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4

Your intuition that If, instead, my measurement is only partly accurate and says that the momentum of the particle is in a set $\Delta =(a_x,b_x)\times(a_y,b_y)\times(a_z,b_z)$, will the measurement collapse the wave function into $P\Psi$ (where $P$ is the spectral projector of the momentum operator on the set $\Delta$)? is exactly correct. ...


1

The statement is simply false as it stands when adopting the standard Hilbert space formulation of QM. The true statement is that a self-adjoint operator with pure point spectrum admits a Hilbert basis made of eigenvectors. (It happens in particular, but not only, when either the operator is compact or its resolvent is.) The proof is not so simple and is a ...


0

“Always a polytope” – definitely not. Moreover, in certain situation $Ω$, if a closed set, may not change at all; I mean product with the 0-dimensional set of states $Ω_{\rm id} = \{1\}$ (one point), considered as a subset of 1-dimensional vector space $V_{\rm id} = {\mathbb R}$. It has the only effect, the unit effect, and corresponds to 1-state quantum ...


0

EDIT 2:There is no general procedure. It is not easy to find out how many commuting observables there are. In classical mechanics some systems are integrable. Then they have as many constants of motion as the number of degrees of freedom. For a system of N particles in d spatial dimensions times the number of degrees of freedom is Nxd. One particle and a ...


0

The little group is the subgroup of the Lorentz group that leaves an arbitrary four-momentum vector invariant, i.e. for an element of the group $g$ and momentum $V$ we have $gV=V$. This group is in general different for massive and massless particles. If you now find that the little group of your holomorphic primaries corresponds to that of massive states, ...


1

Hints to the question (v2): First note that the operator norm $||A||=||UA||=||AU||$ of an operator $A$ is invariant if we compose with an unitary operator $U$ from either left or right. Therefore $\dot{\rho}(t)$ is not the zero-operator: $|| \dot{\rho}(t) || = || [H, \rho(0) || \neq 0. $


1

I try to measure the energy and the position of the system simultaneously The states with definite energy are not states with definite position so there is no particle state of both definite energy and definite position.


1

At the first look the question seemed very interesting, but later I found the mistake. You said you are measuring the position of the particle precisely. But how? You can tell that the particle is inside the well but you can not know the exact position of the particle. For more info read ...


2

If you are considering a system of a single particle in a potential well with infinitely high walls and with finite width, the energy operator is $ H = \frac{p^2}{2m} + V(x) $ where $ V(x) $ is the potential energy operator, vanishing inside the well and infinite outside it. Being that $ \frac{p^2}{2m} $ does not commute with $ x $, how are you saying that ...


5

It is possible indeed !! It is called Hilbert Schmidt scalar product, it is defined in a Hilbert space of bounded compact operators including trace class operators. $$\langle A|B\rangle := tr(A^\dagger B)\:.$$ The space of Hilbert Schmidt operators is made of all bounded operators $A$ in the considered Hilbert space, such that $A^\dagger A$ is trace ...


1

It's difficult to answer your question because a collection of eigenvectors $\{ v_i\}$ does not uniquely specify an operator. For example, any two operators that are simultaneously diagonalizable are, by definition, operators that share the same set of eigenvectors. Moreover, an operator (and its matrix representation) can be defined by its action on basis ...


1

All other functions not containing a non-zero factor of $F$ are eigenfunctions with eigenvalue $0$ of this operator, the eigenspace with that value is degenerate with dimension $\mathrm{dim}(\mathcal{H}) - 1$, where $\mathcal{H}$ is the whole space. Obviously, the operator is simply the projection on $\lvert F \rangle$, so it is $P_F = \lvert F \rangle ...


11

In non-relativistic quantum mechanics the mass can, in principle, be considered an observable and thus described by a self-adjoint operator. In this sense a quantum physical system may have several different values of the mass and a value is fixed as soon as one performs a measurement of the mass observable, exactly as it happens for the momentum for ...


6

Of course, mass is an observable, although in simple models it is constant. This is already the case classically. One cannot determine the path of as rocket that burns fuel (which forms a large fraction of its mass) without taking into account that the mass is variable. The same holds in quantum mechanics, whenever the mass is not fixed by the modeling ...


8

Mass-squared is a Hermitian linear operator, it's a Casimir operator $\hat{C}_{1}=\hat{P}_{0}\hat{P}_{0}-\hat{P}_{i}\hat{P}_{i}$ for the Poincare group. It's Hermitian because the translation generators $\hat{P}_{\mu}$ are Hermitian. It commutes with all the generators of the Poincare group and so it's eigenvalues (mass-squared) are constant on each ...


2

The key concept to look for is displaced number states. These are, quite simply, the number states $|n⟩$, moved by the displacement operator $$D(\alpha)=\exp\left[\alpha a^\dagger-\alpha^*a\right]$$ to some point $\alpha=x+ip$ on the complex phase space. The ground state of a harmonic oscillator which has been displaced to a real $\alpha=x$ is, as you ...


0

Ok, eventually I figured out the answer. $$\mathcal{U}=e^{-i\phi a^{\dagger}a}$$


1

I) Right, the operator $$\tag{1} \hat{A}~\equiv~ \hat{a}-\alpha{\bf 1}, \qquad \alpha\in \mathbb{C},$$ satisfies the same commutation relations $$\tag{2} [\hat{A},\hat{A}^{\dagger}] ~=~{\bf 1}$$ as $$\tag{3} [\hat{a},\hat{a}^{\dagger}] ~=~{\bf 1}.$$ (In OP's example the complex number $\alpha=-1$.) II) Define number operator $$\tag{4} ...


0

Mathematically speaking, U(n) is what is known as a unitary group. In particular, when n = 1, as Acuriousmind stated, we get what is known as the "circle group". This group in particular is composed of all numbers on the complex plane {with asb(1) under the multiplication}. However, for all values of n, the unitary groups contain copies of n = 1 (or U(1)). ...


2

I have 5 bags labelled 1 to 5, and I have randomly dropped the letters A to J into the bags. You choose a letter at random and you win as many Francs as the number on the bag containing your letter. If I have distributed the letters evenly, then there should be 2 letters in each bag, so we could say that ψ(bagnumber) = ψ = sqrt(2). But if we want ...


4

The expectation value (of position) represents the average value (position) for the particle (it has units of length in this case) which is different from the actual location of the particle (also units of length). For example, take an electron on a hydrogen atom; the expectation value for all energy levels is at the nucleus even though many of the energy ...


10

In position-space (that is, when your functions are functions of x), the function $\int|\Psi|^2$ gives the probability of finding the particle in a given range. The expectation value of x is where you'd expect to find the particle. It is often essentially the weighted average of all the positions where the probability density, $|\Psi|^2$, is the weighting ...


4

Let $\Omega\subseteq \mathbb{R}^n$; then $\int_\Omega \lvert\psi(x)\rvert^2dx$, for a normalized function $\psi\in L^2(\mathbb{R}^n)$ gives the probability that the particle is in the region of space $\Omega$, but does not give any further information on its position. If you want to obtain a quantitative information on the latter (within the limits of ...


4

Expectation value is a different concept from probability. In fact, you can have an expectation value of energy, angular momentum, etc., not just for position. An expectation value of an observable for a given state $\Psi$ is the average value of a large number of measurements of that observable, assuming each measurement is made on the same state $\Psi$. ...


2

You have to interpret $|\frac{d}{dq} \psi\rangle$. Knowing that decomposition of the basis $|q'\rangle$ gives : $$|\psi\rangle = \int dq' \psi(q') |q'\rangle \tag{1}$$ You have : $$|\frac{d}{dq}\psi\rangle = \int dq' \frac {d\psi(q')}{dq'} |q'\rangle\tag{2}$$ So, applying it to $|\psi\rangle = |q"\rangle = \int dq' \delta(q"-q') |q'\rangle$, you get : ...



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