# Expectation & Uncertainty > A single measurement gives one random eigenvalue; many measurements give a *distribution*. Its first > two moments — the **expectation value** and the **variance** — summarize what an observable does to a > state. The deepest structural fact in this lesson is that you cannot make *two* observables both > sharp at once unless they commute: the **Robertson uncertainty relation**. We derive it from a single > line of Cauchy–Schwarz, and see exactly where Heisenberg's $\Delta x\,\Delta p$ comes from. ## Learning Objectives After this lesson you will be able to: 1. Compute the **expectation value** $\langle A\rangle = \langle\psi|A|\psi\rangle = \sum_k\lambda_k p_k$ and prove the two forms agree. 2. Compute the **variance** $(\Delta A)^2 = \langle A^2\rangle - \langle A\rangle^2$ and interpret $\Delta A = 0$ as "definite value". 3. **Derive** the Robertson uncertainty relation $\Delta A\,\Delta B \ge \tfrac12|\langle[A,B]\rangle|$ via the Cauchy–Schwarz inequality. 4. Apply it to **Pauli observables**, e.g. bound $\Delta X\,\Delta Z$ using $[X,Z]=-2iY$. 5. Distinguish **compatible** (commuting) from **incompatible** observables and state when joint measurement / a common eigenbasis exists. --- ## Intuition Run the same measurement on many identically prepared copies of $|\psi\rangle$ and histogram the eigenvalues you read off. The **expectation value** $\langle A\rangle$ is the mean of that histogram; the **variance** $(\Delta A)^2$ is its spread. If $|\psi\rangle$ happens to be an eigenstate of $A$, the histogram is a single spike — zero spread, a *definite* value. Now the surprise. For *one* observable you can always find states with zero spread (the eigenstates). But you generally **cannot** find a state that is simultaneously spike-sharp for two *different* observables, like $X$ and $Z$. The obstruction is algebraic: $X$ and $Z$ do not commute. The Robertson relation turns "how badly they fail to commute" — the size of $\langle[A,B]\rangle$ — into a hard lower bound on the product of spreads. Incompatibility is not a measurement clumsiness; it is geometry forced by non-commuting operators. --- ## Theory ### Expectation value > **Definition.** For an observable $A=A^\dagger$ and normalized state $|\psi\rangle$, the **expectation > value** is > $$ \langle A\rangle \;:=\; \langle\psi|A|\psi\rangle . $$ This is the probability-weighted mean of the eigenvalues. Insert the spectral decomposition $A=\sum_k\lambda_k P_k$ from [Lesson 1](01-projective-measurement.md): $$ \langle A\rangle = \langle\psi|\Big(\sum_k\lambda_k P_k\Big)|\psi\rangle = \sum_k \lambda_k\,\langle\psi|P_k|\psi\rangle = \sum_k \lambda_k\,p_k , $$ which is exactly the classical mean $\sum_k\lambda_k\,\Pr[\text{outcome }k]$. So "$\langle\psi|A|\psi\rangle$" and "average the eigenvalues against the Born probabilities" are the *same number* — one computed from the operator, one from the experiment. Because $A$ is Hermitian and the $p_k$ are real, $\langle A\rangle\in\mathbb R$. **Properties** (all immediate from linearity of the inner product): - **Linearity:** $\langle aA + bB\rangle = a\langle A\rangle + b\langle B\rangle$ for real $a,b$. - **Reality:** $\langle A\rangle = \overline{\langle A\rangle}$ since $\langle\psi|A|\psi\rangle = \overline{\langle\psi|A^\dagger|\psi\rangle} = \overline{\langle A\rangle}$. - **Trace form:** $\langle A\rangle = \operatorname{Tr}(\rho A)$ with $\rho=|\psi\rangle\langle\psi|$ ([0.1.3 E5](../../term-0-foundations/c01-linear-algebra/03-dirac-notation.md)) — the form that generalizes to mixed states in [Term 1.5](../c05-density-matrices/01-density-operator.md). ### Variance and the standard deviation > **Definition.** The **variance** of $A$ in state $|\psi\rangle$ is > $$ (\Delta A)^2 \;:=\; \big\langle (A - \langle A\rangle I)^2\big\rangle = \langle A^2\rangle - \langle A\rangle^2, $$ > and the **uncertainty** (standard deviation) is $\Delta A = \sqrt{(\Delta A)^2}\ge 0$. The two expressions agree by expanding and using linearity (write $\mu=\langle A\rangle$, a real scalar): $$ \big\langle (A-\mu I)^2\big\rangle = \langle A^2 - 2\mu A + \mu^2 I\rangle = \langle A^2\rangle - 2\mu\langle A\rangle + \mu^2 = \langle A^2\rangle - \mu^2 . $$ Define the **centered operator** $\bar A := A - \langle A\rangle I$ (Hermitian, with $\langle\bar A\rangle=0$). Then a clean restatement we will use in the derivation: $$ (\Delta A)^2 = \langle \bar A^2\rangle = \langle\psi|\bar A^\dagger \bar A|\psi\rangle = \|\bar A|\psi\rangle\|^2 . $$ So **the variance is a squared norm** — automatically $\ge 0$, and **zero iff $\bar A|\psi\rangle = 0$**, i.e. iff $A|\psi\rangle = \langle A\rangle|\psi\rangle$: the state is an eigenstate and the value is definite. This recovers the [Lesson 1](01-projective-measurement.md) observation that eigenstates give deterministic outcomes, now phrased as $\Delta A = 0$. ### The Robertson uncertainty relation > **Theorem (Robertson, 1929).** For any two observables $A=A^\dagger$, $B=B^\dagger$ and any normalized > state $|\psi\rangle$, > $$ \boxed{\;\Delta A\,\Delta B \;\ge\; \tfrac12\big|\langle[A,B]\rangle\big|\;}. $$ **Derivation.** Let $\bar A = A-\langle A\rangle I$ and $\bar B = B-\langle B\rangle I$ (both Hermitian), and define the vectors $$ |f\rangle := \bar A|\psi\rangle, \qquad |g\rangle := \bar B|\psi\rangle . $$ By the variance-as-norm identity, $(\Delta A)^2 = \langle f|f\rangle$ and $(\Delta B)^2 = \langle g|g\rangle$. The **Cauchy–Schwarz inequality** ([0.1.2](../../term-0-foundations/c01-linear-algebra/02-inner-products-norms.md)) gives $$ (\Delta A)^2 (\Delta B)^2 = \langle f|f\rangle\langle g|g\rangle \;\ge\; |\langle f|g\rangle|^2 . \tag{1} $$ Now compute the cross term $\langle f|g\rangle = \langle\psi|\bar A\bar B|\psi\rangle$ (using $\bar A^\dagger=\bar A$). Split any product into its Hermitian (anticommutator) and anti-Hermitian (commutator) parts: $$ \bar A\bar B = \tfrac12\{\bar A,\bar B\} + \tfrac12[\bar A,\bar B]. $$ Take expectations. The anticommutator $\{\bar A,\bar B\}$ is Hermitian, so $\langle\{\bar A,\bar B\}\rangle$ is **real**; the commutator $[\bar A,\bar B]$ is **anti**-Hermitian ($[\bar A,\bar B]^\dagger = -[\bar A,\bar B]$), so $\langle[\bar A,\bar B]\rangle$ is **purely imaginary**. Hence $$ \langle f|g\rangle = \underbrace{\tfrac12\langle\{\bar A,\bar B\}\rangle}_{\text{real}} + \underbrace{\tfrac12\langle[\bar A,\bar B]\rangle}_{\text{imaginary}}, $$ and because the real and imaginary parts add in quadrature, $$ |\langle f|g\rangle|^2 = \tfrac14\big|\langle\{\bar A,\bar B\}\rangle\big|^2 + \tfrac14\big|\langle[\bar A,\bar B]\rangle\big|^2 \;\ge\; \tfrac14\big|\langle[\bar A,\bar B]\rangle\big|^2 . \tag{2} $$ Finally, the constant shifts cancel in the commutator: $[\bar A,\bar B] = [A - \langle A\rangle I,\,B-\langle B\rangle I] = [A,B]$ (the identity commutes with everything). Chaining (1) and (2), $$ (\Delta A)^2(\Delta B)^2 \;\ge\; |\langle f|g\rangle|^2 \;\ge\; \tfrac14|\langle[A,B]\rangle|^2 . $$ Taking nonnegative square roots gives $\Delta A\,\Delta B \ge \tfrac12|\langle[A,B]\rangle|$. ∎ **Reading the bound.** The right-hand side is *state-dependent* — it is $\tfrac12|\langle\psi|[A,B]|\psi\rangle|$, not a universal constant (except in special cases like $[x,p]=iI$, where it becomes the Heisenberg constant $\tfrac12$). Dropping the anticommutator term in (2) is the only inequality besides Cauchy–Schwarz; keeping it gives the tighter **Schrödinger uncertainty relation** (Exercise E5). > **Heisenberg as a special case.** For position and momentum, $[\hat x,\hat p] = i\hbar I$ (here > $\hbar=1$ per [Appendix C](../../appendix/C-notation-glossary.md)), so $\langle[\hat x,\hat p]\rangle = i$ > for *every* state and Robertson collapses to the familiar $\Delta x\,\Delta p\ge \tfrac12$. The qubit > Pauli version below is the finite-dimensional analog. ### Example structure: Pauli observables For the Paulis, $[X,Z] = XZ - ZX$. Using the Pauli algebra ([Appendix E](../../appendix/E-math-identities.md)), $XZ = -iY$ and $ZX = +iY$, so $$ [X,Z] = -iY - iY = -2iY . $$ (Equivalently from $[\sigma_j,\sigma_k]=2i\varepsilon_{jkl}\sigma_l$: $[X,Z]=2i\varepsilon_{xzy}Y = -2iY$.) Robertson then reads $$ \Delta X\,\Delta Z \;\ge\; \tfrac12|\langle -2iY\rangle| = |\langle Y\rangle| . $$ So the product of $X$- and $Z$-spreads is bounded below by the *magnitude of the $Y$-expectation*. On $|0\rangle$: $\langle X\rangle=\langle Y\rangle=0$, $\langle Z\rangle=1$, $\langle X^2\rangle=\langle Z^2\rangle=1$ (since $X^2=Z^2=I$). Thus $\Delta X = 1$, $\Delta Z = 0$, product $=0$, and the bound $|\langle Y\rangle|=0$ holds with equality — $Z$ is sharp, so the relation permits $X$ to be maximally spread. On $|{+}i\rangle = \tfrac1{\sqrt2}(|0\rangle+i|1\rangle)$ (the $+1$ eigenstate of $Y$): $\langle Y\rangle = 1$, $\langle X\rangle = \langle Z\rangle = 0$, so $\Delta X = \Delta Z = 1$ and the bound is again **saturated**: $1\cdot1 = |\langle Y\rangle| = 1$. ### Compatible vs. incompatible observables > **Definition.** Observables $A,B$ are **compatible** if $[A,B]=0$, else **incompatible**. **Theorem (simultaneous diagonalization).** Two Hermitian operators commute **iff** they share a common orthonormal eigenbasis. *Sketch.* ($\Leftarrow$) If both are diagonal in $\{|v_k\rangle\}$ they trivially commute. ($\Rightarrow$) If $[A,B]=0$ and $A|v\rangle=\lambda|v\rangle$, then $A(B|v\rangle) = B(A|v\rangle) = \lambda(B|v\rangle)$, so $B$ preserves each $\lambda$-eigenspace of $A$; restrict $B$ (still Hermitian) to that eigenspace and diagonalize it there, repeating per eigenspace. The combined eigenvectors diagonalize both. ∎ Consequences for measurement: - **Compatible** ($[A,B]=0$): there is a common eigenbasis $\{|v_k\rangle\}$ with $A|v_k\rangle=a_k|v_k\rangle$ and $B|v_k\rangle=b_k|v_k\rangle$. A single projective measurement in that basis returns *both* values $(a_k,b_k)$ at once — they are **jointly measurable**, the order does not matter, and Robertson's bound is vacuous ($\langle[A,B]\rangle=0$, allowing $\Delta A=\Delta B=0$ simultaneously, e.g. on a common eigenstate). - **Incompatible** ($[A,B]\ne0$): no common eigenbasis, measuring one disturbs the other, and Robertson forces $\Delta A\,\Delta B > 0$ on any state where $\langle[A,B]\rangle\ne0$. $X$ and $Z$ are the canonical example. This is the precise statement of "you can't measure both at once": not a limit of apparatus, but the non-existence of a basis in which both are diagonal. Generalized (non-projective) measurements can *trade off* information about incompatible observables — the subject of [Lesson 3](03-povm-generalized.md). --- ## Worked Examples ### Example 1 — Mean and variance of $Z$ on a general qubit Let $|\psi\rangle = \cos\tfrac\theta2|0\rangle + e^{i\phi}\sin\tfrac\theta2|1\rangle$ (Bloch angles; [Term 1.2](../c02-qubits-bloch/02-bloch-sphere.md)). Then with $Z=\operatorname{diag}(1,-1)$, $$ \langle Z\rangle = \cos^2\tfrac\theta2 - \sin^2\tfrac\theta2 = \cos\theta . $$ Since $Z^2 = I$, $\langle Z^2\rangle = 1$, so $$ (\Delta Z)^2 = \langle Z^2\rangle - \langle Z\rangle^2 = 1 - \cos^2\theta = \sin^2\theta, \qquad \Delta Z = |\sin\theta| . $$ The spread vanishes at the poles ($\theta=0,\pi$, the eigenstates $|0\rangle,|1\rangle$) and is maximal on the equator ($\theta=\pi/2$, e.g. $|+\rangle$), where $\Delta Z = 1$ — matching Example 1 of [Lesson 1](01-projective-measurement.md). ### Example 2 — Saturating $\Delta X\,\Delta Z \ge |\langle Y\rangle|$ Take $|\psi\rangle = |{+}i\rangle = \tfrac1{\sqrt2}(|0\rangle + i|1\rangle)$, the $+1$-eigenstate of $Y$. **Expectations.** $\langle X\rangle = \langle{+}i|X|{+}i\rangle = 0$ and $\langle Z\rangle = 0$ (the Bloch vector points along $+y$, so $x$- and $z$-components vanish), while $\langle Y\rangle = 1$. **Variances.** $X^2=Z^2=I\Rightarrow\langle X^2\rangle=\langle Z^2\rangle=1$, so $(\Delta X)^2 = 1 - 0 = 1$ and likewise $(\Delta Z)^2 = 1$; hence $\Delta X=\Delta Z=1$. **Bound.** $\Delta X\,\Delta Z = 1$ and the Robertson right-hand side is $\tfrac12|\langle[X,Z]\rangle| = \tfrac12|\langle -2iY\rangle| = |\langle Y\rangle| = 1$. The inequality is an **equality** — $|{+}i\rangle$ is a *minimum-uncertainty state* for the pair $(X,Z)$. (Geometrically, maximizing $|\langle Y\rangle|$ while keeping $\langle X\rangle=\langle Z\rangle=0$ pushes both spreads to their largest equal value, and Cauchy–Schwarz is tight because $|f\rangle$ and $|g\rangle$ are proportional — see the NumPy check below.) --- ## Hands-on (Python) We compute expectations and variances exactly, then **numerically verify the Robertson bound over many random states** — and find the states that saturate it. ```python import numpy as np I = np.eye(2, dtype=complex) X = np.array([[0, 1], [1, 0]], dtype=complex) Y = np.array([[0, -1j], [1j, 0]], dtype=complex) Z = np.array([[1, 0], [0, -1]], dtype=complex) def expectation(A, psi): """ = , returned as a real float (A is Hermitian).""" return np.real(psi.conj() @ (A @ psi)) def variance(A, psi): """(ΔA)^2 = - ^2.""" return expectation(A @ A, psi) - expectation(A, psi) ** 2 def uncertainty(A, psi): return np.sqrt(max(variance(A, psi), 0.0)) # clip tiny negatives from roundoff def random_qubit(rng): """Haar-ish random pure qubit: normalize a complex Gaussian vector.""" v = rng.normal(size=2) + 1j * rng.normal(size=2) return v / np.linalg.norm(v) # Commutator [X, Z] = -2 i Y (sanity check) comm_XZ = X @ Z - Z @ X print("[X,Z] == -2iY :", np.allclose(comm_XZ, -2j * Y)) # True ``` ```python # Verify Robertson: ΔA ΔB >= (1/2)|<[A,B]>| for many random states and Pauli pairs rng = np.random.default_rng(1) pairs = {"(X,Z)": (X, Z), "(X,Y)": (X, Y), "(Y,Z)": (Y, Z)} worst_slack = {} for name, (A, B) in pairs.items(): comm = A @ B - B @ A min_slack = np.inf for _ in range(100_000): psi = random_qubit(rng) lhs = uncertainty(A, psi) * uncertainty(B, psi) rhs = 0.5 * np.abs(psi.conj() @ (comm @ psi)) # (1/2)|<[A,B]>| slack = lhs - rhs assert slack >= -1e-9, f"Robertson violated for {name}: {slack}" min_slack = min(min_slack, slack) worst_slack[name] = min_slack print("Min (LHS - RHS) over 100k random states (>=0 confirms the bound):") for name, s in worst_slack.items(): print(f" {name}: {s:.2e}") # ~0 -> the bound is tight (saturated somewhere) ``` ```python # Saturation: |+i> is a minimum-uncertainty state for (X, Z). plus_i = np.array([1, 1j], dtype=complex) / np.sqrt(2) dX, dZ = uncertainty(X, plus_i), uncertainty(Z, plus_i) rhs = abs(expectation(Y, plus_i)) # (1/2)|<[X,Z]>| = || print(f"ΔX·ΔZ = {dX*dZ:.4f}, || = {rhs:.4f} -> equality:", np.isclose(dX * dZ, rhs)) # True # Why it saturates: f = (X-)|ψ> and g = (Z-)|ψ> are proportional (Cauchy–Schwarz tight). f = (X - expectation(X, plus_i) * I) @ plus_i g = (Z - expectation(Z, plus_i) * I) @ plus_i ratio = f / g # constant (up to roundoff) where g != 0 print("f ∝ g :", np.allclose(ratio[0], ratio[1])) # True ``` > The empirical loop is a *property test*: we never construct an adversarial state, we sample broadly > and assert the inequality holds (with `min_slack ≈ 0` proving it is tight, not loose). This is the > standard way to gain confidence in an analytic bound before trusting a derivation in code. --- ## Exercises **E1 (easy).** For $|\psi\rangle = |+\rangle$, compute $\langle X\rangle$, $\langle Z\rangle$, and the variances $(\Delta X)^2$, $(\Delta Z)^2$.

$|+\rangle$ is the $+1$-eigenstate of $X$, so $\langle X\rangle = 1$ and (since $X^2=I$) $(\Delta X)^2 = 1 - 1 = 0$. For $Z$: $\langle Z\rangle = \tfrac12(1) + \tfrac12(-1) = 0$, $\langle Z^2\rangle=1$, so $(\Delta Z)^2 = 1$. The $X$ value is sharp, the $Z$ value maximally spread — consistent with $|+\rangle$ being an $X$-eigenstate.

**E2 (easy).** Show that $\Delta A = 0$ if and only if $|\psi\rangle$ is an eigenstate of $A$.

Solution

**E3 (medium).** Verify the Robertson bound for $A=X$, $B=Z$ on the state $|0\rangle$, and explain why it holds with equality despite $X$ and $Z$ not commuting.

Solution

On $|0\rangle$: $\langle X\rangle=0\Rightarrow(\Delta X)^2 = \langle X^2\rangle = 1$, so $\Delta X=1$; $\langle Z\rangle = 1\Rightarrow(\Delta Z)^2 = 1-1 = 0$, so $\Delta Z=0$. Product $=0$. RHS: $\tfrac12|\langle[X,Z]\rangle| = |\langle 0|Y|0\rangle| = 0$. Equality holds because the *bound itself* vanishes here: $\langle Y\rangle = 0$ on $|0\rangle$. Non-commutativity forbids both being sharp *only where $\langle[A,B]\rangle\ne0$*; on $|0\rangle$ the commutator's expectation is zero, so one observable ($Z$) is allowed to be perfectly sharp.

**E4 (medium).** Prove that if $[A,B]=0$ (with $A,B$ Hermitian) then there exists a state with $\Delta A = \Delta B = 0$ simultaneously, and conversely give a pair where no such state exists.

Solution

If $[A,B]=0$ they share an orthonormal eigenbasis $\{|v_k\rangle\}$ (simultaneous-diagonalization theorem). Take $|\psi\rangle=|v_k\rangle$: it is an eigenstate of both, so $\Delta A=\Delta B=0$ by E2. Conversely, for $A=X$, $B=Z$ ($[X,Z]=-2iY\ne0$): a common-zero-spread state would be a simultaneous eigenstate, hence a common eigenvector, which would force $X,Z$ to commute — contradiction. So no qubit state has $\Delta X=\Delta Z=0$ together. ∎

**E5 (hard).** Derive the **Schrödinger uncertainty relation** by keeping the anticommutator term: $$ (\Delta A)^2(\Delta B)^2 \ge \Big|\tfrac12\langle\{A,B\}\rangle - \langle A\rangle\langle B\rangle\Big|^2 + \Big|\tfrac1{2i}\langle[A,B]\rangle\Big|^2 . $$

Solution

From the derivation, $\langle f|g\rangle = \tfrac12\langle\{\bar A,\bar B\}\rangle + \tfrac12\langle[\bar A,\bar B]\rangle$ with the first term real and the second imaginary. Cauchy–Schwarz gives $(\Delta A)^2(\Delta B)^2 \ge |\langle f|g\rangle|^2 = \big(\tfrac12\langle\{\bar A,\bar B\}\rangle\big)^2 + \big(\tfrac1{2i}\langle[\bar A,\bar B]\rangle\big)^2$ (real$^2$ + imag$^2$, writing the imaginary part as $\tfrac1{2i}\langle[\bar A,\bar B]\rangle\in\mathbb R$). Now $[\bar A,\bar B]=[A,B]$, and the anticommutator expands as $\langle\{\bar A,\bar B\}\rangle = \langle\{A,B\}\rangle - 2\langle A\rangle\langle B\rangle$ (the cross terms; check by expanding $\bar A = A-\langle A\rangle I$). Substituting yields the stated relation. Dropping the (nonnegative) anticommutator/covariance term recovers Robertson. ∎