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

Showing changes from revision #24 to #25: Added | Removed | Changed

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

guest post by Ville Bergholm

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 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.|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.

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

iddt|ψ(t)=H|ψ(t). 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 |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.

|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 So, expected we value call of a self-adjoint operatorn×nn \times n matrix an observable. The expected value of an observable OO in the a state|ψ|\psi \rangle is

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

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

This expected value is constant in time for all states if and only if OO is a constant of the motion if and only if itcommutes 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.

[O,H]=0ddtO ψ(t)=0|ψ 0,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 OO 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|ψ(t)|\psi(t)\rangle is a probability distribution: a vector with nn nonnegative entries that sum to one. And 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|nψ×(nt) |\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:

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

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

HHIf we start with a probability distribution generates a continuous-time Markov process |ψ 0|\psi_0 \ranglee tHe^{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:

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

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

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

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,

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

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

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

O ψ= iO iiψ i \langle O\rangle_{\psi} = \displaystyle{ \sum_i O_{i i} \psi_i }

where ψ i\psi_i is the iith component of the vector |ψ|\psi\rangle.

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

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

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

if and only if

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

and

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

for all tt and all ψ(t)\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[O,H]=0 still the leads expected to value ofOO representing will a be conserved:conserved quantity, meaning that

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

for any ψ 0\psi_0 and all tt . However, John and Brendan showed that—unlike in quantum mechanics—you need more than just the expectation value of the observableOO to be constant to obtain the equation [O,H]=0[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 ddtO ψ(t)=0 \displaystyle{\frac{d}{d t}} \langle O\rangle_{\psi(t)} = 0 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

ddtdtO 2 ψ(t)=0 \displaystyle{\frac{d}{dt}} \displaystyle{\frac{d}{d t}} \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.

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 ifHH 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 Operators. If HH is an infinitesimal stochastic operator with H=H H = H^\top, and OO is an observable, then

[O,H]=0ddtdtO ψ(t)=0|ψ 0,t0. [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 HH is symmetric self-adjoint, we may use a spectral decomposition decomposition:

H= k kE k|ϕ kϕ k|E k|υ kυ k| 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

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 where ϕ k\phi_k are an orthonormal basis of eigenvectors, and E kE_k are the corresponding eigenvalues. We then have:

ddtO ψ(t)=O^|He tH|ψ 0=0|ψ 0,t0O^|He Ht=0t0 \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

kO^|υ kHe tHE ke tE kυ k|=0t0 \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

kO^| υ ϕ kE ke tE kϕ k|=0t0 \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

|O^Span{| υ ϕ k|E ke tE k=0} = ker tH0, \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 |O^Span{|ϕ k|E k=0}=kerH, \iff \quad |\hat{O} \rangle \in \Span\{|\phi_k \rangle | E_k = 0\} = \ker \: H, {|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

where the third equivalence is due to the vectors |ϕ k|\phi_k \rangle being linearly independent. 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 basis of eigenvectors ofOO to be {| υ ϕ k} k \{|\upsilon_k \{|\phi_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, implies that

where we have used the above sequence of equivalences backwards. Now, using John and Brendan’s original proof, we can obtain |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}{d t}} \langle O^2 \rangle_{\psi(t)} = 0 \quad \forall \: |\psi_0 \rangle, \forall t \ge 0, [O,H]=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,H]=0 O [O, H] = 0 . that   remain invariant under a symmetricHH 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 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 are equal equal: (graphs graphs with this property are calledbalanced ). .

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

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

iU(t) ij= jU(t) 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 generatorU(t)U(t) are nonnegative for all t0t \ge 0, if and only if the matrix HH obeys

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

iH ij= jH 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.HH are nonnegative.

Puzzle 4. 3. Let Prove that any doubly stochastic Markov semigroup A U(t) A U(t) be is a of possibly the time-dependent form stochastic observable, and writee tHA \langle e^{t A\rangle H} for where its expected value with respect to some initial stateψ 0H \psi_0 H evolving is as the graph Laplacian of a balanced graph.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

Puzzle 4. Let O(t)O(t) be a possibly time-dependent observable, and write O(t) ψ(t)\langle O(t) \rangle_{\psi(t)} for its expected value with respect to some initial state ψ 0\psi_0 evolving according to the master equation. Show that

Using this (or some other method), prove a stochastic version of the ddtO(t) ψ(t)=[O(t),H]+Ot ψ(t) \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.

categories: blog