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16

I understand the definition of a Hilbert space. But I do not understand why non-commutativity compels us to use Hilbert spaces. It doesn't, but that's not what Scrinzi is saying. The reason is doesn't is because we could work, for example, in Wigner quasiprobability representation: $$\rho\mapsto W(x,p) = \frac{1}{\pi\hbar}\int_{-\infty}^\infty\langle ...


7

For a general and brief overview of the mathematical framework of Quantum Mechanics, see this answer. In a nutshell, Hilbert spaces arise from the representation theory of C*-algebras, which are postulated to be the relevant mathematical object that describes a quantum theory (because it contains observables in its self-adjoint part, and states as special ...


5

If $A$ is self-adjoint, you can define $f(A)$ as a complex-valued observable, where $f: \mathbb R \to \mathbb C$ is a measurable complex-valued function: $$f(A) := \int_{\sigma(A)} f(x) dP^{(A)}(x)\:,$$ $P^{(A)}$ being the spectral measure (projector-valued) of $A$. $N= f(A)$ is a closed normal operator and admits a spectral decomposition $P^{(N)}$ ...


3

The expansion formally works for any operator. It's breaking the exponential as $$ \exp(-x)=\frac{\exp(-x/2)}{\exp(x/2)} $$ and then expanding the numerator and denominator as $e^x\approx1+x$. However, since the exponential term in the Cayley expansion comes from the time-evolution of a wave-function: $$ \psi(x,t+\delta t)=e^{-iH\delta t}\psi(x,t) $$ which ...


3

For that to be true you will have to go beyond the Hilbert space and consider the rigged Hilbert space, that is the Hilbert space plus distributions. Hence this can be made formal if you identify Dirac's bras and kets with the (anti)linear rigged Hilbert space. In more intuitive terms, the delta function $\delta(\mathbf r)$ can be realised as limits of a ...


3

The formula that Ref. 1 uses is $$\tag{*} \exp\left(-\sum_j \eta_j a_j^{\dagger} \right) ~=~ \prod_j\exp\left( - \eta_j a_j^{\dagger} \right) ~=~\prod_j \left(1- \eta_j a_j^{\dagger}\right). $$ Ref. 1 correctly applies [the Hermitian conjugate of] eq. (*) to the bra in answer (a) on p. 181. There is no mistake on p. 181. Ref. 1 does not write a ...


2

($\hbar$ omitted in the following.) That is not weird, it is one of the crucial properties of the Fourier transform $F(\bar{})$ that $$ F(\partial_x f) = \mathrm{i}p F(f)$$ i.e. differentiation by one variable becomes multiplication with the Fourier conjugate variable and vice versa. Because of this, Fourier transformation is a powerful tool to solve ...


2

In fact $xp$ is not self-adjoint, it can have non-real expectation values. But its symmetrized form $D=(1/2)(xp+px)$ is better behaved (it has a self-adjoint extension). It is the generator of dilatations which scales momenta and coordinates. The complexification of ${\exp}[iDa]$ (i.e. $a$ becomes complex) is important for the study of the spectrum of a ...


2

Eigenvectors exist only for the point spectrum of an operator. For any other point of the spectrum one can only find a sequence of vectors for which $(A-\lambda I)u_n\to0$, where $A$ is said operator, and $\lambda$ is a point in the spectrum which is not an isolated point. So in this case there is a sequence of approximate eigenvectors. With a bit of extra ...


2

A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has in its spectrum three different kinds of subspectra: A discrete point spectrum, a continuous spectrum, and a singular spectrum. The latter is physically discarded. The point spectrum consists of the eigenvalues of $T$, that is, the spectral values for which true eigenvectors in $L^2(\mathbb{R})$, and ...


2

How is this last statement true? On the position basis, the (1-D) position operator is multiplication by $x$. A well known property of the Dirac delta distribution is $$f(x)\delta(x) = f(0)\delta(x)$$ since $\delta(x)$ is zero everywhere except $x=0$ were it has unit area. Thus $$x\delta(x) = 0\delta(x)$$ So, operating on $\delta(x)$ with the ...


2

Let $$\tag{1} \hat{T}_{ik}~:=~\hat{n}_i \hat{n}_k-\frac{1}{3}\delta_{ik}\hat{\bf 1}.$$ The phrasing of the problem in Ref. 1 is indeed not the clearest, but by comparing with the given solution, it seems that Ref. 1 is performing a partial averaging over the Hilbert space of states with fixed value of the orbital angular momentum quantum number $\ell$ and ...


2

TL;DR: The property bounded, bounded from above, and bounded from below are different things, cf. Wikipedia. In detail, consider a densely defined symmetric linear operator $A:D\subseteq H \to H$ in a complex Hilbert space $H$. Let $$\langle A \rangle_{\psi}~:=~ \frac{\langle \psi, A\psi\rangle}{||\psi||^2}$$ for $\psi\in D\backslash\{0\}$. It follows ...


2

Comments to the question (v1): We will not discuss tachyonic states here, because they are pathological and signal an instability of the theory. Then $$\tag{1} p^{\pm}~\equiv~\frac{p^0 \pm p^1}{\sqrt{2}}~\geq~0 $$ is manifestly non-negative, since the energy $p^0\geq |p^1|$. In the light-cone formalism $p^{+}>0$ is strictly positive, since the special ...


1

If the operator is not self-adjoint then this is a possibility. If you search on phys.SE you will find questions about the expectation value of xp in the case of the QHO, and this turns out to be imaginary


1

The trace class operators form a Banach space. There is a concept of (countable) basis for Banach spaces that is called Schauder basis. Not every Banach space has a Schauder basis, but it is true e.g. for the space of compact operators (the case of $\mathcal{K}(l^2)$ is given explicitly in the wikipedia article). Since the trace class operators are ...


1

in short the first way, but you can see this via using the position basis, $$ <x|\psi> = \psi(x) , \ 1 = \int |x><x| \ dx, \ p|x> = |x> (-i \hbar \partial_x), \ <x|x'> =\delta(x-x') \ \rightarrow \\ <\psi|xp|\psi> = \int dx \ dx' <\psi|x><x| x p | x'><x'|\psi> = \int dx \ dx' \psi^*(x) x <x|p | x'> ...


1

You are touching on the subject of relativistic quantum mechanics where time and space $(t,x)$ are handled on the same footing as operators. The accepted description is to not use quantum wavefunctions as describing one particle but rather the state of a quantum field. Doing this turns into the subject of quantum field theory and is the basis of modern ...


1

No, the Hilbert space is not spanned by continuous "eigenfunctions" because they are not eigenfunctions at all! A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has a point spectrum, a continuous spectrum and a singular spectrum. The latter is physically irrelevant. The point spectrum consists of the values $\lambda_i$ for which a true eigenvector ...


1

I am not giving a mathematical solution to your problem, as that can easily be found by a quick google search. Instead, I prompt you to imagine the following, in hopes that it furthers your intuitive understanding of the problem: Picture in your head a three-dimensional space. Take any one plane, e.g. the $x$-$y$-plane. Let, as an example, $P$ be the ...


1

You do not obtain the rules for the infinite-dimensional case by "proving" them from the finite-dimensional rules. Rather, you know that you need to have a Hilbert space, which is a complex vector space with an inner product, essentially. If you now search for infinite-dimensional Hilbert spaces that could possibly be used in quantum mechanics, you find ...


1

The problem of including time as an operator rather than a parameter in Quantum Mechanics is what led to the development of Quantum Field Theory. I.e., the position operator was demoted to a parameter rather than promoting time to an operator. The two uncertainty principles you quote are entirely different. The first (position/momentum) principle is the ...


1

This is exactly analogous to the procedure for finding matrix elements of normal operators. Let's first recall how this works in the familiar case. You choose an orthonormal basis of vectors, say $|n\rangle$, with $n = 1,2,\ldots D$, where $D$ is the dimension of the Hilbert space, such that $\langle n\rvert m\rangle = \delta_{mn}$. Now the matrix elements ...


1

If you want to write a super-operator representing left- or right-multiplication, there is a distinct method which is simpler and more elegant. Let us define the left-multiplication superoperator by $$ \mathcal{L}(A)[\rho] = A\rho,$$ and the right-multiplication superoperator by $$ \mathcal{R}(A)[\rho] = \rho A.$$ It should be clear that these operations ...


1

Here we want to craft a counterexample. Let $A$ be a $2\times2$ diagonal matrix $\text{diag}(\lambda_1,\lambda_2)$. Then its exponential is $e^A = \text{diag}(e^{\lambda_1},e^{\lambda_2})$, whereas its trace is $e^{\lambda_1}+e^{\lambda_2}$. If $B$ is another diagonal matrix then it commutes with $A$ and therefore $e^Ae^B = e^{A+B}$. If ...


1

Nothing will help, a plane wave occupies all the space, and the mean value of the position doesn't make much sense, but in your case is zero, because the integrand in the 1st calculus is anti-symmetrical. But this, on one condition, namely if we write the integral as $lim _{a \to \infty} f^2(p_0) \int _a^a x d^3 x$ Otherwise it's hard to say what is the ...


1

First of all, I would encourage you to think of position of acting on momentum states to the left, that is, to commute them with the bra: $$ ⟨\mathbf p|\hat x=-i\hbar \frac{\partial}{\partial p_x}⟨\mathbf p|, $$ where the differentiation is over anything to the right of it, so for instance $$⟨\mathbf p|\hat x|\psi⟩=-i\hbar \frac{\partial}{\partial ...


1

To say that something is a (linear) operator you have to specify the space where it acts. You may say that, for example, wavefunctions of quantum mechanics are maps: $t\to \psi(t)$ that are continuous in $t$ with values in $L^2(\mathbb{R}^d)$. If we restrict to compact time intervals $[0,T]$, we may denote the space of these maps by ...


1

The anticommutation rules for creation/annihilation fermionic operators are what defines these operators. The "proof" that they are correct is that they produce a theory that is compatible with the antisymmetric nature of fermions (and with all the other experimental results of course). For example you can check that they produce the expected result for ...


1

So, the problem is that you've got to enforce Fermionic antisymmetry, but Fock space tries to make things easier by making that invisible. So if we've got two electrons in a box in a definite Fock state, the electrons definitively occupy some single-particle states which we can just call $1, 2$. The actual state that is being occupied is therefore: ...



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