Is a bra the adjoint of a ket?

Solution 1:

$\newcommand{\ip}[1]{\left\langle{#1}\right\rangle}$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 $\bra{a} = \ket{a}^\dagger$.

Solution 2:

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 $$ \langle v | M^{\dagger} $$