# Is every bra vector the complex conjugate of ket vector?

$\renewcommand{\ket}{\lvert #1 \rangle}\renewcommand{\bra}{\langle #1 \rvert}$Suppose we are taking the inner product of two vectors, say $a$ and $b$ as $$\bra{a}b\rangle$$ where $\bra{a}$ is a bra vector and $\ket{b}$ is a ket vector. Is it important that every time we do inner product we do it for a vector and its complex conjugate or we can take inner product of any two vectors? Does it always mean that $\bra{a}$ is complex conjugate of $\ket{b}$ whenever we are taking an inner product?

## 2 Answers

The inner product is always between two (ket) vectors. However, let $\mathcal H$ be the space in which they live. This is a complex vector space with a Hermitian inner product. The inner product defines a map $\mathcal H\to\mathcal H^\vee$, where $\mathcal H^\vee$ is the dual of $\mathcal H$, and consists of linear functionals on $\mathcal H$. The map is given by $a\mapsto a^\ast$ which is defined by $a^\ast(b) = \langle a,b\rangle$.

In fact the inner product defines a metric on $\mathcal H$ and the postulates of quantum mechanics state that this state space is in fact complete, making it a Hilbert space, and it can be seen that the element $a^\ast = \langle a,\,\cdot\,\rangle$ is continuous. Now denote by $\mathcal H^\ast$ the topological dual of $\mathcal H$, consisting only of continuous linear functionals. The Riesz representation theorem asserts that $a\mapsto a^\ast$ is an isometric isomorphism between $\mathcal H$ and $\mathcal H^\ast$. You should see the latter as the space of kets, and so we see that indeed every bra vector is the conjugate of a ket vector. Starting from that point, in physics one forgets about the distinction between $\langle a,b\rangle$ and $a^\ast(b)$, which are both denoted $\langle a|b\rangle$. It is even customary for this reason to write $\langle a|$ for $a^\ast$ and $|b\rangle$ for $b$.

ADDED IN EDIT

As commented by ACuriousMind, it can be enlightening to make this concrete for the finite dimensional case. In the finite dimensional case, if we fix a basis, $a$ and $b$ can be written as

$$a = \begin{pmatrix}a_1\\ \vdots \\ a_n\end{pmatrix},\ \ \ b = \begin{pmatrix}b_1\\ \vdots \\ b_n\end{pmatrix}.$$

The inner product is $\langle a,b\rangle = a_1^\ast b_1 + \cdots + a_n^\ast b_n$, where $a_i^\ast$ is the complex conjugate of $a_i$. The linear functional $a^\ast$ is represented by a matrix in this basis, namely

$$a^\ast = \left(a_1^\ast\ \cdots\ a_n^\ast\right).$$

Clearly we have $a^\ast(b) \equiv a^\ast b = \langle a,b\rangle$.

• Since OP is talking about complex conjugation, you could mention that in finitely many dimensions, or if we take $L^2(\mathbb{C})$, this dual indeed is closely related to complex conjugation. – ACuriousMind Jun 28 '15 at 9:32
• @ACuriousMind Thanks, done! (if this is what you had in mind) – doetoe Jun 28 '15 at 16:15

You can take inner product of any 2 vectors if they have same dimension. However, we consider the physics problem in quantum mechanics, where inner product of bra and ket vector (of course is complex conjugate of bra vector) of wave function means probability density of finding particle. Inner product of 2 arbitrary vectors (with same dimension) is no meaning in quantum mechanics.

• Of course the inner product of two arbitrary vectors has meaning. If $|a>$ is some arbitrary ket state, and $|b>$ is an eigenvector of some Hermitian operator (every vector is an eigenvector of SOME Hermitian operator!) then $<b|a>$ is the squared probability of measuring $|b>$ when applying the Hermitian operator. That's a pretty important meaning! – Jahan Claes Jun 28 '15 at 16:57