# Inner Products & Norms > The inner product is how quantum mechanics measures *overlap* between states — and overlap is > probability. The Born rule, orthogonality of distinguishable outcomes, normalization of states, > and the geometry of the Bloch sphere all reduce to the single structure we build here. Get the > **conjugate-linear** convention right and everything downstream is clean. ## Learning Objectives After this lesson you will be able to: 1. State the axioms of a complex inner product and compute $\langle u, v\rangle$ on $\mathbb{C}^n$. 2. Prove and apply the **Cauchy–Schwarz** and **triangle** inequalities. 3. Test vectors for **orthogonality** and **normalize** them. 4. Run **Gram–Schmidt** to produce an orthonormal basis. 5. Expand a vector in an orthonormal basis via projection, and state **Parseval's identity**. --- ## Intuition In $\mathbb{R}^n$ the dot product $u \cdot v = \sum_k u_k v_k$ measures how much two vectors "agree": it's largest when they point the same way, zero when perpendicular. It also defines length: $\|u\| = \sqrt{u\cdot u}$. Over $\mathbb{C}$ we need a twist. If we naïvely used $\sum_k u_k v_k$, the "length squared" of $v = (i, 0)^T$ would be $i^2 + 0 = -1 < 0$ — nonsense for a length. The fix is to **conjugate one factor**: $\langle v, v\rangle = \sum_k v_k^* v_k = \sum_k |v_k|^2 \ge 0$. That single conjugation is the whole story of complex inner products, and it's exactly what makes $|\langle i\text{-state}\rangle|^2$ a valid probability later. --- ## Theory ### Definition: complex inner product An **inner product** on a complex vector space $V$ is a map $\langle\cdot,\cdot\rangle : V\times V \to \mathbb{C}$ satisfying, for all $u,v,w \in V$ and $a \in \mathbb{C}$: $$ \begin{aligned} &\textbf{(I1) conjugate symmetry:} && \langle u, v\rangle = \overline{\langle v, u\rangle} \\ &\textbf{(I2) linearity in the 2nd argument:} && \langle u,\, a v + w\rangle = a\langle u,v\rangle + \langle u,w\rangle \\ &\textbf{(I3) positive-definiteness:} && \langle v, v\rangle \ge 0,\quad \langle v,v\rangle = 0 \iff v = 0 \end{aligned} $$ > **Convention (physics / this program).** We make the inner product **linear in the second > argument** and **conjugate-linear in the first**: $\langle a u, v\rangle = \overline{a}\langle u,v\rangle$, > which follows from (I1)+(I2). This matches Dirac notation $\langle\phi|\psi\rangle$ (next lesson) and > [NC]. Many *mathematics* texts (e.g. [Axl]) put linearity in the first argument instead — be alert > when reading across sources. See [Appendix C](../../appendix/C-notation-glossary.md). A vector space with an inner product is an **inner product space**. (Finite-dimensional inner product spaces over $\mathbb{C}$ are automatically complete, hence are **Hilbert spaces** — the term you'll hear constantly. Completeness only bites in infinite dimensions, which we avoid.) ### The standard inner product on $\mathbb{C}^n$ $$ \langle u, v\rangle := \sum_{k=1}^{n} \overline{u_k}\, v_k = u^\dagger v, $$ where $u^\dagger = \overline{u}^{\,T}$ is the **conjugate transpose** (a row vector). Check the axioms: (I1) $\overline{\langle v,u\rangle} = \overline{\sum \overline{v_k} u_k} = \sum v_k \overline{u_k} = \langle u,v\rangle$; (I2) is linearity of the sum in $v$; (I3) $\langle v,v\rangle = \sum |v_k|^2 \ge 0$ with equality iff every $v_k = 0$. > ⚠️ **NumPy:** `np.vdot(u, v)` conjugates its **first** argument, matching our convention exactly. > `np.dot` does **not** conjugate — wrong for complex vectors. ### Norm The inner product induces a **norm** (length): $$ \|v\| := \sqrt{\langle v, v\rangle} = \sqrt{\textstyle\sum_k |v_k|^2}. $$ A vector with $\|v\| = 1$ is a **unit vector**; **normalizing** means $v \mapsto v / \|v\|$. **Physical quantum states are unit vectors** — this is the normalization condition $\langle\psi|\psi\rangle = 1$ you'll impose in [Term 1](../../term-1-quantum-mechanics/c01-postulates/01-state-postulate.md). ### Cauchy–Schwarz inequality For all $u, v$ in an inner product space, $$ \boxed{\,|\langle u, v\rangle| \le \|u\|\,\|v\|\,} $$ with equality iff $u$ and $v$ are linearly dependent. **Proof.** If $v = 0$ both sides are $0$. Otherwise, for any $\lambda \in \mathbb{C}$, $0 \le \|u - \lambda v\|^2 = \langle u-\lambda v,\, u-\lambda v\rangle$. Expand using the axioms: $$ 0 \le \langle u,u\rangle - \overline{\lambda}\langle v,u\rangle - \lambda\langle u,v\rangle + |\lambda|^2\langle v,v\rangle. $$ Choose $\lambda = \dfrac{\langle v, u\rangle}{\langle v, v\rangle}$ (the projection coefficient). Using $\langle v,u\rangle = \overline{\langle u,v\rangle}$, the cross terms combine and simplify to $$ 0 \le \|u\|^2 - \frac{|\langle u,v\rangle|^2}{\|v\|^2} \;\;\Longrightarrow\;\; |\langle u,v\rangle|^2 \le \|u\|^2\|v\|^2 . $$ Taking square roots gives the result. Equality holds iff $u - \lambda v = 0$, i.e. $u,v$ are dependent. ∎ Cauchy–Schwarz is *the* workhorse inequality of quantum information — it bounds state overlaps, underlies the uncertainty principle (Term 1.3), and guarantees fidelities lie in $[0,1]$. ### Triangle inequality $\|u + v\| \le \|u\| + \|v\|$. **Proof:** $\|u+v\|^2 = \|u\|^2 + 2\,\mathrm{Re}\langle u,v\rangle + \|v\|^2 \le \|u\|^2 + 2|\langle u,v\rangle| + \|v\|^2 \le \|u\|^2 + 2\|u\|\|v\| + \|v\|^2 = (\|u\|+\|v\|)^2$, using $\mathrm{Re}(z)\le|z|$ then Cauchy–Schwarz. ∎ ### Orthogonality and orthonormal bases - $u \perp v$ (**orthogonal**) means $\langle u, v\rangle = 0$. Physically, orthogonal states are **perfectly distinguishable** by a measurement. - A set $\{e_1,\dots,e_n\}$ is **orthonormal** if $\langle e_i, e_j\rangle = \delta_{ij}$ (orthogonal and each of unit norm). - Orthonormal sets are automatically independent, so an orthonormal spanning set is an **orthonormal basis (ONB)**. **Expansion in an ONB.** If $\{e_k\}$ is an ONB, then for any $v$, $$ v = \sum_k \langle e_k, v\rangle\, e_k . $$ The coordinate is just the projection $\langle e_k, v\rangle$ — *no linear system to solve.* This is why orthonormal bases are so convenient, and why the computational basis $\{|0\rangle,\dots\}$ is chosen orthonormal. The squared coordinates obey **Parseval's identity**: $$ \|v\|^2 = \sum_k |\langle e_k, v\rangle|^2 . $$ For a normalized state and a measurement basis $\{e_k\}$, the terms $|\langle e_k, v\rangle|^2$ are exactly the **Born-rule probabilities** and Parseval says they sum to $1$. ### Gram–Schmidt orthonormalization Any basis can be turned into an orthonormal one. Given independent $\{v_1,\dots,v_n\}$, define $$ u_k = v_k - \sum_{j 1e-12: # skip vectors that are (numerically) dependent basis.append(w / n) return basis E = gram_schmidt([np.array([1, 1]), np.array([1, 0])]) for e in E: print(np.round(e, 3)) # [0.707+0.j 0.707+0.j] -> |+> # [ 0.707+0.j -0.707+0.j] -> |-> # Verify orthonormality: the Gram matrix G[i,j] = ⟨e_i, e_j⟩ should be the identity. G = np.array([[inner(a, b) for b in E] for a in E]) print(np.allclose(G, np.eye(len(E)))) # True ``` ```python # Expansion in an ONB via projection (no linear solve needed): v = np.array([2, 3j], dtype=complex) coords = [inner(e, v) for e in E] # ⟨e_k, v⟩ recon = sum(c * e for c, e in zip(coords, E)) print(np.allclose(recon, v)) # True print(np.isclose(norm(v)**2, sum(abs(c)**2 for c in coords))) # Parseval: True ``` --- ## Exercises **E1 (easy).** Compute $\langle u, v\rangle$, $\|u\|$, and $\|v\|$ for $u = (1, i)^T$, $v = (2, -i)^T$, and verify Cauchy–Schwarz numerically.
Solution $\langle u,v\rangle = \overline{1}\cdot 2 + \overline{i}\cdot(-i) = 2 + (-i)(-i) = 2 + i^2 = 2 - 1 = 1$. $\|u\| = \sqrt{1+1} = \sqrt2$, $\|v\| = \sqrt{4+1} = \sqrt5$. Then $|\langle u,v\rangle| = 1 \le \sqrt{10} = \|u\|\|v\|$. ✓
**E2 (easy).** Show $|+\rangle = \tfrac1{\sqrt2}(1,1)^T$ and $|-\rangle = \tfrac1{\sqrt2}(1,-1)^T$ form an ONB, and expand $|0\rangle = (1,0)^T$ in it.
Solution $\langle +|-\rangle = \tfrac12(1\cdot1 + 1\cdot(-1)) = 0$ and both have norm $1$, so they're an ONB. Coordinates: $\langle +|0\rangle = \tfrac1{\sqrt2}$, $\langle -|0\rangle = \tfrac1{\sqrt2}$. Hence $|0\rangle = \tfrac1{\sqrt2}|+\rangle + \tfrac1{\sqrt2}|-\rangle$. (Born check: each outcome probability is $\tfrac12$, summing to 1.)
**E3 (medium).** Prove the **parallelogram law** $\|u+v\|^2 + \|u-v\|^2 = 2\|u\|^2 + 2\|v\|^2$ from the inner-product axioms.
Solution Expand both: $\|u\pm v\|^2 = \|u\|^2 \pm 2\,\mathrm{Re}\langle u,v\rangle + \|v\|^2$. Adding the $+$ and $-$ versions cancels the cross terms, leaving $2\|u\|^2 + 2\|v\|^2$. ∎
**E4 (medium).** When does equality $|\langle u,v\rangle| = \|u\|\|v\|$ hold? Prove your claim and give the physical meaning for normalized quantum states.
Solution Equality iff $u - \lambda v = 0$ for the $\lambda$ in the Cauchy–Schwarz proof, i.e. $u,v$ are linearly dependent ($u = \lambda v$). For normalized states $|\langle\psi|\phi\rangle| = 1$ means $|\psi\rangle = e^{i\theta}|\phi\rangle$ — they are the *same physical state up to global phase* and are **indistinguishable** by any measurement.
**E5 (hard).** Let $\{e_k\}$ be an ONB and define the **Gram matrix** of vectors $\{v_i\}$ by $G_{ij} = \langle v_i, v_j\rangle$. Prove $\{v_i\}$ is linearly independent iff $G$ is invertible.
Solution Suppose $\sum_i c_i v_i = 0$ for some coefficients $c=(c_i)$. Take $\langle v_j, \cdot\rangle$ of both sides: $\sum_i \langle v_j, v_i\rangle c_i = 0$ for all $j$, i.e. $G^\top$... more carefully, with our convention $\langle v_j, \sum_i c_i v_i\rangle = \sum_i c_i G_{ji} = (Gc)_j = 0$, so $Gc = 0$. If $G$ is invertible the only solution is $c=0$ ⇒ independence. Conversely if the $v_i$ are dependent, a nonzero $c$ gives $\sum c_i v_i = 0$, hence $Gc = 0$ with $c \neq 0$, so $G$ is singular. ∎ (This is exactly the `matrix_rank` independence test, recast through the inner product.)
--- ## Checkpoint 1. Why must a complex inner product conjugate one argument? What goes wrong otherwise? 2. State Cauchy–Schwarz and its equality condition; what does equality mean for quantum states? 3. What does it mean physically for two states to be orthogonal? 4. Why is expanding a vector in an *orthonormal* basis easier than in a general basis? 5. State Parseval's identity and connect it to the Born rule summing to $1$.
Answers 1. Without conjugation $\langle v,v\rangle$ could be negative/complex (e.g. $v=(i,0)$ gives $-1$), so it couldn't define a length or a probability. Conjugating gives $\sum|v_k|^2 \ge 0$. 2. $|\langle u,v\rangle| \le \|u\|\|v\|$, equality iff $u,v$ dependent; for normalized states equality means equal up to global phase — physically identical, indistinguishable. 3. Orthogonal states are perfectly distinguishable: a measurement can tell them apart with certainty. 4. Coordinates are projections $\langle e_k, v\rangle$ — no linear system to solve — and Parseval gives the norm instantly. 5. $\|v\|^2 = \sum_k |\langle e_k,v\rangle|^2$. For a normalized state the right side is the sum of Born probabilities, which equals $\|v\|^2 = 1$.
--- ## Further Reading - **[Axl]** Axler, *Linear Algebra Done Right*, Ch. 6 — inner product spaces, Gram–Schmidt (note: linear in the *first* argument there). - **[NC]** Nielsen & Chuang, §2.1.4 — inner products, Gram–Schmidt, our conjugate convention. - **[Pre]** Preskill, *Ph219*, Ch. 2 — Hilbert spaces. --- ← Prev: [Complex Vector Spaces](01-complex-vector-spaces.md) · Up: [Term 0](../README.md) · Next: [Dirac Notation](03-dirac-notation.md) →