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The mean and spread of position from a wave function, the momentum operator $-i\hbar\,d/dx$, the sandwich rule for any observable, and the kinetic energy of a confined particle.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute expectation values and standard deviations of position, use the momentum and energy operators, find the average kinetic energy of a confined particle, and explain what an expectation value does and does not predict.
You can normalize a wave function and compute the probability of finding a particle in an interval. From statistics you know the mean and variance of a distribution. This lesson computes those for the distribution a wave function describes, and discovers that momentum and energy, which are not simply functions of position, need operators.
| Term | What it means |
|---|---|
| Expectation value | $\langle Q\rangle$, the average of the results of measuring $Q$ on many identically prepared systems. |
| Operator | A rule that acts on a wave function, such as $\hat{p} = -i\hbar\,d/dx$; each observable has one. |
| Momentum operator | $\hat{p} = -i\hbar\,\dfrac{d}{dx}$. |
| Hamiltonian | $\hat{H} = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2} + V(x)$, the energy operator. |
| Sandwich rule | $\langle Q\rangle = \int\Psi^*\hat{Q}\Psi\,dx$. |
| Variance | $\sigma_Q^2 = \langle Q^2\rangle - \langle Q\rangle^2$. |
| Standard deviation | $\sigma_Q$, the spread of the distribution of results. |
| Ehrenfest's theorem | Expectation values obey classical equations: $d\langle p\rangle/dt = \langle -dV/dx\rangle$. |
The wave function gives the probability density of position, so the average position found by measuring many identically prepared systems is
$$\langle x\rangle = \int_{-\infty}^{\infty}x|\Psi(x, t)|^2dx,$$
the expectation value of $x$. Any function of position works the same way: $\langle x^2\rangle = \int x^2|\Psi|^2dx$, and the spread of the results is the standard deviation, from the variance
$$\sigma_x^2 = \langle x^2\rangle - \langle x\rangle^2.$$
Momentum is not a function of $x$, so it needs something new. Writing $\langle p\rangle = m\,d\langle x\rangle/dt$ and using the Schrödinger equation gives
$$\langle p\rangle = \int\Psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi\,dx.$$
The quantity in brackets is the momentum operator, $\hat{p} = -i\hbar\,\partial/\partial x$. Every observable has an operator, and its expectation value is the "sandwich" $\langle Q\rangle = \int\Psi^\hat{Q}\Psi\,dx$. Position's operator is multiplication by $x$; kinetic energy's is $\hat{p}^2/2m = -(\hbar^2/2m)\,\partial^2/\partial x^2$; total energy's is the Hamiltonian*, $\hat{H} = \hat{p}^2/2m + V(x)$.
For a real wave function that vanishes at the boundaries, two shortcuts follow. $\langle p\rangle = 0$, since the integrand is a total derivative times $-i$, and the answer must be real. And integrating by parts,
$$\langle p^2\rangle = -\hbar^2\int\psi\psi''dx = \hbar^2\int(\psi')^2dx,$$
which is always positive and often easier to integrate.
Another way: picture
Picture a thousand identical boxes, each prepared with the same electron state, and a thousand experimenters each measuring position once. They write down their results and average them: that is $\langle x\rangle$. They compute how widely the results scatter: that is $\sigma_x$. A second thousand, prepared the same way, measure momentum instead; their average is $\langle p\rangle$. The wave function predicts both columns of numbers before anyone opens a box.
Another way: steps
Checks. A variance can never be negative; if $\langle x^2\rangle$ comes out smaller than $\langle x\rangle^2$, an integral is wrong. The mean must lie inside the region the particle can occupy. The average kinetic energy of any state confined to a region of length $a$ is at least $\pi^2\hbar^2/(2ma^2)$, the true ground-state energy of a box; a smaller answer is impossible. And every expectation value of an observable must be real.
For $\psi = \sqrt{30/a^5}\,x(a - x)$ the calculations are exact. Symmetry gives $\langle x\rangle = a/2$. The mean square is
$$\langle x^2\rangle = 30a^2\int_0^{1}u^4(1 - u)^2du = 30a^2\left(\frac{1}{5} - \frac{1}{3} + \frac{1}{7}\right) = \frac{2}{7}a^2,$$
so the variance is $\tfrac{2}{7}a^2 - \tfrac{1}{4}a^2 = a^2/28$ and $\sigma_x = a/\sqrt{28} \approx 0.19a$.
For momentum, $\psi' = \sqrt{30/a^5}(a - 2x)$ and
$$\langle p^2\rangle = \hbar^2\frac{30}{a^5}\int_0^a(a - 2x)^2dx = \hbar^2\frac{30}{a^5}\cdot\frac{a^3}{3} = \frac{10\hbar^2}{a^2}.$$
The average kinetic energy is $5\hbar^2/(ma^2)$. The true ground state of a particle in a box, found in the next unit, has energy $\pi^2\hbar^2/(2ma^2) = 4.935\hbar^2/(ma^2)$. The parabola, a guess with no sines in it, comes within $1.3$ percent. That is not luck: it is the variational principle, a later lesson's subject, which guarantees that any trial state's average energy is at least the ground-state energy, and is close when the trial is close.
Differentiating expectation values in time with the Schrödinger equation gives two results that look like Newton's laws:
$$\frac{d\langle x\rangle}{dt} = \frac{\langle p\rangle}{m}, \qquad \frac{d\langle p\rangle}{dt} = \left\langle -\frac{\partial V}{\partial x}\right\rangle.$$
Expectation values move like classical quantities, with one subtlety: the force is the average of $-\partial V/\partial x$ over the wave packet, not the force at the average position. For a packet narrow compared with the distance over which the force changes, the two agree, and the center of the packet follows Newton's second law. That is why a baseball, whose wave packet is unimaginably narrow, obeys classical mechanics perfectly, while an electron in an atom, spread over the whole region where the force varies, does not.
An expectation value is an average, and averages need not be possible values. A fair die has an expectation of $3.5$ though no face shows it. A particle in a symmetric double well, with equal peaks either side of a barrier, has $\langle x\rangle$ exactly at the barrier, where it is almost never found. An electron in a superposition of two energy levels has an average energy between them that no energy measurement ever returns.
What an expectation value always is, is a prediction for a long run of measurements on identically prepared systems. It is also what classical physics sees when quantum spreads are too small to notice, which is why the averages obey classical-looking equations. The spread $\sigma$ tells how reliable any single measurement is as a guide to that average — and the next lesson shows that for position and momentum the two spreads cannot both be small.
Position has a simple meaning at each point of a wave function: the particle might be found there. Momentum does not, because momentum describes how the wave function changes from point to point — its wavelength — and no single point has a wavelength. A plane wave $e^{ipx/\hbar}$ has momentum $p$ everywhere, and differentiating it, $-i\hbar\,d/dx$, returns $p$ times the wave. That is why the momentum operator is a derivative: it reads off the local wavelength.
For a wave function that is a mixture of wavelengths, the sandwich $\int\psi^*(-i\hbar\,\psi')\,dx$ averages the momentum over the whole mixture. The alternative, which the formalism unit develops, is to write $\psi$ as a sum of plane waves by a Fourier transform; the squared amplitudes of that sum are the momentum probability distribution, and it gives the same averages. Either way, momentum is a property of the whole shape of the wave function, not of its value at a point — which is the root of the uncertainty principle of the next lesson.
A quantum dot is a crystal of semiconductor a few nanometers across, small enough that an electron inside is confined like a particle in a box. The confinement adds kinetic energy of order $\hbar^2/(ma^2)$, and it is this energy that sets the color of light the dot emits: smaller dots confine more tightly, raise the energy of the emitted photon, and glow bluer.
The parabolic-state estimate shows the scale. In cadmium selenide the electron's effective mass is about $0.13m_e$, so $\hbar^2/m = 0.0762/0.13 = 0.586$ eV nm², and for a dot $4$ nm across, $\langle K\rangle \approx 5 \times 0.586/16 = 0.18$ eV. Added to the material's band gap of $1.74$ eV, and with the hole contributing a similar share, that shifts the emission from deep red toward green; halving the size would quadruple the shift.
These are the dots in QLED televisions, where arrays of cadmium-free quantum dots convert blue backlight into pure red and green, and the 2023 Nobel Prize in Chemistry went to Bawendi, Brus and Ekimov for discovering and synthesizing them. The color is chosen by choosing the size, a direct application of the expectation value of $p^2$ for a confined particle.
The same expectation value explains the gulf between chemistry and nuclear physics. An electron confined to an atom about $0.2$ nm across has kinetic energy of order $5\hbar^2/(m_ea^2) = 5 \times 0.0762/0.04 = 9.5$ eV: the scale of chemical bonds, and of the photons of visible and ultraviolet light.
A proton confined to a nucleus about $5$ fm across is $1836$ times heavier but confined to a region forty thousand times smaller. Its kinetic energy is of order $5 \times 41.5/25 = 8.3$ MeV, about a million times larger. Nuclear binding energies are indeed around $8$ MeV per nucleon, and nuclear reactions release millions of times more energy per atom than chemical ones — a uranium fission about $200$ MeV against a few electronvolts for burning a carbon atom.
Nothing but the expectation value of $p^2$ for a confined particle, and the ratio of masses and sizes, sets that factor of a million, which is why a nuclear power plant runs a year on a few truckloads of fuel while a coal plant burns a trainload a day.
The name suggests the value you should expect to find, and it misleads. The expectation value is the mean of many measurements on identically prepared systems, and in general no single measurement need return it. For a particle in a symmetric double well, $\langle x\rangle$ sits at the barrier where the particle is almost never found; for a state that is an equal mix of two energy levels, $\langle E\rangle$ lies between them and no energy measurement ever gives it.
It is equally wrong to think of $\langle x\rangle$ as the result of measuring one particle many times. The first position measurement changes the particle's state, and an immediate repeat finds it in the same place again. The statistics that $\langle x\rangle$ and $\sigma_x$ describe come from an ensemble: many systems, prepared the same way, each measured once. A related slip is to compute $\langle x\rangle^2$ where $\langle x^2\rangle$ is needed; their difference is the variance, and it is never negative.
For $\psi = \sqrt{30/a^5}\,x(a - x)$, write the expectation value of $x$.
$\langle x\rangle = \dfrac{30}{a^5}\int_0^a x \cdot x^2(a - x)^2dx$
Weight each position by the probability density.
Substitute $x = ua$.
$\langle x\rangle = 30a\int_0^{1}u^3(1 - u)^2du$
One more power of $a$ than the normalization integral, so one factor of $a$ survives.
Expand and integrate.
$\int_0^1(u^3 - 2u^4 + u^5)du = \dfrac{1}{4} - \dfrac{2}{5} + \dfrac{1}{6} = \dfrac{1}{60}$
Over the common denominator sixty: $15 - 24 + 10 = 1$.
Multiply the factors.
$\langle x\rangle = 30a \times \dfrac{1}{60} = \dfrac{a}{2}$
The center of the region.
Check with symmetry.
$|\psi(a - x)|^2 = |\psi(x)|^2$
The density is symmetric about $a/2$, so the mean had to be $a/2$.
Write the mean square position for the same state.
$\langle x^2\rangle = 30a^2\int_0^{1}u^4(1 - u)^2du$
Weight $x^2$ by the density and substitute $x = ua$.
Expand the integrand.
$u^4(1 - u)^2 = u^4 - 2u^5 + u^6$
Square $(1 - u)$ and multiply.
Integrate term by term.
$\dfrac{1}{5} - \dfrac{2}{6} + \dfrac{1}{7} = \dfrac{42 - 70 + 30}{210} = \dfrac{2}{210} = \dfrac{1}{105}$
A common denominator of $210$.
Multiply by $30a^2$.
$\langle x^2\rangle = \dfrac{30}{105}a^2 = \dfrac{2}{7}a^2$
Simplify the fraction.
Subtract the square of the mean.
$\sigma_x^2 = \dfrac{2}{7}a^2 - \left(\dfrac{a}{2}\right)^2 = \dfrac{8 - 7}{28}a^2 = \dfrac{a^2}{28}$
Variance is mean square minus square of the mean.
Take the root.
$\sigma_x = \dfrac{a}{\sqrt{28}} \approx 0.189a$
Most measurements land within a fifth of the region of the center.
Write $\langle p^2\rangle$ for a real wave function that vanishes at the walls.
$\langle p^2\rangle = -\hbar^2\int_0^a\psi\psi''dx = \hbar^2\int_0^a(\psi')^2dx$
Integrating by parts; the boundary term vanishes because $\psi = 0$ at the walls.
Differentiate the wave function.
$\psi' = \sqrt{\dfrac{30}{a^5}}(a - 2x)$
The derivative of $ax - x^2$.
Square the derivative.
$(\psi')^2 = \dfrac{30}{a^5}(a^2 - 4ax + 4x^2)$
Expand $(a - 2x)^2$.
Integrate from $0$ to $a$.
$\int_0^a(a^2 - 4ax + 4x^2)dx = a^3 - 2a^3 + \dfrac{4}{3}a^3 = \dfrac{a^3}{3}$
Term by term.
Assemble $\langle p^2\rangle$.
$\langle p^2\rangle = \hbar^2 \cdot \dfrac{30}{a^5} \cdot \dfrac{a^3}{3} = \dfrac{10\hbar^2}{a^2}$
Multiply the pieces.
Find the average kinetic energy.
$\langle K\rangle = \dfrac{\langle p^2\rangle}{2m} = \dfrac{5\hbar^2}{ma^2}$
Kinetic energy is $p^2/2m$.
Evaluate for an electron in a $1$ nm region.
$\langle K\rangle = \dfrac{5 \times 0.0762\ \text{eV nm}^2}{(1\ \text{nm})^2} = 0.381\ \text{eV}$
Using $\hbar^2/m_e = 0.0762$ eV nm².
Compare with the exact ground state of a box.
$E_1 = \dfrac{\pi^2\hbar^2}{2ma^2} = 4.935 \times 0.0762 = 0.376\ \text{eV}$
The parabola's energy is only $1.3$ percent higher, and never lower, as the variational principle requires.
Find the mean from symmetry.
$|\psi(-x)|^2 = |\psi(x)|^2 \quad\Rightarrow\quad \langle x\rangle = 0$
The density is even.
Set up the mean square.
$\langle x^2\rangle = \dfrac{2}{a}\int_0^{\infty}x^2e^{-2x/a}dx$
Twice the integral over positive $x$.
Evaluate the expression.
An electron is in a state whose density has two equal peaks, at $x = -3$ nm and $x = +3$ nm, and is nearly zero at $x = 0$. Its expectation value of position is $\langle x\rangle = 0$. What does that mean?
Complete the worked solution: measurements find a particle at $7$, $14$, $21$ or $28$ nm with probabilities $0.1$, $0.2$, $0.3$ and $0.4$. Find its mean position, mean square position and variance.
Weight each position by its probability and add.
$\langle x\rangle = \sum x_iP_i =$ m nm
The mean of the distribution.
Weight each squared position by its probability and add.
$\langle x^2\rangle = \sum x_i^2P_i =$ s nm²
The mean of the squares.
Subtract the square of the mean.
$\sigma^2 = \langle x^2\rangle - \langle x\rangle^2 =$ v nm²
The variance, never negative.
In the position representation, match each quantity for a particle of mass $m$ in a potential $V$ ($1$ spatial dimension) to its operator or formula.
| multiply by $x$ | $-i\hbar\,d/dx$ | $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V$ | $\langle x^2\rangle - \langle x\rangle^2$ | |
|---|---|---|---|---|
| position | ||||
| momentum | ||||
| energy | ||||
| the variance of position |
For $\psi = \sqrt{30/a^5}\,x(a - x)$ with $a = 3$ nm, fill in $\langle x\rangle$ in nm and $\langle x^2\rangle$ and $\sigma_x^2$ in nm², to four decimal places.
| value | |
|---|---|
| $\langle x\rangle$ (nm) | |
| $\langle x^2\rangle$ (nm²) | |
| $\sigma_x^2$ (nm²) |
Position measurements on many identically prepared particles give $9$, $18$, $27$ and $36$ nm with probabilities $0.1$, $0.2$, $0.3$ and $0.4$. What is the variance of position, in nm²?
Answer: nm²
An electron confined to a region of length $a = 0.2$ nm has wave function $\psi = \sqrt{30/a^5}\,x(a - x)$. What is its average kinetic energy, in eV? Use $\hbar^2/m_e = 0.07620$ eV nm².
Answer: eV
Model a proton confined within a nucleus as a particle in a region of length $a = 8$ fm, in the state $\psi = \sqrt{30/a^5}\,x(a - x)$. What is its average kinetic energy, in MeV? Use $\hbar^2/m_p = 41.50$ MeV fm².
Answer: MeV
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $\psi = \sqrt{30/a^5}\,x(a - x)$ with $a = 2$ nm, fill in $\langle x\rangle$ in nm and $\langle x^2\rangle$ and $\sigma_x^2$ in nm², to four decimal places.
| value | |
|---|---|
| $\langle x\rangle$ (nm) | |
| $\langle x^2\rangle$ (nm²) | |
| $\sigma_x^2$ (nm²) |
You can compute averages and spreads of position, momentum and energy from a wave function. Explain to someone why the expectation value of position can be a place where the particle is never found.
16. Your turn: for $\psi = e^{-|x|/a}/\sqrt{a}$, find $\langle x\rangle$ and $\langle x^2\rangle$., step 3
$\int_0^{\infty}x^2e^{-2x/a}dx = 2\left(\dfrac{a}{2}\right)^3 = \dfrac{a^3}{4} \quad\Rightarrow\quad \langle x^2\rangle = \dfrac{a^2}{2}$
Using $\int_0^\infty x^ne^{-x/b}dx = n!\,b^{n + 1}$.