Tobias J. Osborne's research notes

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Stochastic Davidenko equation

Fri, 17 Apr 2009 15:27:00 GMT

Introduction

The Davidenko equation is a system of differential equations which track the zeros of parameter-dependent system of polynomials. This note is aimed at understanding what happens when the polynomials depend on one or more continuous-time martingales, such as simple brownian motions.

Simple example

(This needs to be fixed: the second order terms aren't correct yet!)
To understand what I'm talking about let's derive the stochastic Davidenko equation in the case of a parameter-dependent polynomial in one variable. Note that this discussion is at the level of physical rigour.
Let p(x) = \sum_{j=0}^d a_j(B_t, t) x^j be a polynomial of degree d whose coefficients a_j are functions of B_t and t, where B_t is a Wiener process, i.e., a standard Brownian motion. We want to understand the zeros z_l(B_t, t), l = 1, 2, \ldots, d of p(x). To do this we mimic the derivation of the Davidenko equation and set up a set of Stochastic differential equations for z_l(B_t,t). As with the Davidenko equation this can be done in a variety of ways.
Consider a small change in p(x):

dp(x) = \sum_{j=0}^d da_j(B_t,t) x^j + ja_j(B_t,t)x^{j-1} dx + jx^{j-1}da_j(B_t,t)dx + j(j-1)a_j(B_t,t)x^{j-2}(dx)^2.

(Here we are foreshadowing the use of Itō's rule.)
Now use Itō's rule to write

da_j(B_t,t) = \frac{\partial a_j(y,t)}{\partial t}\Bigg|_{y=B_t} dt + \frac{\partial a_j(y,t)}{\partial y}\Bigg|_{y=B_t} dB_t + \frac12\frac{\partial^2 a_j(y,t)}{\partial^2 y}\Bigg|_{y=B_t}dt

for the SDE satisfied by a_j(B_t,t). Then we have that

dp(x) = \sum_{j=0}^d \left(\frac{\partial a_j(y,t)}{\partial t}\Bigg|_{y=B_t} dt + \frac{\partial a_j(y,t)}{\partial y}\Bigg|_{y=B_t} dB_t + \frac12\frac{\partial^2 a_j(y,t)}{\partial^2 y}\Bigg|_{y=B_t}dt\right) x^j + ja_j(B_t,t)x^{j-1} dx + j(j-1)a_j(B_t,t)x^{j-2}(dx)^2.

Now if z_l(B_t, t) is a zero of p(x) then we must have that 0 = dp(z_l(B_t, t)), i.e.,

0= \sum_{j=0}^d \left(\frac{\partial a_j(y,t)}{\partial t}\Bigg|_{y=B_t} dt + \frac{\partial a_j(y,t)}{\partial y}\Bigg|_{y=B_t} dB_t + \frac12\frac{\partial^2 a_j(y,t)}{\partial^2 y}\Bigg|_{y=B_t}dt\right) z_l(B_t, t)^j + ja_j(B_t,t)z_l(B_t, t)^{j-1} dz_l(B_t, t) + j(j-1)a_j(B_t,t)z_l(B_t,t)^{j-2}(dz_l(B_t,t))^2.

This is the stochastic Davidenko equation.

A small example polynomial

Suppose that

p(x) = x^2+B_tx-1.

Thus, a_0 = -1, a_1 = B_t, and a_2 = 1. In this case the stochastic Davidenko equation becomes

z_l(B_t,t)dB_t + (2z_l(B_t,t) + B_t)dz_l(B_t,t) = 0,

dz_l(B_t,t) = - \frac{z_l(B_t,t)}{2z_l(B_t,t) + B_t} dB_t.

Draft blog posts

Tue, 14 Apr 2009 16:13:00 GMT

Discrete phase space
Random QSAT
Toward efficient quantum circuits for the Laughlin wavefunction
On open science
Hamiltonian complexity
What is a quantum phase transition?

MainMenu

Mon, 13 Apr 2009 22:18:00 GMT

About
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Blog

Mon, 13 Apr 2009 22:13:00 GMT

The central place to find out about my current research is my blog. You can find links to the rss updates from my twitter and del.icio.us accounts and from this TiddlyWiki there.

Draft posts can be found here.

Jordan's lemma

Fri, 10 Apr 2009 17:11:00 GMT

From Oded Regev's lecture notes:

As we all know, for any two lines in \mathbb{R}^n that go through the origin (i.e., one-dimensional subspaces), one can define the angle between them. If we take a line and a plane, we can again define the angle between them in a natural way. But what happens if we take two planes? Here our three-dimensional intuition is no longer good enough. Indeed, in three-dimensions, two two-dimensional subspaces always intersect in a line, and orthogonal to that line we find the angle between the two subspaces. This is no longer true in higher dimensions: Starting from four dimensions, two two-dimensional subspaces generally have a trivial intersection, and instead of forming an angle, they form two angles!
In more generality, the question we consider in this section is how two subspaces interact. This question turns out to have a very elegant answer, as we shall soon see. This answer, which was first given in a remarkable paper of C. Jordan in 1875, was since rediscovered many times by mathematicians, statisticians, physicists, and computer scientists. In addition to being a crucial component in witness-preserving amplification of QMA, this question also plays an important role in many recent results in quantum computation, often in an implicit way... This topic is also covered in Chapter VII of the book by
Bhatia.

The lemma itself is as follows.

Lemma 1 (Jordan, 1875). For any two Hermitian projectors \Pi_0 and \Pi_1, there exists an orthogonal decomposition of the Hilbert space into one dimensional and two dimensional subspaces that are invariant under both \Pi_0 and \Pi_1. Moreover, inside each two-dimensional subspace, \Pi_0 and \Pi_1 are rank-one projectors. (In other words, inside each two-dimensional subspace there are two unit vectors |v\rangle and |w\rangle suchthat \Pi_0 projects on |v\rangle and \Pi_1 projects on |w\rangle.)

The proof of this lemma is actually elementary:

Proof. The idea is to look at the hermitian matrix H = \Pi_0 + \Pi_1. It turns out that the eigenvectors of H can be partitioned into sets of size one or two, and these sets in fact span the subspaces mentioned in the lemma.
Let |\phi\rangle be an eigenvector of H, of unit length. Let \lambda be the corresponding eigenvalue. Then
\Pi_0|\phi\rangle + \Pi_1|\phi\rangle = \lambda|\phi\rangle.
Assume first that \Pi_0|\phi\rangle lies in the subspace spanned by |\phi\rangle. By the eigenvector equation above, we have that \Pi_1|\phi\rangle lies in the same subspace. This means that the subspace spanned by |\phi\rangle is invariant under \Pi_0 and \Pi_1, hence it is an eigenvector of both projectors. Hence \Pi_0|\phi\rangle is 0 or |\phi\rangle, and similarly for \Pi_1.
So we now assume that \Pi_0|\phi\rangle is not in the subspace spanned by |\phi\rangle. So consider, instead, the two-dimensional subspace S spanned by |\phi\rangle and \Pi_0|\phi\rangle. Evidently S is invariant under \phi_0 because
\Pi_0(\alpha |\phi\rangle + \beta\Pi_0|\phi) = (\alpha+\beta)\Pi_0|\phi\rangle \in S.
It is also invariant under \Pi_1 thanks to the eigenvector equation:
\Pi_1|\phi\rangle = \lambda|\phi\rangle - \Pi_0|\phi\rangle \in S
so that
\Pi_1\Pi_0|\phi\rangle = \Pi_1(\lambda|\phi\rangle - \Pi_1|\phi\rangle) = (\lambda-1)\Pi_1|\phi\rangle \in S.
Because S is invariant under both \Pi_0 and \Pi_1 it is invariant under H. Hence, the vector orthogonal to |\phi\rangle in S is actually another eigenvector of H, and so S is actually spanned by a pair of eigenvectors of H, as claimed. To conclude, note that inside S the operators \Pi_0 and \Pi_1 are rank-one projectors.

Mathematics notes

Fri, 10 Apr 2009 15:56:00 GMT

Jordan's lemma

Notes

Fri, 10 Apr 2009 15:53:00 GMT

Mathematics notes

About

Fri, 10 Apr 2009 15:53:00 GMT

I am a researcher in quantum information theory based at Royal holloway, University of London. I have worked in applied mathematics, quantum information theory, and condensed matter physics for the past eight years or so.

Inspired by the recent discussions (see, eg., Michael Nielsen’s blog postings) surrounding open science I have decided to make my research notes open. This TiddlyWiki is one attempt to make my notes accessible to a wider audience, and it is hoped that it can be as current a snapshot of my research notes as possible. The content is a little sparse at the moment, but I hope to add more as time passes.

If you make any progress on the ideas or open problems discussed on this site then please let me know: I am more than happy for anything here to become a joint project, the more the merrier! (After all, it might be easier to write a joint paper with the collaborators on this site rather than rewriting everything yourself…)

If you want to look at any of my notes which include many formulae then I'd strongly suggest installing the jsMath fonts (they are 4 ttf files).

Quantum circuits for the Laughlin wavefunction

Thu, 09 Apr 2009 10:25:00 GMT

Introduction

In a recent paper arXiv:0902.4797 Riera, Pico, and Latorre describe a quantum circuit which efficiently prepares a quantum register in the integer quantum Hall effect ground state.
An open problem from this paper is to describe quantum circuits which efficiently prepare the Laughlin wavefunction for the fractional quantum Hall effect. This problem is significantly more complicated than the integer quantum Hall effect case because there are delicate cancellations that can occur and must be accounted for.

Sequential generation

One might hope to build a quantum circuit for the Laughlin wavefunction sequentially. To do this one needs to have some kind of recursion relation.

Representing the Laughlin function in a quantum register

The way we represent a multiparticle wavefunction in a quantum register is to use a basis where a monomial multiplied by a gaussian weighting:

z_1^{d_1}z_2^{d_2}\cdots z_n^{d_n} e^{-\frac12\sum_{j=1}^n |z_j|^2}

is represented as

|d_1d_2\cdots d_n\rangle.

Because the gaussian weighting is the same for all terms in the wavefunctions we wish to represent we typically drop it from now on and just speak of the monomial z_1^{d_1}z_2^{d_2}\cdots z_n^{d_n} as being represented by |d_1d_2\cdots d_n\rangle.

A simple recursion relation for the Vandermonde determinant

A nice recursion relation is described in the paper of Dunne, who notes the following properties of the Vandermonde determinant: let

V_n(z_1, z_2, \ldots, z_n) = \prod_{1\le j<k}^n(z_j-z_k).

Note that

V_n(z_1, z_2, \ldots, z_n) = V_{n-1}(z_2, z_3, \ldots, z_n)\prod_{j=2}^n(z_1-z_j).

We now write

\prod_{j=2}^n(z_1-z_j) = \sum_{k=0}^{n-1} (-1)^kz^{n-1-k}e_k,

where e_0 = 1, e_1 = \sum_{k=2}^n z_k, \ldots, e_{n-1} = z_2z_3\cdots z_{n}, i.e., e_l is the sum of all products of l distinct variables z_2, z_3, \ldots, z_n.
This observation now allows us to write out the desired recursion:

(V_n(z_1, z_2, \ldots, z_n))^{2m+1} = \left(\sum_{k=0}^{n-1} (-1)^kz^{n-1-k}e_k\right)^{2m+1}(V_{n-1}(z_2, z_3, \ldots, z_n))^{2m+1}.

An example for m=1

Let's consider the simplest case, namely V_2(z_1, z_2): this is represented as

V_2(z_1, z_2) = (z_1-z_2) \mapsto \frac{1}{\sqrt{2}}(|10\rangle-|01\rangle).

Now the Laughlin wavefunction with m=1 is represented as

(V_2(z_1, z_2))^3 = (z_1-z_2)^3 \mapsto \frac{1}{\sqrt{20}}(|30\rangle - 3|21\rangle + 3|12\rangle - |03\rangle).

Introducing the linear operator

L_{j,k}|x_1x_2\cdots x_n\rangle = |x_1x_2\cdots x_{j-1} (x_j+k) x_{j+1} \cdots x_n\rangle

allows us to also write

(V_2(z_1, z_2))^3 \mapsto (L_{1,2} - 2L_{1,1}L_{2,1} + L_{2,2}) \times \frac{1}{\sqrt{2}}(|10\rangle-|01\rangle).

It is convenient to adopt a multi-index notation and define

L_{\mu}|x_1x_2\cdots x_n\rangle = |(x_1+\mu_1)(x_2+\mu_2)\cdots(x_n+\mu_n)\rangle

where \mu = (\mu_1, \mu_2, \ldots, \mu_n). Thus

(V_2(z_1, z_2))^3 \mapsto (L_{(2,0)} - 2L_{(1,1)} + L_{(0,2)}) \times \frac{1}{\sqrt{2}}(|10\rangle-|01\rangle).

A nonunitary(!) quantum circuit for the Laughline wavefunction

The recursion relation for the Vandermonde determinant can now be represented, in quantum notation, as

|V_n\rangle = \sum_{k=0}^{n-1} (-1)^k|n-1-k\rangle E_k|V_{n-1}\rangle,

where

|V_n\rangle = \frac{1}{\sqrt{n!}} \sum_{\pi\in S_n} \epsilon(\pi)|\pi^{-1}(0)\pi^{-1}(1)\cdots \pi^{-1}(n-1)\rangle

is the quantum representation of the Vandermonde determinant, and

E_k = \sum_{\mu\subset \{0,1,\ldots, n-1\}, |\mu| = k} L_\mu

Cubing this recursion in the obvious way gives a nonunitary circuit for the m=1 Laughlin wavefunction. Unfortunately the operators E_k aren't unitary, so this is why the problem is hard.

Computer science problems

Thu, 09 Apr 2009 09:25:00 GMT

The simulation problem

Problems

Thu, 09 Apr 2009 09:25:00 GMT

Computer science problems
Mathematics problems
Physics problems
Miscellaneous problems

Physics problems

Thu, 09 Apr 2009 09:24:00 GMT

Solving a simple QSAT instance
Quantum circuits for the Laughlin wavefunction

Davidenko equation

Mon, 06 Apr 2009 11:30:00 GMT

Introduction

The Davidenko equation is a system of differential equations which track the zeros of a parameter-dependent system of polynomials.

Simple example

Let p_t(x) = \sum_{j=0}^d a_j(t) x^j be a parameter-dependent polynomial of degree d in the variable x. Write z_j(t), j = 1, 2, \ldots, d for the (possibly compex) zeroes of p_t(x). We want to set up a system of ordinary differential equations for the parameter-dependent zeros of p_t(x). This can be done in several ways. One way is to suppose that we know z_j(t), j = 1, 2, \ldots, d at a time t and to consider a small change in H(x,t) = p_t(x):

dH(x,t) = \frac{\partial H(x,t)}{\partial x} dx + \frac{\partial H(x,t)}{\partial t} dt.

Now, substituting x = z_j(t) we obtain

dH(z_j(t),t) = \frac{\partial H(z_j(t),t)}{\partial x} dz_j(t) + \frac{\partial H(z_j(t),t)}{\partial t} dt.

Now, because z_j(t) is supposed to be a zero of p_t(x) for all time we must have that

0 = \frac{\partial H(z_j(t),t)}{\partial x} dz_j(t) + \frac{\partial H(z_j(t),t)}{\partial t} dt

so that

\frac{dz_j(t)}{dt} = - \left(\frac{\partial H(z_j(t),t)}{\partial x}\right)^{-1} \frac{\partial H(z_j(t),t)}{\partial t}.

This is the basic Davidenko equation.

A small example polynomial

Suppose that

p_t(x) = x^2 + (1+t)x - 1.

In this case the Davidenko equation becomes

\frac{dz_j(t)}{dt} = - \frac{z_j(t)}{2z_j(t) + t - 1}, \quad j = 1, 2.

Full derivation

Ideas

Mon, 06 Apr 2009 10:45:00 GMT

Stochastic Davidenko equation

The simulation problem

Tue, 31 Mar 2009 21:18:00 GMT

The simulation problem aims to formalise in a concrete way what a theoretical physicist does when they make a prediction about a physical system. I would argue that the simulation problem is the central problem of the newly emerging field of hamiltonian complexity.

Entanglement remote-control and deciding QSAT

Mon, 30 Mar 2009 15:46:00 GMT

In this note I will describe how to construct all of the ground states for a class of interacting quantum spin systems. The systems considered here pertain to n qudits (with local dimension d) arranged in a line, and have the following form (they are a subclass of the family of quantum satisfiability instances).

\sum_{j=1}^{n-1} |\phi_{j,j+1}\rangle\langle\phi_{j,j+1}|\otimes \mathbb{I}_{[n]\setminus\{j,j+1\}},

where the pure states |\phi_{j,j+1}\rangle have Schmidt decomposition

|\phi_{j,j+1}\rangle = \sum_{k=1}^d\sqrt{q_k^{(j)}}|u_kv_k\rangle,

with q_j>0. The pure states |\phi_{j,j+1}\rangle are otherwise arbitrary.

The way to solve these systems is to exploit Proposition 1: we first write

|\phi_{j,j+1}\rangle = A_j\otimes \mathbb{I}_{j+1}|\Psi^+\rangle,

where |\Psi^+\rangle = \frac{1}{\sqrt{d}}\sum_{j=1}^d |jj\rangle. The next step is to take a ground state |\Omega\rangle of

K = \sum_{j=1}^{n-1} |\Psi_{j,j+1}^+\rangle\langle\Psi_{j,j+1}^+|\otimes \mathbb{I}_{[n]\setminus\{j,j+1\}}.

and construct

|\Gamma\rangle = T_1\otimes T_2\otimes \cdots \otimes T_{n-1}\otimes \mathbb{I}_n|\Omega\rangle,

where, for j even (need to check these formulae!),

T_{n-j} = (A_{j-1}^\dagger A_{j-2}^*\cdots A_2^* A_1^\dagger)^{-1},

and for j odd,

T_{n-j} = (A_{j-1}^* A_{j-2}^\dagger\cdots A_2^* A_1^\dagger)^{-1}.

We now have the following

Proposition 1. |\Gamma\rangle is a ground state of H.

Proof.

Entanglement remote-control

Mon, 30 Mar 2009 15:46:00 GMT

In the (now foundational) paper quant-ph/9707038 Lo and Popescu describe protocols involving local operations and classical communication on quantum bipartite entangled states. There are several tricks that one can play with entangled states in this setting and I just want to note down some of the more basic and fundamental results. All of the results here pertain to two d-dimensional quantum systems (qudits).

Write |\Psi^+\rangle = \frac{1}{\sqrt{d}}\sum_{j=1}^d |jj\rangle. Then we have the following

Lemma 1. Let M be a d\times d matrix. Then
M\otimes \mathbb{I}|\Psi^+\rangle = \mathbb{I}\otimes M^T|\Psi^+\rangle.

Proof.
M\otimes \mathbb{I}|\Psi^+\rangle = \frac{1}{\sqrt{d}}\sum_{j,k=1}^d m_{k,j} |kj\rangle = \frac{1}{\sqrt{d}}\sum_{j,k=1}^d m_{j,k} |jk\rangle = \mathbb{I}\otimes M^T|\Psi^+\rangle.

The second result tells us how to prepare essentially any two-qudit state from |\Psi^+\rangle via a local rotation and a post-selected measurement on just one half of |\Psi^+\rangle. This is the remote-control of entanglement trick. It is also known as the Reeh-Schlieder theorem in algebraic quantum field theory (where it applies in a vastly more general setting) which is also called the Taj-Mahal theorem (because someone could create the Taj-Mahal on the other side of the universe by simply measuring one half of a sufficiently entangled state and post selecting). The precise result is

Proposition 1. Let |\phi\rangle = \sum_{j=1}^d \sqrt{p_j}|u_jv_j\rangle be the Schmidt decomposition of an arbitrary pure state of two qudits. Suppose that p_j>0. Then
|\phi\rangle = A\otimes \mathbb{I}|\Psi^+\rangle,
where A = MUV^T with U|j\rangle = |u_j\rangle, V|j\rangle = |v_j\rangle, and M|j\rangle = \sqrt{dp_j}|j\rangle.

Proof. Follows directly after applying Lemma 1.

Solving a simple QSAT instance

Mon, 30 Mar 2009 15:21:00 GMT

The approach to deciding QSAT instances via entanglement remote-control reduces the problem of first finding the ground-eigenspace of hamiltonians of the form

H = \sum_{e\in E} |\Psi^{+}\rangle_e\langle \Psi^{+}|\otimes \mathbb{I}_{[n]\setminus e},

where |\Psi^+\rangle = \frac{1}{\sqrt{d}}\sum_{j=1}^d |jj\rangle and E is the edge set of some finite tree-graph G = (V=[n], E) with [n]=\{1,2,\ldots, n\}. The second step is to add in extra constraints to form loops.

I'll focus on solving the first problem. In fact, I'll just try and describe the ground eigenspace for the simplest possible example, namely, the following hamiltonian on 3 qudits

H = |\Psi^{+}\rangle_{12}\langle \Psi^{+}|\otimes \mathbb{I}_3 + \mathbb{I}_1\otimes |\Psi^{+}\rangle_{23}\langle \Psi^{+}|.

It's pretty easy to find quite a lot of states in the ground eigenspace: any state of the form |jkl\rangle with j\not=k and k\not=l, j,k,l \in [d] will do. How many of these are there? Well, there are d^3-2d^2+d such states. Are these all the ground states? Well, no, there are a couple more, namely, any state of the form

|\eta(k)\rangle = \frac{1}{\sqrt{d}}\sum_{j=1}^d e^{\frac{2\pi i}{d}jk} |jjj\rangle,

with k\in [d-1] is also a ground state. There are d-1 such states. Thus there are at least d^3-2d^2+1 orthogonal ground states. Is this all of them? There should be more because the smallest the ground eigenspace can be is d^3-2d-dimensional.

Tobias J. Osborne

Thu, 26 Mar 2009 23:11:00 GMT

I am a lecturer in the department of mathematics at Royal Holloway, University of London.

I am currently exploring the possibilities offered by open science via this TiddlyWiki and my blog.