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Blog - Noether's theorem: quantum vs stochastic (Rev #24)

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.

Photo of Emmy Noether

The theorem and its generalizations have found particularly wide use in quantum theory. Those of you following the network series 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

In standard (closed system) quantum theory, the unitary time evolution of a state |ψ(t)|\psi(t)\rangle is generated by a self-adjoint matrix HH which is called the Hamiltonian. So, |ψ(t)|\psi(t)\rangle satisfies Schrödinger’s equation:

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

The state of a system starting off at time zero in the state |ψ 0|\psi_0 \rangle and evolving for a time tt is then given by

|ψ(t)=e itH|ψ 0. |\psi(t) \rangle = e^{-i t H}|\psi_0 \rangle.

The observable properties of a quantum system are associated with self-adjoint operators. The expected value of a self-adjoint operator OO in the state |ψ|\psi \rangle is

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

OO is a constant of the motion if and only if it commutes with the Hamiltonian HH:

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

The stochastic case

In stochastic mechanics, the story changes a bit. Now the Hamiltonian HH 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 |ψ(t)|\psi(t)\rangle is a normalized probability distribution obeying the master equation:

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

HH generates a continuous-time Markov process e tHe^{t H}, which maps the space of normalized probability distribution to itself. So, a probability distribution evolves as follows:

|ψ(t)=e tH|ψ 0.|\psi(t)\rangle = e^{t H} |\psi_0 \rangle.

In this context an observable OO is a real diagonal matrix, and its expected value is given by

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

The version of Noether’s theorem for stochastic systems is stated as:

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

[O,H]=0 [O,H] =0

if and only if

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

and

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

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

From this theorem, we can see immediately that every symmetry in stochastic mechanics given by [O,H]=0[O,H]=0 still leads to OO representing a conserved quantity, meaning that

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

for any ψ 0\psi_0 and all tt.

However, John and Brendan showed is that—unlike in quantum theory—you need more than just the expectation value of the observable OO to be constant to obtain the equation [O,H]=0[O,H]=0. You really need both

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

together with

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

for all initial data ψ 0\psi_0 to be sure that [O,H]=0[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 process is also self-adjoint? In other words, what if HH is both an infinitesimal stochastic matrix but also its own transpose: H=H 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 HH is an infinitesimal stochastic operator with H=H H = H^\top, and OO is an observable, then

[O,H]=0ddtO ψ(t)=0|ψ 0,t0. [O, H] = 0 \quad \iff \quad \displaystyle{\frac{d}{dt}} \langle O \rangle_{\psi(t)} = 0 \quad \forall \: |\psi_0 \rangle, \forall t \ge 0.

Proof. Since HH is symmetric we may use a spectral decomposition

H= kE k|υ kυ k|H = \sum_k E_k |\upsilon_k \rangle \langle \upsilon_k |. As in John and Brendan’s proof, the \implies direction is easy to show. In the other direction, we have

ddtO ψ(t)=O^|He Ht|ψ 0=0|ψ 0,t0 \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

O^|He Ht=0t0 \iff \quad \langle \hat{O}| H e^{Ht} = 0 \quad \forall t \ge 0

kO^|υ kE ke tE kυ k|=0t0 \iff \quad \sum_k \langle \hat{O} | \upsilon_k \rangle E_k e^{t E_k} \langle \upsilon_k| = 0 \quad \forall t \ge 0

O^|υ kE ke tE k=0t0 \iff \quad \langle \hat{O} | \upsilon_k \rangle E_k e^{t E_k} = 0 \quad \forall t \ge 0

|O^Span{|υ k|E k=0}=kerH, \iff \quad |\hat{O} \rangle \in \Span\{|\upsilon_k \rangle | E_k = 0\} = \ker \: H,

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

|O^kerH|\hat{O} \rangle \in \ker \: H

implies that we can choose the eigenbasis of OO to be {|υ k} k\{|\upsilon_k \rangle\}_k, which implies

[O,H]=0[O, H] = 0.

Alternatively,

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

|O 2^kerHddtO 2 ψ(t)=0|ψ 0,t0, |\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,

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

In summary, by restricting ourselves to the intersection of quantum and stochastic generators, we have essentially recovered the quantum version of Noether’s theorem. However, this simplification comes at a cost. We find that the only observables OO that remain invariant under a symmetric HH 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 HH generates a 1-parameter Markov semigroup as follows:

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

defined for all non-negative times tt.

Puzzle 1. Suppose that also H=H H = H^\top, so that HH is a Dirichlet operator and hence iHi H generates a 1-parameter unitary group. Show the following. Let nn label any node of the underlying adjacency matrix corresponding to HH. Then the indegree and outdegree of any node nn is equal (graphs with this property are called balanced).

Puzzle 2. Now assume further that U(t)U(t) is doubly stochastic for all times tt, e.g.

iU ij= jU ij=1 \sum_i U_{ij} = \sum_j U_{ij} = 1

and show that this is equivalent to the condition on the generator

iH ij= jH ij=0 \sum_i H_{ij} = \sum_j H_{ij} = 0

Puzzle 3. 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.

Puzzle 4. Let AA be a possibly time-dependent stochastic observable, and write A\langle A\rangle for its expected value with respect to some initial state ψ 0\psi_0 evolving as e tHψ 0e^{t H}\psi_0. Show that

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

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

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