# The Azimuth Project Blog - Noether's theorem: quantum vs stochastic (Rev #25, changes)

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This blog article in progress, written by Ville Bergholm. See also Blog - quantum network theory (part 1) and Blog - quantum network theory (part 2). To see the discussion of the article being written, visit the Azimuth Forum. If you want to write your own article, please read

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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 Network series Theory series here on Azimuth might recallPart 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

In I’ll standard assume (closed you’re system) familiar with quantum theory, the but unitary let’s time start evolution with of a state few reminders.$|\psi(t)\rangle$ is generated by a self-adjoint matrix $H$ which is called the Hamiltonian. So, $|\psi(t)\rangle$ satisfies Schrödinger’s equation:

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

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

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

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

The observable properties of a quantum system are associated with self-adjoint operators. The So, expected we value call of a self-adjoint operator$n \times n$ matrix an observable. The expected value of an observable $O$ in the a state$|\psi \rangle$ is

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

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

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

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

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

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

### The stochastic case

In stochastic mechanics, the story changes a bit. Now the a Hamiltonian state$|\psi(t)\rangle$ is a probability distribution: a vector with $n$ nonnegative entries that sum to one. And now the Hamiltonian $H$ is an infinitesimal stochastic operator, i.e. a real-valued square matrix with non-negative off-diagonal entries and columns summing to zero. And now |\psi(t)\rangle n \times n is matrix a with normalized non-negative probability off-diagonal distribution entries obeying and columns summing to zero. Schrödinger’s equation gets replaced by the master equation:master equation:

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

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

$H$If we start with a probability distribution generates a continuous-time Markov process $|\psi_0 \rangle$$e^{t H}$ at time zero and evolve it according to the master equation, at some later time we’ll have the probability distribution, which maps the space of normalized probability distribution to itself. So, a probability distribution evolves as follows:

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

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

In So, this context anobservable$e^{t H}$ maps the space of probability distributions to itself, and we call it a$O$continuous-time Markov process , is or more precisely a real diagonal matrix, and its expected value is given byMarkov semigroup.

$\langle O\rangle_{\psi} = \sum_{i \in X} \langle i | O |\psi\rangle = \langle \hat{O} | \psi \rangle,$

In this context an observable $O$ is a real diagonal $n \times n$ matrix, and its expected value is given by

The version of Noether’s theorem for stochastic systems is stated as:$\langle O\rangle_{\psi} = \langle \hat{O} | \psi \rangle$

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

$[O,H] =0$

$[O,H] =0$

if and only if

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

and

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

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

From So, this just theorem, as we can see immediately that every symmetry in stochastic quantum mechanics mechanics, given whenever by$[O,H]=0$ still the leads expected to value of$O$ representing will a be conserved:conserved quantity, meaning that

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

for any $\psi_0$ and all $t$ . However, John and Brendan showed 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

However, John and Brendan showed is that—unlike in quantum theory—you need more than just the expectation value of the observable $\displaystyle{\frac{d}{d t}} \langle O\rangle_{\psi(t)} = 0$$O$ to be constant to obtain the equation $[O,H]=0$. You really need both

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

together with

 \displaystyle{\frac{d}{dt}} \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.

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 process 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 Operators. If $H$ is an infinitesimal stochastic operator with $H = H^\top$, and $O$ is an observable, then

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

Proof.Proof. Since The$\implies$ 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 symmetric self-adjoint, we may use a spectral decomposition decomposition:

 H = \displaystyle{ \sum_k E_k |\upsilon_k |\phi_k \rangle \langle \upsilon_k \phi_k | |}. As in John and Brendan’s proof, the $\implies$ direction is easy to show. In the other direction, we have

$\displaystyle{\frac{d}{dt}} \langle O \rangle_{\psi(t)} = \langle \hat{O} | H e^{Ht} |\psi_0 \rangle = 0 \quad \forall \: |\psi_0 \rangle, \forall t \ge 0$where $\phi_k$ are an orthonormal basis of eigenvectors, and $E_k$ are the corresponding eigenvalues. We then have:

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

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

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

 \iff \quad |\hat{O} \langle \rangle \hat{O} \in \Span\{|\upsilon_k \rangle | \phi_k \rangle E_k e^{t E_k} = 0\} 0 = \quad \ker \forall \: t H, \ge 0

where the third equivalence is due to $\iff \quad |\hat{O} \rangle \in \Span\{|\phi_k \rangle | E_k = 0\} = \ker \: H,$$\{ |\upsilon_k \rangle \}_k$ being a linearly independent set of vectors. For any infinitesimal stochastic operator $H$ the corresponding transition graph consists of $m$ connected components iff we may 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

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 may 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 eigenbasis basis of eigenvectors of$O$ to be  \{|\upsilon_k \{|\phi_k \rangle\}_k, which implies

$[O, H] = 0$.

Alternatively,

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

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

where we have used the above sequence of equivalences backwards. Now, using John and Brendan’s original proof, we can obtain $|\hat{O^2} \rangle \in \ker \: H \quad \iff \quad \ldots \quad \iff \quad \displaystyle{\frac{d}{d t}} \langle O^2 \rangle_{\psi(t)} = 0 \quad \forall \: |\psi_0 \rangle, \forall t \ge 0,$$[O, H] = 0$.

In where summary, by restricting ourselves to the intersection of quantum and stochastic generators, we have essentially used recovered the quantum above version sequence of Noether’s equivalences theorem. backwards. However, Now, this using simplification John comes and at Brendan’s a original cost. proof, We we find can that obtain the only observables O [O, H] = 0 . that   remain invariant under a symmetric$H$ are of the very restricted type described above, where the observable has to have the same value in every state in a connected component.

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 $O$ that remain invariant under a symmetric $H$ are of the very restricted type described above, where the observable has to have the same value in every state in a connected component.

### Puzzles

Suppose that a graph Laplacian matrix $H$ generates a 1-parameter Markov semigroup as follows:

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

defined for all non-negative times $t$.

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 the following. Let $n$ label any node of the underlying adjacency matrix corresponding to $H$. Then the indegree and outdegree of any node $n$ is are equal equal: (graphs graphs with this property are calledbalanced ). .

Puzzle 2. Now Show assume further that U(t) = e^{t H} is doubly stochastic for Markov all semigroup, times e.g.$t$, e.g.

$\sum_i U_{ij} = \sum_j U_{ij} = 1$

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

and show all that this is equivalent to the condition matrix on entries the of generator$U(t)$ are nonnegative for all $t \ge 0$, if and only if the matrix $H$ obeys

$\sum_i H_{ij} = \sum_j H_{ij} = 0$

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

Puzzle 3.and all the off-diagonal entries of Prove that a doubly stochastic continuous-time Markov semigroup is generated by a balanced graph. Observe further that symmetric graphs are a strict subclass of balanced graphs.$H$ are nonnegative.

Puzzle 4. 3. Let Prove that any doubly stochastic Markov semigroup A U(t) be is a of possibly the time-dependent form stochastic observable, and write \langle e^{t A\rangle H} for where its expected value with respect to some initial state \psi_0 H evolving is as the graph Laplacian of a balanced graph.$e^{t H}\psi_0$. Show that

$\frac{d}{d t}\langle A\rangle = \langle [A, H] \rangle+ \left\langle \frac{\partial A}{\partial t}\right\rangle$

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

Using this (or some other method), prove a stochastic version of the $\displaystyle{ \frac{d}{d t}\langle O(t)\rangle_{\psi(t)} = \langle [O(t), H] \rangle + \left\langle \frac{\partial O}{\partial t}\right\rangle_{\psi(t)} }$Ehrenfest theorem.

Using this or some other method, prove a stochastic version of the Ehrenfest theorem.

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