int_0^L \varphi_1(x)\bigl(x - L/2\bigr)\varphi_3(x)dx = 0$ and $\langle\varphi_1|\hat x|\varphi_3\rangle = (L/2)\langle\varphi_1|\varphi_3\rangle = 0$. Rules of this kind, in which symmetry makes particular matrix elements vanish, are called selection rules, and they determine between which levels an atom can emit light.

7. Exercises

Exercise 7.1Standard

Let $\lambda > 0$ be a constant and consider $\psi(x) = A\,e^{-|x|/\lambda}$ with $A > 0$.

1. Normalise $\psi$ and find $A$.
2. Find $\langle \hat x\rangle$, $\langle \hat x^2\rangle$ and $\Delta x$.
3. Find the probability of locating the particle in $|x| \le \lambda$.

Solution

1. Since $|\psi|^2 = A^2 e^{-2|x|/\lambda}$ is even,

$$\int_{-\infty}^{\infty}A^2e^{-2|x|/\lambda}dx = 2A^2\int_0^{\infty}e^{-2x/\lambda}dx = 2A^2\cdot\frac{\lambda}{2} = A^2\lambda .$$

Setting this equal to $1$ gives $A = 1/\sqrt{\lambda}$, that is, $\psi(x) = \lambda^{-1/2}e^{-|x|/\lambda}$.

2. The function $x|\psi(x)|^2$ is odd (an odd function $x$ times the even function $|\psi|^2$) and the integral converges absolutely, so $\langle \hat x\rangle = 0$. Next, substituting $\beta = 2/\lambda$ into the gamma-function formula $\int_0^\infty x^2 e^{-\beta x}dx = 2/\beta^3$ gives $\int_0^\infty x^2e^{-2x/\lambda}dx = 2\lambda^3/8 = \lambda^3/4$, so

$$\langle \hat x^2\rangle = \frac{1}{\lambda}\cdot 2\int_0^\infty x^2 e^{-2x/\lambda}dx = \frac{2}{\lambda}\cdot\frac{\lambda^3}{4} = \frac{\lambda^2}{2} .$$

Hence $\Delta x = \sqrt{\langle \hat x^2\rangle - \langle \hat x\rangle^2} = \lambda/\sqrt2$.

3.

$$P(|x|\le\lambda) = \frac{2}{\lambda}\int_0^{\lambda}e^{-2x/\lambda}dx = \frac{2}{\lambda}\cdot\frac{\lambda}{2}\left(1 - e^{-2}\right) = 1 - e^{-2} \approx 0.865 .$$

Exercise 7.2Standard

For the probability current density $j = (\hbar/m)\,\mathrm{Im}(\overline{\psi}\,\partial_x\psi)$ defined in Theorem 4.2, show the following.

1. If $\psi(x,t) = e^{i\theta}f(x,t)$ with $\theta$ a real constant and $f$ real-valued, then $j \equiv 0$. In particular the stationary states of Example 6.4 carry no flow of probability.
2. For a plane wave $\psi(x,t) = A e^{i(kx - \omega t)}$ with $A$ a complex constant, find $\rho$ and $j$ and compare with the classical picture.

Solution

1. Since $\overline{\psi} = e^{-i\theta}f$ and $\partial_x\psi = e^{i\theta}\partial_x f$,

$$\overline{\psi}\,\partial_x\psi = e^{-i\theta}e^{i\theta}f\,\partial_x f = f\,\partial_x f ,$$

which is real because $f$ is real-valued. The imaginary part of a real number is $0$, so $j = 0$. The stationary states of Example 6.4 are $\psi = \varphi_n(x)e^{-iE_nt/\hbar}$ with $\varphi_n$ real-valued; the phase factor $e^{-iE_nt/\hbar}$ depends on $t$, so this is not literally of the form above. But at each fixed time the phase is constant, so the same computation gives $j = 0$. Physically, a stationary state confined in the well is a standing wave in which the right-moving and left-moving waves balance, so there is no net flow.

2. We have $\rho = |\psi|^2 = |A|^2$, which is uniform. Since $\partial_x\psi = ik\psi$,

$$\overline{\psi}\,\partial_x\psi = ik|\psi|^2 = ik|A|^2 \quad\Longrightarrow\quad \mathrm{Im}\bigl(\overline{\psi}\partial_x\psi\bigr) = k|A|^2 .$$

Hence

$$j = \frac{\hbar k}{m}|A|^2 = \frac{p}{m}|A|^2 = v\,\rho$$

where $v = p/m$ is the particle’s velocity. This has exactly the form “flux = density × velocity” of a classical fluid, and it supports reading $j$ as “the probability passing through a unit cross-section per unit time”. Note that a plane wave has divergent $\int|\psi|^2dx$, so strictly speaking it is not a state in the sense of Definition 2.1 (see the Appendix).

Exercise 7.3Hard

Consider Theorem 5.5.

1. For the harmonic oscillator $V(x) = \tfrac12 m\omega^2x^2$, show that $\langle \hat x\rangle(t)$ satisfies the classical equation of motion of a harmonic oscillator exactly, and solve it in terms of the initial values $\langle\hat x\rangle_0$ and $\langle\hat p\rangle_0$.
2. Assuming $V'$ is continuous, show that a necessary and sufficient condition for ”$\langle V'(\hat x)\rangle = V'(\langle \hat x\rangle)$ for every normalised state” is that $V'$ be affine (that is, that $V$ be at most quadratic).

Solution

1. Since $V'(x) = m\omega^2 x$ is linear in $x$, linearity of the expectation value gives

$$\langle V'(\hat x)\rangle = m\omega^2\langle \hat x\rangle$$

with no approximation. The two identities of Theorem 5.5 become

$$\frac{d\langle \hat x\rangle}{dt} = \frac{\langle\hat p\rangle}{m}, \qquad \frac{d\langle \hat p\rangle}{dt} = -m\omega^2\langle \hat x\rangle ,$$

and differentiating the first in $t$ and substituting the second,

$$\frac{d^2\langle \hat x\rangle}{dt^2} = \frac{1}{m}\frac{d\langle\hat p\rangle}{dt} = -\omega^2\langle \hat x\rangle .$$

This is precisely the classical harmonic-oscillator equation. Solving it with initial conditions $\langle\hat x\rangle(0) = \langle\hat x\rangle_0$ and $(d\langle\hat x\rangle/dt)(0) = \langle\hat p\rangle_0/m$,

$$\langle \hat x\rangle(t) = \langle \hat x\rangle_0\cos\omega t + \frac{\langle \hat p\rangle_0}{m\omega}\sin\omega t .$$

So for the harmonic oscillator the centroid of the wave packet traces the classical orbit exactly, no matter how badly the shape of the packet is distorted.

2. (Sufficiency.) If $V'(x) = \alpha x + \beta$, then by linearity of the expectation value and $\langle 1\rangle = \|\psi\|^2 = 1$,

$$\langle V'(\hat x)\rangle = \alpha\langle \hat x\rangle + \beta = V'(\langle \hat x\rangle).$$

(Necessity.) Take any $x_0 \in \mathbb{R}$ and $d > 0$, and consider the state obtained by superposing, with equal weights, two normalised states concentrated within a width of order $\epsilon$ near $x_0 - d$ and near $x_0 + d$. In the limit $\epsilon \to 0$ we have $\langle \hat x\rangle \to x_0$ and $\langle V'(\hat x)\rangle \to \tfrac12\bigl(V'(x_0-d) + V'(x_0+d)\bigr)$ (using the continuity of $V'$). By hypothesis these two must satisfy

$$\frac{V'(x_0-d) + V'(x_0+d)}{2} = V'(x_0) ,$$

and this holds for all $x_0$ and all $d$. This says that $V'$ satisfies the midpoint Jensen equation with equality, and a continuous $V'$ satisfying it must be affine, $\alpha x + \beta$. Indeed, putting $g(x) := V'(x) - V'(0) - (V'(1)-V'(0))x$, the function $g$ satisfies the same midpoint condition and $g(0) = g(1) = 0$; iterating the midpoint condition gives $g = 0$ on the dyadic rationals, and continuity gives $g = 0$ for all $x$. That $V'$ is affine is equivalent to $V$ being at most quadratic.

References

• E. Schrödinger, “Quantisierung als Eigenwertproblem (Erste Mitteilung)”, Annalen der Physik 384 (1926), 361–376. doi:10.1002/andp.19263840404 — the paper in which the wave equation was first proposed.
• M. Born, “Zur Quantenmechanik der Stoßvorgänge”, Zeitschrift für Physik 37 (1926), 863–867. doi:10.1007/BF01397477 — the scattering-theory paper that introduced the probability interpretation.
• P. Ehrenfest, “Bemerkung über die angenäherte Gültigkeit der klassischen Mechanik innerhalb der Quantenmechanik”, Zeitschrift für Physik 45 (1927), 455–457. doi:10.1007/BF01329203 — the original paper for Theorem 5.5.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020 — Chapters 1 and 2. A standard text taking Dirac notation as its starting point.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018 — Chapters 1 and 2. A careful treatment of the wave function and one-dimensional problems.
• Keiji Igi and Hikaru Kawai, Ryoshi Rikigaku I, Kodansha Scientific, 1994 (in Japanese) — Chapters 2 and 3. A Japanese standard, written out to the last detail of every computation.

Appendix: plane waves cannot be normalised

The plane wave $\psi_p(x) = A\,e^{ipx/\hbar}$ used in §3.1 has constant $|\psi_p|^2 = |A|^2$, so

$$\int_{-\infty}^{\infty}|\psi_p|^2dx = |A|^2\int_{-\infty}^{\infty}dx = \infty$$

and it does not belong to $L^2(\mathbb{R})$ as long as $A \ne 0$. It is therefore not a state in the sense of Definition 2.1, and the probability interpretation of Definition 4.1 does not apply to it directly. The eigenvalue equation $\hat p\,\psi = p\,\psi$ for $\hat p$ has no solution inside $L^2$. This is the common situation for operators with continuous spectrum, and it means that the hypothesis of Theorem 6.3 that the eigenfunctions form a complete orthonormal system fails for the free particle.

In practice there are two ways to proceed.

Delta-function normalisation. Choosing the normalisation constant so that $\psi_p(x) = (2\pi\hbar)^{-1/2}e^{ipx/\hbar}$, one has, in the sense of distributions,

$$\langle \psi_{p'}\mid\psi_p\rangle = \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}e^{i(p - p')x/\hbar}dx = \frac{1}{2\pi\hbar}\cdot 2\pi\,\delta\!\left(\frac{p-p'}{\hbar}\right) = \delta(p - p')$$

(using $\int e^{ikx}dx = 2\pi\delta(k)$ and $\delta(ax) = \delta(x)/|a|$). The Kronecker delta $\delta_{mn}$ is replaced by the Dirac delta $\delta(p-p')$, and the sum in Theorem 6.3 becomes an integral (a Fourier transform). The framework that makes this rigorous is the theory of rigged Hilbert spaces.

Putting the particle in a box. Imposing periodic boundary conditions $\psi(x+L) = \psi(x)$ on an interval of length $L$ discretises the allowed wave numbers to $k_n = 2\pi n/L$ with $n \in \mathbb{Z}$, and $\psi_n(x) = L^{-1/2}e^{ik_nx}$ is a genuinely normalised function. One then takes $L \to \infty$ at the end. This method is common in solid-state physics.

Either way, the states physically realised are always normalisable wave packets, and plane waves should be regarded as a convenient tool for expanding them. Scattering states and bound states in concrete one-dimensional problems are examined in detail in Simple one-dimensional systems.