Expectation Value

In one sentence: The expectation value is the average result you would get if you repeated a measurement many times on identical particles.

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

Read it out loud: “The expectation value of x, written with angle brackets, equals the integral from minus infinity to plus infinity of x times the squared size of psi of x, times d x. The angle brackets around x mean the average of x. The integral sign means add up many thin slices. Psi (said SY) is the wave function, and the vertical bars mean squared size. The result is the average position, in meters.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Average and weighted average: An average adds up values and divides by how many there are. A weighted average gives each value a weight, such as how common it is. Each value is multiplied by its weight before the adding. When the weights are probabilities, they must add up to 1. The expectation value is a weighted average, where each possible result is weighted by how likely it is.
Probability, from 0 to 1: A probability is a chance written as a number from 0 to 1. Zero means never and 1 means always. A probability of 0.5 means that something happens half of the time, on average, such as heads for a fair coin. All the probabilities for the possible results of one measurement add up to 1, because one of those results must happen.
Angle brackets, like $\langle x \rangle$: The angle brackets $\langle\ \rangle$ around a letter mean the expectation value of that quantity. So $\langle x \rangle$ is read ‘the expectation value of x’, which is the average position. The brackets do not mean multiplication. They are a short label that says ‘average of this quantity’.
The sum sign $\sum$ and subscripts like $a_i$: The capital Greek letter sigma, $\sum$ , means add up. Its limits, written under and over it, say which items to add. The little $i$ in $a_i$ is a counting label, like a number on a list of results, so $a_1$ is the first result and $a_2$ the second. So $\sum_i a_i P(a_i)$ means multiply each result by its probability, then add the products.
Integrals and $dx$: An integral, written $\int$ , is the continuous version of a sum. It adds up many thin slices, each of width $dx$ . When positions form a continuous line, there is no list of separate results, so the sum becomes an integral. The integral from minus infinity to infinity covers the whole line, from far left to far right.
Probability density $|\psi(x)|^2$: The wave function $\psi$ is the mathematical wave of a particle, and it can be a complex number. The vertical bars mean the size, and the small 2 means squared, so $|\psi|^2$ is a positive number at each position. It is a probability density, a chance per meter, and it acts as the weight for each position in the average.

What each symbol means

Symbol	Meaning
$\langle x \rangle$	The expectation value of position, the average position over many measurements on identical particles. It is measured in meters (m). It may be a value that no single measurement gives, because it is an average of many results.
$x$	The position along a line, in meters (m). Inside the integral it is the value being averaged, weighted by the probability density at each spot along the line.
$\|\psi(x)\|^2$	The probability density at position $x$ , which is the squared size of the wave function. Its unit is $\text{m}^{-1}$ , a chance per meter. It acts as the weight for each position in the average.
$\psi(x)$	The wave function at position $x$ , a complex number. Its unit is $\text{m}^{-1/2}$ in one dimension. Only its squared size gives a probability density, so $\psi$ itself is not used as the weight.
$dx$	A thin slice of position, of width $dx$ , in meters (m). The integral adds these slices together over the whole line.
$\int_{-\infty}^{\infty}$	The integral over all positions, from far left to far right. It is the continuous version of a sum, and it adds up many thin slices of the line.
$a_i$	The possible result number $i$ of a measurement, such as one energy value. In the list version of the rule, these are the only values that can come out. For energy the unit is electron volts (eV).
$P(a_i)$	The probability of result $a_i$ , a number from 0 to 1. All the probabilities add up to 1, because one of the possible results must happen.

The formula piece by piece

$x$: Start with a single position $x$ , in meters. It is the value being averaged. A position of 1.0 m, for example, is a spot one meter from the start of the line.
$|\psi(x)|^2$: Next, the weight of that position is $|\psi(x)|^2$ , the probability density there. A spot where the density is high counts more in the average, and a spot where the density is zero does not count at all.
$x\, |\psi(x)|^2$: Multiply the position by its weight. The product $x\,|\psi(x)|^2$ is a position counted by how likely it is, so the average leans toward the positions the particle is most likely to be found at.
$\langle x \rangle = \int_{-\infty}^{\infty} x\, |\psi(x)|^2\, dx$: The whole formula in one sentence: the average position equals the integral, over all positions, of each position times its probability density, with the results added up in thin slices.
$\langle a \rangle = \sum_i a_i P(a_i)$: Another standard form, for separate results instead of a continuous line: the expectation value of a quantity $a$ equals the sum, over every possible result $a_i$ , of that result times its probability $P(a_i)$ .

What the formula is doing, and why

The expectation value $\langle A \rangle$ of a quantity is the average you get when you measure it many times on identical systems. It is not one measured result. What goes in is the wave function, or a list of results with their probabilities. What comes out is the average. Each possible result counts in proportion to how likely it is, so the average leans toward the common outcomes.

Why a weighted sum? When the possible results are separate values, the rule is $\langle a \rangle = \sum_i a_i P(a_i)$ . Multiply each possible result $a_i$ by its probability $P(a_i)$ , then add. The probabilities must add to 1, so each weight is a fair share of the whole. Likely results pull the average toward them.

Why an integral? Position is continuous, so there is no list of separate results. The sum becomes an integral, and the density $|\psi(x)|^2$ plays the role of the probabilities. In $\langle x \rangle = \int x\, |\psi(x)|^2\, dx$ , each position $x$ is weighted by the density at that spot, and the thin slices $dx$ add up over the whole line.

Why symmetry helps. For the lowest state of a box of width $L$ , the density is the same on the left and on the right, centered in the middle of the box. The average position is therefore $\langle x \rangle = \frac{L}{2}$ , the center of the box. Symmetry gives the answer without doing the integral.

What an average does not tell you. An expectation value is an average, so it does not show how far single results scatter. The spread of the results, often measured by the standard deviation, tells how far single measurements typically land from the average. The average can also be a number that no single measurement can give, as the example below shows.

An everyday picture

Think of a class test. If half the students score 1, 30 percent score 2 and 20 percent score 4, the class average is 1.9, even though no student scored 1.9. Each score counts in proportion to how many students got it. The expectation value works the same way, weighting each possible result by its probability.

Worked example: Average energy from a measurement

A measurement of the energy of a particle gives one of three values. The value 1 eV has probability 0.5, the value 2 eV has probability 0.3, and the value 4 eV has probability 0.2. These are made-up numbers for practice. We find the expectation value of the energy, $\langle E \rangle$ .

Check the probabilities add up to 1. Every measurement gives one of the three values, so the probabilities must total 1: $0.5 + 0.3 + 0.2 = 1.0$ . If they did not, the weights would be wrong and the average would be off.
Multiply each value by its probability. Each value is multiplied by how likely it is: $1\ \text{eV} \times 0.5 = 0.5$ eV for the first, $2\ \text{eV} \times 0.3 = 0.6$ eV for the second, and $4\ \text{eV} \times 0.2 = 0.8$ eV for the third. The unit stays as eV, because the probabilities have no unit.
Add the three products. $\langle E \rangle = 0.5 + 0.6 + 0.8 = 1.9$ eV. This is the weighted average. The answer sits between the smallest and largest results, and it is closer to 1 eV, the most likely result.
Compare with a plain average. Averaging the three values without weights gives $(1 + 2 + 4)/3 \approx 2.33$ eV. The weights pull the answer down to 1.9 eV, because the 1 eV result is the most likely and the 4 eV result is the least likely.
Imagine 1,000 measurements. About 500 would give 1 eV, 300 would give 2 eV and 200 would give 4 eV. Their total is $500 \times 1 + 300 \times 2 + 200 \times 4 = 1900$ eV. Dividing by the 1000 measurements gives 1.9 eV per measurement, which is the expectation value.
Check that the result is not one of the values. No single measurement gives 1.9 eV, because the only possible results are 1, 2 and 4 eV. The expectation value is an average, so it can fall between results that can actually occur.

Answer: The expectation value of the energy is 1.9 eV. No single measurement gives 1.9 eV, because the possible results are only 1, 2 and 4 eV. Over many measurements, though, the average of the results gets close to 1.9 eV.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Averaging the possible results without weights. The plain average of 1, 2 and 4 is 2.33, which is the wrong answer here. Each result must be weighted by its probability.
Expecting the answer to be one of the possible results. The value 1.9 eV is not a result that a single measurement can give. It is an average.
Using $\psi$ in place of $|\psi|^2$ in the integral. The probability density is the squared size of $\psi$ , and $\psi$ itself can be negative or complex.
Leaving out the factor $x$ and integrating only $|\psi|^2$ . That gives the total probability, which is 1, not the average position.

Where you will meet it

Quantum chemistry, where the expectation value of the energy for a trial wave function gives an upper bound on the lowest possible energy.
Quantum computing, where algorithms such as the variational quantum eigensolver estimate expectation values of molecular energies.
Experiments with atoms, photons and electron spins, where many repeated measurements are averaged and compared with a calculated expectation value.

Questions people ask

Is the expectation value the most likely result?

Not always. The most likely result is the highest point of the probability picture. The expectation value is the average, which can sit away from that peak, especially when the distribution is lopsided.

Does one measurement give the expectation value?

No. One measurement gives one of the possible results. The expectation value shows up as the average of many measurements on identical systems.

How many measurements does the average need?

The more measurements you average, the closer the result gets to the expectation value. The random scatter of the average shrinks roughly as $\frac{1}{\sqrt{N}}$ for $N$ measurements.

Born Rule: The Born rule turns a quantum wave into chances: the probability of a result is the squared size of the wave for that result, so bigger waves mean more likely results.
Wave Function Normalization: Normalizing a quantum wave means scaling it so the chances of finding the particle, added over every spot, total exactly 1.
Heisenberg Uncertainty Principle: The more tightly you pin down where a tiny particle is, the more its speed must spread out, and no instrument can beat that trade-off.