68

How is the claim "information is indestructible" compatible with "information is lost in entropy"? Let's make things as specific and as simple as possible. Let's forget about quantum physics and unitary dynamics, let's toy with utterly simple reversible cellular automata. Consider a spacetime consisting of a square lattice of cells with a trinary (3-...


46

(The answers by Mark H and B.fox were posted while this one was being written. This answer says the same thing in different words, but I went ahead and posted it anyway because sometimes saying the same thing in different words can be helpful.) The key is to appreciate the difference between losing information in practice and losing information in principle....


37

1) If you want a Noether theorem for information, there is no such thing. Trying to obtain it from a symmetry law, by Noether's theorem can't work, simply because information is not a quantity that can be obtained for instance by the derivative of the Lagrangian with respect to some variable. Information is not scalar, vector, tensor, spinor etc. 2) ...


22

When Dr. Hawking talks about information being destroyed, he is talking about the erasure of all evidence that the information ever existed. In the case of burning a written letter, you could track the trajectory and composition of every smoke particle as the book burned. Since ink and paper generate different kinds of smoke and start at different locations, ...


21

Let $U$ be an unitary operator. Write $$ U=\mathbb I+\epsilon A $$ for some $\epsilon\in\mathbb C$, and some operator $A$. Unitarity means $U^\dagger U=\mathbb I$, i.e., $$ U^\dagger U=(\mathbb I+\epsilon^* A^\dagger)(\mathbb I+\epsilon A)=I+\epsilon^*A^\dagger+\epsilon A+\mathcal O(\epsilon^2) $$ Therefore, if $U^\dagger U=\mathbb I$, we must have $$ \...


21

Is it true that two different states cannot evolve into the same final state? That depends on exactly what you mean. If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time. You may have learned that quantum states evolve with a unitary transformation, i.e. $$ \lvert ...


20

I don't know in which context Susskind mentioned this, but he probably meant time evolution is unitary. That means, among other things, that it's reversible, ie no information can ever get lost because you can essentially, starting from any time (time-like slice), run time backwards (theoretically) and compute what happened earlier. If black hole evolution ...


19

As Lubos Motl and twistor59 explain, a necessary condition for unitarity is that the Yang Mills (YM) gauge group $G$ with corresponding Lie algebra $g$ should be real and have a positive (semi)definite associative/invariant bilinear form $\kappa: g\times g \to \mathbb{R}$, cf. the kinetic part of the Yang Mills action. The bilinear form $\kappa$ is often ...


18

One can indeed motivate the Schrödinger equation along the lines you suggest. The crucial point you are missing is time shift invariance of your quantum system. It is this that lets you write down $U = \exp(i\alpha\,H\,t)$. To explain further: The fact of $\psi(t) = U(t) \psi(0)$ is simply a linearity assumption. The evolution wrought by your state ...


17

We don't have to modify the basic laws of quantum mechanics to describe unstable particles. The full state of the system is includes the state of the decay products, and what you really have is a coupling from one state to another. No imaginary energies are required to describe this, but you do need to include the states of the decay products in your ...


14

CPT seems to imply it. You can reverse the system evolution by applying charge, parity and time conjugation, so the information about the past must be contained in the present state. That implies conservation of information by the evolution. This may not be the answer you wanted, because it does not imply unitarity, but it is the only relationship between ...


13

I'll forewarn that I'm no string theorist and Susskind's work is not therefore fully wonted to me (and likely I couldn't understand it if it were) so I do not fully know the context (of the supposed quote that entropy is hidden information). But what he maybe means by "hidden" information is one or both of two things: the first theoretical, the second ...


13

No, the ground state is not well-defined because the energy is unbounded below. To see this, switch back to the variables $x$ and $p$ using $a \sim x + ip$ to find $$H \sim p^2 - x^2.$$ This is the Hamiltonian for a particle in a potential that just pushes it further away from the origin, so you can make the energy as negative as you want, and there's no ...


12

(1) The completeness relationship for a basis of vectors orthonormal with respect to $\eta_{\mu\nu}$ is \begin{equation} \eta_{ij}\epsilon^{(i)}_\mu \epsilon^{(j)}_\nu = \eta_{\mu\nu} \end{equation} This normalization convention is picked for Lorentz invariance... I know you said you didn't want that answer but the point is that the normalization of these ...


11

I recommend that you to read the chapter 15.2 in "The Quantum Theory of Fields" Volume 2 by Steven Weinberg, he answers precisely your question. Here a short summary In a gauge theory with algebra generators satisfying $$ [t_\alpha,t_\beta]=iC^\gamma_{\alpha\beta}t_\gamma $$ it can be checked that the field strength tensor $F^\beta_{\mu\nu}$ transforms as ...


11

1. How can we show that $\partial\cdot j\equiv 0$ at the quantum level? For example, by showing that the Ward Identity holds. It should be more or less clear that the WI holds if and only if $\partial\cdot j=0$. There are multiple proofs of the validity of the WI; some of them assume that $\partial\cdot j=0$, and some of them use a diagrammatic analysis to ...


10

It's because you want the kinetic part of the Yang Mills action $$ \int Tr({\bf{F^2}}) dV$$ to be positive definite. To guarantee this, the Lie algebra inner product you're using (Killing form) needs to be positive definite. This is guaranteed if the gauge group is compact and semi-simple. (I'm not sure if it's only if G is compact and semi simple though. ...


10

To expand on @user26374's answer a little, the phrase "A QFT is unitary" comes from the requirement that the $S$-matrix is unitary, i.e. $S S^\dagger = S^\dagger S = 1$ which is equivalent to the statement that sum of probabilities is 1. Unitarity implies several serious constraints on how a QFT can be formulated. For example, unitarity implies the Froissart ...


10

Short answer since I'm on mobile: No, it's not a postulate but rather a theorem. First, clear things up a little. We want a symmetry of the theory to act as an arbitrary transformation which conserves the unitarity of our theory. A transformation which does not act as a symmetry of our system need not to be a linear transformation. Now, for symmetries, ...


9

I) We are given an angular momentum operator $\vec{S}$ in an (unitary, finite-dimensional, irreducible) spin $s$-representation $$\tag{1} \vec{S}^2~=~s(s+1){\bf 1}, \qquad s\in \frac{1}{2}\mathbb{N}_0, $$ $$\tag{2} [S_i,S_j]~=~i\sum_{k=1}^3\epsilon_{ijk} S_k, \qquad i,j,k\in\{x,y,z\}, \qquad S_i^{\dagger}~=~ S_i,$$ or in terms of raising and lowering ...


9

I think it may also be fruitful to not think of the Hamiltonian as the energy. The Hamiltonian is the generator of time translations, and so an eigenvalue of the Hamiltonian that is complex tells you that there is some decay, as was pointed out by Edoot. This is an important distinction. When we compute the "energy spectrum" of a Hamiltonian, what we are ...


9

For Poincaré algebra there are (as far as I know) two different approaches to find its representations. In the first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on the space of some fields on Minkowski space. Representation so obtained is usually not irreducible ...


9

Let me first make a general remark about internal symmetry groups, unrelated to our problem of the correct symmetry group for QCD. The symmetry must act on Hilbert space as a unitary operator for the conservation of probability. Now let us turn to the strong interaction. The most important experimental facts were that Observed hadron spectrum was ...


9

We know, from Schrodinger's equation, that $$ \frac{d\psi}{dt}=\left(\frac{i\hbar}{2m}\frac{d^2}{dx^2}-\frac{i}{\hbar}V(x)\right)\psi $$ Taking the conjugate tells us how $\psi^*$ evolves. $$ \frac{d\psi^*}{dt}=\left(-\frac{i\hbar}{2m}\frac{d^2}{dx^2}+\frac{i}{\hbar}V(x)\right)\psi^* $$ So let's calculate $\frac{d}{dt}|\psi|^2$. \begin{align} \frac{d}{...


9

Proposition. Let $G$ be a connected non-compact Lie group that is a semisimple Lie group and $$U: G \ni g \mapsto U_g \in B(H)$$ ($B(H)$ being the set of bounded operators $A:H \to H$) a continuous unitary representation over the finite-dimensional Hilbert space $H$. The following facts hold. (a) $U$ cannot be faithful. (b) If $G$ is a simple group or, ...


9

Well, the point is not continuity of the representation (if continuity is referred to the representation, see the final comment), but a related deeper fact concerning the group itself (though some vague argument based on continuity also of the representation is recurrent in the literature). One has also to remember that (Wigner's theorem) symmetries are ...


9

The Stone's theorem proves the following. Consider a group of unitary operators $(U(t))_{t\in\mathbb{R}}$ acting on a Hilbert space $\mathscr{H}$ (i.e. satisfying $U(t+s)=U(t)U(s)$, in more mathematical terms this is a unitary representation of the abelian group $\mathbb{R}$ on $\mathscr{H}$). If in addition such group is strongly continuous, namely is such ...


9

I must admit, I'm not the most qualified person to answer here. If you place the burning letter into a sealed box (a closed system) and allow the system to carry itself to its final state, with a perfect model of the physics involved, you can trace the end state back to its original state if you know the properties of the end state perfectly because the ...


9

Every representation $(D,V)$ of a finite group $G$ is equivalent to a unitary representation. It is often termed as Weyl's unitary trick. This works by simply redefining your inner product by averaging over on the space $V$. This smoothening trick works precisely because of finite number of elements and invariance of sum of finite elements (i.e. $\sum_{g \in ...


8

$S^{-1}=S^*$ is just the condition for unitarity. It is usually written as $S^*S=1$ (together with invertibility) and means that $\psi^*\psi$ doesn't change when $\psi$ is replaced by $S\psi$: $(S\psi)^*(S\psi)=\psi^*S^*S\psi=\psi^*\psi$ Therefore probability is conserved, a must for a good scattering matrix. In general, unitarity of the S-matrix is a ...


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