Dual Vectors

We have been treating a bra $\langle\psi|$ as "the conjugate transpose of the ket." That is correct in $\mathbb{C}^n$ , but the deeper truth is that bras are elements of a different space — the dual space — and the correspondence between kets and bras is a theorem, not a definition.

The dual space

Given a Hilbert space $\mathcal{H}$ , a linear functional is a map $f : \mathcal{H} \to \mathbb{C}$ that respects linear combinations:

f(\alpha|u\rangle + \beta|v\rangle) = \alpha\, f(|u\rangle) + \beta\, f(|v\rangle).

The set of all (continuous) linear functionals on $\mathcal{H}$ is itself a complex vector space, called the dual space $\mathcal{H}^*$ . Bras live here: $\langle\phi|$ is the functional that sends a ket $|\psi\rangle$ to the number $\langle\phi|\psi\rangle$ .

Every ket induces a bra

Fix a ket $|\phi\rangle$ . Define a functional by "take the inner product with $|\phi\rangle$ ":

f_\phi(|\psi\rangle) = \langle \phi | \psi \rangle.

This is linear in $|\psi\rangle$ by the inner-product axioms, so $f_\phi \in \mathcal{H}^*$ . We give it the name $\langle\phi|$ . Thus every ket produces a bra. Concretely, in $\mathbb{C}^n$ ,

|\phi\rangle = \begin{pmatrix} \gamma_1 \\ \vdots \\ \gamma_n \end{pmatrix} \quad\Longrightarrow\quad \langle\phi| = \begin{pmatrix} \overline{\gamma_1} & \cdots & \overline{\gamma_n} \end{pmatrix},

and applying the bra to a ket is row-times-column matrix multiplication.

The Riesz representation theorem

Does every bra come from some ket — is the correspondence onto? Yes. The Riesz representation theorem states that for any continuous linear functional $f$ on a Hilbert space there exists a unique ket $|\phi\rangle$ such that

f(|\psi\rangle) = \langle \phi | \psi \rangle \quad \text{for all } |\psi\rangle.

So the dual space $\mathcal{H}^*$ is in one-to-one correspondence with $\mathcal{H}$ : kets and bras are two faces of the same information. This is what justifies Dirac's notation, in which a bra is written by simply "flipping" its ket.

The correspondence is conjugate-linear (antilinear)

There is one subtlety that trips up newcomers. The map $|\psi\rangle \mapsto \langle\psi|$ is not linear — it is conjugate-linear. Scaling a ket by $\alpha$ scales its bra by $\overline{\alpha}$ :

|\psi\rangle \to \alpha|\psi\rangle \quad\Longrightarrow\quad \langle\psi| \to \overline{\alpha}\,\langle\psi|.

Likewise the dual of a sum is the sum of duals, but with conjugated coefficients:

\big(\alpha|u\rangle + \beta|v\rangle\big)^\dagger = \overline{\alpha}\,\langle u| + \overline{\beta}\,\langle v|.

This is exactly the conjugate that appears in the first slot of the inner product, and it is why taking the dual is the same operation as Hermitian conjugation (the dagger $\dagger$ ).

Worked example

Let $|\psi\rangle = (1 + i)|0\rangle + 2|1\rangle$ . Its dual bra is obtained by conjugating each coefficient:

\langle\psi| = (1 - i)\langle 0| + 2\langle 1|.

Then the norm-squared is the bra acting on the ket,

\langle\psi|\psi\rangle = (1-i)(1+i) + (2)(2) = (1 + 1) + 4 = 6,

using $\langle i|j\rangle = \delta_{ij}$ . The conjugation turned $(1+i)$ paired with $(1+i)$ into the real number $|1+i|^2 = 2$ , as positive-definiteness requires.

The takeaway

Bras are not "row-vector versions" of kets by fiat — they are elements of the dual space, the space of linear functionals. The Riesz theorem guarantees a one-to-one, conjugate-linear correspondence between kets and bras, which is the rigorous backbone of Dirac notation and the origin of the conjugation in the inner product.