Noether’s Theorem: Quantum vs Stochastic

In 1915 Emmy Noether discovered an important connection between the symmetries of a system and its conserved quantities. Her result has become a staple of modern physics and is known as Noether’s theorem.

The theorem and its generalizations have found particularly wide use in quantum theory. Those of you following the Network Theory series here on Azimuth might recall Part 11 where John Baez and Brendan Fong proved a version of Noether’s theorem for stochastic systems. Their result is now published here:

• John Baez and Brendan Fong, A Noether theorem for stochastic mechanics, J. Math. Phys. 54:013301 (2013).

One goal of the network theory series here on Azimuth has been to merge ideas appearing in quantum theory with other disciplines. John and Brendan proved their stochastic version of Noether’s theorem by exploiting ‘stochastic mechanics’ which was formulated in the network theory series to mathematically resemble quantum theory. Their result, which we will outline below, was different than what would be expected in quantum theory, so it is interesting to try to figure out why.

Recently Jacob Biamonte, Mauro Faccin and myself have been working to try to get to the bottom of these differences. What we’ve done is prove a version of Noether’s theorem for Dirichlet operators. As you may recall from Parts 16 and 20 of the network theory series, these are the operators that generate both stochastic and quantum processes. In the language of the series, they lie in the intersection of stochastic and quantum mechanics. So, they are a subclass of the infinitesimal stochastic operators considered in John and Brendan’s work.

The extra structure of Dirichlet operators—compared with the wider class of infinitesimal stochastic operators—provided a handle for us to dig a little deeper into understanding the intersection of these two theories. By the end of this article, astute readers will be able to prove that Dirichlet operators generate doubly stochastic processes.

Before we get into the details of our proof, let’s recall first how conservation laws work in quantum mechanics, and then contrast this with what John and Brendan discovered for stochastic systems. (For a more detailed comparison between the stochastic and quantum versions of the theorem, see Part 13 of the network theory series.)

The quantum case

I’ll assume you’re familiar with quantum theory, but let’s start with a few reminders.

In standard quantum theory, when we have a closed system with $n$ states, the unitary time evolution of a state $|\psi(t)\rangle$ is generated by a self-adjoint $n \times n$ matrix $H$ called the Hamiltonian. In other words, $|\psi(t)\rangle$ satisfies Schrödinger’s equation:

$i \hbar \displaystyle{\frac{d}{d t}} |\psi(t) \rangle = H |\psi(t) \rangle.$

The state of a system starting off at time zero in the state $|\psi_0 \rangle$ and evolving for a time $t$ is then given by

$|\psi(t) \rangle = e^{-i t H}|\psi_0 \rangle.$

The observable properties of a quantum system are associated with self-adjoint operators. In the state $|\psi \rangle,$ the expected value of the observable associated to a self-adjoint operator $O$ is

$\langle O \rangle_{\psi} = \langle \psi | O | \psi \rangle$

This expected value is constant in time for all states if and only if $O$ commutes with the Hamiltonian $H$ :

$[O, H] = 0 \quad \iff \quad \displaystyle{\frac{d}{d t}} \langle O \rangle_{\psi(t)} = 0 \quad \forall \: |\psi_0 \rangle, \forall t.$

In this case we say $O$ is a ‘conserved quantity’. The fact that we have two equivalent conditions for this is a quantum version of Noether’s theorem!

The stochastic case

In stochastic mechanics, the story changes a bit. Now a state $|\psi(t)\rangle$ is a probability distribution: a vector with $n$ nonnegative components that sum to 1. Schrödinger’s equation gets replaced by the master equation:

$\displaystyle{\frac{d}{d t}} |\psi(t) \rangle = H |\psi(t) \rangle$

If we start with a probability distribution $|\psi_0 \rangle$ at time zero and evolve it according to this equation, at any later time have

$|\psi(t)\rangle = e^{t H} |\psi_0 \rangle.$

We want this always be a probability distribution. To ensure that this is so, the Hamiltonian $H$ must be infinitesimal stochastic: that is, a real-valued $n \times n$ matrix where the off-diagonal entries are nonnegative and the entries of each column sum to zero. It no longer needs to be self-adjoint!

When $H$ is infinitesimal stochastic, the operators $e^{t H}$ map the set of probability distributions to itself whenever $t \ge 0,$ and we call this family of operators a continuous-time Markov process, or more precisely a Markov semigroup.

In stochastic mechanics, we say an observable $O$ is a real diagonal $n \times n$ matrix, and its expected value is given by

$\langle O\rangle_{\psi} = \langle \hat{O} | \psi \rangle$

where $\hat{O}$ is the vector built from the diagonal entries of $O.$ More concretely,

$\langle O\rangle_{\psi} = \displaystyle{ \sum_i O_{i i} \psi_i }$

where $\psi_i$ is the $i$ th component of the vector $|\psi\rangle.$

Here is a version of Noether’s theorem for stochastic mechanics:

Noether’s Theorem for Markov Processes (Baez–Fong). Suppose $H$ is an infinitesimal stochastic operator and $O$ is an observable. Then

$[O,H] =0$

if and only if

$\displaystyle{\frac{d}{d t}} \langle O \rangle_{\psi(t)} = 0$

and

$\displaystyle{\frac{d}{d t}}\langle O^2 \rangle_{\psi(t)} = 0$

for all $t \ge 0$ and all $\psi(t)$ obeying the master equation. █

So, just as in quantum mechanics, whenever $[O,H]=0$ the expected value of $O$ will be conserved:

$\displaystyle{\frac{d}{d t}} \langle O\rangle_{\psi(t)} = 0$

for any $\psi_0$ and all $t \ge 0.$ However, John and Brendan saw that—unlike in quantum mechanics—you need more than just the expectation value of the observable $O$ to be constant to obtain the equation $[O,H]=0.$ You really need both

$\displaystyle{\frac{d}{d t}} \langle O\rangle_{\psi(t)} = 0$

together with

$\displaystyle{\frac{d}{d t}} \langle O^2\rangle_{\psi(t)} = 0$

for all initial data $\psi_0$ to be sure that $[O,H]=0.$

So it’s a bit subtle, but symmetries and conserved quantities have a rather different relationship than they do in quantum theory.

A Noether theorem for Dirichlet operators

But what if the infinitesimal generator of our Markov semigroup is also self-adjoint? In other words, what if $H$ is both an infinitesimal stochastic matrix but also its own transpose: $H = H^\top$ ? Then it’s called a Dirichlet operator… and we found that in this case, we get a stochastic version of Noether’s theorem that more closely resembles the usual quantum one:

Noether’s Theorem for Dirichlet Operators. If $H$ is a Dirichlet operator and $O$ is an observable, then

$[O, H] = 0 \quad \iff \quad \displaystyle{\frac{d}{d t}} \langle O \rangle_{\psi(t)} = 0 \quad \forall \: |\psi_0 \rangle, \forall t \ge 0$

Proof. The $\Rightarrow$ direction is easy to show, and it follows from John and Brendan’s theorem. The point is to show the $\Leftarrow$ direction. Since $H$ is self-adjoint, we may use a spectral decomposition:

$H = \displaystyle{ \sum_k E_k |\phi_k \rangle \langle \phi_k |}$

where $\phi_k$ are an orthonormal basis of eigenvectors, and $E_k$ are the corresponding eigenvalues. We then have:

$\displaystyle{\frac{d}{d t}} \langle O \rangle_{\psi(t)} = \langle \hat{O} | H e^{t H} |\psi_0 \rangle = 0 \quad \forall \: |\psi_0 \rangle, \forall t \ge 0$

$\iff \quad \langle \hat{O}| H e^{t H} = 0 \quad \forall t \ge 0$

$\iff \quad \sum_k \langle \hat{O} | \phi_k \rangle E_k e^{t E_k} \langle \phi_k| = 0 \quad \forall t \ge 0$

$\iff \quad \langle \hat{O} | \phi_k \rangle E_k e^{t E_k} = 0 \quad \forall t \ge 0$

$\iff \quad |\hat{O} \rangle \in \mathrm{Span}\{|\phi_k \rangle \, : \; E_k = 0\} = \ker \: H,$

where the third equivalence is due to the vectors $|\phi_k \rangle$ being linearly independent. For any infinitesimal stochastic operator $H$ the corresponding transition graph consists of $m$ connected components iff we can reorder (permute) the states of the system such that $H$ becomes block-diagonal with $m$ blocks. Now it is easy to see that the kernel of $H$ is spanned by $m$ eigenvectors, one for each block. Since $H$ is also symmetric, the elements of each such vector can be chosen to be ones within the block and zeros outside it. Consequently

$|\hat{O} \rangle \in \ker \: H$

implies that we can choose the basis of eigenvectors of $O$ to be the vectors $|\phi_k \rangle,$ which implies

$[O, H] = 0$

Alternatively,

$|\hat{O} \rangle \in \ker \, H$

implies that

$|\hat{O^2} \rangle \in \ker \: H \; \iff \; \cdots \; \iff \; \displaystyle{\frac{d}{d t}} \langle O^2 \rangle_{\psi(t)} = 0 \; \forall \: |\psi_0 \rangle, \forall t \ge 0,$

where we have used the above sequence of equivalences backwards. Now, using John and Brendan’s original proof, we can obtain $[O, H] = 0.$ █

In summary, by restricting ourselves to the intersection of quantum and stochastic generators, we have found a version of Noether’s theorem for stochastic mechanics that looks formally just like the quantum version! However, this simplification comes at a cost. We find that the only observables $O$ whose expected value remains constant with time are those of the very restricted type described above, where the observable has the same value in every state in a connected component.

Puzzles

Suppose we have a graph whose graph Laplacian matrix $H$ generates a Markov semigroup as follows:

$U(t) = e^{t H}$

Puzzle 1. Suppose that also $H = H^\top,$ so that $H$ is a Dirichlet operator and hence $i H$ generates a 1-parameter unitary group. Show that the indegree and outdegree of any node of our graph must be equal. Graphs with this property are called balanced.

Puzzle 2. Suppose that $U(t) = e^{t H}$ is doubly stochastic Markov semigroup, meaning that for all $t \ge 0$ each row and each column of $U(t)$ sums to 1:

$\displaystyle{ \sum_i U(t)_{i j} = \sum_j U(t)_{i j} = 1 }$

and all the matrix entries are nonnegative. Show that the Hamiltonian $H$ obeys

$\displaystyle{\sum_i H_{i j} = \sum_j H_{i j} = 0 }$

and all the off-diagonal entries of $H$ are nonnegative. Show the converse is also true.

Puzzle 3. Prove that any doubly stochastic Markov semigroup $U(t)$ is of the form $e^{t H}$ where $H$ is the graph Laplacian of a balanced graph.

Puzzle 4. Let $O(t)$ be a possibly time-dependent observable, and write $\langle O(t) \rangle_{\psi(t)}$ for its expected value with respect to some initial state $\psi_0$ evolving according to the master equation. Show that

$\displaystyle{ \frac{d}{d t}\langle O(t)\rangle_{\psi(t)} = \left\langle [O(t), H] \right\rangle_{\psi(t)} + \left\langle \frac{\partial O(t)}{\partial t}\right\rangle_{\psi(t)} }$

This is a stochastic version of the Ehrenfest theorem.

This entry was posted on Saturday, May 3rd, 2014 at 4:38 pm and is filed under mathematics, networks, physics, probability. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

22 Responses to Noether’s Theorem: Quantum vs Stochastic

stasheff says:

3 May, 2014 at 4:42 pm

She had two theorems about symmetries of a system. Best not to perpetuate the myth that there is THE Noether Theorem.

jim

Reply
linasv says:

4 May, 2014 at 3:55 am

In the quantum case, the self-adjoint operators generate unitary transforms viz elts in U(n). What would the equivalent theorems look like if instead one considered the (real, orthogonal) generators of O(n)? How about O(1,n)? Or the euclidean group? (asks someone too lazy to try to do it himself … but surely this can’t be hard… there must surely be some ‘obvious’ generally applicable version, but without working out the various special cases, I don’t see it.)

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- John Baez says:
  
  4 May, 2014 at 9:14 am
  
  People have spent a lot of time studying real and quaternionic quantum mechanics, where the unitary group U(n) gets replaced by O(n) or Sp(n). I wrote a paper about this stuff:
  
  • Quantum theory and division algebras.
  
  arguing that we should think of all three flavors of quantum mechanics as part of a unified structure. The math here gets quite interesting. The Jordan–Wigner–von Neumann classification of formally real Jordan algebras plays an important role (I explain that in my paper), and the groups O(1,n) show up in this theorem too, though in a way that’s still mysterious.
  
  All these flavors of quantum mechanics are quite different from stochastic mechanics.
  
  Reply
John Baez says:

4 May, 2014 at 9:26 am

I like your Ehrenfest theorem for stochastic mechanics:

$\displaystyle{ \frac{d}{d t}\langle O(t)\rangle_{\psi(t)} = \left\langle [O(t), H] \right\rangle_{\psi(t)} + \left\langle \frac{\partial O(t)}{\partial t}\right\rangle_{\psi(t)} }$

It looks exactly like the quantum version, but all the expressions mean something a bit different! And here’s an interesting twist.

In quantum mechanics if $H$ and $O(t)$ are observables (self-adjoint matrices), so is $[O(t),H],$ at least after we divide it by $i.$

But in stochastic mechanics, while $O(t)$ is an observable in the above formula, and so is $\partial O(t)/\partial t,$ the Hamiltonian $H$ is not, and neither is $[O(t),H].$

At least this is true with the definition we’re currently using of observables: diagonal matrices. At first glance this suggests maybe we should generalize our definition of observable. But actually I think the concepts of observable (like $O(t)$ ) and symmetry generator (like $H$ ) become distinct in stochastic mechanics. I don’t think there’s an observable deserving the name ‘energy’ associated to the symmetry generator $H.$ And there may even be other concepts, waiting to be named and understood!

What sort of thing is $[O(t),H],$ for example? It’s not an observable according to our current defintion (a diagonal matrix), nor is it a symmetry generator (an infinitesimal stochastic matrix). Yet it’s still important—we need to take its expected value! It seems to be something new.

Reply
- domenico says:
  
  5 May, 2014 at 9:58 am
  
  It may be too simple, but in classical mechanics exist Poisson bracket that give invariants, so that this can be an invariant in stochastical mechanics (if the matrix zero is an observable).
  
  I am thinking, that if this is true, then two invariants $O_1$ and $O_2$ given another invariant $[O_1,O_2],$ and so on.
  
  Reply
  - John Baez says:
    
    6 May, 2014 at 3:17 pm
    
    In this respect stochastic mechanics is very much like quantum mechanics. Let’s call an operator $O$ a conserved quantity if $[O,H] = 0;$ this implies that
    
    $\displaystyle{ \frac{d}{dt} \langle O \rangle_{\psi(t)} = 0 }$
    
    whenever $\psi(t)$ is a solution of the master equation
    
    $\displaystyle{ \frac{d}{dt} \psi(t) = H \psi(t) }$
    
    and $H$ is infinitesimal stochastic.
    
    It then follows by simple algebra that if $O_1$ and $O_2$ are conserved quantities then so is $[O_1, O_2].$
    
    So, that’s nice. The problem I’m concerned with is that in stochastic mechanics it’s tempting to say $O$ is an observable iff it’s diagonal in the standard basis. This is based on the idea that in probability theory, an observable is a function on the probability measure space.
    
    But with this definition, the commutator of two observables is always zero. Worse, if $O$ is an observable, its time-evolved version
    
    $O(t) = \exp(-tH) O \exp(tH)$
    
    is not an observable according to this definition! So, we don’t get a ‘Heisenberg picture’.
    
    But if we want to change our definition of ‘observable’ in stochastic mechanics, the question becomes: what definition should we use?
    
    One option is to say any operator counts as an observable. However, then it’s possible for two different observables to have the same expected value in every state. Maybe this is okay, but it goes against the standard philosophy about observables.
    
    Also, if we expand our definition of observable to include non-diagonal matrices, then observables don’t commute. I already said this… but now I want to point out that this introduces noncommuting observables, normally considered a feature of quantum theory, into probability theory! Maybe that’s a good thing. But it’s worth thinking about.
    
    Reply
    - Graham Jones says:
      
      7 May, 2014 at 5:21 pm
      
      I only understand quantum theory in a vague sort of way, but it seems to me that it would be ‘quantum-ish’ to define the value of an operator O in the stochastic case as $s^T O s$ where s is the (element-wise) square root of $\psi$ . I don’t know if this helps…
    - John Baez says:
      
      9 May, 2014 at 2:56 pm
      
      I’ve thought about this a bit, but because this ‘square root map’ from probability distributions to wavefunctions is nonlinear, while time evolution in both quantum and stochastic mechanics is linear, I don’t see how to use it to turn stochastic systems into quantum ones.
      
      The Quantum Techniques for Stochastic Mechanics book is based on another idea, which is that it might be interesting to copy some ideas from quantum mechanics to stochastic mechanics while deliberately ignoring the fact that the usual relation between probability distributions and amplitudes is nonlinear. While this is weird, it seems to work.
lee bloomquist says:

5 May, 2014 at 12:56 am

Wow, it seems like the matrix in both quantum and stochastic accounts is information being created and information being destroyed. One vector is destroyed by the operation of the matrix, and from that operation either a same or different vector is created.

Are there conservation laws at work here?

Reply
John Baez says:

5 May, 2014 at 10:57 am

I posted this on Google+:

I’m in Erlangen, where the great German mathematician Emmy Noether was born in 1882. She was the daughter of the well-known mathematician Max Noether—but as a woman, she was only allowed to audit courses at the university here. Somehow she finished a PhD thesis in 1907. She then worked here without pay for 7 years, since women were excluded from academic jobs.

Her thesis advisor, Paul Gordan, specialized in doing complicated calculations to find all the polynomials that were unchanged by certain symmetries. Around this time David Hilbert proved a powerful general theorem that said all these polynomials could be gotten by adding, subtracting and multiplying a finite set of them, called ‘generators’. But he didn’t say how to find these generators! Gordan said

This is not mathematics; this is theology.

Noether did her thesis, On Complete Systems of Invariants for Ternary Biquadratic Forms, in the style of Gordan’s work. It was well received, but she later said it was “crap”. While working without pay, she learned Hilbert’s ideas and started revolutionizing the subject of algebra.

In 1915 she was invited to the University of Göttingen by David Hilbert and Felix Klein. Their attempt to recruit her was fought by the philologists and historians, who didn’t want a woman on the faculty. Hilbert fought back, saying

After all, we are a university, not a bath house.

It took years for her to actually get paid, but she started working at Göttingen and soon proved the theorem physicists remember her for, relating symmetries and conservation laws. They call it Noether’s Theorem.

In fact she proved two important theorems on this subject, but the easier one is more famous: Leon Lederman later said it’s

certainly one of the most important mathematical theorems ever proved in guiding the development of modern physics, possibly on a par with the Pythagorean theorem.

At the time, Einstein wrote:

Yesterday I received from Miss Noether a very interesting paper on invariants. I’m impressed that such things can be understood in such a general way. The old guard at Göttingen should take some lessons from Miss Noether! She seems to know her stuff.

Her theorem applies to classical mechanics and classical field theory, but there’s also a quantum version, and more recently Brendan Fong and I proved a ‘stochastic’ version, which applies to random processes. The stochastic version is weirdly different from the quantum version, but Ville Bergholm has just written a nice article discussing this issue, and some results he discovered with Jacob Biamonte and Mauro Faccin. Check it out!

Emmy Noether finally started getting a salary in 1923, sixteen years after finishing her thesis. If anyone asks why there are fewer famous women mathematicians than men, consider pointing this out!

Noether did extraordinary work until 1933, when the Nazis kicked her out of the University of Göttingen. She wound up in Bryn Mawr College, a women’s college near Philadelphia. She died of complications from surgery in 1935.

But here are some of the wonderful things she did:

In 1921 she stated the general definition of ‘ring’ and ‘ideal’, and proved that in a ring where every increasing sequence of ideals stops growing after finitely many steps, every ideal has finitely many generators. Such rings are now called Noetherian.

In 1927 she gave a massive generalization of the fundamental theorem of arithmetic, about unique factorization into primes. She characterized commutative rings in which the ideals have unique factorization into prime ideals as the integral domains that are Noetherian, 0- or 1-dimensional, and integrally closed in their quotient fields. Sorry—this sounds technical, and it is! But everyone who studies modern number theory takes this result as basic: such rings are now called Dedekind domains, but Noether discovered them.

Even more important than either of these massive results are the beautifully simple ‘Noether isomorphism theorems’ that everyone learns near the start of a course on group theory.

And perhaps even more important was her discovery of ‘homology groups’ while attending lectures by the famous topologists Alexandrov and Hopf. Other people would have made a whole career out of this discovery, which utterly revolutionized topology. But she only gave it a tiny mention in one of her works on group theory! She was truly a fountain of new ideas.

I now have an office in the Emmy-Noether-Zentrum für Algebra at the university in Erlangen.

For more, try:

• Emmy Noether, Wikipedia.

Reply
nad says:

6 May, 2014 at 8:15 am

Wikipedia says:

During her first years teaching at Göttingen she did not have an official position and was not paid; her family paid for her room and board and supported her academic work. Her lectures often were advertised under Hilbert’s name, and Noether would provide “assistance”.

So in some sense her parents paid for her “position”. Wikipedia says about her father Max Noether:

In 1880 he married Ida Amalia Kaufmann, the daughter of another wealthy Jewish merchant family.

I don’t know how wealthy they were. Part of her “payment” was eventually also coming from the salary of her mathematician father. So in some sense he “shared his salary”, which e.g. David Hilbert didn’t. But then David Hilbert had to care for his son Franz who probably was amongst others in need for therapy:

His son Franz suffered throughout his life from an undiagnosed mental illness…

The interesting question here is of course, whether her parents paid for her for the reason of loving paternal care or whether they felt the moral obligation to pay for their daughter or whether they actively wanted to support her in her rather “crazy” fight for equal rights. I write “crazy” because in particular by looking at her life I am not sure whether one can even speak about “women’s liberation” in this context .

Reply
- John Baez says:
  
  6 May, 2014 at 11:57 am
  
  It’s indeed interesting to wonder what Emmy Noether’s parents thought about her, and what she thought about her own situation. I don’t know what’s recorded about this—I’ve never read a detailed biography of her. Clearly she thought about these issues, e.g. after she visited the Institute for Advanced Studies at Princeton she wrote that she was not welcome at this “men’s university, where nothing female is admitted”.
  
  Reply
  - nad says:
    
    7 May, 2014 at 8:02 am
    
    they actively wanted to support her in her rather “crazy” fight for equal rights.
    
    I would like to add that her main motivation for working were probably the subjects she was working on, that is she probably found that it was important to do this work and the concrete environment she was working in was eventually only of secondary importance.
    Or in other words – I don’t know how good the access to literature was in these days, but in principle she eventually could have kept way less contact with those people there,
    So I regard the fact that she was teaching and actively contributing as a “fight for her rights”. In particular she could have thought that at one point that this apparent imbalance of treatment is noticed and that she would get more support. It is however not clear to me how many collegues were convinced of the importance of her work at that time and if they were, why they wouldn’t try to change the way things were. May be because she was not making enough fuzz and trouble? But in fact in the end she was at least supported in fleeing from the Nazi’s.
    
    Clearly she thought about these issues, e.g. after she visited the Institute for Advanced Studies at Princeton she wrote that she was not welcome at this “men’s university, where nothing female is admitted”.
    
    I find it problematic to sort out typical “female” or “male” features that is on average I think there are differences, but the distribution is often rather individual and similar to racial differences a lot of differences, if not most, are due to cultural differences. But yes as said on average there seem to be differences. So if there are way more males that females in an institution there may be a noticable dfifference in atmosphere, but that depends a bit also on what kind of males/females. I went from an all girls school to almost basically all boys math-physics school and I found the atmossphere quite different. I think I understand what she could have meant.
    
    It’s indeed interesting to wonder what Emmy Noether’s parents thought about her, and what she thought about her own situation.
    
    I couldn’t find any personal recollections of her or her parents. That seems her writings seem to have been mostly of a mathematical nature.
    
    Reply
Graham Jones says:

6 May, 2014 at 11:12 am

Is there a connection with time reversibility here?

A stationary Markov chain is reversible if and only if the matrix of transition probabilities can be written as the product of a symmetric and a diagonal matrix. So says Exercise 4, page 10 of Frank Kelly’s book Reversibility and Stochastic Networks.

If exp(tH) has this property for all t, does that mean H does? I was thinking that if H=DS where D is diagonal and S symmetric, and assuming D is invertible, you’d have OH=HO iff ODS=DSO iff DOS=DSO iff OS=SO.

Reply
- John Baez says:
  
  6 May, 2014 at 11:59 am
  
  I love thinking about these things but it always takes me a while to remember all the relationships between reversible Markov chains, doubly stochastic operators, Dirichlet operators, etc. So, I’ll hope that Jacob Biamonte or Ville Bergholm can field your question!
  
  Reply
  - Jacob Biamonte says:
    
    8 May, 2014 at 11:20 am
    
    Here in Torino, we have Jacob Turner from Penn State visiting for a while and in collaboration with his PhD advisor, Jason Morton we’ve been thinking a lot these days about time-reversal symmetry in quantum mechanics. We have thought about how this relates to the more familiar notions of reversing Markov processes. I want to think more about what you’re saying.
    
    Roughly stated, I think of the difference in probability between starting at network node $n$ and going to $m$ and the opposite as being a sort of ‘probability current’. The reason I think of this is that it actually relates to time-reversal symmetry in quantum mechanics. So if you let $L$ be a valid stochastic generator and define $U(t) := e^{t L}$ for all non-negative times $t$ . Then you can define the ‘stochastic probability current’ as
    
    $\hat{s}_{nm}(t) = U_{nm}(t) - U_{mn}(t)$
    
    This definition is relevant to this post since the following are all equivalent.
    
    (i) $\hat s = 0$ $\forall t$
    (ii) $L = L^\top$
    (iii) $L$ is a Dirichlet operator (e.g. under the image of the exponential map $iLt$ generates a 1-parameter unitary group)
    
    Also note that $\hat s = 0$ $\forall t$ implies the standard definition. I was actually going to ask and see if John was interested in having us create a little mini-series on
    time-reversal symmetry breaking. (Full disclosure, as John knows, I’m supposed to be making a post about how some of the ideas in the Network Theory Series can be used to study infectious disease propagation—applied to a long chain of rabbits. This is a missing chapter from our course notes/book).
    
    Reply
ixxra says:

6 May, 2014 at 10:04 pm

Reblogged this on Homepage of Rene Garcia.

Reply
ixxra says:

6 May, 2014 at 10:21 pm

The way I learned about Noether’s theorem was in quantum field theory, when studying Lagrangian symmetries and conserved currents. Also, I know that in Lie group theory there is another formulation of Noether’s in terms of the moment map. Is there something similar to a Lagrangian formulation for stochastic mechanics?

Reply
- John Baez says:
  
  9 May, 2014 at 2:50 pm
  
  Stochastic mechanics is a lot like quantum mechanics. In particular, the usual path integral description of a free point particle in quantum mechanics is a lot like the path integral description of Brownian motion in stochastic mechanics. In quantum mechanics the amplitude for the free particle to take a path is proportional to the exponential of $i$ times its kinetic energy integrated over time. In stochastic mechanics the probability for Brownian motion to take a path is proportional to the exponential of $-1$ times its kinetic energy integrated over time. Starting from this idea, we should be able to take some ideas about Noether’s theorem for the quantum mechanics of a free point particle and translate them into ideas that apply to Brownian motion. It becomes a bit more interesting if we allow the particle to move around on a Riemannian manifold that’s not necessarily Euclidean $\mathbb{R}^n.$
  
  There should also be a version relating quantum field theory to stochastic field theory, but if you want to study this I suggest starting with particles instead of fields.
  
  There is already some work on this, but I don’t understand it very well, and not merely because it’s in languages other than English:
  
  • A. Barros and D. Torres, Teorema de Noether no cálculo das variacoes estocástico, available as arXiv:1208.5529.
  
  • J. Cresson and S. Darses, Plongement stochastique des systémes lagrangiens, C. R. Math. Acad. Sci. Paris 342 (2006), 333–336. Also available as arXiv:math/0510655.
  
  Reply
  - ixxra says:
    
    11 May, 2014 at 11:49 pm
    
    Thanks Prof. Baez!
    
    Reply
radamion says:

18 May, 2014 at 11:23 am

This is a small, maybe trivial, point but it’s something that I have been thinking about recently in regard to my own research. It seems to me that your master equation has the wrong sign. As I said, a small point and probably doesn’t change anything but let me ask anyway.

If we are thinking about $H$ as something like a Hamiltonian, that is, not just a time-evolution operator but an energy-like quantity then it seems to me that the master equation should be $\frac{d\psi}{dt} = -H\psi$ . My reasoning is that when you add a potential (one that is not balanced by viscous forces) in, for example, a diffusion (or Fokker-Planck) equation, it enters with a minus sign;

$\frac{\partial\psi}{\partial t} =D\frac{\partial^2\psi}{\partial x^2} -V(x)\psi$ ,

and this suggests that either the right-hand side is $-H\psi$ or $L\psi$ , where $L$ is a Lagrangian-like operator. Personally, I think the former makes more sense.

Reply
- John Baez says:
  
  18 May, 2014 at 9:11 pm
  
  If you want $H$ to be not only infinitesimal stochastic but self-adjoint, it’s common in quantum mechanics to assume $H$ is a positive self-adjoint operator, e.g.
  
  $H = - \nabla^2$
  
  Then the right equation for stochastic time evolution, e.g. the heat equation, is
  
  $\displaystyle{ \frac{d \psi}{d t} = - H \psi }$
  
  On the other hand, in my book with Jacob Biamonte I got tired of all these minus signs and used a different convention, namely
  
  $\displaystyle{ \frac{d \psi}{d t} = H \psi }$
  
  which works well when
  
  $H = \nabla^2$
  
  It’s all just a matter of different conventions; the physics is the same either way.
  
  Reply

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