$\renewcommand{\ket}[1]{\lvert #1 \rangle}\renewcommand{\bra}[1]{\langle #1 \rvert}$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?
[Physics] Is every bra vector the complex conjugate of ket vector
hilbert-spacequantum mechanics
Related Solutions
The confusion is coming from the fact that you're thinking in terms of the bra-ket physics notation without understanding how the underlying vector spaces are constructed.
"Kets" are vectors in a vector space, i.e. a set of objects on which vector-vector addition and vector-scalar multiplication is defined (for some field of scalars). "Bra"s are covectors, aka one-forms, defined as linear functions from a vector space to its field of scalars. They also form a vector space, and they exist even if we don't define an inner product on the set of kets.
Since the space of bras is a vector space, it can be tensored with another vector space such as the space of kets. This is defined just like any other tensor product of two vector spaces (which is the Cartesian product, equipped with an intuitive definition of addition and multiplication). You could write this down in two equivalent ways: $\vert{\uparrow}\rangle \otimes \langle{\downarrow}\vert$ or $\langle{\downarrow}\vert \otimes \vert{\uparrow}\rangle$. They are structurally the same thing, just with a different convention for the ordering. (Likewise, coordinates in 3D can be ordered $(x,y,z)$ or $(z,y,x)$ without changing anything.)
The first case is often abbreviated $\vert{\uparrow}\rangle \langle{\downarrow}\vert$ since there's no risk of confusing it with anything else. (Of course, that notation is a bit dangerous since it suggest we have a way of associating a bra $\langle x \vert$ with every ket $\vert x \rangle $, but that would be a mistake because we haven't defined an inner product.)
On the other hand, if you tried to write the second case as $\langle{\downarrow}\vert \vert{\uparrow}\rangle$ or $\langle{\downarrow}\vert{\uparrow}\rangle$, it could be misinterpreted as letting the bra (which is a linear functional from the space of kets to the scalars) act on the ket, which produces a scalar. That's a very different mathematical object. If the bras and kets have vector space dimension $N$, then the object $\langle{\downarrow}\vert{\uparrow}\rangle$ has dimension 1 while the object $\langle{\downarrow}\vert \otimes \vert{\uparrow}\rangle$ has dimension $N^2$.
Let me work in mathematicians' notation for a bit and then switch back to Dirac notation.
Suppose you start with a Hilbert space $\mathscr H$, which you can understand as a space of functions from some coordinate space $S$ into $\mathbb C$, i.e. if $f\in\mathscr H$ then $f:R\to \mathbb C$, and that you have some suitable notion of inner product $(·,·):\mathscr H\times \mathscr H\to\mathbb C$, like e.g. an integral over $R$. (Note that here $(·,·)$ should be linear on the second argument.)
Given this structure, for every vector $f\in\mathscr H$ you can define a linear functional $\varphi_f:\mathscr H\to \mathbb C$, i.e. a function tha takes elements $g\in \mathscr H$ and assigns them complex numbers $\varphi_f(g)\in \mathbb C$, whose action is given specifically by $\varphi_f(g) = (f,g)$. As such, $\varphi_f$ lives in $\mathscr H^*$, the dual of $\mathscr H$, which is the set of all (bounded and/or continuous) linear functionals from $\mathscr H$ to $\mathbb C$.
There's plenty of other interesting functionals around. For example, if $\mathscr H$ is a space of functions $f:R\to \mathbb C$, then another such functional is an evaluation at a given point $x\in R$: i.e. the map $\chi_x:\mathscr H\to\mathbb C$ given by $$\chi_x(g) = g(x).$$ In general, this map is not actually bounded nor continuous (w.r.t. the topology of $\mathscr H$), but you can ignore that for now; most physicists do.
Thus, you have this big, roomy space of functionals $\mathscr H^*$, and you have this embedding of $\mathscr H$ into $\mathscr H^*$ given by $\varphi$. In general, though, $\varphi$ may or may not cover the entirety of $\mathscr H^*$.
The correspondence of this into Dirac notation goes as follows:
$f$ is denoted $|f\rangle$ and it's called a ket.
$\varphi_f$ is denoted $\langle f|$ and it's called a bra.
$\chi_x$ is denoted $\langle x|$, and it's also called a bra.
Putting these together you start getting some of the things you wanted:
2. $\langle x |f\rangle$ is $\chi_x(f) = f(x)$, i.e. just the wavefuntion.
6. $\langle f | g \rangle$ is $\varphi_f(g) = (f,g)$, i.e. the iner product of $f$ and $g$ on $\mathscr H$, as it should be.
Note in particular that these just follow from juxtaposing the corresponding interpretations of the relevant bras and kets.
7. Somewhat surprisingly, $\langle f | x\rangle$ is actually defined - it just evaluates to $f(x)^*$. This is essentially because, in physicists' brains,
9. $|x\rangle$ is actually defined. It's normally understood as "a function that is infinitelly localized at $x$", which of course takes a physicist to make sense of (or more accurately, to handwave away the fact that it doesn't make sense). This ties in with
8.' $\langle x' | x\rangle$, the braket between different positions $x,x'\in R$, which evaluates to $\delta(x-x')$. Of course, this then means that
8. $\langle x | x\rangle$, with both positions equal, is not actually defined.
If this looks like physicists not caring about rigour in any way, it's because it mostly is. I should stress, though, that it is possible to give a rigorous foundation to these states, through a formalism known as rigged Hilbert spaces, where you essentially split $\mathscr H$ and $\mathscr H^*$ into different "layers". On balance, though, this requires more functional analysis than most physicists really learn, and it's not required to successfully operate on these objects.
Having done, that, we now come to some of the places where you've gone down some very strange roads:
3. $\langle x| O\rangle$ does not mean anything. Neither does "operator on an eigenvalue of $O$ that produces the corresponding eigenfunction under a position basis".
4. $|O\rangle$ is not a thing. You never put operators inside a ket (and certainly not on their own).
Operators always act on the outside of the ket. So, say you have an operator $O:\mathscr H\to\mathscr H$, which in mathematician's notation would take a vector $f\in \mathscr H$ and give you another $O(f)\in \mathscr H$. In Dirac notation you tend to put a hat on $\hat O$, and you use $\hat O|f\rangle$ to mean $O(f)$.
In particular, this is used for the most fundamental bit of notation:
- $\langle f |\hat O|g\rangle$, which a mathematician would denote $\varphi_f(O(g)) = (f,O(g))$, or alternatively (once you've defined the hermitian conjugate $O^*$ of $O$) $\varphi_{O^*(f)}(g) = (O^*(f),g)$.
This includes as a special case
5. $\langle f |G(\hat x)|f\rangle$. This is sometimes abbreviated as $\langle G(\hat x)\rangle$, but that's a good recipe for confusion. In this case, $G:R \to \mathbb C$ is generally a function, but $G(\hat x)$ is a whole different object: it's an operator, so e.g. $G(\hat x)|f\rangle$ lives in $\mathscr H$, and its action is such that this vector has wavefunction
$$ \langle x| G(\hat x) | f \rangle = G(x) f(x).$$
The general matrix element $\langle g |G(\hat x)|f\rangle$ is then taken to be the inner product of $|g\rangle$ with this vector, i.e. $\int_R g(x)^*G(x) f(x)\mathrm dx$, and similarly in the special case $g=f$.
Finally, this brings us to your final two questions:
10. The statement that "the bra portion of the bra-ket is always a dummy variable" is false. As you have seen, $\langle f|$ is perfectly well defined. (Also, $x$ and $p$ are not "dummy" variables, either, again as you have seen above.)
11. Similarly, the statement that "the ket portion is always a function/operator" is also false. You never put operators inside a ket (you put them to the left), and it's generally OK to put $x$'s in there (though, again, this does require either more work to bolt things down, or a willingness to handwave away the problems).
I hope this is enough to fix the problems in your understanding and get you using Dirac notation correctly. It does take a while to wrap one's head around but once you do it is very useful. Similarly, there's plenty of issues in terms of how we formalize things like position kets like $|x\rangle$, but they're all surmountable and, most importantly, they make much more sense once you've been using Dirac notation correctly and comfortably for a while.
Best Answer
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 $b$.
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 $a^\ast(b) \equiv a^\ast b = \langle a,b\rangle$.
ADDED IN SECOND EDIT
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 $\mathscr F^\vee$.