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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 ...


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 ...


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 ...


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.) ...


5

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; ...


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

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 $E= \hbar |\vec{k}| = \hbar \omega = h\nu$ ...


4

Just a few pointers for you to explore more on this. Check out Aharonov's paper the time symmetric formulation of quantum mechanics: http://arxiv.org/abs/quant-ph/9501011 Tony Leggett talks about this: http://www.youtube.com/watch?v=IGim9uzcumk It's a nice video and quite simple to understand.


4

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 ...


4

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 ...


3

$PT-, T-, P-$ transformations refer to subgroup of discrete transformations of the Lorentz group. They transform connected components of the Lorentz group between each other ($PT$ transformation transforms $L^{\uparrow}_{+}$ representation to $L^{\downarrow}_{+}$). In general, they can't be represented as the special case of rotation, which refer to subgroup ...


3

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 ...


2

There are two possible answers to why $T^2=-1$: a) Why not. The total phase of a quantum state is unphysical. So a symmetry may be realized as a projective representation. Here T may be viewed as a projective representation of time reversal $T_{phy}$ which satisfy $T^2_{phy}=1$. b) If we define the time reversal symmetry to be realized as a regular ...


2

|Dear Mr Student, the time-reversal symmetry in conventional QM at best (e.g. no magnetic fields) only applies to the unitary evolution of a quantum system; the measurement process is not time-reversal symmetric. Also, the second law of Thermodynamics says (very roughly speaking) the Entropy should decrease if you go back in time; again there is no ...


2

If I'm reading your question correctly, it is at least in part whether the nominal event of "wave collapse" (please note that different schools of thought describe that event differently!) is reversible in time. I won't try to address the schools, but rather whether what you ask has any experimental meaning. This is not a complete answer, but the concept of ...


2

Let us here just consider the classical case ($\hbar=0$). I) Noether's theorem does not work for discrete symmetries like time reversal symmetry, $$\tag{1} T: t ~\longrightarrow~ -t, $$ cf. e.g. this Phys.SE post. II) Instead, energy conservation follows from time translation symmetry $$\tag{2} t ~\longrightarrow~ t + a, \qquad a~\in~ \mathbb{R}, $$ ...


2

Again, thanks to the $SU(2)$ PSG proposed by prof.Wen, I can answer my question now, $THT^{-1}$ is in fact $SU(2)$ gauge equivalent to $H$, and the statement "$H$ is also not SU(2) gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$" in my question is wrong. Let's rewrite the Hamiltonian as ...


2

I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with $$H_{\pi}=t\left(\begin{array}{cccc} 0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\ e^{-\mathbf{i}\pi/4} & 0 & ...


2

I think the answer should be 'no'. Because when we introduce the antiunitary time-reversal(TR) opeartor $T$ for spin-system, it should satisfy $T\mathbf{S}_iT^{-1}=-\mathbf{S}_i$ since angular-momentum should be sign reversed under TR(due to the classical correspondence). Thus, spin-spin interactions like $\mathbf{S}_i\cdot\mathbf{S}_j$ are invariant under ...


2

The laws of electromagnetism are indeed time-reversal invariant, but this can sometimes be in a fairly restricted sense, particularly when dealing with causality as regards the Liénard-Wiechert potentials. Consider the following situation. A charged particle is travelling under the influence of some force $\mathbf F_\text{ext}(t)$ which is due to a ...


2

The the easiest way to see that time reversal transforms electrons into positrons relies on the fact that PCT (parity, charge conjugation and time reversal) combined are a symmetry of every Lorentz-invant QFT. Using $P^{-1} = P$, $C^{-1} = C$, $T^{-1} = T$, i.e. a parity transformation is undone by a second parity transformation etc. you can see that $$1 = ...


1

Based on the article by Kupervasser suggested in the comments by Ben Crowell, I suspect the answer is that my hypothetical situation is impossible: in general, for a complex enough dynamical system, there is no solution to the equations is which two universes with opposite arrows of time can interact. In order to have that, your system should have to satisfy ...


1

In real world there is always dissipation and therefore there is an "arrow of time" and time-reversal symmetry does not hold (right?). There are many phenomena in physics that we understand by invoking the time reversal symmetry of microscopic interactions. This is distinct from the "arrow of time" and the fact that an out-of-equilibrium system ...


1

You are just wrong. 1) The time reversal symmetry you are speaking of is not the time reversal symmetry which is considered when topological isulators are discussed. In the latter case just no magnetic field (both external and internal) is enough. In that case effective Hamiltonian of the system allows for the time inversion symmetry. Formally you may ...


1

I) First of all, one should never use the Dirac bra-ket notation (in its ultimate version where an operator acts to the right on kets and to the left on bras) to consider the definition of adjointness, since the notation was designed to make the adjointness property look like a mathematical triviality, which it is not. See also this Phys.SE post. II) OP's ...


1

Feynman studied the relation between negative energy, antimatter, and particles moving backward in time. Let me quote him [1]: "The fundamental idea is that the 'negative energy' states represent the states of electrons moving backward in time [...] reversing the direction of proper time s amounts to the same as reversing the sign of the charge so that the ...


1

Remember that for crystalline materials, we usually assume an infinite number of particles, and that electrons do not interact. This allows us to Fourier transform and see that each pseudo-momentum $k$ is independent --- essentially to consider a single unit cell. In this context, Kramer's theorem states that if there is an odd number of electrons per unit ...


1

Consider derivative at $t=0$; denote $\Psi(0)$ as $\Psi_{0}$ $(\partial /\partial t)T\Psi(t)\big|_{t=0}=\displaystyle lim_{h\rightarrow0}((T\Psi_{0})(h)-T\Psi_{0})/h$ Since $T\Psi$ evolves according to $i(\partial /\partial (-t))T\Psi(t)=HT\Psi(t)$ So $(T\Psi_{0}) (h)= exp(ihH)T\Psi_{0}$. Hence we have : $\displaystyle ...



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