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14

There are a few ways to answer your question, and I will try to list some of them. According to Quantum Mechanics, and due to the Heisenberg Uncertainty Principle, we cannot predict the future state (position and momentum) of any system. Given the state of a system in classical phase space $(\textbf{r}(t_0), \textbf{p}(t_0))$, we cannot determine the state ...


14

A deterministic universe need not be predictable. And even a deterministic universe not hampered by any limits to observability need not be predictable. As an example take a toy universe consisting of an infinite chain of $0$'s and $1$'s. This 1D cellular universe evolves according to cellular automata rule-110: the state of a cell becomes $1$, unless the ...


5

Saying that things "are both waves and particles" is a vestige of the 18th century way of thinking, and really ought to be done away with. Everything is described by a wavefunction. Period. What is a wavefunction? It is a complex-valued function. If you are interested in an electron's position, it is easy to think of it as a complex-valued function of ...


4

Radioactivity is the result from a confluence of special relativity and quantum mechanics. Special relativity introduces the generalized energy, E=m*c^2 , which allows the energy conservation to count in the sum the rest masses of the particles which comprise a nucleus. In this relativistic energy conservation we find some nuclear isotopes which are at a ...


3

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...


3

Isn't it just generally true that in the absence of any potential, the momentum eigenfunctions are also energy eigenfunctions? In other words, when there is no potential, (in the right units) $$ H = p^2/2m + V(x) = p^2/2m $$ Since the Hamiltonian is proportional to the momentum operator squared, it's easy to see that any eigenket of the momentum operator ...


3

These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions ($\Delta x$ and $\Delta p$) is of the same order as a typical position or momentum within those distributions ($r$ and $p$).


3

The magnetic field inside of a cylindrical solenoid of radius $R$ is given by $$ \textbf{B} = \mu_0 n I \hat{z}, $$ where n is the turn density in turns/m and I is the current. The field everywhere outside the solenoid is zero. Let's assume that the current $I(t)$ is linearly increasing, so $$ I(t) = at . $$ For a cylinder of radius $r \ge R$, the flux ...


3

In quantum field theory, you change the space of wavefunctions w.r.t. quantum mechanics. The space is still a Hilbert space, but it is called Fock space, and it takes into account the possibility of having any number of identical "particles" (or excitations of the field). A vector of such Fock space is of this form: $$(\psi_0,\psi_1,\psi_2,\psi_3,\dotsc)$$ ...


3

In the representation given by you, the $V(x,t)$ is a scalar function depending on $x$ and $t$. However, you already have choses a basis (the $x$-basis). In general, the $\hat{V}$ or rather $\hat{H}$ in the Schödinger Equation is an operator. This operator can then be evaluated in a basis of your choice. If you are familliar to the dirac notation one can ...


3

The potential is, without any ambiguity, an operator. There is no other way, because it changes the spatial dependence of the wavefunction: $$ \Psi(x,t)\text{ changes to }V(x,t)\Psi(x,t). $$ It's unclear to me what you mean by "scalar", though, because this word can have multiple meanings. For example, it is indeed a scalar operator - as opposed to vector ...


3

$$D(\hat{X}_i):= \left\{ \psi \in L^2(K, d^nx)\:\left|\: \int_K x_i^2|\psi(x)|^2 d^nx< \right.+\infty \right\} = L^2(K, d^nx)$$ where the last identity holds true if $K$ is bounded (in particular compact) because $x_i^2$ is bounded as well thereon (in this case the operator is bounded, too). Moreover $$D(\hat{J}^2):= \left\{\psi \in L^2(\mathbb S^2, ...


3

An electron can interfere with itself. there have been experiment of interference with single electron. Saying that an electron is alway a particule is then wrong. The wave function is the "best" way we have find to describe electrons and other quantons. object at the quantum level are not wave or particles. They just follow quantums rules and waves or ...


3

Even in a quantum universe, all evolution is deterministic if interpreted under Many-Worlds interpretation. So all possible futures could be "already determined", but you would still be unable to know which of those futures will be directly experienced by your qualia, since qualia experiences are always described by non-unitary probabilistic projection ...


2

In the special case where $\ell^2$ or $s^2$ has eigenvalue zero, then $j^2$ is fixed. Otherwise you must know the projections $m_\ell,m_s$ to find $j$.


2

The way this is justified is as follows: We start with the uncertainty principle, which can be roughly stated as $$\Delta x \Delta p \geq \hbar$$ For this rough estimate, we will ignore some factors of perhaps $2$ or $\pi$, but we're interested in some order of magnitude, not the exact result. Now, our second assumption will be that the ground state of the ...


2

Perhaps you are starting by the wrong end. Your concern seems to be related in the first term with the totally misleading notation of integrals in quantum mechanics, and this is more related with the spectral theorem than with distributions itself. Distributions only appear in Quantum mechanics when certain operators has empty spectrum in the usual Hilber ...


2

There seems to be no way to proceed unless more information is given. In fact, from what you have above (correcting the typo pointed out by Bernhard and ticster): $$ \vec{j}=\vec{l}+\vec{s} \quad \Rightarrow \quad \vec{j}^2=\big(\vec{l}+\vec{s}\big)^2 = \vec{l}^2+\vec{s}^2 + 2\, \vec{l}\cdot\vec{s}\,, $$ meaning that knowledge of the eigenvalues of ...


2

I) Quantum mechanically, the Lie group associated with rotational symmetry is $$G~:=~Spin(3)~\cong~ SU(2),$$ which is a double cover of $SO(3)$, and has a 3-dimensional real Lie algebra $$L~:=~so(3)~\cong~ su(2)$$ with generators $J_i$ satisfying $$[J_i,J_j]~=~i\hbar\sum_{k=1}^3\epsilon_{ijk} J_k.$$ II) In quantum mechanics, we are interested in Lie ...


2

It is not surprising that you found one of the Pauli matrices as the generator of your rotation. Let us see how it can be seen algebraically that it is to be expected: Observe that 2D rotations embed naturally into 2D unitary matrices, as $\mathrm{SO}(2) \subset \mathrm{SU}(2)$, corresponding to the subgroup of real matrices. Now, as we know, the ...


2

The quantum state and the state vector is exactly the same thing. The word "vector" is meant to emphasize that quantum states form a vector space, the so-called Hilbert space. Because they're synonyma, the question in the first paragraph is a meaningless talkative tautology, like "Is a car meant to denote just one vehicle or are many cars meant to be many ...


2

Since you want a bit of mathematical rigor: A quantum state is a self-adjoint positive trace class operator on a Hilbert space with trace 1. This is called density matrix $\rho$. In its simplest form, given $\psi\in \mathscr{H}$, $\rho$ is the orthogonal projector on the subspace spanned by $\psi$. Let $E_\rho(\cdot):D_\rho\subset\mathcal{A}(\mathscr{H})\to ...


2

$$ <\hat{A}> = \int \psi^*(x)\hat{A}\psi(x) dx $$ now $\hat{A}\psi(x)=a(x)\psi(x)$ so, $$<\hat{A}>=\int a(x)|\psi(x)|^2dx$$ Let $|\psi(x)|^2dx = d \mu$ now $\int|\psi(x)|^2dx=\int d\mu = 1$ $$ <\hat{A}> = \int a(\mu)d\mu $$


2

This wavefunction is an idealization of a wave function that has very very steep, but not discontinuous, behavior at $0$ and $a/2$. You are right, the wavefunction as written is not a proper wave function. That's a good observation. Often in physics the math is much easier if a real situation is modeled by one that is close to it, but mathematically more ...


2

There are many ways to define the Wigner characteristic function corresponding to a density operator $\rho$. In the case of a single degree of freedom (i.e. one optical mode, or one point particle moving in one dimensions), one way is $$\chi (\xi) = \mathrm{Tr}[\rho \hat{D}_{\xi}]$$ where $$D_{\xi} =\mathrm{exp}(i \hat{X}^{\mathrm{T}} \! \omega \, \xi)$$ ...


2

$Q$ is not a functional, but a linear operator. Since it is linear, there are no problems in using the chain rule. I will pretend there are no domain problems here, and that you can exchange limits and integrals as you wish (e.g. I'm supposing you can use the dominated convergence theorem), and obviously that each quantity is differentiable. ...


2

You seem to be assuming that there is a measurement process that can be applied to the electron on a single, isolated hydrogen atom twice within somewhere under 10–19 seconds, and that because an electron in a hydrogen atom's ground state (or any $s$ state) is equally likely to be found along any direction from the nucleus, there's some chance that the ...


1

Actually, "unitary representation" is meant with respect to the spinors, which do not form a finite-dimensional space and therefore allow a unitary representation of the proper Lorentz group. The action is defined by $D(\Lambda)\psi(x)=U(\Lambda)\psi(\Lambda^{-1}x)$, and you can simply calculate that this is unitary on your spinor space. However, this does ...


1

There are two answers to this. One answer simply points out that the probability of the jth outcome specified by the Born rule $p_j = tr(\rho\hat{P}_j)$, where $\hat{P}_j$ is the projector onto the jth outcome, satisfy the axioms of probability: http://mathworld.wolfram.com/ProbabilityAxioms.html. Another answer is that the Born rule can be explained ...


1

In addition to Qmechanic's interesting answer, you might want to explicitly see that $\mathrm{SU(2)}$ only has three independent variables. Let us start by writing the $U \in \mathrm{SU(2)}$ as: \begin{equation} U=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{equation} with $a,b,c,d \in \mathbb{C}$. Since we are looking at the special unitary ...



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