The instructor in my quantum computation course sometimes uses the equivalence $$(\left|a\right>)^\dagger\equiv\left<a\right|$$ I understand that this is true for the typical matrix implementation of Dirac's bra-ket notation, and taking $x^\dagger\equiv\bar{x}^T$; but does it follow from the definition of an inner product space and the adjoint? Should I be able to derive this as a fact from definition of an inner product space and the adjoint of linear operators on that space or is it simply a notational convention?


3 Answers 3


$\newcommand{\ip}[1]{\left\langle{#1}\right\rangle}$Actually, you can view a bra vector as the honest-to-goodness adjoint of the corresponding ket vector, but it requires a trick.

So, let $V$ be a finite-dimensional complex inner product space. The trick is that you have a canonical isomorphism $\Phi : V \to L(\mathbb{C},V)$ given by $\Phi(v) : \lambda \mapsto \lambda v$ for all $v \in V$; indeed, one has that $\Phi^{-1}(s) = s(1)$. Now, since $\mathbb{C} = \mathbb{C}^1$ is also an inner product space, for any $v \in V$ we can form the adjoint $\Phi(v)^\ast \in L(V,\mathbb{C}) = V^\ast$ of $\Phi(v)$, and lo and behold, for any $w \in V$, $$ \Phi(v)^\ast(w) = \ip{1,\Phi(v)^\ast(w)}_{\mathbb{C}} = \ip{\Phi(v)(1),w}_V = \ip{v,w}_V, $$ so that $\Phi(v)^\ast : w \mapsto \ip{v,w}_V$, as required. Thus, up to application of a canonical isomorphism, a bra vector really is the adjoint of the corresponding ket vector.

$\newcommand{\ket}[1]{\left|{#1}\right\rangle} \newcommand{\bra}[1]{\left\langle{#1}\right|} $ADDENDUM: In more physics-friendly notation, here's what's going on. Let $H$ be your (finite-dimensional) Hilbert space, and let $\ket{a} \in H$. You can interpret $\ket{a}$, in a completely natural way, as defining a linear transformation $\Phi[\ket{a}] : \mathbb{C} \to H$ by $\Phi[\ket{a}](\lambda) := \lambda \ket{a}$. The Hermitian conjugate of $\Phi[\ket{a}]$, then, is a linear transformation $\Phi[\ket{a}]^\dagger : H \to \mathbb{C}$, so that $\Phi[\ket{a}]^\dagger$ is simply a bra vector. The computation above then shows that $\Phi[\ket{a}]^\dagger = \bra{a}$. Thus, as long as you're fine with identifying $\ket{a}$ with $\Phi[\ket{a}]$ (which is actually completely rigorous), you do indeed have that $\bra{a} = \ket{a}^\dagger$.

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    $\begingroup$ That sounds great. I wish I understood better what it means. $\endgroup$
    – orome
    Commented Mar 27, 2013 at 19:47
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    $\begingroup$ The thing that's great about this trick is that it naturally relates back to the notation: if you regard everything as a map, the pointiness of something's right-hand symbol tells you its domain and the pointiness of its left-hand symbol tells you its range. (Though it's one of those things that'd be even better if we wrote function application on the right...) $\endgroup$
    – Micah
    Commented Mar 27, 2013 at 19:57
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    $\begingroup$ @raxacoricofallapatorius I've added an addendum, giving a summary in Dirac notation. $\endgroup$ Commented Mar 27, 2013 at 19:59
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    $\begingroup$ Only if you restrict to continuous linear functionals---this is the analytic content of the Riesz representation theorem. That said, if you fix a dense subspace $\mathcal{S}$ of your Hilbert space $H$ and equip it with a finer topology (e.g., the Schwarz space $\mathcal{S}(\mathbb{R})$ inside $L^2(\mathbb{R})$), you can now conjugate-linearly identify the continuous dual $\mathcal{S}^\prime$ of $\mathcal{S}$ with an enlargement of $H$ containing additional "generalised vectors"---this is the famous formalism of "rigged Hilbert spaces". $\endgroup$ Commented Mar 23, 2023 at 14:04
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    $\begingroup$ Correct. Note that any linear map from a finite-dimensional Hilbert space to any Hilbert space is automatically bounded. $\endgroup$ Commented Mar 23, 2023 at 14:11

Given a complex Hilber space $H$ we can define the complex conjugate of $H$, $\overline{H}$, which is all the same than $H$ except that the scalar multiplication by $z\in \mathbb{C}$ is changed by the conjugate $\overline{z}$. Obviously, there is a conjugate linear isomorphism between $H$ and $\overline{H}$

On the other hand, we can define the continuous dual $H^*$ of $H$ like the vector space of all continuous and linear maps from $H$ to $\mathbb{C}$. The inner product gives rise to a morphism from $H$ to his dual $H^*$ that is conjugate-linear:

$$ \begin{array}{rcc} \Phi: H & \longmapsto & H^*\\ v & \longmapsto & \langle v,\cdot \rangle \end{array} $$

The Riesz representation theorem tell us that $\Phi$ is an isomorphism (this is only evident in the finite dimensional case). Moreover, it is an isometry respect to the norm.

In conclusion, $H^* \cong\overline{H}$, since the composition of two conjugate-linear maps is a genuine linear map. This conclusion is what justifies the bra-ket notation: For an element $v\in H$ we will write $|v \rangle$ and for $\Phi(v)\in H^*$ we will write $\langle v|$. This way, for $v,w\in H$, you know that $\Phi(v)(w)=<v,w>$, but in the new notation is, directly $$ \langle v|w \rangle=\langle v,w \rangle $$

Observe that to $z\cdot| v \rangle$ we associate $\langle v |\cdot\overline{z}$. The scalar is at the right because is an habit of physicist.

A linear map between two complex Hilbert spaces $f: H_1\mapsto H_2$ gives rise to other linear map $$ \overline{f}: \overline{H_1} \longmapsto \overline{H_2} $$

Let us identify $\overline{H_i}$ with $H_i^*$ and call $\Phi_i$ to the conjugate linear isomorphism from $H_i$ to $H_i^*$ (for example, $\Phi_1(v)=<v,\cdot >$). We will take $\overline{f}=\Phi_2\circ f \circ \Phi_1^{-1}$.

We are going to study this in the bra-ket notation. Suppose $|v\rangle \in H_1$ such that $f(|v\rangle)=|w \rangle$, what is $\overline{f}(\langle v |)$? Obvisouly, is $\langle w |$. But let $M$ be the matrix of $f$ respect to the basis $\{|e_i\rangle\} \subset H_1$, $\{|g_j\rangle\} \subset H_2$. What is the marix of $\overline{f}$ respect to $\{\langle e_i|\} \subset H_1^*$, $\{\langle g_j|\} \subset H_2$?

If we take, say for example, $\langle e_1|$ and follow the compositions that produce $\overline{f}$ we obtain $$ \overline{f}(\langle e_1|)=\langle \sum_j m_{j1}g_j|= \sum_j \langle g_j|\cdot \overline{m}_{j1} $$

But wait a moment. Physicist has the habit that represent $|v\rangle$ like a column vector, and $\langle v|$ like a row vector; and, in the latter case, apply the matrix of linear maps multiplying by the right. Putting all together we obtain that the matrix of $\overline{f}$ is $$ M^{\dagger}:=\overline{M}^T $$ which is calle the Hermitian conjugate of $M$.

And we can write that to $M|v\rangle$ we associate $$ \langle v | M^{\dagger} $$


If $L:V\to W$ is linear then the adjoint of $L$ is a linear map from the dual of $W$ to the dual of $V$. So I would argue that linear maps have adjoints, but vectors do not. I do not think either bra or ket is a linear map.

The correspondence that sends $|a\rangle$ to $\langle a|$ is an isomorphism from a vector space into its dual space. In projective geometry this might be called a correlation (but I have not seen the term used outside this context). Any inner product on a vector space $V$ determines an isomorphism for $V$ to its dual.

It is important to bear in mind that physicists have their own way of thinking about linear algebra, and this can be quite different from what you will meet in mathematics courses. (For example, one physics text writes: "a vector in $\mathbb{R}^3$ is not just a triple of numbers, it has a meaning".)

  • $\begingroup$ So the notation above is (as used here, for example, in section 7) is nothing more than a kind of consistent shorthand for "do whatever is necessary to generate the dual", right? $\endgroup$
    – orome
    Commented Mar 27, 2013 at 19:19
  • $\begingroup$ Yes, I think you are correct. One problem for a mathematician in all this is that, in those notes, a ket is defined as a string of three symbols with certain rules of syntax that produce a bra from ket. So a ket is something on a page, which is just bizarre mathematically. $\endgroup$ Commented Mar 27, 2013 at 20:43

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