doublefelix
• Member for 7 years, 2 months
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• New York, Munich

Good question. The textbook formalism in Quantum Mechanics & QFT just doesn't deal with this problem (as well as a few others). It deals with cases where there is a well-defined moment of ...

Good question. They actually do not all make the same predictions. This is something that people say which isn't quite true. Take, for example, spontaneous collapse models (like GRW). They have ...

I just finished a thesis on this subject and I'm happy to share. None of the linked papers are my own. The time of arrival in quantum mechanics is actually a subject of ongoing research. It is ...

Because when you push on the chair, you're also pulling on the chair in the opposite direction without realizing it. For example, I just tried pushing the back of the chair I'm sitting in away, but ...

I think the other answers are pretty complicated. Doing two consecutive measurements of spin always gives the same result. That is a clear reason why we collapse the wave function - so that after the ...

If you have $n$ input parameters in a deterministic theory you can perfectly fit at most $n$ data points just by adjusting those parameters. In a probabalistic theory that is more subtle, but there is ...

People call it an operator. It is a thing which would map a function to another function. It's a lot like how a matrix maps a vector to another vector. You can invent lots of operators, like $$2$$ or $... View answer Accepted answer 5 votes There were certainly no sound waves before the universe began. The article was talking about a period of time after the big bang during which the "stuff" in the universe was so hot that no atoms were ... View answer Accepted answer 4 votes The point at which boundary conditions are specified is subtle. You can find this point by adding a constant vector to$\vec{E}$in each step until it varies the equation: it should not change any ... View answer Accepted answer 4 votes The usually-unstated premise for "translational symmetry implies momentum conservation" is that the laws of physics of the system in question follow the Euler-Lagrange equation of motion ... View answer 4 votes To put it concretely: In my course on QED, I can remember two conditions which were imposed due to relativity. The fields transform via a Lorentz transformation, and the Lagrangian should not change ... View answer 3 votes I'm going to give a bit more literal of an answer with a bit less story since that's my taste. What we call a quantum field is an operator as a function of position and/or time, i.e. at every$x, t$... View answer 3 votes Here the author strongly implicates the momentum of photons as being responsible for the collapse. But is that correct to say? The fact that photons have momentum does not make them cause things to ... View answer 3 votes I know that there are theories that include multiple dimensions of time (this would be required for a direction other than forward/backward). People have gotten so used to being surprised by ... View answer 3 votes Light DOES have lag time, and it affects how we see things. It's just so fast that humans don't notice without careful experiment. One of the most interesting results of this is that when we look ... View answer Accepted answer 3 votes Object A increases in mass, and so increases in volume I'm going to make the assumption that we are adding mass to A by providing more material of the same density$\rho_{A}$, rather than exchanging ... View answer Accepted answer 2 votes Since the commutation relation can be used to derive$\hat{p}$, and the explicit action of$\hat{p}$in the x basis can be used to verify the commutation relation, you are using an equivalent set of ... View answer Accepted answer 2 votes The usual way to do this is to average over the possible incoming states, but sum over the outgoing states. In that case the 1/2 is correct as there are 2 possible incoming states. This would be ... View answer 2 votes Yes, QFT books are unfortunately often vague about the connection to reality until the chapters where they discuss scattering, in my experience. The measurement-related postulates in QFT are the same ... View answer 2 votes $$e^x = 1 + x + x^2 + ...$$ Therefore if$x$has units,$e^x$doesn't even make sense by dimensional analysis. The argument of log, and any trigonometric function is also unitless for the same reason.... View answer 2 votes It's non-local in the sense that you said: Bob's wave function must be updated as soon as Alice makes a measurement, by setting$\psi=0$in Alice's detection region and renormalizing it to 1 ... View answer 2 votes The integral form to the Schrödinger equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi$$ is:$$\psi(\vec{x})=\psi_0(\vec{x})-\frac{m}{2\pi\hbar^2}\int \frac{e^{ik|\vec{x}-\vec{x}_0|}}{|\vec{x}... View answer 2 votes The answers in the comments are gone and so in case someone else has the same question I will type it here but not accept the answer as it is mostly not my own. Benefits of defining$D_\mu = \...

One thing missing in the other answers... people actually do discuss eigenstates of the field operator, or at least they are important in QFT. A complete set of field eigenstates are used to prove ...

Infrared = Low-energy/momentum Ultraviolet = High-energy/momentum You're right, it's not only used to describe EM radiation in this context. So "infrared physics" will be physics which is only valid ...

I have understood your question as two separate questions. What does it mean that a state has definite momentum? A general state $|\psi \rangle$in quantum mechanics is a superposition of ...

Here is a method that should work for any hamiltonian which is a polynomial of degree two or less in $x_1, ...x_N$, $p_1, ...p_N$, so it can have stuff like $x_1 p_2$ or $p_1 p_3$, but not $x_1 p_2^2$...

To be perfectly honest - physics does not know the answer to your question. It seems mysterious to me, too, that the particle would have information on how fast it moved in the past. There is no ...

Yes such Hamiltonians are considered equivalent. And, two different hamiltonians which only produce a global phase difference must differ by a real constant $H_2 = H_1 + C$. You can have the same for \$...