$\renewcommand{\ket}[1]{\lvert #1 \rangle}\renewcommand{\bra}[1]{\langle #1 \rvert}$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 2


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 bras, 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 $b$.


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


The above argument is a strictly mathematical one, in which the (common) assumption is made that kets are the elements of some Hilbert space and bra's are defined to be elements of its continuous dual. This definition is mathematically convenient but not physicaly unavoidable.

First of all, only kets directly relate to physical reality, bra's are just there for mathematical convience. The convenience, and the whole usefulness of the Dirac formalism disappears when you abandon the duality between bra's and kets.

ZeroTheHero remarked that in the book of Cohen-Tannoudji e.a. the definition of bra's and kets (for a single spinless particle) are such that this duality doesn't strictly hold. The authors do acknowledge the computational desirability of this duality, and propose a formal solution, in which the duality is supposedly restored, but with the understanding that not all elements are physical (only aproximately so), and without going into details. I do think they take great care to indicate possible pain points in the formalism, and probably they do the right thing by not going into distracting detail.

My guess is that their intention was to give a pragmatic approach in which physical motivation and intuition are preferred over mathematical rigor. To be specific, first of all they define the Hilbert space of kets $\mathscr F$ to be a subspace of $L^2$ of "sufficiently regular functions" of which they "shall not try to give a precise, general list of [...] supplementary conditions".

Then they define the space of bra's as the space of all linear functionals on $\mathscr F$, the space that in my earlier notation would have been $\mathscr F^\vee$. I don't think it is ever useful to consider the entire (algebraic) dual in the infinite-dimensional case, since this space is huge, and full of highly pathological elements. They'd better have "defined" it as an otherwise unspecified subspace of "sufficiently regular" linear functionals.

Then they construct a very reasonable bra (namely evaluation at a point), and show that it is not associated to any ket, and finally they note that this can be physically resolved by going to generalized kets to restore the duality, even though one should not attribute a physical meaning to them. Many reasonable bra's correspond to a generalized ket, but this is still not the case for the large majority of the elements of $\mathscr F^\vee$.

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    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Commented Jun 28, 2015 at 9:32
  • $\begingroup$ @ACuriousMind Thanks, done! (if this is what you had in mind) $\endgroup$
    – doetoe
    Commented Jun 28, 2015 at 16:15
  • $\begingroup$ According to the seminal book on QM by Claude Cohen-Tannoudji et al, there are bras to which correspond no kets. See Chap.2, section B.2.c.$\delta$. $\endgroup$ Commented Nov 25, 2019 at 22:53
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    $\begingroup$ @ZeroTheHero I am saying that if we assume the state space to be a complex Hilbert space, a common assumption made in QM, then the space of bras and the space of kets, where bras are taken as elements of the continuous dual, are isomorphic. This is a mathematical theorem, hence non-negotiable. $\endgroup$
    – doetoe
    Commented Nov 25, 2019 at 23:38
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    $\begingroup$ @ZeroTheHero (continued) In the book you are referring to, the set of bra's is defined as the vector space of all linear functionals on $\mathcal H$, a space that I referred to as $\mathcal H^\vee$. This space is incredibly much larger than $\mathcal H^\ast$, mostly full of meaningless elements, but indeed some elements, like the one constructed by Cohen-Tannoudji, are not all that pathological. $\endgroup$
    – doetoe
    Commented Nov 25, 2019 at 23:39

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.

  • $\begingroup$ 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! $\endgroup$ Commented Jun 28, 2015 at 16:57

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