Meaning of $\langle q_n | E \rangle$ when $E$ is energy but $q_n$ is not In this lecture of Prof. Binney (go to 15:40), he is explaining that if we have a system with a state of constant energy, then the expectation value of any observable of that system remains unchanged with respect to time. He writes the expression $$\langle q_n | E \rangle$$ where $q_n$ is an observable. It is not stated that the observable $q_n$ has to be some energy. Now, with my limited knowledge of linear algebra, I understand  $\langle \phi | \psi \rangle$ as 


*

*an inner product of $| \phi \rangle$ and $|\psi \rangle$ and

*$\langle \phi | \psi \rangle = f(|\psi\rangle)$ where $f$ is the linear functional associated with $| \phi \rangle$ which resides in the dual space of the space in which $| \phi \rangle $ and $| \psi \rangle$ reside.


Now, if $q_n$ is not energy, then how can $| q_n \rangle$ reside in the space in which $| E \rangle$ resides? If not, then how does $\langle q_n | E \rangle$ make sense, in the context of 1? 
Ref: Binney, James; Skinner, David The Physics of Quantum Mechanics, Oxford University Press, 2014.
Edit:
I think that I'm confusing the meaning of state here. If I'm right, then $| E_1 \rangle$ means the state in the system has definite energy $E_1$, $| p_1 \rangle$ means the state in which the system has a momentum $p_1$ and $| E_1, p_1 \rangle$ means the state in which the system has both energy $E_1$ and momentum $p_1$. Now, I want to ask if $| E_1 \rangle$, $| p_1 \rangle$  and $| E_1, p_1 \rangle$ lie in the same space or not.
 A: Since the $|E\rangle$s form a basis, any vector $|q_n\rangle$ maybe expressed as a linear combination of them. 
Then the meaning of the inner product would be the amplitude of the state $q_n$ having an energy $E$ or vice versa. 

Having read your edit, your confusion does indeed lie in what a state is. In quantum mechanics, with each system (Hamiltonian) there is an associated Hilbert space which contains all the possible states of your system. Each (normalised) vector $|\psi\rangle$ of the Hilbert space corresponds to a possible state of your system. 
Now if you want to measure an observable with an operator representation of $A$ with eigenstates $|a_i\rangle$ of your system, then:
$$|\psi\rangle=\sum_i|a_i\rangle\langle a_i|\psi\rangle$$
This is because the eigenstates form a complete basis. Thus to label the state with an observable, we express it in the basis (eigenstates) of that observable by projecting the state.
Now in cases where two observable commute, it is possible to label the state with the two observable simultaneously. For example if $[H,L]=0$ then it is possible to find express states in the basis of $|E_n,l\rangle$. In fact we do so in the case of Hydrogen atom. 
A: I suspect $\langle q_n\vert E\rangle$ might be a typo.  First it is not an expectation value: the expectation value of the observable should be written $\langle E\vert \hat q\vert E\rangle$ or 
$\langle E_n\vert \hat q\vert E_n\rangle$ if the system is prepared to have the $n$'th possible value of energy.  Now, the time evolution of an energy eigenstate is $\vert E(t)\rangle = e^{-iEt/\hbar}\vert E\rangle$ so that
\begin{align}
\langle q\rangle = \langle E(t)\vert \hat q\vert E(t)\rangle = e^{iEt/\hbar}\langle E\vert \hat q\vert E\rangle e^{-iEt/\hbar} = \langle E\vert \hat q\vert E\rangle
\end{align}
does not depend on time.  Alternatively, in terms of wavefunctions in the position representation:
\begin{align}
\langle q\rangle &=\int dx \Psi(x,t)^* \hat q \Psi(x,t)\, ,\\
&=\int dx \psi^*(x) e^{iEt/\hbar} \hat q \psi(x) e^{-i Et/\hbar} \, ,\\
&= \int dx \psi^*(x)  \hat q \psi(x) 
\end{align}
also time independent, and where $\Psi(x,t)=\psi(x) e^{-i Et/\hbar}$ has been used.
If $\hat q$ is an observable, then it has a complete set of eigenstates (in the same space as the Hilbert space spanned by the set $\{\vert E_i\rangle\}$ of energy eigenstate of $\hat H$).  Denoting by $\vert q_n\rangle$ the $n$'th eigenstate of $\hat q$ then
$\langle q_n\vert E\rangle$ is in general a complex number but 
$\vert \langle q_n\vert E\rangle\vert^2$ is the probability of getting the eigenvalue $q_n$ (i.e. probability of outcome $q_n$) when measuring the observable $\hat q$ when the system is prepared in the state $\vert E\rangle$. 
