quantum mechanics - How to arrive on the diffraction pattern for the double slit experiment using path integrals for the Gaussian slit case?

I wish to take the path integral route to derive the diffraction pattern for the double slit experiment using the Gaussian slits as the nature of the slits. The kernel looks like: \begin{equation} K((x,L+D),T;(0,0),0)=\\\int_{t=0}^{T} \int_{y= -b+a}^{-b-a} K((0,0),0;(y,L),T)*K((y,L),t;(x,L+D),T)dy *dt+\\\int_{t=0}^{T} \int_{y= b+a}^{b-a} K((0,0),0;(y,L),T)*K((y,L),t;(x,L+D),T)dy *dt \end{equation} if we approximate the slit to be of a Gaussian distribution, we can say the slits are represented by $G_1$ and $G_2$:

\begin{equation} G_1 = \frac{1}{\sqrt{2\pi}}\exp[{(y-d)^2/8b^2}]\\ G_2 = \frac{1}{\sqrt{2\pi}}\exp[{(y+d)^2/8b^2}].\\ \end{equation} If we assign the values: \begin{equation} K1=\int_{t=0}^{T} \int_{y= -b+a}^{-b-a} K((0,0),0;(y,L),T)*K((y,L),t;(x,L+D),T)dy *dt\\ K2=\int_{t=0}^{T} \int_{y= b+a}^{b-a} K((0,0),0;(y,L),T)*K((y,L),t;(x,L+D),T)dy *dt \end{equation} Then the kernel becomes: \begin{equation} K=\int_{t=0}^{T} \int_{y= \infty}^{-\infty}K1*G1*dy*dt +\int_{t=0}^{T} \int_{y= \infty}^{-\infty}K2*G2*dy*dt \end{equation} But if I assume: \begin{equation} K((0,0),0;(y,L),t)=(\frac{m}{2\pi i \hbar t})\exp[\frac{i}{\hbar}\frac{m}{2}\frac{y^2+L^2}{T}]\\ K((x,L+D),T;(y,L),t)=(\frac{m}{2\pi i \hbar (T-t)})\exp[\frac{i}{\hbar}\frac{m}{2}\frac{y^2+L^2}{(T-t)}] \end{equation} which is the kernel for free particles, I get a solution of the form: \begin{equation} K= \int_{t=0}^T \delta* \sqrt{\frac{\pi}{\alpha}}\exp[\beta^2/4\alpha]dt\\ where:\\ \delta = (\frac{m}{i \hbar})^2*\frac{1}{(2\pi)^{5/2}}\frac{1}{t(T-t)} \exp(\frac{im}{2\hbar}[\frac{L^2}{t}+\frac{D^2}{t'}+\frac{x^2}{t'}])\exp[\frac{d^2}{8b^2}] \\ \alpha = \exp[\frac{im}{2\hbar}[\frac{1}{t}+\frac{1}{t'}]+\frac{1}{8b^2}]\\ \beta = \exp[\frac{-2d}{8b^2}-\frac{2x}{t'}]\\ t'=T-t \end{equation} Now the problem is, I don't know how to integrate this over the time domain. Can someone tell me if I'm on the right track, and what I need to do to arrive at the right interference pattern on the screen, but finally taking the square of the thusly obtained amplitude? And possibly what the final what the final expression looks like?

Pressure difference along horizontal in accelerated fluids

In accelerated fluids, fluid in a container can orient itself in a direction due to acceleration. In that case, pressure at different heights (at the surface) is same (atm). Then at the same height, pressure is different. How does this pressure increase along the horizontal?

gravity - Why should mass be attractive in nature?

Why does a mass attract all the masses around it? Why should't it repel or just stay calm? Why should it be like that?

Answer

Gravity may be treated as a quantum field theory. In this kind of theory, interactions are represented by field correlations, more known as "virtual particles", "virtual gravitons" in the case of gravity. The fact that two charges (more precisely, in the case of the gravitation, $2$ positive energy densities) attract each other is due to the sign associated to a quantity named propagator which describes these field correlations. The sign of the propagator depends on the relation between the Lagrangian and an invariant LorentZ quantity, described below, with the constraint that the time derivative of the spatial components of the fields get a positive sign in the Lagrangian.

Take a metric $g_{ij} = Diag (1,-1,-1,-1)$. This metrics is used to raise or lower indices.

With a scalar field (spin 0), the invariant lorentz quantity is $\partial_\mu \phi \partial^\mu \phi = (\partial_0 \phi)^2 - \sum\limits_{i=1}^3(\partial_i \phi)^2$, the Lagrangian is then $L = + \partial_\mu \phi \partial^\mu \phi $

With a spin one field (electromagnetic field), the invariant Lorentz quantity is proportionnal to $(\partial_\mu A_\nu - \partial_\nu A_\mu) (\partial^\mu A^\nu - \partial^\nu A^\mu) = - 2 \sum\limits_{i=1}^3 (\partial_0 A_i)^2 + ....$

The minus sign here comes from $g^{ii}=-1$, for $i=1;2,3$. We must have a positive sign for the time derivatives of the spatial components, so we have to put a minus sign, that is $L $ proportionnal to $- (\partial_\mu A_\nu - \partial_\nu A_\mu) (\partial^\mu A^\nu - \partial^\nu A^\mu)$

With the graviton, which has spin $2$, and is representing gravity interactions, the interesting part of the invariant Lorentz quantity is : $(\partial_\mu h_{\nu \rho} \partial^\mu h^{\nu \rho}) = + \sum\limits_{i,j=1}^3 (\partial_0 h_{ij})^2 + ....$

The $+$ sign comes from : $g^{ii} g^{jj} = (-1)^2= +1$, so the Lagrangian is $L = \partial_\mu h_{\nu \rho} \partial_\mu h^{\nu \rho}$ + other terms

This game with the sign has a direct physical consequence. When you have a $+$ sign, charges of same nature attract, and when you have a $-$ sign, charges of same nature repell. So, positive energy densities ("masses") attract each other by gravitational interaction.

This is the general idea. Well, I made some simplifications... , for a complete discussion, see : Zee (Quantum Field Theory in a nutshell), Chapter 1.5.

quantum mechanics - Distinguishing identical particles

I've been going through Shankar's Principles of Quantum Mechanics. In the section of the system of identical particles, he uses an example of billiards to illustrate the difference between identical particles in classical vs. quantum mechanics.

He argues that in classical mechanics, we can track the history of a particle (a billiard ball) to distinguish it from another particle with no intrinsic differences. In quantum mechanics, however, he argues since continual observation is not possible, we can't use the same method to distinguish identical particles.

A potential counter-example I thought was...suppose we have two non-interacting particles in a same square well. And at the end of some measurement, we find that one particle 1 is in a stationary state $\psi_1$, and particle 2 in $\psi_2$. We measure the system again after some time t, then we know whichever particle that's in $\psi_1$ must be particle 1 from the previous measurement. And the same goes for particle 2. Thus we can distinguish the two "identical" particles.

What conceptual mistakes am I making here?

Answer

Let me try to understand what you are proposing: If there are two different stationary states, then they must have different energies, unless they are degenerate states. You stated that the two identical particles are non-interacting, thus they cannot exchange energy. This would mean that each particle would be stuck in their stationary states and we'll be able to distinguish them apart.

So I see that there's no other energy involved, both particles have different energies and therefore will be in different stationary states. I think the problem is that you state they are non-interacting. The two particles are distinguishable by construction.

At first I thought you were talking about position measurements, if so read the following:

You can see the problem clearly when their probability density overlaps in the same region of space. Notice that if you draw out the stationary state of both particles (no matter which states you choose), you have to overlap them because both of these are in the same well! There will be overlap for certain places. So, if you find out that one particle's position is where the overlap is, you won't know if that came from the first stationary state or the second.

Try this example: Take an O2 molecule. How can you tell which electron belongs to which nuclei?

quantum mechanics - Localization of Electron Matter Field Excitation in Simple Electron QFT Model

I believe QFT represents a single free (stationary) electron as a an excitation of the electron matter field which then couples to the EM field to create a local 'attached' EM field - if this is accurate, what keeps the electron matter field excitation localized instead of dispersing out through the extended field. In an atom one can imagine that the electron matter field excitation (and associated attached EM field) stays put as a standing wave pattern due to attraction between the attached EM field and the positively charged nucleus. But for a single free electron what mechanism keeps the electron matter field excitation localized?

Answer

There's three levels of approximations that eventually allow for the QFT electron to be described by the point particle. I will give them in order, but my suspicion is that you're looking for the 3rd one (coherent states).

Approximation #1: the adiabatic hypothesis, interacting QFT $\rightarrow$ free QFT. In the regime of weak interactions, an isolated system of electrons can be described by perturbation theory. The space of states of the QFT then is approximately described by the space of states of the corresponding free QFT, and the matrix elements of the Hamiltonian are given by a formal series in the coupling constant $\alpha$ (the fine structure constant).

This hypothesis is actually dead wrong, an important result known as the Haag's theorem. However, it turns out that adiabatic hypothesis works extremely well in practice, provided an important caveat: the interacting QFT and the corresponding free QFT have different values of field normalization constants and masses. Moreover, usually one of the two values is actually divergent.

The correct way to apply approximation #1 is therefore to:

1. Build the Hilbert space of the QFT in question (QED in your example) as the Fock space of the corresponding free QFT with different "renormalized" values of particle masses and wavefunction normalization constants.


2. Express physical observables such as e.g. the scattering matrix elements in terms of formal power series in the coupling constants (in your example, the fine structure constant $\alpha$).

This means that the leading (in $\alpha$) contribution to the behavior of an isolated electron comes from the free QFT with a renormalized (physical) electron mass.

Approximation #2: the non-relativistic limit, QFT $\rightarrow$ quantum mechanics. Now that we've approximated the original problem by a free QFT of non-interacting electrons and photons with the physical electron mass, let's consider its nonrelativistic limit.

One characteristic quantity that depends on $c$ is the (reduced) Compton wavelength of the electron

$$ \lambda = \frac{\hbar}{m c}. $$

In the formalism of the free QFT, the Compton wavelength plays a specific role. Say we've confined a particle into a region (in one dimension) of length $\Delta x$. Due to the uncertainty principle, this implies an uncertainty in the momentum. Due to the kinematic relation between energy and momentum, this implies uncertainty in energy. The characteristic scale of $\Delta x$ for which the uncertainty in energy is comparable to the rest energy of the electron is exactly $\lambda$.

In other words, try confining an electron to a region smaller than $\lambda$, and you'll get enough energy coming from the quantum uncertainty to create more electrons and positrons, so you don't have an isolated electron anymore.

All of the above was just a back-of-the-envelope calculation, of course. To describe what happens in the free QFT we should look at the behavior of the propagator for small spacelike separation. It turns out that the propagator decays as (numeric coefficients dropped for simplicity)

$$ S(x, y) \sim e^{-\left| x - y \right| / \lambda}, $$

which means that the quantum field values taken at spacelike separated points aren't orthogonal in the Hilbert space, unless the separation is much larger than $\lambda$, in which case they are orthogonal.

That is exactly the regime of approximation #2. It is easy to see that $\lambda \rightarrow 0$ corresponds to $c \rightarrow \infty$, that's why it is also the non-relativistic approximation.

Approximation #3: coherent states, quantum mechanics $\rightarrow$ classical mechanics. So now we've reduced our initial QED to a theory of a single isolated electron, and we now think of the delta-function basis of wavefunction as of an orthogonal basis in the (appropriate extension of the) Hilbert space, because in the non-relativistic limit $S(x, y)$ vanishes.

This is actually exactly equivalent to Pauli's theory of a non-relativistic electron. We've lost antimatter and other relativistic effects on approximation 2.

But even here the electron is just a mist of quantum magic, the wavefunction. What ensures that it doesn't spread?

Indeed, there are quantum-mechanical states for which the wavefunction rapidly spreads in time. Think delta-function-like distributions (but keep in mind that they have to be at least as wide as the Compton wavelength, or else the electron will explode in jets of QFT particles).

There's however a special subset of states which are states of minimal spread, or semiclassical states, aka coherent states. These are wave packets, given by Gaussian wavefunctions peaked at the classical position of the packet multiplied by a phase given by the classical value of the (non-relativistic) momentum.

Whenever the system is in a coherent state, the peak's spread in time is minimal (for free noninteracting theories like the idealisation here it vanishes). The classical values of position and momentum (determining the shape of the wavepacket) obey the classical equations of motion, which is the real reason why classical mechanics is a limiting case of quantum mechanics.

This approximation corresponds to $\hbar \rightarrow 0$, because the width of the wavepacket is proportional to $\hbar^{1/2}$. Hence, the classical approximation is valid when the scale of observation is large compared to the width of the wavepacket.

Conclusion. It is when the three approximations above are applied in sequential order that the elementary particles can be thought of as billiard balls moving through space. Note that this only works for massive particles, because for massless particles there’s no Compton wavelength and approximation #2 isn’t valid. Instead, one would typically take the classical limit of the relativistic QFT which leads to the classical field theory, eg electromagnetism. There's another notion of coherent states of the relativistic QFT, which is the eigenstate of the annihilation operator and describes the behavior of the classical field (and not a particle), which is the tool that highlights this correspondence.

quantum information - Types of photon qubit encoding

How many types of qubit encoding on photons exist nowadays? I know only two:

• 
  

  Encoding on polarization: $$ \lvert \Psi \rangle = \alpha \lvert H \rangle + \beta \lvert V \rangle $$ $$ \lvert H \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{H}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$ $$ \lvert V \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{V}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$

  
  


• 
  

  Time-bin: $$ \lvert \Psi \rangle = \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle $$ $$ \lvert 0 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$ $$ \lvert 1 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c+\tau}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c+\tau)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$

  
  

Is there anything else?

general relativity - Do photons bend spacetime or not?

I have read this question:

Electromagnetic gravity

where Safesphere says in a comment:

Actually, photons themselves don't bend spacetime. Intuitively, this is because photons can't emit gravitons, because, as any massless particles not experiencing time, photons can't decay by emitting anything. The latest theoretical results show that the gravitational field of a photon is not static, but a gravitational wave emanating from the events of the emission and absorption of the photon. Thus the spacetime is bent by the charged particles emitting or absorbing photons, but not by the photons themselves.

If photon can bend spacetime how does it exchange graviton?

Is there experimental evidence that massless particles such as photons attract massive objects?

where John Rennie says:

as far as I know there has been no experimental evidence that light curves spacetime. We know that if GR is correct it must do, and all the experiments we've done have (so far) confirmed the predictions made by GR, so it seems very likely that light does indeed curve spacetime.

Now this cannot be right. One of them says photons do bend spacetime, since they do have stress-energy, but it is hard to measure it since the energy they carry is little compared to astronomical body's stress-energy. So they do bend spacetime, it is just that it is hard to measure it with our currently available devices.

Now the other one says that photons do not bend spacetime at all. It is only the emitting charge (fermion) that bends spacetime.

Which one is right? Do photons bend spacetime themselves because they do have stress-energy or do they not?

Answer

Classical electromagnetic fields carry energy and momentum and therefore cause spacetime curvature. For example, the EM field around a charged black hole is taken into account when finding the Reissner-Nordstrom and Kerr-Newman metrics.

The question of whether photons cause spacetime curvature is a question about quantum gravity, and we have no accepted theory of quantum gravity. However, we have standard ways of quantizing linear perturbations to a metric, and reputable journals such as Physical Review D have published papers on graviton-mediated photon-photon scattering, such as this one from 2006. If such calculations are no longer mainstream, it is news to me. Given that photons have energy and momentum, it would surprise me if they do not induce curvature.

I also note that the expansion of the "radiation-dominated" early universe was caused by what is generally described as a photon gas and not as a classical electromagnetic field. So the idea that photons bend spacetime is part of mainstream cosmology, such as the standard Lambda-CDM model.

Finally, the idea of a kugelblitz makes no sense to me unless photons bend spacetime.

So in Rennie v. Safesphere, I am on the Rennie side, but I look forward to Safesphere defending his position in a competing answer.

Addendum:

Safesphere declined to answer; in a now-removed comment, he said that knzhou’s answer explains the disagreement. I don’t agree. I disagree with knzhou that “bends spacetime” is vague. It is commonly understood by most physicists to mean “contributes to the energy-momentum tensor on the right side of the Einstein field equations”. And most physicists believe that real photons do exactly this, for the reasons that Ben Crowell and I have stated.