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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?

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$\newcommand{\ip}[1]{\left\langle{#1}\right\rangle}$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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That sounds great. I wish I understood better what it means. –  raxacoricofallapatorius Mar 27 '13 at 19:47
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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...) –  Micah Mar 27 '13 at 19:57
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@raxacoricofallapatorius I've added an addendum, giving a summary in Dirac notation. –  Branimir Ćaćić Mar 27 '13 at 19:59
    
Thanks. That's cool. –  raxacoricofallapatorius Mar 27 '13 at 20:12

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".)

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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? –  raxacoricofallapatorius Mar 27 '13 at 19:19
    
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. –  Chris Godsil Mar 27 '13 at 20:43

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