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If your desired basis is the set ${|n\rangle}$, then the completeness relation tells you: $\hat{O} = \sum_a \sum_b \langle a|\hat{O}|b \rangle |a \rangle \langle b|$. Ideally, we prefer to do this in the orthonormal basis in which the operator $\hat{O}$ is diagonal, in which case this becomes $\hat{O} = \sum_a \langle a|\hat{O}|a \rangle |a \rangle \langle ... 2 One gets there by noting that$\langle x | p \rangle = e^{i p x/\hbar}$is a plane wave, and you have to throw on the test wave function to talk about the derivative operation. So, you are worried about $$\langle x| P | \Psi \rangle = \int dp ~p~ e^{i p x/\hbar} ~\langle p | \Psi \rangle$$ From there you note that you can get the ... 6 If$\cal H$is a complex Hilbert space, and$A :D(A) \to \cal H$is linear with$D(A)\subset \cal H$dense subspace, there is a unique operator, the adjoint$A^\dagger$of$A$satisfying (this is its definition) $$\langle A^\dagger \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \phi \in D(A)\:,\forall \psi \in D(A^\dagger)$$ with: ... 0$ | x \rangle $is a position eigenstate, the state for a particle with definite location$x$. This is an abstract vector.$\delta (x - x_0)$is a wavefunction (or distribution) for a particle with definite location$x_0$. It is the state$| x_0 \rangle$on the position basis $$\langle x | x_0 \rangle = \delta (x - x_0)$$ If$\hat x$is the position ... 0 If you put$\chi_{r_0}(r)= \delta (r-r_0 )$then$[ \chi_{r_0}(r)]$forms a basis. In Dirac's notation:$\chi_{r_0}(r) \rightarrow |r_0 \rangle$and you can verify that this set is a basis because it satisfy: Orthonormality:$\langle r_0 | r'_{0} \rangle = \delta (r_0 - r'_0 )$Closure relation:$\int d^3 r_0 \ \ |r_0 \rangle \langle r_0|=$1 where 1 ... 1 The latter description is correct (as is described in Sakurai, Gasiorowicz, Griffiths, and probably some other books that I don't own). What it is saying is that the inner product between$|x\rangle$and$|x_0\rangle$is either 0 if$x\neq x_0$or 1 if$x=x_0$. That is, the states are orthogonal. The momentum space description $$\langle ... 3 Comments to the question: Under the ordering symbol (such as, e.g. normal ordering :\ldots:, time ordering T(\ldots), radial ordering {\cal R}(\ldots), etc) all the operators (super)commute, e.g.$$ : \hat{A}\hat{B}: ~=~ (-1)^{|\hat{A}||\hat{B}|}: \hat{B}\hat{A}:, $$even if the (super)commutator [\hat{A},\hat{B}]\neq 0 is non-vanishing. Ordering ... 4 The eigenvalue equation$$\tag{1} \hat{x}\psi(x)~=~x_0\psi(x)$$in the standard Schrödinger position representation$$\tag{2} \hat{x}~=~x, \qquad \hat{p}~=~-i\hbar\frac{\partial}{\partial x},$$reads$$\tag{3} (x-x_0)\psi(x)~=~0,$$which has general solution$$\tag{4} \psi(x) ~\propto~ \delta(x-x_0). $$3 Just a remark from the physical side. If a physical system is described in a non-separable Hilbert space whatever Hamiltonian operator one chooses, thermal (Gibbs canonical or grand canonical) states cannot be defined as density matrices (mixed states) in the given Hilbert space. So, if one wants to describe thermodynamics of that system he/she ... 3 A linear operator A: D(A) \to {\cal H} with D(A) \subset {\cal H} a subspace and {\cal H} a Hilbert space (a normed space could be enough), is said to be bounded if:$$\sup_{\psi \in D(A)\:, ||\psi|| \neq 0} \frac{||A\psi||}{||\psi||} < +\infty\:.$$In this case the LHS is indicated by ||A|| and it is called the norm of A. Notice that, ... 1 This is much to do with the possible eigenvalues of the operators. Normal operators on a Hilbert space are closely analogous to complex numbers, with the adjoint taking the role of the conjugate; these relations are typically inherited directly to the operator's eigenvalues. Thus, if a linear operator L has an eigenfunction f with eigenvalue \lambda, ... 4 I do not know it this is an answer, since I am not sure to have understood your question. The structure of the equation is formally hyperbolic:$$\frac{\partial^2 \psi}{\partial t^2} - A\psi = S\quad (1)$$where$\psi =(p,q)^t$. If$A$were self-adjoint and non-negative (or non positive, changing a sign and inserting a further$i$in front of$\sqrt{-A}$as ... 10 On the actual Hilbert space of a consistent relativistic quantum mechanical system, the Lorentz transformations including boosts actually are unitary – which also means that the generators$J_{0i}$are as Hermitian as the generators of rotations$J_{ij}$. We say that the Hilbert space forms a unitary representation of the Lorentz group. What the OP must be ... 2 Infinite matrices, if properly handled, are nothing but linear operators (either bounded or unbounded) on the Hilbert space$\ell^2(\mathbb N)$. So they can have point spectrum, continuous spectrum, residual spectrum just in view of the general theory of operators in general Hilbert spaces. 8 What is your Hilbert space? In$L^2(\mathbb R)$your eigenfunction would have infinite norm. If you dealt instead with a bounded set$L^2([a,b])\$, your operator would not be Hermitian unless you impose suitable boundary conditions to discard boundary terms. These boundary conditions, however, would rule out your candidate eigenvector!