Axioms of Quantum Mechanics - Lec01 - Frederic Schuller

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Recommends Johnny Longsom · 2020-04 · Physics Stack Exchange · on “Why do we take the wave function to be zero at the edges of the box when solving the Schrödinger Equation for particle in a box?”

Ok, the answer from a physical perspective was already given by others, however it can also be reasond from the mathematical perspective. One of the axioms of Quantum mechanics says that observables are discribed by self-adjoint operators. Note that a self-adjoint operator $A$ satisfies $\langle \psi | A\varphi\rangle = \langle A \psi | \phi\rangle$ , it is symmetric. Now considering the momentum operator $$P: \psi \mapsto (-i)\psi^\prime$$ of a particle living on the intervall $I=[a,b]$ we find that for $\psi$ and $\varphi$ in the domain of $P$ that $$\langle \psi | P\varphi\rangle = \int_a^b dx \bar\phi(x) (-i \varphi^\prime(x)) = - i\int_a^b dx \bar\phi^\prime(x)) \varphi(x) + \left[\bar\

Cites Max Borovkov · 2017-09 · Physics Stack Exchange · on “Calculation of exponential of momentum operator”

I'm puzzling out the way one can determine $exp(iA)$ operator for an unbounded $A$. More precisely I would like to know, how could I deal with the momentum operator $p_x = -i\hbar\partial_x$. I'm trying to follow the instruction which I learned from Frederic Schullers lectures ( [this video] lecture 10 and 11) The instruction is as following: 1) Construct the positive real-valued measure $\mu_{\psi}$ using Stieltjes inversion formula : $\mu_{\psi}((-\infty,\ \lambda]) = \lim_{\delta\to 0^{+}}\lim_{\epsilon\to 0^{+}}\frac{1}{\pi} \int\limits_{-\infty}^{\lambda+\delta}dt \ Im<\psi, R_p(t + i\epsilon)\psi> $, where $R_p(z)$ is a resolvent operator. 2)Construct complex-valued measure $\mu_{\psi, \phi

Cites j4nd3r53n · 2018-08 · Physics Stack Exchange · on “Wave-functions in strong gravitational fields?”

When I was introduced to quantum mechanics at university, a wave function was described as a function $\Psi: \mathbb{R}^4 \rightarrow \mathbb{C}$, that is space + time. General relativity, meanwhile, is modeled as a 4-manifold. According to Frederic Schuller's lecture series on Youtube, when fitting the idea of the wave function to a manifold, it should more correctly be implemented as a section of the $\mathbb{C}$-bundle over the manifold, the main difference being that in the "classical" view (if you will excuse the label), $\Psi$ maps $\mathbb{R}^4$ to the same copy of $\mathbb{C}$, whereas in the manifold implementation, each point in space-time maps into a 'private' copy of $\mathbb{C}$

Cites user199113 · 2019-05 · Physics Stack Exchange · on “Why can't every quantum state be expressed as a density matrix/operator?”

It was my previous impression that all quantum states in a Hilbert space can be represented using density matrices † and that's already the most general formulation of a quantum state. Then I came across yuggib [LINK] 's comment here [LINK] : Everything would be so easy if there was the one-to-one correspondence you are describing. Sadly, there are many very strong suggestions that this should not be the case. The existence of uncountably many inequivalent irreducible representations of the canonical commutation relations for quantum fields is one of such suggestions. Another is the fact that not every quantum state can be represented, in a given (irreducible) representation, as a ray in Hil

Cites Johnny Longsom · 2020-09 · Physics Stack Exchange · on “Meaning of Eigenvalues for position operator”

It is sometimes said that the delta distributions $\delta(x-x_0)$ are the eigenstates of the position operator $\hat x$ . However, as its name may imposes, those are not classical functions and precisely not elements of the Hilbert space $x$ is defined on. The same is true for the momentum operator and the functions $e^{i\langle x,k\rangle}$ . They are not square integrable and, thus, no elements of the Hilbert space. Clearly, an eigenvector has to be an element of the Hilbert space, so what's the matter? The problem we're dealing with here is the infinite dimension of our Hilbert space. Things get more sophisticated then in the finite dimensional scenario where we can just compute eigenstat

Cites Clemens Bartholdy · 2023-04 · Physics Stack Exchange · on “Continuity of measurement of observables in Classical Mechanics”

In these lecture notes on QM by Simon Rhea & Richie Dadhley based on the first lecture Frederic Schuller [this video] , the following is mentioned on page-2: Recall that in classical mechanics an observable is a map $F : Γ → R$ , where $Γ$ is the phase space of the system, typically given by the cotangent space $T^{*}Q$ of some configuration manifold $Q$ . The map is taken to be at least continuous with respect to the standard topology on $R$ and an appropriate topology on $Γ$ , and hence if $Γ$ is connected, we have $F (Γ) = I ⊆ R$ . What is the basis of wanting the Observable to be continous w.r.t standard topology on $R$ and approproriate topology on $\Gamma$ ?