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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 ...


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 ...


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 ...


7

There's a trick to proving this result which you would certainly be forgiven for not spotting! Consider the quantity $\exp(\mu X) \exp(\mu Y)$ which appears on the right hand side. Now differentiate this with respect to $\mu$. Say what!? Yeah, bear with me, just try it: $$\frac{d}{d \mu} \exp(\mu X) \exp(\mu Y) = X\exp(\mu X) \exp(\mu Y) + \exp(\mu X) Y ...


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 ...


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 ...


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. ...


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 $\Psi$. ...


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

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 ...


3

You have obtained several interesting answers. Here is the one I prefer, which actually is a variation of gj255's answer. Let us attack the problem by computing the function $$f(\mu) := e^{\mu X} Y e^{-\mu X}\:.$$ This function verifies $$f'(\mu) = e^{\mu X} X Y e^{-\mu X} - e^{\mu X} YX e^{-\mu X} = e^{\mu X} [X, Y] e^{-\mu X} = \lambda I\:.$$ ...


3

This is a basic example of a BCH formula. There are many ways to prove it. For example, write the exponential as $$ \exp(\mu X + \mu Y) = \lim_{N\to \infty} \left(1 + \frac {\mu X+ \mu Y}N\right)^N = \dots $$ Because the deviations from $1$ scale like $1/N$, it is equal to $$ = \lim_{N\to \infty} \left[\left(1 + \frac {\mu X}N\right)\left(1 + \frac {\mu ...


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 ...


2

In classical physics, quantities are ordinary, commuting $c$-numbers. The order in which we write terms in expressions is of no consequence. In quantum field theory (QFT), on the other hand, quantities are described by operators that, in general, don't commute. Classical physics is a low-energy approximation of quantum physics - the road from quantum to ...


2

The choice of normal ordering prescription $:~:$ is typically adjusted to the choice of vacuum state $|\Omega\rangle$ so that the bra-ket-sandwich of normal-ordered operators $$\langle \Omega|:\hat{\cal O}_1\ldots \hat{\cal O}_n : |\Omega\rangle~=~0 $$ vanishes. The relation of normal ordering prescription to Wick theorem and other operator ordering ...


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 ...


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 ...


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 ...


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

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

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 ...


1

Here are some hints. use the definition of $\exp(x) = 1+x+\frac{1}{2}x^2+...$ on both sides. Now compare coefficients of $X^mY^n$ on both sides. Start with the $XY$ term as a warm up. Be careful to introduce commutators if you are swapping $X$s and $Y$s (Lie algebras are non-commutative in general). You should be able to write down the proof now!


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 ...


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

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} ...


1

First of all, note that the radial operator ordering ${\cal R}$ is implicitly implied in many textbooks of CFT (e.g. Ref. 1). Polchinski is in his eq. (2.2.7) discussing Wick's theorem between two operator ordering prescriptions. In this case between normal ordering $:~:$ and radial ordering ${\cal R}$. See also e.g. this Phys.SE post. The basic 2-point ...



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