Bosons occupy the same quantum states, even if fermions obey Pauli's Exclusion Principle. Heisenberg's Uncertainty Principle rules out determining the quantum states occupied by two close fermions, etc....  I mean to say is it impossible in Quantum Mechanics to trace the history, or follow it, of a boson or fermion? There are inevitable errors. Why are we concerned about the loss of information in a black hole, where degeneracy breaks down? I mean, how are two mysteries different, with respect to conservation of information? We have laws of black hole thermodynamics, which prohibit the loss of energy...
3519
I really can't understand what Leonard Susskind means when he says in the video  that information is indestructible.  Is that information that is lost, through the increase of entropy really recoverable?   He himself said that entropy is hidden information. Then, although the information hidden has measurable effects, I think information lost in an irreversible process cannot be retrieved. However, Susskind's claim is quite the opposite. How does one understand the loss of information by an entropy increasing process, and its connection to the statement “information is indestructible”.  Black hole physics can be used in answers, but, as he proposes a general law of physics, I would prefer an answer not involving black holes.
53723
We know that there is uncertainty principle, so question: can we ever measure wavefunction of particles? I do not think this is possible, but I am not sure. I guess that everything is probabilistic. (that's why, I believe, we have sigma level when we say we discovered some particles...)
97127
What is the underlying reason that the same pairs of conjugate variables (e.g. energy &amp; time, momentum &amp; position) are related in  (e.g. time symmetry implies energy conservation) and likewise in QM (e.g. $\Delta E \Delta t \ge \hbar$)?
97475
Consider the  of Bennett et. al. How does one prove that this protocol would never violate the conservation of energy? At the face of it, it doesn't seem to be something obvious, as the various measurements, communications, rotations etc. don't seem to be able to account for the differences in the Hamiltonians of Alice's and Bob's labs. The state $\psi$ might have an expected energy $E_1$ in Alice's lab and a completely different expected energy $E_2$ in Bob's. Where does the difference come from?
69284
When quantum entanglement is explained in "layman's terms", it seems (to me) that the first premise, that we have to accept on faith, is that a particle doesn't have a certain property (the particle is not IN one state or another) until that property is measured.    For example, I have read that a particle may have "spin up" or "spin down" but -- here's where I get lost -- it's not that we don't KNOW which spin the particle has until we measure it, but actually the particle has NEITHER spin until we measure it.  Now, if we accept that premise, then we can see how spooky it is, when two particles are created with correlated spins -- if one has spin up, the other has spin down, but neither particle "has" either spin until one of them is measured.  At that instant, the other particle "assumes" the other spin.   At least that's how I interpret things I have read.  Certainly, if we accept that a particle doesn't actually HAVE the property until it is measured, but rather the particle exists in a superposition of states until the measurement is made, then we "regular citizens" can understand what is meant by "spooky action at a distance".  (As soon as one particle is measured, the other particle assumes the "other" spin, no matter how far away it is.  How did the second particle know the first particle was measured?  That, I think, is the spooky action at a distance (and the information transfer can be faster than light).)  However, from a layman's perspective, I want to cry out "but the particle does have a definite spin, we just don't KNOW what it is, until it is measured!  Duh!"  Now, I can't believe that all of the physics community hasn't thought of that objection -- but -- here's my point -- although the spooky action at a distance can be explained to us, once we accept the premise of superposition -- Why isn't the fact of superposition explainable also, in terms mortals can understand?  I have looked, but I haven't found a layman's explanation of why we should accept the fact that a particle doesn't have a particular spin, or other property, until it's measured.  It sure seems that the particles "have" the properties, even though we haven't measured them yet.    I know that much of quantum mechanics is not "intuitive".  If anyone can explain why particles don't have a definite property, even before we measure them, I would be grateful.
105458
is formed when particle, anti particle pairs formed by the uncertainty principle are separated by the event horizon of a black hole.        It seems like an equal amount of particles and anti-particles should end up on each side of the event horizon. So why don't the particles and anti-particles annihilate with a new partner once separated from their original partner by the event horizon? Thus canceling out any radiation released.
78045
It seems to be related to , but I can't penetrate the Wikipedia article. What has the Pauli exclusion principle to do with indistinguishability?
94957
The way I learned it from practicing Fourier analysis and signal processing besides quantum mechanics, is that Energy conservation cannot be achieved in short time scales, and that limits energy conservation in Quantum mechanics.  In other words: Energy conservation is limited by the Heisenberg uncertainty principle in our universe.  $$\Delta E \cdot \Delta t \geq \frac{\hbar}{2}$$  As I posted this answer somewhere, someone said this is wrong. So I posted this issue here, for maybe I don't understand it properly and you guys could tell me why this is wrong.  Example:  A beta decay produces a W-boson, which is ~85 times larger in mass than the initial particles (which decays eventually to neutrino and some beta particle), and this is possible due to the very short time scale, during which the W-boson is produced. I.e, energy is not conserved in very short time scales.  Do I understand this correctly? Thank you.
19301
I don't understand how quantum mechanics (and therefore also quantum computers) can work  given that while we work with quantum states, particles that this quantum state consist of cannot be observed, which is the most fundamental requirement.  If I am not mistaken, by "observed" we mean interaction with any other particle (photon, gluon, electron or whatever else). So my very important questions:   Aren't the particles this quantum state consists of interacting with each other? Why doesn't that cause the state to collapse? Aren't all particles in the universe interacting with Higgs field and gravitons etc? Why doesn't that cause every quantum state to collapse?   I feel there is something very fundamental in quantum mechanics that I am not aware of, hence I would be very pleased to have these questions answered.
62401
In Lawrence Krauss's book "A Universe From Nothing"; page 62 mentions that for a very short period of time, so small it cannot be measured, an electron due to the uncertainty principle can appear to be moving faster than the speed of light; another way to interpret this is that it's moving back in time.  It further proceeds to say that Feynman used this to suggest that what happens was that an electron and a positron were created, the positron will annihilate with the original electron while the new electron will continue on its merry way.  It finally says that this behavior is confirmed by the spectrum of the Hydrogen atom, among other things.   My question is, how does this not violate special relativity? How is it possible to get an electron moving faster than light? And how did we say it's okay to assume that and simply consider it moving back in time?   (Note, I do not have a degree in Physics so please take that into account while explaining the basic idea).
119493
Can someone summarize as to what are the problems and/or the open questions with the ? (in the context of understanding black-holes or as a computer science question)It gives a framework to see how the black-hole can thermalize in a $log (entropy)$ time scale. What next?   To start off,   The sort of problem with the Hayden-Preskill circuit is that it works via random unitary transformations on disjoint pairs of qubits and hence doesn't have an Hamiltonian interpretation. I believe one has always believed that matrix models (of matrix size $\sqrt{entropy}$) can saturate this logarithmic thermalization bound via Hamiltonian evolution.    Has this above belief about the matrix models been proven yet?  Is there an intuitive explanation as to why a theory with finite dimensional matrices should behave like an infinite dimensional system?
43286
I Appreciate the statement of Heisenbergs uncertainty principle however am a bit confused as to how exactly it applies to the Quantum Mechanical situation of an infinite square well.  I understand how to apply schrodingers equation and appreciate that energy Eigenvalues can be deduced  as   $$E_n=\frac{n^2\hbar^2\pi^2}{2mL^2}$$  However I have read somewhere that the reason that the quantum particle can not have $n = 0$ in other words $E = 0$ is because by having zero energy we also have a definite momentum with no uncertainty and by the Heisenberg uncertainty principle this should lead to an infinite uncertainty in the position of the particle however this cannot be in an infinite well as we know it should be somewhere in the box by definition. Therefore $n$ may be another greater or equal to one.  Surely when $n = 1$ we have the Energy as $\hbar^2 \pi^2 / (2mL^2)$ which is also a known energy and so why does this (as well as the other integer values of $n$) not violate the uncertainty principle.
68703
I've always hit two big stumbling blocks in conceiving of the proof or disproof of  as being even valid idea, let alone an answerable question... I feel I must be misunderstanding some very fundamental things.   Quantum mechanics is deterministic, ignoring the bit where we take our complex number and fudge it into a probability. We use computers to make calculations to provide the predictions of quantum mechanics - how is this not itself a hidden variable theory?   From what I understand the famous Bell theorem and work around it doesn't disprove hidden variable theories, just one specific flavour of them, and even then I am not convinced. I feel like I must be missing something again, because this result seems to me in no way profound... although it certainly fits into the picture of the EPR 'paradox' and is a good demonstration of the validity of QM.  I guess the crux of my problem is that this 'area' really addresses wave function collapse and the problems it causes, but I don't see the need for some collapse mechanism to begin with... again I feel like I am missing something really fundamental.  Apologies if this is a bad question. I have had a look around and there are similar questions, but I don't feel that either of my points are addressed...
49222
In quantum mechanics, one makes the distinction between mixed states and pure states.  A classic example of a mixed state is a beam of photons in which 50% have spin in the positive $z$-direction and 50% have spin in the positive $x$-direction.  Note that this is not the same as a beam of photons, 100% of which are in the state $$ \frac{1}{\sqrt{2}}[\lvert z,+\rangle + \lvert x,+\rangle]. $$ It seems, however, that at least in principle, we could describe this beam of particles as a single pure state in a 'very large' Hilbert space, namely the Hilbert space that is the tensor product of all the $\sim 10^{23}$ particles (I think that's the proper order of magnitude at least).  So then, is the density operator a mathematical convenience, or are there other aspects of quantum mechanics that truly require the density operator to be a 'fundamental' object of the theory?  (If what I mean by this is at all unclear, please let me know in the comments, and I will do my best to clarify.)
48267
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 the 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 &quot;identical&quot; particles. What conceptual mistakes am I making here?
79853
I read that the non-commutativity of the quantum operators leads to the . What I don't understand is how both things hang together. Is it that when you measure one thing first and than the other you get a predictably different result than the measuring the other way round? I know what non-commutativity means (even the minus operator is non-commutative) and I think I understand the uncertainty principle (when you measure one thing the measurement of the other thing is kind of blurred - and vice versa) - but I don't get the connection. Perhaps you could give a very easy everyday example with non-commuting operators (like subtraction or division) and how this induces uncertainty and/or give an example with commuting operators (addition or multiplication) and show that there would be no uncertainty involved.
65939
If photons are spin-1 bosons, then doesn't quantum mechanics imply that the allowed values for the z-component of spin (in units of $\hbar$) are -1, 0, and 1?  Why then in practice do we only use the $\pm 1$ states?  I have been told that this is directly related to the two polarizations of the photon.  This seems to be more of a classical argument however, arising from the fact that Maxwell's equations do not permit longitudinal EM waves in a vacuum.  I have also heard that it is related to the fact that photons have no rest mass, although I understand far less of this reasoning.  What I'm looking for are elaborations on these two arguments (if I have them correct), and perhaps an argument as to how these two are equivalent (if one exists).