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Hermitian operators and their physical meaning

If I understand your question correctly, you are asking what is the intuitive meaning of the complex numbers appearing in quantum mechanics, when the classical world appears to deal only in real ...
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How to apply anti-unitary symmetry operators?

You forgot the crucial fact that the observables are selfadjoint and that the relevant matrix elements have the same entries. Only those matrix elements are physically measurable since they are ...
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What is a singular continuous spectrum?

I know it's an old question, but it's a topic I love. ;) I always recommend this paper which goes into details in a very clear way. Let's try to sum up the most important parts for your question. Step ...
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Hermitian operators and their physical meaning

In QM we still want expectation values of our observables to be real numbers, i.e. we want $\langle\psi|\hat A|\psi\rangle$ to be real for observables. To see what this condition imposes on operators ...
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Prove that every component of angular momentum commutes with $f$

Commuting with $L$ is equivalent to being invariant under rotations. The quantities $r^2$, $r\cdot p$ and $p^2$ are all rotationally invariant, as is any function of them. That is all that is needed.
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What does $\dot x$ mean as an operator in quantum mechanics?

In the Heisenberg picture operators evolve in time, and dot denotes a time derivative $$ \dot{\hat{x}}^j~:=~\frac{d\hat{x}^j}{dt}, \qquad \ddot{\hat{x}}^j~:=~\frac{d^2\hat{x}^j}{dt^2},\tag{A}$$ cf. ...
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Conditions for the Hamiltonian's spectrum to be discrete

(I propose a second answer because the first was affected by a trivial but devastating mistake). First of all Definition. A selfadjoint operator $A: D(A) \to H$, where $H$ is a complex Hilbert space ...
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What does $\dot x$ mean as an operator in quantum mechanics?

$\dot{\hat{x}}$ means the same thing it does in Newtonian mechanics. $$\dot{\hat{x}} = \frac{d\hat{x}}{dt}.$$ The trick is this is happening in the Heisenberg picture, not the Schrodinger picture that ...
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Why is Dirac's Phase Operator Non-Hermitian?

There is an even shorter way than the one pointed out by @ACuriousMind. Suppose that there is a unitary $V$ such that $$a= Vn^{1/2}\:, \quad a^* = n^{1/2}V^*$$ on the corresponding domains. As a ...
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Integration and differentiation of an operator in Quantum mechanics

Yes it is true under a suitable choice of the topology used to define the integral of an operator-valued function. First of all we have to define the integral of a function that is operator-valued $I\...
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Hadamard formula in quantum mechanics

Making the answer of @onetoinfinity more explicit I proceed as follows. First to group the terms (commutators) in the right way, one should know explicitly how $[A,[A,[...[A,B]...]]$ look like, which ...
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Point splitting in bosonization

There are two "cutoff" (for lack of a better word) scales here, $\Delta$ and $\alpha$. $\Delta$ is inherent to the fermionic theory, it is introduced to remove the divergence of the ...
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How can an operator be proportional to a scalar?

There is the momentum operator $\hat p$, which when acting in the $x$-representation is given by the equation you cited. The eigenstates of the $\hat p$ operator can be labeled by eigenvalue $p$, ...
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Tong QFT Problem set 2, question 6: Normal ordering of angular momentum operator

Okay, here is what I get $$\int d^3\vec{x}\ x^jT^{0k}=\int d^3\vec{x}\int \not{d}^3\vec{p}\int \not{d}^3\vec{p}' \omega_p p^{'k}\Big[ a_px^je^{-i(p+p')\cdot x}a_{p'}- a_px^je^{-i(p-p')\cdot x}a_{p'}^{...
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About obtaining eigenvalues for angular momentum with ladder operators

I am not sure to understand what you mean by non-normalisable. However, the only non-normalisable vector in a vector space with scalar product is the zero vector. So what it is actually obtained is ...
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How is anything *not* ultimately a position measurement?

You're close, but you still have a misconception. We don't actually measure the position of anything, because position is an abstract concept that would require infinite precision to measure. What we ...
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How is anything *not* ultimately a position measurement?

Fundamentally, position and time intervals are the ultimate geometrical measurements. But we also have to take into account statistics: events counting. This would include other kind of measurements ...
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How is anything *not* ultimately a position measurement?

In classical physics all objects have a position (or they are extended) in space. In that sense every measurement is however associated to a location in space. A more delicate situation takes place ...
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Particle in a box with absolutely continuous spectrum

The answer is negative for smooth bounded below potentials: there is no continuous part of the spectrum. That is because, as is well known, the heat semigroup $e^{-tH}$, $t\geq 0$, for strictly ...
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Dependence of the Eigenvalues of the Angular Momentum Operator on the Mass, Energy and the Reference point wrt which the Angular Momentum is Measured

I think I found where the dependence on the reference point in the quantum case should be. This was a deceptive question to myself. Well this dependence has moved to the expansion coefficients and to ...
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4 votes

I'm getting imaginary eigenvalues of $X^2-P^2$

The question is formulated in a such vague manner that it is not possible to properly answer. This is a mathematical issue and to answer it is necessary to fix all mathematical hypotheses. First of ...
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Confusion regarding Heisenberg Uncertainty Principle

The operators are just that- operators. Their relationship with observable properties (according to the interpretation of QM I studied in the 1980s) is that the properties are eigenvalues of the ...
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Confusion regarding Heisenberg Uncertainty Principle

My problem with this statement is that it seems to suggest that there exist things like "position" and "momentum" that can be measured. I'm supposing that what's being referred to ...
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Confusion regarding Heisenberg Uncertainty Principle

The operators X and P in quantum mechanics are postulated to lead to probabilities for real-life measurements of position and momentum. In real experiments, those quantities are measured in a pretty ...
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What is the angular momentum operator?

Use the chain rule for partial derivatives to show that with $$ x=r \cos\theta\\ y= r \sin\theta $$ you have $$ -i \hbar (x\partial_y - y\partial_x)= -i \hbar \partial_ \theta. $$ Your two formulae ...
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How to calculate the commutator between Hamiltonian and momentum operator squared?

$$[H,p^2]= [p^2/2m+ kx^2, p^2]\\ = k[x^2,p^2]=k\{p,[x^2,p]\}\\ = i\hbar k \{p, 2x\}\equiv 2ik\hbar (xp+px). $$
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Equal-time Canonical Commutation Relation for a scalar field

You can write the integrals in question as $$\int_{\mathbb{R}^3} f(\vec{p}) d^3 p = \int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty\!\! f(p_x,p_y,p_z) dp_x dp_y,dp_z$$ where ...
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2 votes
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Product notation for operators

Short answer: yes. Long answer: yes.
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Expectation value of $\langle H^n \rangle_{t=0}$

$$| \psi, 0 \rangle = \int | E\rangle \langle E| \psi, 0 \rangle dE\qquad$$ $$ \langle H^n \rangle(0) = \langle \psi,0 | \psi, 0 \rangle = \int \langle E | H^n | E \rangle \langle E| \psi,0 \rangle ...
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What's the contraction for non-adjacent fields?

Non-adjacent operators have similar contractions. This can be seen as follows. Recall that everything (super)commutes under a operator ordering symbol, such as, time ordering $T$, normal ordering $:~:$...
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The composition property of the time-evolution operators

Since proving it requires knowing facts that have been mentioned in answers to some other questions, I decided to write a single answer which includes everything relevant to the proof. For any three ...
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Cross Product of two Hermitian Operators

Is the cross product between p and L also Hermitian? No it is not. But on using the vector triple product relation, I see that it's Hermitian indeed. No, even using a triple product relation, it's ...
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Expression of Klein-Gordon field in Heisenberg picture

One easy to deal with these difficulties is to just imagine the integral is discretized, then the measure even more clearly is just a c-number multiplying the annihilation and creation operators, $H$ ...
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Normal ordered exponential of one-body operators

Let us prove OP's claim for a single bosonic mode: Proposition: $$ e^{ta^{\dagger}a}~=~:e^{(e^t-1)a^{\dagger}a}: \qquad t~\in~\mathbb{C}.\tag{A}$$ Sketched proof of eq. (A): Let's call the LHS for $...
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Normal ordered exponential of one-body operators

Hints: First try to show it for a single bosonic oscillator (for fermions this was done by the OP already). To this end, define the following functions: \begin{align} f(M)&:=\exp{a^\dagger a M} \...
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Decomposing a coherent state?

You are effectively asking for a change of basis from oscillator number states to position eigenstates $|x\rangle$, essentially the wavefunction of the Schrödinger wavepacket. (Momentum eigenstates ...
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The composition property of the time-evolution operators

Yes, the group property (3) holds for any order of $t_1,t_2,t_3$. The proof follows from $$U(t_2,t_1)^{-1}~=~ U(t_1,t_2) \tag{A}$$ and the time-ordered version of eq. (3). Proof of eq. (A): We may ...
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In what way are eigenfunctions of an observable operator complete?

Here, "wave function" refers to any possible wave function for the system. A wave function that is invalid for the system for which you found those eigenfunctions $f$ cannot, in general, be ...
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Srednicki's QFT: Why $\langle p|\phi(0)|0\rangle$ in the interacting theory is Lorentz invariant?

This is why we use the LIPS (Lorentz Invariant Phase Space) normalization $$ \langle {\bf p}|{\bf p}'\rangle = (2\pi)^3 2E_{\bf p}\delta^3({\bf p}-{\bf p}') $$ for the single particle states. Without ...
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Srednicki's QFT: Why $\langle p|\phi(0)|0\rangle$ in the interacting theory is Lorentz invariant?

A function is Lorentz invariant if $f(p) = f(\Lambda p)$. Consider the function $$ f(p) = \langle p| \phi(0) | 0 \rangle = \langle p| U(\Lambda)^{-1} U(\Lambda) \phi(0) U(\Lambda)^{-1} U(\Lambda) | 0 \...
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Calculating $\langle p | [x,p] | \psi \rangle $ using Dirac notation

That commutator is just a constant! Since the canonical commutation relation is $$[x,p]=i\hbar,$$ this means that $$\langle p|[x,p]|\psi\rangle=\langle p|i\hbar|\psi\rangle=i\hbar\langle p|\psi\rangle....
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What is the Hamiltonian operator, and is it unique?

In the axiomatic formulation of quantum theory, there is an axiom that is used to define the time evolution of quantum states in an isolated system, \begin{equation} |\psi(t)\rangle = \hat{U}(t) |\psi(...
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When does an operator in Quantum Mechanics have a discrete spectrum?

Intuitively, you can see that the spectrum is continuous using the right conjugations. From $[x,p]=i$, you can formally deduce from solving Heisenberg's equations: $$ e^{iup}xe^{-iup} = x+u\\ e^{-itx}...
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Calculating $\langle p | [x,p] | \psi \rangle $ using Dirac notation

The first term on the RHS of your second equation is not equal to your last equation. But you're right that we first apply $P$ and then $X$. More concretely, it might help to denote $P|\psi\rangle =:|\...
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Weinberg Chapter 10: Sign convention for momentum operator

Weinberg says that translational invariance produces a conserved momentum, i.e., $P^\mu$, such that (Eq. 10.1.1): \begin{align*} [P_\mu, O(x)] = +i\hbar \frac{\partial}{\partial x^\mu} O(x). \tag{...
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Hamiltonian Open String

The orthogonality relations for sine and cosine functions should be proportional to the Kronecker $\delta$ symbol. Namely $$\int d\sigma \cos(m\sigma)\cos(n\sigma)\sim \delta_{mn}$$ $$\int d\sigma \...
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Conservation of momentum in quantum mechanics

The potential energy of the particles is only dependend on their relative location to each other and therefore only on $x_{12}=x_1-x_2$. We have: \begin{align*} [P_x,x_{12}] =[P_{x1}+P_{x2},x_1-x_2] =[...
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Proper notation and definition of the Hamilton operator

You are not pedantic, you are just wrong! $$ \hat H= \hat p^2 /2m+ V(\hat x). $$ Recall $$ \hat p= -i\hbar \int\! dx ~|x\rangle \partial_x \langle x| ,\\ \hat x= \int\! dx ~|x\rangle x \langle x|~, \...
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