New answers tagged

0

I'm not an expert in quantum Hall physics, and this may not be a comprehensive answer, but let me point out an interesting connection. In arXiv:1712.09904 the Read-Green model is restudied at the special point $\mu=\frac{m\Delta^2}{2}$ where the Hamiltonian can be written in a factorized form \begin{eqnarray}\label{eq:K_bb} \hat{K}&\equiv&\int_{S}\...


5

The delta function here can be thought of a position eigenstate, but as it is not square-integrable, it cannot be an actual wave function. Hence, "unrealistic". An actual, square-integrable wave function can be thought as being made of a superposition of dirac deltas, however: $$ \psi(x) = \int \psi(x_0) \delta(x - x_0) dx_0. $$ You can think of this as a ...


3

Let $|0\rangle$ be the vacuum state, and let $q_C(f)$ denote the operator that, when applied to any state, adds another quark with color $C$ and "spatial wavefunction" $f$. The statement that quarks are fermions means that these operators anticommute with each other. Now consider the three-quark state $$ |f,g,h\rangle := \sum_{\pi}(-1)^\pi q_{\pi(R)}(f)q_{...


0

No. 2 because the frequency of the wave is greater between a and b and it increases from a to b. The wavefunction on No. 2 is also identical to the left and right of the potential well as it should be since the potential is the same and unchanging there (it's flat). The reason for all this is because energy of the particle is proportional to the frequency ...


3

First of all, notice that from the general theory of angular momentum, the eigevalues $m$ of $J_z$ are integer if and only if $j$ is integer because $$m = -j, -j+1,\ldots, j-1, j\:.$$ At this juncture notice that, passing from Cartesian coordinates to spherical ones, you find $$L_z= -i \partial_\phi\:.$$ From direct inspection, using this expression, you ...


1

I'm not convinced that your transformations are correct. This is one place where it is important, if $x=x'-vt$, to distinguish between $$ \left(\frac{\partial \psi}{\partial t}\right)_x $$ and $$ \left(\frac{\partial \psi}{\partial t}\right)_{x'}. $$ They are not the same thing. You notation is uclear as to what you are keeping fixed in your partials. I ...


1

This is but a sum of cardinal sine functions with peaks at $\omega = E_n/\hbar$, for all n. Absorb $\hbar$ into the Es out of respect for sanity. Recall $$ |g\rangle=\sum_n |\psi_n\rangle \langle \psi_n|g\rangle, \quad \Longrightarrow \quad |g(t)\rangle=\sum_n |\psi_n\rangle e^{-iE_nt}\langle \psi_n|g\rangle , $$ so that $$ (1/2\pi)\int_{-T}^T \!\!dt ~ e^{...


2

Depends what you mean by "measuring". Obviously, if the photon is absorbed, then nothing at all will happen afterwards, as pointed out in the other answer. You can however also imagine a way to know which slit the photon went through without absorbing it. For example, you can make it so that when the photon passes through the upper slit its polarization is $...


3

Both plane waves have time dependence, so the question is which one satisfies the Time-Dependent Schrodinger Equation (TDSE) in one dimension, $$i\hbar\frac{\partial}{\partial t}\psi(x,t)=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right)\psi(x,t).$$ The answer is that the complex one does (for $V=0$, since we’re talking about a free ...


0

Regardless of what it means, carrying the imaginary part and using Euler's formula actually makes the algebra much less cumbersome. Once the legwork is done, just extract the real part and you'll be back in familiar territory (a real plane wave). Later on, you'll learn more about how the imaginary part is otherwise useful. Hope this helps.


0

I think you would not. In the wave theory photon which is moving from you represents as a wave which E-vector oscillates in the flat perpendicular to the you-detector line. And another photon, which is moving from left to right represents as a wave moving from left to right with E-vector oscillates in the flat perpendicular to the line left-right but ...


1

If you measure a photon then it is fully absorbed and the wave function is collapsed and it does not go onwards at all. Physicists try use special materials that can make a higher energy photon produce 2 lower energy photons that are entangled, and these 2 photons will proceed onwards (the first photon is collapsed). The 2 photons will proceed onwards in ...


0

This "collapse"language is completely navel gazing as far as measurements go. In this answer I show a position measurement, which I copy: This event was "measured" a few decades ago, by being immortalized in a picture. The data accompanying the picture, the magnetic field ,allow to measure the momentum of the electron, (again and again if one wants) and ...


0

As far as I know, the "collapse" (or the environmental decoherence that imitates a collapse) is always to the position basis. I think that a lot of the confusion surrounding this issue comes from the fact that there is a symmetry between position and momentum in the Hamiltonian formalism of QM, so it looks as though wavefunction collapse should have no ...


1

From eq. (4) to eq. (5): The Fourier transform of $1$ is the $\delta$ function, i.e. $\int \mathrm{e}^{\mathrm{i}px}\mathrm{d}p = \delta(x)$, cf. e.g. Wikipedia. From eq. (5) to eq. (6): This is just the characteristic property of the $\delta$-function, i.e. $\int f(x) \delta(x - y) \mathrm{x} = f(y)$.


1

The 1D TISE is a second-order homogeneous ODE. The complete solution span a 2-dimensional vector space parametrized by 2 constants. It is not hard to see that if a solution and its derivative vanish in the same point, it must be the zero-solution. The zero-solution is unacceptable as a wave-function.


2

You skipped the "either" option. Let's work in one dimension, and recall your unlocalized plane wave wavefunctions $\langle x| p\rangle\propto \exp(ixp/\hbar)$ and $\langle x|- p\rangle\propto \exp(-ixp/\hbar)$ are pointless for your free particle example: why don't you work in momentum space, instead? So, let's work in momentum space, now ultralocalized, ...


15

The system doesn't "know" anything. The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $\delta$-function, if we ...


7

The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement. Always remember the wavefunction isn't physical. It'...


5

This question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. There are two aspects to the measurement problem: If we adopt the collapse postulate, then for any given measurement-like interaction, how is the measurement basis determined? What is the nature of the evolution of the system ...


4

Quick idea: the Hamiltonian is the time-translation operator. That means that a state $\Psi(t_0)$ at time $t_0$ can be evaluated at time $t$ like $\Psi(t) = e^{iH(t-t_0)}\Psi(t_0)$. If you have $\Psi(t_0 = 0) = \Psi_1 + \Psi_2$, then $\Psi(t) = e^{iE_1t}\Psi_1 + e^{iE_2t}\Psi_2$. So each $\Psi_1$ and $\Psi_2$ evolve differently depending on the eigenvalue $...


3

There is no bound state for the positive Dirac potential. Look at the derivation of the bound state in e.g. these lecture notes. Formula (11.15) for the wave function of the bound state $$\Phi(x,t)=\frac{\sqrt{am}}{\hbar}e^{-\frac{am|x|}{\hbar^2} + i \frac{a^2 m t}{2\hbar^3} }$$ is derived without assuming $a>0$. But if a negative $a$ is inserted in the ...


1

What you have is called a functional derivative. The notation is usually somewhat different. It is defined by the relationship $$ \frac{\delta \psi(q)}{\delta\psi(q')} = \delta(q-q') . $$ In your example, we have $$ \langle\psi|\hat{O}|\psi\rangle = \int \psi^*(q)O(q,q')\psi(q') dqdq' . $$ Note that I have changed the way the $q$'s appear to make the ...


-4

Some processes are entangled but not suitable to show the uncertainty of entanglement. Using an electric circuit you may separat ions and electrons [QM objects) on the plates of a capacitors. Later you disconnect the plates and take them into different places. As long as you don’t know the direction of the potential difference of the circuit, your knowledge ...


3

Due to the diversity in the sets of orthogonal functions, one cannot make any statements about the overlap of the intensities (or densities) of two mutually orthogonal functions. It suffice to illustrate this by examples. As a first example, consider two mutually orthogonal plane waves: $\phi_{1,2}=\exp(i\mathbf{k}_{1,2}\cdot\mathbf{x})$, for different wave ...


5

Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ \psi = ae^{ikx} + be^{-ikx} $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as ...


5

Using your parameterization, the wave is $ae^{-\kappa x}+be^{\kappa x}$. Note that this particular wavefunction blows up at $x=+\infty,-\infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$. If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize ...


2

In the mainstream standard model of particle physics, all matter is made up of point particles with a fixed mass which we measure as best as we can within our experimental errors. There is no width in this masses at the table. Protons and (and neutrons bound in a nucleus) are stable composite particles , made up by a great multitude of quarks antiquarks ...


0

Antisymmetrization is usually done w/r to coordinates so that $$ \Phi(r_1)\Psi(r_2)-\Phi(r_2)\Psi(r_1) $$ is antisymmetric. In your case $\Phi\to Y_{\ell_1m_1}$ and $\Psi\to Y_{\ell_2m_2}$. This can be done systematically using Young’s symmetrizers; normalization is done at the end.


1

If you want the states themselves, It's essentialy a question about solving PDEs - you're meant to solve the Schrödinger equation for that potential, i.e. $$\frac{-\hbar^2}{2m}\nabla^2 \psi + \left[\frac{-A}{y^2+a^2}+\frac{-A}{(y-r)^2+a^2} \right]\psi = E \psi$$ That particular choice of potential doesn't look very solvable analytically though. If you ...


2

Huygens' principle really just says the space a wave travels through is homogeneous. The wave equation $\partial_t^2u(x,t)-$ $c^2\partial_x^2u(x,t)=0$ is from Maxwell's equations. We can just we well consider the Schroedinger equation for a wave $\psi(x,t)=\psi(x)e^{-iEt/\hbar}$ with $i\hbar\partial_t\psi(x,t)=$ $E\psi$. The Schroedinger wave equation $i\...


1

$\sqrt{\frac{2}{a}} \sin (\frac{n\pi}{a}x)$ is just the spatial dependency. Recall that all stationary states have a temporal factor of $e^{-i E t/ \hbar }$. Multiply these to obtain $\psi(x,t)= e^{-i E t/ \hbar }\sqrt{\frac{2}{a}} \sin (\frac{n\pi}{a}x) $. Edit: why the downvote? OP clearly thought the ISQ wavefunction was "all real" in the sense of "...


2

is there fundamental initial state that we can use? No. That's a ridiculous question, because the initial state depends on the conditions at which the system was started, and it is generally determined by the preparation procedure. Put it this way: if you transplant the question to newtonian mechanics you're basically asking something like "I know that $F=...


3

$\hat O$ is a linear operator. That means that $\hat O(\alpha \psi) = \alpha \hat O(\psi)$ for all scalar $\alpha$, and the factor $e^{-iEt / \hbar}$ is scalar. "$\mathrm i\hbar \partial_t$" is not an operator on the Hilbert space, because time is a parameter in quantum mechanics, see e.g. here. If $\hat O = \hat O(t)$ has an explicit time-dependence, like ...


1

The strict answer to the question in the title is: no, isolated systems evolving unitarily do not increase their entropy (if they are isolated, where would the information go?). The subtleties around this issue have to do with the question of "what is entropy anyway". Entropy is always defined by postulating some ignorance about the complete knowledge of the ...


1

In the classical case, the energies are related to the eigenvalues of the square root of the 1D Laplace operator, so that they are proportional to the numbers $E_n^{1/2}$ which are equally spaced as functions of the natural $n$. In the quantum case, they are instead related to the eigenvalues $E_n$ of Laplace operator. These differnces can be traced back ...


0

The square well (harmonic) potential can be viewed as a power law potential $|x|^p$, where the power $p$ is $p=\infty$ ($p=2$), respectively. See e.g. this related Phys.SE post.


1

If I understood correctly, you want to postulate a more general Born rule while continuing to require phase invariance. As a matter of fact, this won't give you any advantage due to Gleason's theorem Theorem. Suppose $H$ is a separable Hilbert space. A measure on $H$ is a function $f$ that assigns a nonnegative real number to each closed subspace of $H$ ...


0

The Born rule is a statement of the conserved nonrelativistic Noether charge distribution, divided by e. I don't consider it a postulate. Since as far we know the electron is a point charge and its charge is immutable, the charge distribution tells us where the electron is. You may consider other conserved quantities such as energy-momentum but these ...


3

We are only interested in solutions $\psi(x)$ which are normalizable, i.e. $$ \int_{-\infty}^{+\infty} |\psi(x)|^2 \text{d}x = 1.$$ If $\psi(-\infty)$ or $\psi(+\infty)$ diverge, then this normalization would not be possible. Edit: Actually the above is nearly true (for practical physical purposes), but not completely true (in strict mathematical sense). ...


1

Relativistic quantum mechanics (RQM) constructs single-particle wave equations that are consistent with special relativity (SR). However SR allows to create particles out of energy, while quantum mechanics is based on the conservation of probability. Nevertheless at energies low compared to the masses involved a single-particle description is a good ...


Top 50 recent answers are included