Expectation Value of an Observable

Equivalent forms

\langle x\rangle=\int x\,|\psi(x)|^2\,dx

One integral turns the probability cloud |ψ|² into a single number — the bridge from quantum randomness to a measurable average.

Symbol	Name	SI unit	Dimension	Range
$\langle\hat{A}\rangle$	Expectation valueoutput Ensemble-average of observable A (here position)	m	L	—
$\psi$	Wavefunction Normalized 1D probability amplitude	m^{-1/2}	L^-1/2	—
$\hat{A}$	Observable operator Hermitian operator (e.g. position x̂)	m	L	—
$x$	Position Integration variable	m	L	-5 – 5

Where it holds

Requires a normalized state

(\int |\psi |^{2}=1)

. For unnormalized states divide by ⟨

\psi |\psi

⟩. Â must be Hermitian for the result to be real.

Dimensional analysis

⟨x̂⟩

\to [L]

\int x|\psi |^{2} dx \to [L][L^-1][L] = [L]

|

\psi |^{2} has

dimension L^-1 in 1D; times x dx gives length.

\checkmark

Discovery

Max Born / Paul Dirac · 1926

Once Born's rule identified |ψ|² as a probability density (1926), the average of any observable followed as the probability-weighted mean. Dirac's bra-ket formalism recast it as the elegant ⟨ψ|Â|ψ⟩, making expectation values the workhorse of quantum predictions.

Research status: stable

Real-world applications

Predicting mean energy, position, and momentum of quantum states
Computing observables in quantum chemistry
Defining variance $(\Delta A)^{2}$ = ⟨Â $^{2}$ ⟩ $-$ ⟨Â⟩ $^{2} for$ uncertainty

Common misconceptions

⟨Â⟩ need not be an allowed eigenvalue — the average $of \pm 1$ spin results can be 0
Operator order matters: ⟨ $\psi$ |Â| $\psi$ ⟩ is $not \int$ Â $\psi *\cdot \psi in$ general
It describes an ensemble average, not the outcome of one measurement

Experimental verification

Averaging repeated measurements (e.g., electron position in a double-slit build-up, or spin projections) reproduces the computed ⟨Â⟩; Ehrenfest's theorem shows ⟨x⟩ and ⟨p⟩ obey Newton's laws on average.

Derivation

By the Born rule the probability of finding the particle in [x, x+dx] is |

\psi (x)|^{2} dx

.

The average of any function f(x) is then

\int f(x)|\psi |^{2} dx

.

Promoting f to a general operator Â that may involve derivatives, the operator must sit between

\psi

*

and \psi

(order matters): ⟨Â⟩

= \int \psi

* Â

\psi dx

.

In Dirac notation this is ⟨

\psi

|Â|

\psi

⟩.

What if…

What

if \psi is

an eigenstate of Â?

Then ⟨Â⟩ equals that eigenvalue exactly and the variance is zero — a definite, repeatable measurement.

What if the state isn't normalized?

Divide by ⟨ $\psi |\psi$ ⟩: ⟨Â⟩ = ⟨ $\psi$ |Â| $\psi$ ⟩/⟨ $\psi |\psi$ ⟩, so probabilities still sum to one.

1

⟨x⟩ of a symmetric state

Given ·

\psi (x)

real and even about

x=0

Find · ⟨x⟩

Steps

| $\psi |^{2} is$ even, x is odd, so $x|\psi |^{2} is odd$
Integral of an odd function over symmetric limits is 0
⟨x⟩ $= 0$ — the packet is centered

Result · ⟨x⟩

= \int x|\psi |^{2} dx = 0 (odd

integrand)

Born Rule→Wavefunction Normalization→Ehrenfest Theorem→