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The question asks about the time dependence of the function $$f(t) := \langle\psi(t)|(\Delta \hat{x})^2|\psi(t)\rangle \langle\psi(t)|(\Delta \hat{p})^2|\psi(t)\rangle,$$ where $$\Delta \hat{x} := \hat{x} - \langle\psi(t)|\hat{x}|\psi(t)\rangle, \qquad \Delta \hat{p} := \hat{p} - \langle\psi(t)|\hat{p}|\psi(t)\rangle, \qquad ... 38 The direction of the gravitational force would not change under time reversal. Your object would feel a force downward, just as it does usually. It might be easier to imagine you had a movie of an object under the influence of gravity. Drop the ball from rest some distance above the floor. You'll see it move downward and speed up. You'd interpret this as a ... 31 Good question. Let's first consider the ball falling immediately before it hits the table. Neglect friction with the air for simplicity. The ball has a velocity in the downward direction. If we reverse time, the ball is in the same position right above the table, but now it has an upward velocity. A ball with an upward velocity will rise with a negative ... 18 The Schrodinger equation is time-symmetric. The answer is therefore No. From all of the comments, I feel like I must be oversimplifying or missing something, but I can't see what. 16 The simplest example in condensed matter physics that spontaneously breaks time reversal symmetry is a ferromagnet. Because spins (angular momentum) change sign under time reversal, the spontaneous magnetization in the ferromagnet breaks the symmetry. This is a macroscopic example. The chiral spin liquid (Wen-Wilczek-Zee) mentioned in the question is a ... 14 I would say that it is a result of time reversal symmetry. If you consider the projectile at the apex of its trajectory then all that changes under time reversal is the direction of the horizontal component of motion. This means that the trajectory of the particle to get to that point and its trajectory after that point should be identical apart from a ... 11 One of the problems you will encounter is causality. Imagine you have a ball resting on the ground. Without already knowing how it behaved in the past you cannot uniquely define the next frame of your game. You cannot tell if the ball should: move upwards vertically. move upwards in any direction. roll on the ground towards any direction. do nothing. ... 9 If \Psi(x,t) solves the Schrodinger equation, so does \Psi^*(x,-t) , so no, there is nothing at all that must increase. 9 The point is that the symmetries in QM (bijective operations sending states to states preserving the transition probability) can be represented by either unitary or antiunitary operators. This is the statement of a famous theorem due to Wigner. It is possible to prove also that, if the Hamiltonian of a system is bounded below, time reversal must be ... 8 There are numerous research groups engaged in a search for an electric dipole moment of the electron, which, if it exists, would violate time-reversal symmetry. You can see this because any dipole moment the electron might have would need to be parallel to the spin (or anti-parallel). When you reverse time, the spin necessarily flips, but the electric dipole ... 8 I would consider that since acceleration is a constant vector pointing downward, that the time the projectiles downward component takes to accelerate from V(initial) to 0 would be the same as the time it takes to accelerate the object from 0 to V(final) 7 As the neutron is not point-like, consider it has a continuous distribution of charge \rho(\mathbf{r}) confined in a volume \Omega. The dipole electric moment is then given by \mathbf{D}(\mathbf{r})=\int_\Omega \rho(\mathbf{r}')\delta(\mathbf{r}-\mathbf{r}')d^3r' where the coordinates are measured from the centre of mass of the ... 7 You have all the elements in your question, your difficulty is about what is meant by "time reversal symmetry". Time reversal symmetry holds if, when "playing backwards", the motion observed obeys the same law. With friction it is not the case : friction opposes movement, when playing backwards it (seemingly) promotes it. You can also go to equations for ... 7 Is time reversible? Look at the stroboscopic photograph. Is the ball "falling up" or falling down? The answer is surely we don't know! A motion picture of this sort of sequence of the event could be run backward & would inevitably be impossible for the viewer to detect any violation of Newton's laws. A time-reversal changes both t,v \to -t,-v ... 6 No. Here's a simple example where it shrinks: You have a particle that has a 50% chance of being on the left going right, and a 50% chance of being on the right going left. This has a macroscopic error in both position and momentum. If you wait until it passes half way, it has a 100% chance of being in the middle. This has a microscopic error in position. ... 6 The microscopic laws of physics are reversible or, to say the least, CPT-symmetric (processes are invariant if they're run backwards in time, in mirror, and with antiparticles). The CPT symmetry follows from the Lorentz symmetry. Langton's ant as well as pretty much any other Turing machine or cellular automaton fails to be microscopically reversible; ... 6 Conservation of energy follows from invariance under translation in time, not inversion. This symmetry states that no matter when you do your experiment, it will give the same results. All isolated systems obey this symmetry (and therefore conserve energy) and no violation of it has ever been detected. (Needless to say, it would be a huge event if it were.) ... 6 The English Wikipedia article on magnetic monopoles has the following equation for the 'extended' Lorentz-Force of a magnetic field on a electrically and magnetically charged particle:$$ \vec{F}=q_{\mathrm e}\left(\vec{E}+\vec{v}\times\vec{B}\right) + q_{\mathrm m}\left(\vec{B}-\vec{v}\times\frac{\vec{E}}{c^2}\right) $$Under time reversal (t is ... 5 The functions -iEt and -i(-E)(-t) are exactly the same so they obviously correspond to the same sign of energy if they appear in the exponent defining |\psi\rangle. It seems that you think that you may freely replace t by -t and change nothing else. However, this operation isn't a symmetry of the laws of physics, as you have actually demonstrated ... 5 At research level, you might be interested in the PDG review on conservation laws: http://pdg.lbl.gov/2009/reviews/rpp2009-rev-conservation-laws.pdf Also, the review about CPT invariance gives information about tests of CPT violation in neutral kaons, but I can apparently only post one hyperlink (reputation too low). Note that CP violation itself is still ... 5 You ask: From my studies in quantum mechanics, I don't remember any postulates stating anything like this, but this all makes sense to me. Are there any theories out there that go along these lines? Indeed there is such a theory. It's called decoherence. You mention the comparison with thermodynamics, and this is basically the same way decoherence ... 5 Reference (page 13, formula 17.71) The time-reversal operator is \Theta = Ke^{-i\pi S_y/\hbar}, where K is the complex conjugation operator. Taking a spin 1/2, we have a wavefunction which is a 2- component spinor \psi(x) = \begin{pmatrix} \psi_+(x) \\ \psi_-(x) \end{pmatrix}, Note that, for spin 1/2, e^{-i\pi S_y/\hbar} = e^{-i \large ... 5 Energy, frequency and period are all scalars. It's correct to write:$$ E = \frac{h}{\tau} $$where \tau is the period, but the sign of \tau does not change if you run time backwards so the sign of energy doesn't change in QM either. 5 As John Rennie wrote in his answer, what one should consider is not a generic \Delta t but the period \tau which is a positive number. However positivity also arises form coherence with other relations. In particular, in relativistic quantum mechanics E= \sqrt{\hbar^2 \vec{k}^2 + m^2c^4}. For m=0 you have E= \hbar |\vec{k}| = \hbar \omega = h\nu ... 5 I think its because both halves of a projectile's trajectory are symmetric in every aspect. The projectile going from its apex position to the ground is just the time reversed version of the projectile going from its initial position to the apex position. 5 That argument is correct, but it applies to a trajectory, which is a function mapping times to position vectors. A single position vector itself, something like (-5,2,3), is not associated with time in any way, and when you reverse time, it doesn't change the point that is labeled by those coordinates. Maybe this simplified example will help: consider a ... 5 The Poincare algebra implies$$ T ( i H ) T^{-1} = - i H $$where T is the time-reversal operator. (Can you prove this?) Now, suppose T is a linear operator, then T H T^{-1} = - H. This implies that if |\Psi\rangle is an eigenstate of the Hamiltonian with energy E, then T^{-1} | \Psi \rangle has energy -E. This implies that the Hamiltonian is ... 5 One argument is based on the preservation of the CCR$$\tag{1} [x,p]~=~i\hbar~{\bf 1}.$$Let T be an invertible \mathbb{R}-linear operator with the usual time-reversal properties:$$\tag{2} TxT^{-1}~=~x\quad\text{and}\quad TpT^{-1}~=~-p.$$Then$$\tag{3} Ti\hbar T^{-1}~=~ T[x,p]T^{-1}~=~[TxT^{-1},TpT^{-1}]~=~-[x,p]~=~-i\hbar~{\bf 1}, i.e. $T$ ...
The transformation $t \mapsto -t$ transforms an ascending function into a decreasing one. It thus inverts the flow of time. You have to think of it as a transformation on the variable $t$ which is an input in observables : each observable $O(t)$ will see time flowing backwards. Assume you want to apply the time-inverting transformation at some time $t_0$. ...