TITLE: Integration using substitution
QUESTION [0 upvotes]: Can we substitute a linear function while integrating a function which is discontinous.
If not then what can be the complications when we get wrong answer doing that.
I have observed that many students don't check for continuity and differentiability of function while substitution. they take it for granted that it will be true. Isn't it wrong.

REPLY [2 votes]: The only complications to an integration by substitution come from irregularities in the function being substituted. What's irregular here? Any discontinuities - that's pretty much automatically fatal. Zero derivative somewhere - because we might have to divide by it. Not being monotone - because then the interval folds back onto itself. Not so bad if we have the derivative of the substitution there, but serious trouble if we don't. An inverse we can't calculate cleanly - this doesn't really impact the theory, but it'll likely ruin our chance of getting a nice formula for the answer.
If that's a linear (affine) function we're substituting, none of these irregularities apply. Affine substitutions always work. They're worth using even if only to gain some minor convenience.
As for the function we're integrating having discontinuities - that's not going to matter. They'll still be there after the substitution, just the same.


TITLE: What is the value of $-(-6)$?
QUESTION [0 upvotes]: I know this is not absolute value, so what do the parentheses mean? If it was absolute value, it would have lines so I can rule that out.
And I guess I should mention, the question I was asked is "Enter the opposite of this number in the box." So, that's the context.
$-(-6)$

REPLY [1 votes]: Remember that a negative number times a negative number is negative.
If you see $-(x)$ that's a short way of writing $(-1) \times x$, that is, "minus one" or "negative one" multiplied by $x$. In this particular case we have $x = -6$, so then $(-1) \times (-6) = 6$.
The parentheses in this case are for clarity: compare $--6$. Without the proper typesetting, that can look like one very long dash, instead of two minus signs. (In other cases, parentheses override operator precedence, plus they also have other mathematical uses that are not relevant to your question here--as for "minus one" versus "negative one", that's another can of worms altogether).


TITLE: Inverse Fourier Transform of the Hdamard's finite part - 2 dimensions
QUESTION [1 upvotes]: I'm trying to compute the Fourier transform of the Hadamard's finite part distribution in two dimensions, and it's giving me a little trouble.
In two dimensions, we defined the finite part $\text{PF}\left(1/k^2\right)$, where $k\equiv|\vec{k}|=\sqrt{k_x^2+k_y^2}$, as the distribution such that for each test function $f$
\begin{equation}
\left\langle \text{PF}\left(\frac{1}{k^2}\right),f\right\rangle =\text{PF}\iint\!d^2\vec{k}\frac{f(\vec{k})}{k^2} =\iint_{k<1}\!\!\!d^2\vec{k}\frac{f(\vec{k})-f(0)}{k^2}+\iint_{k>1}\!\!\!d^2\vec{k}\frac{f(\vec{k})}{k^2}
\end{equation}
Quoting my professor:

Sloppily then we can compute the inverse Fourier transform as
  \begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{(2\pi)^2}\text{PF}\iint\!d^2\vec{k}\frac{e^{i\vec{k}\cdot\vec{x}}}{k^2} = \frac{1}{(2\pi)^2}\text{PF}\int_0^\infty\!dk \frac{k}{k^2}\int_0^{2\pi}d\vartheta\,e^{ikr\cos{\vartheta}}
\end{equation}
  Where we used spherical coordinates in two dimensions, and we called $r\equiv |\vec{x}|=\sqrt{x^2+y^2}$. Using Bessel $J_n$ functions, one eventually get
  \begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{(2\pi)^2}\text{PF}\int_0^\infty\!dk\, (2\pi)\frac{J_0(kr)}{k} \tag{1}
\end{equation}
  Using now the definition of finite part, and changing variable $u\equiv rk$, it can be showed that
  \begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{2\pi}\left\{-\log(r)+\int_0^1\!du\frac{J_0(u)-J_0(0)}{u}+\int_1^\infty\!du\frac{J_0(u)}{u}\right\} \tag{2}
\end{equation}

First of all, I can't understand very well how do we get to equation (1), in the sense that it's seems that going in spherical coordinates allows us to "delete" one dimension, in the sense that equation (1) seems the Hadamard's finite part of the one dimensional variable $k$. I really don't understand the meaning of this equation.
Then, if we accept that, I cannot understand how to get to equation (2). Indeed if we rename $u=kr$ we get
\begin{equation}
\frac{1}{(2\pi)^2}\left\langle \text{PF}\left(\frac{1}{k^2}\right),e^{i\vec{k}\cdot\vec{x}}\right\rangle = \frac{1}{2\pi}\int_0^{1/r}\!du\,\frac{J_0(u)-J_0(0)}{u} +\frac{1}{2\pi}\int_{1/r}^\infty\!du\,\frac{J_0(u)}{u}
\end{equation}
And i really can't understand how to go on, and get the $\log(r)$ term.

REPLY [2 votes]: I'll change the notation slightly and use $k$ and $w$ for the vectors. Write the action of $1/|k|^2$ on $e^{i k \cdot w}$ by definition as if $e^{i k \cdot w}$ were a test function. In the resulting integral, make the change of variables $k_x = \rho \cos \phi, k_y = \rho \sin \phi$:
$$\left< \frac 1 {|k|^2}, e^{i k \cdot w} \right> =
\iint_{|k| < 1} \frac {e^{i k \cdot w} - 1} {|k|^2} dk_x dk_y +
 \iint_{|k| > 1} \frac {e^{i k \cdot w}} {|k|^2} dk_x dk_y = \\
\int_0^1 \int_0^{2\pi}
  \frac {\exp(i \rho (w_x \cos \phi + w_y \sin \phi)) - 1}
  \rho d\phi d\rho + \\
 \int_1^\infty \int_0^{2\pi}
  \frac {\exp(i \rho (w_x \cos \phi + w_y \sin \phi))} \rho d\phi d\rho.$$
We know that
$$w_x \cos \phi + w_y \sin \phi =
r \left( \frac {w_x} r \cos \phi + \frac {w_y} r \sin \phi \right) =
r \cos(\phi + \phi_0)$$
for some $\phi_0$, where $r = |w|$; since we're integrating over a complete period, the result does not depend on $\phi_0$, and
$$\int_0^{2\pi} e^{i \rho r \cos \phi} d\phi =
2 \pi J_0(\rho r) \Rightarrow \\
\frac 1 {2\pi} \left< \frac 1 {|k|^2}, e^{i k \cdot w} \right> =
\int_0^1 \frac {J_0(\rho r) - 1} \rho d\rho +
 \int_1^\infty \frac {J_0(\rho r)} \rho d\rho = \\
\int_0^r \frac {J_0(u) - 1} u du +
 \int_r^\infty \frac {J_0(u)} u du = \\
\left( \int_0^1 \frac {J_0(u) - 1} u du +
 \int_1^r \frac {J_0(u) - 1} u du \right) + \\
 \left( \int_1^\infty \frac {J_0(u)} u du +
 \int_r^1 \frac {J_0(u)} u du \right) = \\
-\ln r + \int_0^1 \frac {J_0(u) - 1} u du +
 \int_1^\infty \frac {J_0(u)} u du = \\
-\ln r - \gamma + \ln 2.$$
The last step is for extra points.


text:
\begin{document}
\maketitle

\begin{abstract}

The shifted boundary method (SBM) is an approximate domain method for boundary value problems, in the broader class of unfitted/embedded/immersed methods. It has proven to be quite efficient in handling problems with complex geometries, ranging from Poisson to Darcy, from Navier-Stokes to elasticity and beyond.

The key feature of the SBM is a {\it shift} in the location where Dirichlet boundary conditions are applied - from the true to a surrogate boundary - and an appropriate modification (again, a {\it shift}) of the value of the boundary conditions, in order to reduce the consistency error.

In this paper we provide a sound analysis of the method in smooth and non-smooth domains, highlighting the influence of geometry and distance between exact and surrogate boundaries upon the convergence rate.

Without loss of generality, we consider the Poisson problem with Dirichlet boundary conditions as a model and we first detail a procedure to obtain the crucial shifting between the surrogate and the true boundaries. Next, we give a sufficient condition for the well-posedness and stability of the discrete problem. The behavior of the consistency error arising from shifting the boundary conditions is thoroughly analyzed, for smooth boundaries and for boundaries with corners and edges. The convergence rate is proven to be optimal in the energy norm, and is further enhanced in the $L^2$-norm.

\end{abstract}

\noindent \keywords{Embedded methods; immersed boundary methods; non-smooth domains; finite element methods; weak boundary conditions; Taylor expansions; consistency error; convergence analysis}

\noindent\amsclass{Primary 65N30; Secondary 65N12, 65N50}

\input{./LaTeX_files/SBM_Poisson_introduction}

\input{./LaTeX_files/SBM_Poisson_method}

\input{./LaTeX_files/SBM_Poisson_stability}

\input{./LaTeX_files/SBM_Poisson_remainder}

\input{./LaTeX_files/SBM_Poisson_convergence}

\input{./LaTeX_files/SBM_Poisson_duality}

\input{./LaTeX_files/SBM_Poisson_numericalTests}

\section*{Acknowledgments}
The support of the Army Research Office (ARO) under Grant W911NF-18-1-0308 is gratefully acknowledged. CC performed this research in the framework of the Italian MIUR Award ``Dipartimenti di Eccellenza 2018-2022" granted to the Department of Mathematical Sciences, Politecnico di Torino (CUP: E11G18000350001), and with the support of the Italian MIUR PRIN Project 201752HKH8-003.
He is a member of the Italian INdAM-GNCS research group.

\bibliographystyle{plain}
\bibliography{./SBM_Poisson}

\end{document}

config: arxiv


TITLE: What is the ratio of Carmichael pseduo-primes to true primes for $1$ to $n$? Or is it known?
QUESTION [7 upvotes]: Let $\pi(n)$ be the prime counting function. And let $\varphi(n)$ be the count of Carmichael pseudo-primes for $1$ to $n$. Is the ratio,
$$\frac{\varphi(n)}{\pi(n)}$$
is known, as $n \to \infty$? I am asking this because of i want to know how much safe is the Fermat Primality Test (the one which only uses Fermat's Little Theorem).
I recently wrote a Python script (initially for RSA) and produced more than 10, random 1024 bit primes with the help of this test. After i used the powerful mathematical software PARI/GP to check if they are really primes or pseudo-primes. In conclusion none of them were pseudo-primes. So, if know this rate, i will know the probability that this test will encounter a pseudo-prime and fail.

REPLY [2 votes]: We have the following result (Knödel, $1953$) :-
$$\varphi(n)<n\exp\big(-k(\log n \log\log n)^\frac{1}{2}\big)$$
By strengthened version of PNT, we will have for sufficiently large $n$:
$$\pi(n)>\frac{n}{\log n}$$
Thus $\forall$ sufficiently large $n$, we have:-
$$\frac{\varphi(n)}{\pi(n)}<\log n \exp\big(-k(\log n \log\log n)^\frac{1}{2}\big)=\exp \big(\log \log n -k(\log n \log\log n)^\frac{1}{2}\big)$$
Note that:-
$$\lim_{n \to \infty} \log \log n -k(\log n \log\log n)^\frac{1}{2}=- \infty \implies \lim_{n \to \infty} \exp \big(\log \log n -k(\log n \log\log n)^\frac{1}{2}\big)=0$$
Hence, we have:-
$$\lim_{n \to \infty}\frac{\varphi(n)}{\pi(n)}=0$$


TITLE: The volume enclosed by a sphere of radius $r$ is $\frac{4}{3} \pi r^3$
QUESTION [0 upvotes]: The volume enclosed by a sphere of radius radius $r$ is $\frac{4}{3} \pi r^3$. The surface area of the same sphere is $4\pi r^2$. You may already have noticed that the volume is exactly $\frac{1}{3}r$ times the surface area. Explain why this relationship should be expected. One way is to consider a billion-faceted polyhedron that is circumscribed about a sphere of radius $r$; how are its volume and surface area related?
I completed a problem similar to this relating area and perimeter, but I can't seem to think of a way that volume and surface area are related... Does anyone have tips?? Thank you!

REPLY [0 votes]: Here is an arguement based on calculus.
If you think of inflating the sphere.  Then the change in volume is surface area times the change in radius (for small changes.)
$dA = \frac {dV}{dr} dr = 4\pi r^2 dr$
Another way to think about it is a multifaceted polyhedra that encloses the sphere, with each face tangent to the sphere.
The polyhedra is then many pyrmids, each with volume equal to $\frac 13$ the base area times the radius.  At the limit the surface area of the polyhedra equals the surface area of the sphere and the volume equals the volume.
Based on classical geometry, Archimedes showed that the surface area of a sphere projects onto a cone of equal radius and height equal to twice the radius.
While Cavileri showed that the volume of a sphere equals the volume of the same cylinder less the volume of a doubled cone that fits inside this cylinder.
https://en.wikipedia.org/wiki/Cavalieri%27s_principle


TITLE: Continuity of a function and its restriction
QUESTION [1 upvotes]: Let $(X,T)$ be a topological space and $A$ subset in $X$, and
$f:X\to (Y,t)$ a function.
If $A$ is open and $g:A\to Y$ is continuous (where $g$ is the restriction of $f$ on $A$) , then so is $f$.
Can someone prove it for me?

REPLY [0 votes]: Under the additional assumption, stated in the comments (by the OP and later deleted), that $X=\cup\{A_i:i\in I\}$ (for some index set $I$) where each $A_i$ is open and each restriction $f:A_i\to Y$ is continuous, then indeed $f:X\to Y$ is continuous$.
Take any open $U\subset Y$ we need to show that $f^{-1}(U)$ is open in $X$. Each $A_i\cap U$ is open, and hence, by continuity of $f$, we have that $f^{-1}(A_i\cap U)$ is open.
Finally $f^{-1}(U)=\cup\{f^{-1}(A_i\cap U):i\in I\}$ is open, as the union of open sets.
Regarding the question as originally stated:
Let $X$ be a topological space and $A$ subset in $X$, and $f:X\to Y$ a function. If $A$ is open and $g:A\to Y$ is continuous (where $g$ is the restriction of $f$ on $A$), then so is $f$. Can someone prove it for me?
The above statement cannot be proved as it is incorrect. If it were correct, then it would imply that all functions between all topological spaces are continuous. Indeed, let $A$ be the empty set $\emptyset$. Then the restriction $g:A\to Y$ is continuous. If the above statement were correct, Then $f$ would also be continuous.
For a specific example, let $X=[0,1]$, $A=[0,1)$ and $Y=\Bbb R$ (or $Y=[0,1]$, or $Y=\{0,1\}$, either way). Let $f(x)=0$ for $x\in A$, and $f(1)=1$. Then $f$ is not continuous (since $\lim\limits_{x\to1^-}f(x)=0\neq1=f(1)$), even if the restriction of $f$ to $A$ is constant (and hence continuous).
For an even better/intuitive understanding take the following point of view.
Start with $X,Y,A$ and continuous $g:A\to Y$. Then extend $g$ on all of $X$, arbitrarily, to obtain some function $f:X\to Y$. There are (in general) zillions of ways to obtain such extensions, and very few of them (under some special/additional conditions) would be continuous. Most of the time such an extension would be discontinuous. (@HennoBrandsma was also alluding to this, in a comment to his answer, I would believe.)


TITLE: Why does variation of parameters work?
QUESTION [7 upvotes]: I was looking at the variation of parameters method, and to be sincere, when I took my differential equations course I felt like too much of it was Hocus Pocus.
For example, there is this
https://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx
I don't know where does (3) comes out from.
It says:
"Now, there is no reason ahead of time to believe that this can be done. However, we will see that this will work out."
My differential equations professor didn't explain it. It wasn't outrageous so I used, simply as a recipe. I would like to have some intuition, some idea behind it.
When you solve some equations in which you need to integrate from both sides, for example
$$
\int f(t) dt = \int g(t) dt
$$
Then you have $F(t)+c_{1}=G(t)+c_{2}$, but you only write $F(t)=G(t)+c$ because $c_{1}$ and $c_{2}$ don't matter, only their difference. I guess something similar is going on there, there are a lots of ways to write that with different functions and you're not interested on the $u_{1}(t)$ and the $u_{2}(t)$ but on a relationship between them, so you choose them accordingly in order to simplify the equation as the relationship will hold.
But I would love to hear some idea, some intuition behind it, as each time that someone ask me for help with differential equations I want to tell them that I know nothing about them, because that's how I really feel.
Thank you for your time.

REPLY [6 votes]: An interesting way of understanding the variation of parameters is as a multidimensional integrating factor method. You can rewrite your linear system of differential equations as first-order linear differential equation with multiple variables, i.e. as
\begin{equation}
x^\prime(t) + A(t)x(t) = f(t),
\end{equation}
where $x(t),f(t)$ are $n$-dimensional vectors and $A(t)$ is an $n\times n$ matrix (in your case $x(t) = (g(t),g^\prime(t),...,)$ if $g$ is the solution to the higher order linear system). In fact this problem is more general.
Now let $B(t)$ be a fundamental matrix for the corresponding homogeneous system, i.e. the columns of $B(t)$ are $n$ linearly independent solutions to the corresponding equation with $f = 0$. This means that $B(t)$ is invertible for all $t$ as long as the coefficient matrix $A$ is continuous (Abel's theorem), and satisfies
\begin{equation}
B^\prime(t) + A(t)B(t) = 0.
\end{equation}
We can calculate $(B^{-1})^\prime(t)$ by differentiating the identity $B^{-1}(t)B(t) = I$ to get
\begin{equation}
(B^{-1})^\prime(t)B(t) + B^{-1}(t)B^\prime(t) = 0.
\end{equation}
Using that $B^\prime(t) = -A(t)B(t)$, we get
\begin{equation}
(B^{-1})^\prime(t)B(t) = B^{-1}(t)A(t)B(t),
\end{equation}
so that (cancelling the $B(t)$ which is invertible)
\begin{equation}
(B^{-1})^\prime(t) = B^{-1}(t)A(t).
\end{equation}
Thus $B^{-1}(t)$ is an integrating factor for the original equation and we get
\begin{equation}
(B^{-1}x)^\prime(t) = B^{-1}(t)(x^\prime(t) + A(t)x(t)) = B^{-1}(t)f(t).
\end{equation}
Solving for $x(t)$, we get
\begin{equation}
x(t) = B(t)\left(\int_0^t B^{-1}(t)f(t) + C\right),
\end{equation}
where the vector $C$ depends on the initial conditions.
This explains why the solution can be written as a linear combination (with time-varying coefficients) of the fundamental homogeneous solutions (the columns of $B(t)$).


TITLE: If A is a subset of a topological space, then $Bd(A) \subseteq Cl(A)$. Prove using elements/ sets.
QUESTION [0 upvotes]: If A is a subset of a topological space, then $Bd(A) \subseteq Cl(A)$. Prove.
I know this statement is true.  I am now trying to prove it.  I am in a basic topology class and to do a lot of set proofs we start by letting an element be included in one side and show it is in the other.
I was given the hint to use this fact $Cl(A)= A \cup Bd(A)$.  However I have to prove that statement first.  For this statement I started with Let $x \in Cl(A)$ then by defintion of closure $x \in A$ since the closure is the smallest closed set containing $A$ and therefore $x \in A \cup Bd(A)$.
Now let $x \in  A \cup Bd(A)$. How do I show $x \in$ the $Cl(A)$ and then use that prove my first statement?

REPLY [0 votes]: I am not sure I understand the hint. If you are allowed to assume it, the question is trivial, but proving it is equivalent to proving the problem...
Let $x\in Bd(A)$. Then by definition,  every open set containing $x$ intersects both $A$ and $X−A$.
If $x\notin Cl(A)$, then there exist on open set, namely $U = X-Cl(A)$, containing $x$ such that $U\cap A = \emptyset$.
We can see this since $A\subset Cl(A)$.
But this contradicts the definition of $x\in Bd(A)$, so we get that $x\in Cl(A)$.
Thus, $Bd(A)\subset Cl(A)$.


TITLE: Smooth map that is a topological submersion but not a smooth submersion
QUESTION [4 upvotes]: This is an exercise from John Lee. I need to find a smooth map that is a topological submersion but not a smooth submersion. I am trying to work with the Euclidean space but cannot find a proper example. Could anyone please help me?

REPLY [6 votes]: Consider the map from $\mathbb{R}^2$ to $\mathbb{R}$ sending $(x,y)$ to $x^3$. This is a topological submersion, as we can choose a chart on $\mathbb{R}$ by having $z = x^3$, and on this chart this map is the projection $(x,y) \mapsto x$. It is not a smooth submersion, as the derivative at $(0,0)$ is $0$.


text:
\begin{document}
\frenchspacing

\title{Locality Bound for Dissipative Quantum Transport}
\author{Xizhi Han and Sean A. Hartnoll \\ {\it Department of Physics, Stanford University,}\\
{\it Stanford, California, USA}
}

\date{}

\begin{abstract}
We prove an upper bound on the diffusivity of a dissipative, local and translation invariant quantum Markovian spin system: $D \leq D_0 + \left(\a \, v_\text{LR} \tau + \beta \, \xi \right) v_\text{C}$. Here $v_\text{LR}$ is the Lieb-Robinson velocity, $ v_\text{C}$ is a velocity defined by the current operator, $\tau$ is the decoherence time, $\xi$ is the range of interactions, $D_0$ is a decoherence-induced microscopic diffusivity and $\a$ and $\b$ are precisely defined dimensionless coefficients. The bound constrains quantum transport by quantities that can either be obtained from the microscopic interactions ($D_0, v_\text{LR}, v_\text{C},\xi$) or else determined from independent local non-transport measurements ($\tau,\a,\beta$). We illustrate the general result with the case of a spin half XXZ chain with on-site dephasing. Our result generalizes the Lieb-Robinson bound to constrain the sub-ballistic diffusion of conserved densities in a dissipative setting.
\end{abstract}

\maketitle

{\it Introduction.---} Quantum transport processes are at the heart of experimental studies of unconventional metals \cite{RevModPhys.75.1085, hussey, Bruin804}, ultracold atomic gases \cite{schneider2012fermionic, PhysRevLett.108.093002, koschorreck2013universal, hild14, chien2015quantum,  PhysRevLett.118.130405} and potential spintronic systems \cite{Wolf1488, RevModPhys.76.323, awschalom2007challenges, bogani2008molecular, han2014graphene}. It is crucial to have theoretical tools that connect transport observables to microscopic processes. In quasiparticle systems such as conventional metals, Fermi liquid theory and Boltzmann equations offer an excellent and well-understood handle on transport \cite{ziman}. For strongly quantum transport regimes, however, there are many fewer tools available. Controlled theoretical work with realistic interactions has largely been restricted to numerics in one spatial dimension  \cite{SCHOLLWOCK2011, kar12, Leviatan:2017, knap17, zal18}.

For general ballistic systems, several important, rigorous bounds on quantum transport have been established. The Mazur-Suzuki inequality bounds the Drude weight in terms of the overlap of currents with conserved charges \cite{MAZUR69, SUZUKI71, pro11}. The Lieb-Robinson velocity $v_\text{LR}$ bounds the propagation of linearly dispersing collective modes, such as spin waves

\cite{lieb1972, PhysRevLett.104.190401}.

Many important quantum transport processes are diffusive rather than ballistic \cite{hartnoll15}. A lower bound on the high-temperature diffusivity has been established for certain systems with integrability or additional symmetries \cite{pro14, PhysRevLett.119.080602}. Recently, it was argued that in general local systems, Lieb-Robinson causality requires that the diffusivity be upper bounded as $D \lesssim v_\text{LR}^2 \tau_\text{th}$ \cite{Hartman:2017hhp}. Here $\tau_\text{th}$ is a `local thermalization time'. This relation usefully identifies key physical ingredients that constrain diffusive transport. However, it is not totally satisfactory because a numerical prefactor is undetermined and furthermore the timescale $\tau_\text{th}$ was not precisely defined.

In this work we prove a rigorous and precise upper bound on the diffusivity of dissipative quantum Markovian spin systems. The full result is given in (\ref{eq:fullbound}) below. In the limit of long decoherence time $\tau$, the bound takes the form $D \leq \alpha v_\text{LR} v_\text{C} \tau$. This expression is the dissipative counterpart of the earlier bound \cite{Hartman:2017hhp}, and all quantities on the right hand side will now be precisely defined. The velocities $v_\text{LR}$ and $v_\text{C}$ are straightforwardly computed given a microscopic Hamiltonian while the dimensionless coefficient $\a$ and decoherence time $\tau$ can be independently and unambiguously determined from local non-transport observables. Therefore, this bound can be precisely verified in experiments. It generalizes the Lieb-Robinson bound to the diffusive behavior of conserved densities, in the context of dissipative quantum Markovian dynamics.

{\it Translation invariant Lindbladian dynamics.---} Non-unitary quantum dynamics describes the quantum evolution of a dissipative system coupled to an external environment. On timescales much longer than the relaxation time of the reservoir, the dynamics can be well approximated as Markovian and hence described by the Lindblad equation \cite{lindblad1976, bre07}. The final state of Lindbladian non-unitary dynamics is expected to be an infinite temperature generalized Gibbs ensemble, so our diffusive dynamics occurs close to this state.

We assume that the external bath couples locally in space to the degrees of freedom of interest, and preserves spatial translation invariance.
In this case, the most general Heisenberg equation of motion for an operator $O(t)$ on a lattice takes the Lindblad form
\be
\dot{O}  = i \sum_{\bx} [H_{\bx}, O] + c \sum_{\bx, \a} \left( 2 L_{\bx}^{\a \dagger} O L_{\bx}^{\a} -  \left\{ L_{\bx}^{\a \dagger} L_{\bx}^{\a}, O\right\} \right), \label{eq:1}
\ee
where $\bx$ is the lattice index and $H_{\bx}$ is a term in the Hamiltonian localized near lattice site $\bx$. The anticommutator $\{A,B\} = AB +BA$. The $L_{\bx}^\alpha$ are decoherence operators localized near site $\bx$ and $c \geq 0$ is the decoherence strength.

It will be important that a Lieb-Robinson velocity exists for such local Lindbladian dynamics \cite{PhysRevLett.104.190401, nachtergaele2011lieb}.

Throughout, we illustrate our general formalism and results with the example of an infinite, spin-half antiferromagnetic XXZ chain with on-site dephasing:
\begin{align}\label{eq:model}
    H_x  = X_x X_{x + 1} + Y_x Y_{x + 1} + \Delta Z_x Z_{x + 1} \,, \quad L_x  = Z_x \,.
\end{align}
Here $X_x, Y_x, Z_x$ are Pauli matrices acting on spin $x \in \mathbb{Z}$ and $\Delta > 0$ is the anisotropy. The dephasing Lindbladian is a common phenomenological description of decoherence due to coupling to a photon or phonon bath \cite{RevModPhys.70.101}. Diffusion in this model was studied numerically in \cite{z, PhysRevLett.106.220601, men13, pro16, ljubotina2017spin}, and we will compare with those results. Our approach, however, is not limited to one dimensional models.

The model (\ref{eq:model}) conserves spin: $\sum_{x} \dot{Z}_x = 0$. More generally, we require a local charge operator $C$ such that
\be\label{eq:conserved}
\sum_{\bx} \dot{C}_{\bx} = 0,
\ee
where $C_{\bx}$ is the operator $C$ translated to site $\bx$. A conserved operator in the sense of (\ref{eq:conserved}) has important consequences for the dynamics on the longest timescales, after all non-conserved operators have decayed. A single, scalar conserved operator is expected to lead to a diffusive mode with long wavelength dispersion
$\omega(\bk) = - i D k^2 + \ldots$, see e.g. \cite{cha95}.  Here $D$ is the diffusivity and $\ldots$ denotes terms of higher order in the wavevector $\bk$. Our objective in the remainder is to connect the microscopic Lindbladian dynamics (\ref{eq:1}) to the long wavelength hydrodynamic mode, and in this way bound the diffusivity $D$ in terms of microscopic quantities.

To exploit the translation invariance of the dynamics, we introduce the linear space of operators $\ocal_{\bk}$ with wavevector $\bk$, defined to be the space of all operators $O$ on an infinite lattice $\Lambda$ such that
\be
\Ts_{\bx}[O] = O\, e^{i \bk \cdot \bx},
\ee
where $\Ts_{\bx}$ translates operators by a vector $\bx$. It will be useful to take the following basis of operators in $\ocal_{\bk}$. Fix an origin of the lattice and a direction $\hat{\bk}$ of the wavevector. We can then write the basis elements of $\ocal_{\bk}$ as
\be
    |O_a) \equiv (O_a)_{\bk} \equiv \sum_{\bx} \Ts_{\bx}[O_a] e^{-i \bk \cdot \bx}, \label{eq:7}
\ee
where $\{O_a\}$ is the set of product operators that are localized in the region $\{\bx \in \Lambda \,|\, \bx \cdot \hat{\bk} \geq 0\}$ and are not the identity at the origin \footnote{Also, the identity operator $I$ itself doesn't contribute to $\ocal_{\bk}$ for ${\bk}\neq 0$ because then $(I)_{\bk} = 0$ in (\ref{eq:7})\label{foot:id}}. We drop the $\bk$ label on the $|O_a)$ to avoid clutter, this basis gives a canonical isomorphism between the different $\ocal_{\bk}$.  For the example of the XXZ chain, the $\{O_a\}$ are strings of Pauli operators starting at the origin:
$X_0, Y_0, Z_0, X_0 X_1, X_0 Y_1, \ldots, X_0 I_1 Y_2, \ldots,$
where subscripts are lattice indices $x \in \mathbb{Z}$. The corresponding basis elements in (\ref{eq:7}) are then operators such as
$(X_0 Y_1)_k = \ldots + X_{-1} Y_0 e^{i k} +  X_0 Y_1 + X_1 Y_2 e^{-i k} + X_2 Y_3 e^{-2 i k} + \ldots$,
from which it is clear that $(X_0 Y_1)_k$ and $(X_1 Y_2)_k$ only differ by a phase prefactor. This is why the operators must be taken to start at $x = 0$.

Translational symmetry implies that the $\ocal_{\bk}$ are preserved by time evolution. Therefore, it is possible to diagonalize $\del_t$ in each $\bk$-sector. An eigenoperator $O_{\bk} \in \ocal_{\bk}$ satisfies
\be\label{eq:oodot}
\dot{O}_{\bk} = - i \w(O, {\bk}) O_{\bk},
\ee
for some $\w(O, \bk) \in \mathbb{C}$ and with $\Im{\w} \leq 0$. Note that $i \pa_t$ is not Hermitian but the negative imaginary part of its eigenvalues means that time evolution is stable. Diffusion is then described by a coarse-grained charge operator $\wtd{C}_{\bk}$ that is an eigenoperator of $\del_t$ with
\be
    \w(\wtd{C}, {\bk}) \equiv \W_{\bk} = - i D k^2 + o(k^2) \,, \label{eq:12}
\ee
which defines the diffusivity $D$ of the conserved charge. More generally $D$ may depend on the direction of $\bk$, and this definition works for any fixed direction of $\bk$. We will obtain the operator $\wtd{C}_{\bk}$ explicitly below.

We are able to discuss diffusion as an operator equation, as in (\ref{eq:oodot}) and (\ref{eq:12}) above, because decoherence causes operator norms to decay. This is a significant technical simplification relative to the case of unitary evolution at finite temperature, where diffusion only occurs within thermal expectation values. In the following section we compute $\W_{\bk}$ in small $k$ perturbation theory. This will give an explicit expression for $D$.

{\it Perturbation theory at small wavevector.---}
At small $k$, we can expand $\del_t|_{\ocal_{\bk}}$ in $k$. Fixing a direction of $\bk$:
\be
\del_t|_{\ocal_{\bk}} = \Lag \equiv \sum_{n \geq 0} k^n \Lag^n, \label{eq:14}
\ee
which defines superoperators $\Lag^n$. For example, in the XXZ chain, the operator $(Z_0)_k \in \ocal_k$ obeys
\begin{align}
(\dot{Z}_0)_k

 = 2 (e^{-i k} - 1) (X_0 Y_1)_k - 2 (e^{-i k} - 1) (Y_0 X_1)_k \,.\label{eq:13}
\end{align}
With respect to the Pauli string basis, $\del_t|_{\ocal_k}$ is represented as a $k$-dependent matrix. Expanding the coefficients of the basis elements in (\ref{eq:13}) at small $k$ we obtain components of the superoperators in (\ref{eq:14}). As expected from conservation of $Z$, $\Lag^0 |Z_0) = 0$, while $\Lag^1 |Z_0) = 2 i |Y_0 X_1) - 2 i |X_0 Y_1)$ and $\Lag^2 |Z_0 ) = |Y_0 X_1) - |X_0 Y_1)$.

The eigenvalue $- i \Omega_{\bk}$ of $\del_t|_{\ocal_{\bk}}$ can be found using
standard second-order perturbation theory in small $\bk$ (cf. the memory matrix formalism \cite{for75}). At $\bk = 0$ we know that the eigenvector is the conserved charge $|C)$, with vanishing eigenvalue. Therefore, up to order $k^2$:
\begin{align}
     - i \Omega_{\bk} & = 	k (C | \mathcal{L}^1 | C) \label{eq:16} \\
    & + k^2 \left[(C | \mathcal{L}^2 | C) - \sum_{E^0_a \neq 0} (C | \mathcal{L}^1 | E^0_a) \frac{1}{E^0_a} (E^0_a| \mathcal{L}^1 | C)\right].  \nonumber
\end{align}
The basis vectors $|E^0_a)$ are given by linear combinations of the $|O_a)$ in (\ref{eq:7}) with ${\mathcal L}^0 |E^0_a) = E^0_a |E^0_a)$. The corresponding eigenoperator $|\wtd{C}) = |C) - k \sum_{E^0_a \neq 0} | E^0_a) \frac{1}{E_a^0} (E^0_a| \mathcal{L}^1 | C)$ is the dressed charge operator to this order. We are assuming that the only operator with $E_a^0 = 0$ is the single conserved charge $C$. It is straightforward to extend our analysis to a finite number of conserved charges. We will be more precise about the absence of additional slow operators in the following section. The superoperator $\del_t$ is not antihermitian in general and the eigenoperators $|E^0_a)$ are not necessarily orthogonal. The above perturbation theory formulae retain their standard form, but $(E^0_a | O)$ is defined to be the coefficient in front of $|E^0_a)$ in the expansion of $|O )$ in the basis $\{|E^0_a)\}$. The $(E^0_a|$ are elements of a dual vector space to that spanned by the $|E^0_a)$, and hence have opposite dimensionality. In the case that the operators $E^0_a$ are orthogonal, i.e. $\tr (E_a^{0 \dagger} E^0_b) = \delta_{a b} \tr( E_a^{0 \dagger} E^0_a)$ for some given $\bk$, then we can write $(E^0_a | O) = \tr (E_a^{0 \dagger} O) / \tr (E_a^{0 \dagger} E^0_a)$, as usual.

Our main objective is to use the expression (\ref{eq:16}) to bound the diffusivity (\ref{eq:12}) in generality. However, in simple models such as the dissipative XXZ chain (\ref{eq:model}) it is possible to compute the diffusivity by evaluating (\ref{eq:16}). The on-site dephasing in that model suppresses Pauli strings with $X$ and $Y$ terms. For example: $\mathcal{L}_{\mathrm{dis}}[X_0] = - 4 X_0$ and  $\mathcal{L}_{\mathrm{dis}}[X_0 Y_1] = - 8 X_0 Y_1$, where $\mathcal{L}_{\mathrm{dis}}[O]$ is the second sum in (\ref{eq:1}).

The explicit computation is easiest in the limit $c \gg 1$, where the Hamiltonian term in (\ref{eq:1}) is negligible compared to the dephasing term. In this limit (\ref{eq:16}) becomes (to leading order in $k$ and $c^{-1}$)
\begin{align}
    - i \Omega_{\bk} = \frac{k^2}{8 c} \sum_{A} (Z_0| \Lag^1 | A) (A| \Lag^1|Z_0) = - \frac{k^2}{c} \,,
\end{align}
where in the sum $A = X_0 Y_1$ and $Y_0 X_1$. The system is diffusive with $D = c^{-1} + O(c^{-2})$ for strong decoherence $c \gg 1$, cf. \cite{PhysRevE.92.042143}. This asymptotic behavior is verified numerically in Fig.\,\ref{fig1}, showing numerical results for finite $c$.
\begin{figure}[!ht]
    \centering
	\includegraphics[width=0.95\columnwidth]{Fig2.pdf}
    \caption{Diffusivity $D$ of the dissipative XXZ model versus dephasing strength $c$, with anisotropies $\Delta = 0.5, 1.0, 1.5$. The asymptotic behavior $D \sim 1 / c$ is also shown. Operator spaces are truncated in numerics so that only Pauli string operators of length at most $n = 7$ are kept. Finite-size effects are strong for small $c$ and indicated by the shaded region, which is estimated from truncations with $n = 6, 8$. \label{fig1}}
\end{figure}

The numerical results are obtained using a truncated space of operators in (\ref{eq:7}) to evaluate (\ref{eq:16}). This is a different method compared to previous work, and is relatively straightforward to implement. It works best for larger values of $c$ where long operators are strongly suppressed by dissipation. These results on the XXZ example agree with those in the literature \cite{z,PhysRevLett.106.220601, men13, pro16, ljubotina2017spin}. In particular, for $0 < \Delta < 1$ the system is known to show ballistic spin transport in the absence of dephasing ($c=0$). Therefore, while transport is diffusive at nonzero dephasing, the diffusivity diverges as $c \to 0$.

{\it Constraints from the Lieb-Robinson bound.---} For diffusive rather than ballistic transport, $(C | \Lag^1 | C)$ must vanish in (\ref{eq:16}). Indeed, $|J) = \Lag^1 |C)$ is the current operator, and it is known from the Mazur-Suzuki bound \cite{MAZUR69, SUZUKI71} that if $(C | J) \neq 0$ at $k=0$, transport is ballistic. We restrict attention to non-ballistic systems \footnote{In the ballistic case, our argument gives a bound on the rate of attenuation of the linearly-dispersing mode}. Then the diffusivity can be rewritten as
\be
D = - \left[ (C| \mathcal{L}^2 |C) + \int_0^\infty dt\, (C| \mathcal{L}^1 e^{\mathcal{L}^0 t} \mathcal{L}^1 |C) \right], \label{eq:23}
\ee
where $\frac{1}{E^0_a} |E^0_a)(E^0_a|$ in (\ref{eq:16}) has been replaced by an integral of $\exp \left(\Lag^0 t\right)$.
$D$ is manifestly real in (\ref{eq:23}) because in a basis of hermitian operators $\mathcal{L}^0,\mathcal{L}^2$ are real matrices and $\mathcal{L}^1$ is imaginary, and furthermore $(C|$ is a real vector.

To isolate the dynamics of the single conserved density we make a physical assumption about the spectrum of the Lindbladian decoherence operators: all local operators other than the charge density decay exponentially, at least as fast as $e^{- t / \tau}$. Here $\tau$ defines the `local decoherence time'.
The local difference in behavior between conserved densities and all other operators will be important for our argument.
Technically, we will require
\begin{itemize}
\item {\bf Single-mode ansatz:} There exist $A, \tau > 0$ such that any local operator $O$ can be decomposed into local operators $O = \gamma I + O_1 + O_2$, where $I$ is the identity operator and $\gamma$ a coefficient, $O_1$ is a sum of $C$'s and $\|O_2(t)\| \leq A \|O\| e^{- t / \tau}$, $\|\dot{O}_2(t)\| \leq A \tau^{-1} \|O\| e^{- t / \tau}$ for $t >0$.
\end{itemize}
We will bound the diffusivity by the Lieb-Robinson velocity and the decoherence time $\tau$.

Let $\|\cdot\|$ be any operator norm contracted by the time evolution (\ref{eq:1}) \footnote{Completely positive unital maps --- such as Lindbladian time evolution --- between C*-algebras contract C*-norms, see Chap 8 of \cite{Nielsen} and Chap 3 of \cite{paulsen_2003}. }. This induces a seminorm (with $\||I)\| = 0$) for $|O) \in \mathcal{O}_0$:
\be\label{eq:induced}
    \||O)\| \equiv \lim_{\bk \to 0} \lim_{N \to \infty} N^{-1} \|\sum_{\bx} \Ts_{\bx}[O] e^{-i \bk \cdot \bx} \|,
\ee
where $N = \sum_{\bx} 1$ is the number of lattice sites.
For example, $\||Z)\| = \lim_{k \to 0} \lim_{N \to \infty} N^{-1} \|\sum_x Z_x e^{-i k x}\| = \|Z\|$,
and generally $\||O)\| \leq \|O\| \,,$ by the triangle inequality. From the definition (\ref{eq:induced}), this seminorm is also contracted by time evolution. As a result of contraction in time combined with the single-mode ansatz:
\be
\||O)\| \geq \lim_{t \to \infty} \||O(t))\|  = |(C | O)| \||C)\|, \label{eq:22}
\ee
bounding the norm of the $\bk=0$ state by its projection onto the conserved charge.

We use (\ref{eq:22}) to bound the two terms in the diffusivity (\ref{eq:23}). For the first term, let $|O) = \mathcal{L}^2 |C) \in \mathcal{O}_0$. Then
\be
|(C | \mathcal{L}^2 | C)| \leq \| \mathcal{L}^2|C) \| / \||C)\| \,. \label{eq:28}
\ee
Given the operator equation of motion, the right-hand side of (\ref{eq:28}) is easily calculable.

To bound the second term in (\ref{eq:23}), take a local operator $O$ such that $|O) = \mathcal{L}^1 |C) + \alpha |C) \in \mathcal{O}_0$, with $\alpha \in \mathbb{C}$. Then:
\begin{align}\label{eq:odot}
    (C| \mathcal{L}^1 & e^{\mathcal{L}^0 t} \mathcal{L}^1 |C) = (C| \mathcal{L}^1 e^{\mathcal{L}^0 t} |O) \\
    & =   (C | \lim_{k\to 0} \partial_k (\mathcal{L} e^{\mathcal{L} t}) |O) =   (C | \lim_{k\to 0} \partial_k  | \dot{O}(t) ). \nonumber
\end{align}
The first equality uses $\mathcal{L}^0 |C) = 0$ and $(C | \mathcal{L}^1 | C) = 0$; the second equality uses $(C | \mathcal{L}^0 = 0$.
In (\ref{eq:odot}), $\partial_k$ is defined to be the $k$-derivative of the components of matrices such as $\mathcal{L}$ or vectors such as $|O)$ in the prescribed basis $|O_a)$ in (\ref{eq:7}). Explicitly, for any local operator $P$ we can uniquely write $P = c I + \sum c^a_{\bx} \Ts_{\bx} [O_a]$ so that in $\mathcal{O}_{\bk}$,
$|P) = \sum c^a_{\bx} e^{i \bk \cdot \bx} |O_a)$ and  $- i \partial_k |P) = \sum c^a_{\bx}  (\hat{\bk} \cdot \bx) |\Ts_{\bx}[O_a])$, which is seen to be the first moment of the operator $P$.

Using (\ref{eq:odot}) in (\ref{eq:22}) gives the bound
\be
\left|\int_0^\infty dt\, (C| \mathcal{L}^1 e^{\mathcal{L}^0 t} \mathcal{L}^1 |C)\right| \leq \int_0^\infty d t \, \frac{\|\partial_k |\dot{O}(t))\|}{\||C)\|} \label{eq:29}.
\ee

Take an operator $J$ (the current) localized near the origin such that $|J) = \mathcal{L}^1 |C)$. According to the single-mode ansatz we can write $J = O + \sum_{\bx} c_{\bx} \Ts_{\bx}[C]$,  where $O$ is also localized near the origin and for $t > 0$
\be
     \|\dot{O}(t)\| \leq A \tau^{-1} \|J\| e^{- t / \tau}. \label{opdecay}
\ee

We can choose this $O$ as the operator in (\ref{eq:odot}).
From the bound (\ref{opdecay}) on $\|\dot{O}(t)\|$ we must now obtain a bound on $\|\partial_k |\dot{O}(t))\|$, that appears in (\ref{eq:29}).

Let $\mathcal{P}_l$ for $l \in \mathbb{R}$ be the projection onto the operator subspace spanned by all product operators  supported on the half-space $\hat{\bk} \cdot \bx \geq l$ and let $\mathcal{Q}_{l} = \text{Id} - \mathcal{P}_l$.
By an adaption \footnote{By \cite{commutator}, for any $l \in \mathbb{R}$ and $t > 0$ there exists $\tilde{O}$ localized in $\hat{\bk} \cdot \bx < l$ such that $\lVert\dot{O}(t) - \tilde{O}\rVert \leq A' \lVert\dot{O}\rVert e^{(v t - l) / \xi}$, hence $\lVert\mathcal{P}_{l}[\dot{O}(t)]\rVert = \lVert\mathcal{P}_{l}[\dot{O}(t) - \tilde{O}]\rVert \leq A' \lVert\dot{O}\rVert e^{(v t - l) / \xi}$. Note $\lVert\mathcal{P}_{l}[\dot{O}(t)]\rVert \leq \lVert\dot{O}(t)\rVert$ as well, hence $\lVert\mathcal{P}_{l}[\dot{O}(t)]\rVert \leq \min\{A' \lVert\dot{O}\rVert e^{(v t - l) / \xi}, \lVert\dot{O}(t)\rVert\}$ and plug in (\ref{opdecay}) to obtain (\ref{eq:30}). Similarly for (\ref{eq:31}) except that $\lVert\mathcal{Q}_{l}[\dot{O}(t)]\rVert = \lVert\dot{O}(t) - \mathcal{P}_{l}[\dot{O}(t)]| \leq 2 \lVert\dot{O}(t)\rVert$.} of the Lieb-Robinson bound \cite{PhysRevLett.104.190401, nachtergaele2011lieb} for (\ref{opdecay}), there exist $A' \geq 1$ and $v, \xi > 0$ such that for all $l \in \mathbb{R}, t > 0$,
\begin{align}
\|\mathcal{P}_{l}[\dot{O}(t)]\| &\leq A \|J\|  \tau^{-1} \min\{e^{-t / \tau}, A' e^{(v t - l) / \xi} \}, \label{eq:30}\\
\|\mathcal{Q}_{l}[\dot{O}(t)]\| &\leq 2 A \|J\|  \tau^{-1} \min\{e^{-t / \tau}, A' e^{(v t + l) / \xi}\}. \label{eq:31}
\end{align}
The length $\xi$ is the range of microscopic interactions. We saw that $\partial_k$ corresponds to taking the first moment. Therefore $\partial_k|\dot{O}(t)) = i |O'(t))$, with
\begin{align}
O'(t) = \int_0^\infty d l\, \mathcal{P}_{l}[\dot{O}(t)] - \int_{-\infty}^0 d l\, \mathcal{Q}_{l}[\dot{O}(t)].
\end{align}
Indeed, from $\mathcal{P}_l[\mathcal{T}_{\bx}[O_a]]=\mathcal{T}_{\bx}[O_a]$ for $\hat{\bk} \cdot \bx \geq l$, and vanishing otherwise, we have $\int_0^\infty d l\, \mathcal{P}_l[\mathcal{T}_{\bx}[O_a]] = \hat{\bk} \cdot \bx\, \mathcal{T}_{\bx}[O_a]$ if $\hat{\bk} \cdot \bx \geq 0$, which is precisely the first moment. The second integral of $\mathcal{Q}_l$ similarly takes care of the $\hat{\bk} \cdot \bx \leq 0$ terms. Now, $\|\partial_k |\dot{O}(t))\| \leq \|O'(t)\|$ and, using (\ref{eq:30}) and (\ref{eq:31}), $\|O'(t)\| \leq 3 A \|J\| e^{-t / \tau} \tau^{-1} [v t +  \xi (1 + t / \tau + \ln A')]$. Hence, substituting into (\ref{eq:29}),
\be
\left|\int_0^\infty dt\, (C| \mathcal{L}^1 e^{\mathcal{L}^0 t} \mathcal{L}^1 |C)\right| \leq 3 A [v \tau + \xi (2 + \ln A')] \frac{\|J\|}{\||C)\|}. \nonumber
\ee

Putting the results together gives the diffusivity bound
\be\label{eq:fullbound}
D \leq D_0 + \left(\a \, v_\text{LR} \tau + \beta \, \xi \right) v_\text{C} \,.
\ee
Here $D_0 = \|\mathcal{L}^2|C)\| / \||C)\|$ is a `microscopic' diffusivity from the dissipative equation of motion. The Lieb-Robinson velocity $v_\text{LR} = v$ and $v_\text{C} = \|J\| / \||C)\|$ is a velocity obtained by dividing the current by the charge. As above $\tau$ is the decoherence time and $\xi$ is the range of microscopic interactions. The dimensionless coefficients $\a = 3 A$ and $\beta = 3 A (2 + \ln A')$.

Equation (\ref{eq:fullbound}) establishes that the diffusivity is bounded by microscopic velocities and time and lengthscales in the system.

The quantities $D_0$, $v_\text{LR}$, $v_\text{C}$ and $\xi$ can be obtained from the equations of motion. The quantities $A$ and $\tau$ are instead best determined experimentally or numerically from the decay of local non-conserved operators. From equation (\ref{opdecay}), $\tau$ determines the late time decay rate of the non-conserved part of the local current and $A = {\max_{t > 0} \| \dot{O}(t)\|/(\tau^{-1} \|J\| e^{-t/\tau})}$.

$A'$ does not have a strong effect as it appears in a logarithm in our bound.

{\it Final comments.---} The bound (\ref{eq:fullbound}) has nontrivial consequences for the dephasing XXZ chain. For $0 < \Delta < 1$ the diffusivity diverges in Fig.\,\ref{fig1} as $c \to 0$. The bound states that $D$ cannot diverge faster than $\tau$.

In the XXZ model $v_\text{C} = 4$ and, from \cite{nachtergaele2011lieb}, $v_{\text{LR}} \leq 2 + \Delta$ are independent of $c$.

Now $\tau = { \max_k 1/(-\text{Re}\, E^1_k)}$, where $E^1_k \in \mathbb{C}$ is the first eigenvalue of $\partial_t|_{\ocal_k}$ above the slow mode. We evaluated this eigenvalue numerically by truncating the operator space as described around Fig.\,\ref{fig1}. At $\Delta = 0.5$ the ratio $D/\tau = 3.8(2)$ indeed remains finite as $c \to 0$.

We end with some broader comments. Firstly, (exponential) locality of interactions and a finite decoherence time are essential, as otherwise there can be superdiffusive transport \cite{PhysRevLett.108.093002, hild14, ljubotina2017spin}, where the perturbation theory (\ref{eq:16}) is no longer valid due to degeneracies or divergences.

The decoherence-induced decay of operators such as long Pauli strings is phenomenologically similar to the decay of the thermal expectation values of those operators.
To obtain a rigorous bound on diffusion in unitary quantum dynamics in a thermal state, however, there will be several challenges to overcome. The diffusivity must be discussed in terms of expectation values rather than operators, and projections with respect to thermal inner products are difficult to evaluate (e.g. \S 5.6 of \cite{har18}). The butterfly velocity may causally constrain finite temperature transport \cite{Hartman:2017hhp, Lucas:2017ibu, Shenker:2013pqa, Roberts:2014isa, PhysRevLett.117.091602}, but a temperature-dependent bound on this velocity has not been established. These interesting problems are left for future work.

If a rigorous bound of the form $D \lesssim v^2 \tau + v \xi$ can indeed be established for diffusion in finite temperature states, it may shed light on the phenomenon of resistivity saturation \cite{RevModPhys.75.1085, hussey}. As temperature is increased $\tau$ will typically descrease, but $\xi$ is a microscopic and temperature-independent lengthscale. Therefore,  the resistivity $\rho \propto 1/D \gtrsim 1/(v^2 \tau + \xi v)$ is able to saturate at high temperatures where $v \tau < \xi$.

\section*{Acknowledgements}

The work of SAH is partially supported by DOE award DE-SC0018134. XH is supported by a Stanford Graduate Fellowship.

\bibliography{decoh}

\end{document}


TITLE: Wave Equation Descending from 2D to 1D
QUESTION [1 upvotes]: I am stuck deriving the familiar d'Alembert formula using the Method of Descent, going from 2D to 1D. After using Kirchhoff's Formula, writing the solution independently of $y$ and some manipulations, I have obtained the following result.
$$\frac{1}{2\pi c} \int_{x-ct}^{x+ct} \frac {g(x)} {({(ct)^2} - x^2)^{(1/2)}}dx $$
Where the initial conditions are $f(x) \equiv 0$ and $u_t = g(x)$.
How do I further reduce the integral to the familiar equation:
$${u(x,t) = \frac{1}{2c} \int_{x-ct}^{x+ct}g(\overline{x}) d\overline{x}} $$
It is very close to the desired result, but the next step escapes me at this moment. Should I consider Polar coordinates or a change of variables?
Thanks in advance.

REPLY [0 votes]: I am unsure if this answers your question, but maybe it might still help. The proof of the wave equation i know uses the descend method only for higher dimensions. The derivation of the 2d wave equation i know (following evans) is rather obtained by a spliting method, rewriting the wave equation into a product of transport equations, that are solved first for the homogenous transport equation and then for the inhomogenous transport equation.


TITLE: Find all integers such that $2 < x < 2014$ and $2015|(x^2-x)$
QUESTION [4 upvotes]: Find all integers, $x$, such that $2 < x < 2014$ and $2015|(x^2-x)$.
I factored it and now I know that $x > 45$ and I have found one solution so far: $(156)(155)= (2015)(12)$. It's just that I don't think I'm approaching it the right way.
I'm new to this kind of stuff and I'm just doing this for fun so any help please?

REPLY [1 votes]: I think the following algorithm can be usefull:

Observe that $x^2-x=x(x-1)$. Now $x$ and $x-1$ are coprime and then, if $2015=ab$, with $(a,b)=1$ you can suppose $a|x$ and $b|x-1$
For every fixed couple $(a,b)$ of coprime divisors of 2015 solve by CRT the system af congruences:
$x \equiv 0 \mod a$     ^      $ x \equiv 1 \mod b$.
CRT gives a unique solution $\mod ab=2015$. We call its class $X$
Now you have just to find the representant of the class of $X$ between $2$ and $2014$, if it exist, to find an $x$ that works.

So now you have just to apllie this algorithm to the 3 possible couple of coprime divisors of 2015, and for every couple there are only two cases to examinate...


TITLE: Interpreting plane waves solutions of the Dirac equation
QUESTION [1 upvotes]: The Dirac equation $(i\gamma^{\mu}\partial_{\mu}-m)\psi = 0$ has plane wave solutions of the form:
$$\psi_{+,k}(t,x) = e^{-iE(k)t+ik\cdot x}u(k) \quad \mbox{and} \quad \psi_{-,k}(t,x) = e^{iE(k)t -ik\cdot x}v(-k)$$
where $k, x \in \mathbb{R}^{3}$.
Question: The above expressions are functions of $t,x$ for each fixed $k$. How do I interpret these solutions? Does it mean that the Dirac equation have solutions with constant well-defined momentum? Each such pair of solutions represent a fermion with constant momentum $k$?

REPLY [1 votes]: I'd just like to say that, in physics and field theory, whenever you have a solution to a linear theory which is proportional to $e^{-i k_\mu x^\mu}$ (with maybe some constant n dimensional vector out front) where $k^2 = m^2$, you always interpret this quantum mechanically as a particle of four momentum $k_\mu$. This is justified when you go to quantum field theory, as you can build an oscillator in the mode of this solution with integer occupation numbers.


TITLE: Representations of $SL(2,3)$
QUESTION [2 upvotes]: I studied first four chapters of Serre's book on linear representations of finite groups, which talks about irreducible representations and their characters, canonical decomposition of a representation, induced representations and their characters (along with obvious big theorems: Schur's lemma and Maschke's theorem).
Then I took the group $SL(2,3)$, a non-abelian group of order $24$, and moved for constructing its representations/characters. But, I could not complete the character table: I found irreducible characters of degree $1,1,1,3$ simply by pulling them via quotient modulo center $\pm I$.
I thought to get remaining by induction; but then after some hitting, I reached to conclusion that it has no subgroups of index $2$.
Then, how should I proceed to construct at least one or all the three irreducibles of degree $2$? (Should I get familiar with some further topics of Serre for this? )
(The existence of three irreducibles of degree $2$ can be guaranted from the rule that $24=1^2+1^2+1^2+3^2+ \sum_i n_i^2$ with $n_i>1$.)

REPLY [0 votes]: I understand the 2D complex irrep best with quaternions.
Every quaternion looks like $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$, in other words the formal sum of a scalar and a 3D vector. The product of any two of $\mathbf{i},\mathbf{j},\mathbf{k}$ in cyclic order is the third, and in reverse order is the opposite. If you check the multiplication table, then, you find if $\mathbf{u},\mathbf{v}$ are 3D vectors then the scalar and vector components (aka real and imaginary parts) of the quaternion product $\mathbf{uv}$ are $-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$ (since it is true for basis vectors). This implies the square roots of $-1$ are the unit vectors, and Euler's formula $\exp(\theta\mathbf{v})=\cos(\theta)+\sin(\theta)\mathbf{v}$ for unit vectors $\mathbf{v}$ follows. Furthermore, every quaternion has a polar form $r\exp(\theta\mathbf{v})$ with $r\ge0$ its magnitude, $\mathbf{v}$ a unit vectors, and $0\le\theta\le\pi$ convex (which is unique for nonreal quaternions, and $\theta=0$ or $\pi$ with $\mathbf{v}$ arbitrary for nonzero reals). If $\mathbf{u}$ is a unit vector, we can extend to an ordered orthonormal basis $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ of the quaternion algebra. If we write $L_q(x)=qx$ and $R_q(x)=xq$ with respect to this basis, where $q=\exp(\theta\mathbf{u})$, we see that the conjugation $C_q=L_q\circ R_q^{-1}$ given by $C_q(x)=qxq^{-1}$ restricts to a rotation of 3D vectors around the oriented axis by an angle of $2\theta$. In this way, $q\mapsto C_q$ is a double covering (i.e. $2$-to-$1$ group homomorphsim) $S^3\to\mathrm{SO}(3)$, where $S^3$ is the group of unit quaternions.
If $T\cong A_4$ is the tetrahedral group (of order $12$), its preimage under $S^3\to\mathrm{SO}(3)$ is the binary tetrahedral group $2T$ of order $24$. It forms the vertices of a $24$-cell, namely $\{\pm1,\pm\mathbf{i},\pm\mathbf{j},\pm\mathbf{k},\frac{1}{2}(\pm1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k})\}$. The group is a semidirect product $2T=Q_8\rtimes C_3$ with $C_3$ generated by any $3$rd root of unity $\frac{1}{2}(-1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k})$, with polar form $\omega=\exp(\frac{2\pi}{3}\frac{\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}}{\sqrt{3}})$. Geometrically, we can place unit diameter $3$-spheres at all integer coordinates in 4D, which leaves just enough space to place unit diameter spheres centered at the hypercube centers to create an optimal sphere packing (the Hurwitz quaternions); the vertices of the spheres surrounding the central sphere forms the $24$-cell.
We can have $2T\curvearrowright\mathbb{H}$ act from the left and complex scalars from $\mathbb{C}=\{a+b\mathbf{i}\}$ act from the right (so it is a "right" vector space). Picking a basis $\{1,\mathbf{j}\}$ for $\mathbb{H}$ as a complex vector space allows us to write the right $\mathbb{C}$-linear transformations from $2T$ as $2\times2$ complex matrices. This is a 2D complex irrep of $2T$. We can check the character table to see tensoring this 2D rep with the nontrivial 1D reps gives the two other 2D irreps.
If we mod these $2\times2$ matrices by $3$ (the smallest modulus where $\pm1$ are distinct), we get $\mathrm{SU}_{1,1}(\mathbb{F}_{3^2})$, where $\mathbb{F}_{3^2}=\mathbb{F}_3(i)$, which is conjugate via the Cayley transform $[\begin{smallmatrix}1&-i\\1&i\end{smallmatrix}]$ to $\mathrm{SL}_2(\mathbb{F}_3)$. I elaborate a little more in my answer here (which shows the binary octahedral group $2O$ is a twisted version of $\mathrm{GL}_2\mathbb{F}_3$).


TITLE: Question about a step in Schur product theorem
QUESTION [0 upvotes]: In the proof of Schur product theorem based on the trace formula (see https://en.wikipedia.org/wiki/Schur_product_theorem), there is one point where it says that
$$A=N^{\frac{1}{2}}diag(a)M^\frac{1}{2}\neq 0 \Leftrightarrow a\neq0$$
I don't get this. Is it a consequence of some property? It doesn't depend on $N,M$?
Thanks

REPLY [2 votes]: Hint: $N$ and $M$ are positive definite. So
$$\mathrm{diag}(a) =  N^{-1/2} A M^{-1/2}.$$


TITLE: What methods can I use to get a rank?
QUESTION [0 upvotes]: I am building a ranking system. To get a rank of an item, I have considered using ratio and method.
Let's say I have an item with 10 wins and 10 losses.
Ratio: 10/20 = 0.5
Point: 10 - 10 = 0
What other options do I have?

REPLY [1 votes]: The ranking problem comes up in many contexts. People who want to rank American college football teams have come up with the BCS system. Part of this involves asking experts for their opinions, but part of it is purely mathematical. Start with the wikipedia page.
Chess players also get ranked, by the Elo system. Here's a wikipedia link.