Download presentation

Presentation is loading. Please wait.

Published byHayden Scott Modified over 3 years ago

1
**Henry Haselgrove School of Physical Sciences University of Queensland**

GRIFFITH QUANTUM THEORY SEMINAR 10 NOVEMBER 2003 Entanglement, correlation, and error-correction in the ground states of many-body systems Henry Haselgrove School of Physical Sciences University of Queensland Michael Nielsen - UQ Tobias Osborne – Bristol Nick Bonesteel – Florida State Today I’ll be talking about some quite general results I have regarding properties of ground states of many-body quantum systems. I should acknowledge my collaborators in this work: first, my supervisor Michael Nielsen. Also Tobias Osborne, who many of you would know as being an ex UQ PhD student. He’s now a postdoc at Bristol. And finally, Nick Bonesteel, who’s a condensed matter physicist at Florida State University. quant-ph/ quant-ph/ – to appear in PRL

2
When we make basic assumptions about the interactions in a multi-body quantum system, what are the implications for the ground state? Basic assumptions --- simple general assumptions of physical plausibility, applicable to most physical systems. Nature gets by with just 2-body interactions Far-apart things don’t directly interact Implications for the ground state --- using the concepts of Quantum Information Theory. This is the main question behind the work I’m talking about today. That is, “when we make basic assumptions about the interactions in a multi-body quantum system, what are the implications for the properties of the ground state?” So, what do I mean by the “basic assumptions” about the interactions in a multi-body quantum system? Well, if you take a many-body quantum system, and you don’t make any physical assumptions about the interactions, then plain old Quantum Theory will allow you to have any Hamiltonian describing the interactions, and will therefore allow any state to be the ground state --- and we can’t say anything smart about its properties. The other extreme is that you know exactly what interactions are going on in the system. Then we can write down an exact Hamiltonian matrix, and diagonalise it to find an exact ground state vector, from which you can infer whatever properties you want. But that just tells you about that specific system and nothing else. So, I’m taking a kind of middle ground, where I’m considering some quite general assumptions of physical plausibility of many-body interactions, which are still strict enough to have interesting effects on the ground state. These implications are going to be described using the concepts of Quantum Information Theory. So, the first notion of physical plausibility is that interactions can be described as being a sum of two-body interactions. It’s true that nature seems to get by almost exclusively with two-body interactions. Quantum-theory allows you to imagine Hamiltonians that are much more complicated, involving interactions which can’t be broken into sums of two-body interactions. So, that particular assumption of physical plausibility is actually quite a strict one, mathematically speaking, that has interesting side-effects for properties of physically plausible ground states. The side-effects I’ll talk about today relate to the error-correcting abilities of the ground state. The other basic assumption I’ll consider, is that there are certain parts within a many-body system that don’t directly interact with each other, although they might indirectly interact via other bodies. The reason that they don’t directly couple might just be that they’re so far apart. It turns out that this type of assumptions has implications for how much entanglement, and other correlations, you can get in the ground state, between those objects that don’t directly couple. Error-correcting properties Entanglement properties

3
**Why ground states are really cool**

Physically, ground states are interesting: T=0 is only thermal state that can be a pure state (vs. mixed state) Pure states are the “most quantum”. Physically: superconductivity, superfluidity, quantum hall effect, … Ground states in Quantum Information Processing: Naturally fault-tolerant systems Adiabatic quantum computing I’d just like to remind you why it’s even interesting to look at properties of ground states. Of course, the ground state is basically what you get when you cool a system down to absolute zero. The interesting thing about the zero-temperature state is that it is the only thermal state that can possibly be a pure quantum state. For nonzero temperatures, you get mixed states, that have classical randomness. Whereas a pure state is in some sense the most quantum. As a result, you see interesting phenomena that are uniquely quantum, like superconductivity, superfluidity, the quantum hall effect, and so on… In Quantum Information Processing, ground states crop up quite a bit too. Here are a couple of the more important places you see them. In a naturally fault-tolerant system, you engineer the system so that its ground state is naturally resistant to noise, making it a good place to store quantum information. Adiabatic Quantum Computing is a particular model for quantum computation, where you efficiently encode a hard computational problem into the Hamiltonian of many-body system, in such a way that the ground state contains the answer.

4
**Part 1: Two-local interactions**

N interacting quantum systems, each d-level … 1 2 3 4 N Interactions may only be one- and two-body Consider the whole state space. Which of these states are the ground state of some (nontrivial) two-local Hamiltonian? Lets look at the first assumption: that a many-body system has only one- and two-body interactions. When the interactions in a system are a sum of one- and two-body interactions, we say that the system is two-local. The question is, what effect does this assumption have on the possible ground states? In other words, first consider the whole set of states on the system. Which of these states represent the ground state of some two-local system? If you pick a particular state, is there some way of fiddling with the actual interactions, so that you get that state as the ground state? I should point out my use of the word nontrivial here. We’re interested in the ground states of all possible two-local systems, with one exception: we’re not interested in the system that has all interactions turned off. In that trivial case, every state belongs to a totally degenerate ground state. So the precise question is, what states can be the ground state of some nontrivial two-local Hamiltonian. I’ll show, that there are holes in this set of possible states. That is, any state inside these regions are provably not the ground state of any nontrivial physically plausible system.

5
**Two-local interactions**

2 1 4 3 Classically: So, classically, a system is two-local if the force on any one particle can be written as a sum of forces due to each of the other particles in turn. Each of those functions in the sum must be a function of the properties of at most two particles. The result is that the total energy of the system must be able to be written as a sum of functions over the various pairs of particles. Quantum-mechanically, its pretty much the same situation. Except now, the energy is a sum of functions of each of the pair-wise reduced density matrices for the state. But of course, the actual form of that function is given by the Hamiltonian of the system. So we need to know what it means for a Hamiltonian to be two-local. Quantum-mechanically:

6
**Two-local Hamiltonians**

N quantum bits, for clarity Any imaginable Hamiltonian is a real linear combination of basis matrices An, {An} = All N-fold tensor products of Pauli matrices, Any two-local Hamiltonian is written as where the Bn are N-fold tensor products of Pauli matrices with no more than two non-identity terms. For the sake of clarity, lets look at a system of N quantum bits. First of all, any imaginable Hamiltonian on that system, physically plausible or not, can be written as a linear combination of basis matrices A_n. For the case of N qubits, these basis matrices consist of all combinations of tensor products of the various Pauli matrices. So why bother expressing a Hamiltonian this way? Well, because for any particular term in that expansion, its easy to tell the number of bodies that the term acts on: you just count the number of the tensor product factors that aren’t the identity. That makes it really easy to write down a completely arbitrary two-local Hamiltonian. You just perform the same sum, but this time you’re restricting the basis matrices to being ones that are at most two-body. I call that subset of the A_n matrices B_n.

7
**Example is two-local, but is not. O(N2) O(2N) parameters**

Why two-locality restricts ground states: parameter counting argument O(N2) So, for an example, consider these two Hamiltonians for a three-qubit system. I’ve left out the tensor product symbols for brevity. The first one is two local, because the there are three two-body terms and one one-body term. The second one is not two-local, because it has a three-body term. It’s not hard to convince yourself, with a crude parameter counting argument, that two-locality should restrict the possible ground states drastically. If you look at the expansion for an arbitrary two-local Hamiltonian on N objects, the number of real numbers that you need to specify is of order N^2, because that’s the number of tensor products of Pauli operators that act on no more than two qubits. However, to specify an arbitrary state on N qubits, you need of order 2^N real numbers. So, there’s simply not enough degrees of freedom in nature to access just any arbitrary ground state. O(2N) parameters

8
**Necessary condition for |> to be two-local ground state**

We have and Take E=0 Not interested in trivial case where all cn=0 So, having written an arbitrary two-local Hamiltonian as a sum of one and two-body Pauli operators, we can start to do some interesting things. First of all, lets simply substitute the expression for H into the eigenvalue equation H-psi-equals-E-psi. We may as well make things simpler by taking the energy eigenvalue to be zero, since we are free to choose the overall energy scale. We are left with this equation. It is saying that for psi to be the ground state (or any eigenstate for that matter) of a two-local system, you must be able to take a linear combination of these vectors, given by the various Bn acting on psi, and make it add up to zero. Plus, you want to do that without simply making the coefficients c_n all zero (because that would give us a trivial Hamiltonians). That means, the set of vectors B_1 acting on psi, B_2 acting on psi, and so on, must be a linearly dependent set, for psi to be the eigenstate of a nontrivial physically plausible system. It’s quite easy numerically to check if a list of vectors is linearly dependent. So, if someone were to give you a state psi and ask you if it could be physically plausible ground state, this is a simple method which you could use to perhaps answer definitely not, if you found these vectors to be linearly independent. So the set must be linearly dependent for |i to be a two-local ground state

9
**Nondegenerate quantum error-correcting codes**

A state |> is in a QECC that corrects L errors if in principle the original state can be recovered after any unknown operation on L of the qubits acts on |> The {Bn} form a basis for errors on up to 2 qubits A QECC that corrects two errors is nondegenerate if each {Bn} takes |i to a mutually orthogonal state So lets talk about an example of a type of state that certainly cannot be a physically plausible ground state. It is a type of state defined in quantum information theory, and it is known as a nondegenerate quantum correcting code state. So first of all, what is QECC, and what does it mean for such a thing to be nondegenerate? Well, a state Psi is said to be part of a QECC that corrects L errors, if in principle the original state can be recovered after any unknown operation (that is, and error) affects some L of the qubits. So how does this relate to the problem at hand? Well, the set Bn of Pauli operators acting on up to two bodies at a time, that we used as a basis for all two-local Hamiltonians, is actually also a basis for all errors on up to two qubits. So, you’ll see that we can make a connection to QECCs that correct up to two errors. In particular, a so-called nondegenerate code. A code Psi is said to be nondegenerate, if each of the basis matrices Bn take psi to a different mutually orthogonal state. The idea behind such a code, is that after a error occurs, you just make a measurement in the orthonormal basis B1|psi> B2|psi> etc. Even though there is a continuous range of errors that could of occurred, the measurement collapses the effective error down to being one of the B_n. Then you know precisely what error occurred, and you can correct for it. But the important thing for our argument, is that the vectors B1|psi>, B2|psi> etc, being orthogonal to each other, are certainly not linearly dependent, and so, from our reasoning before, the state psi can’t be the ground state of any nontrivial physically plausible system. Only way you can have is if all cn=0 ) trivial Hamiltonian

10
**A nondegenerate QECC can not be the eigenstate of any nontrivial two-local Hamiltonian**

In fact, it can not be even near an eigenstate of any nontrivial two-local Hamiltonian Which is exactly what I say there. But what’s more, such a state can’t even be near any eigenstate of any nontrivial two-local Hamiltonian, a fact that I’ll prove on the next slide.

11
**H = completely arbitrary nontrivial 2-local Hamiltonian **

= nondegenerate QECC correcting 2 errors E = any eigenstate of H (assume it has zero eigenvalue) Want to show that these assumptions alone imply that || - E || can never get small Okay, so I’m proving that any state that is within a certain radius of a NGQECC state, cannot possibly be the ground state of a nontrivial two-local Hamiltonian. Let H be a completely arbitrary nontrivial two-local Hamiltonian. Let psi be a nondegenerate QECC state that corrects two errors. Remember, this will mean that each of the basis matrices Bn will take psi to a different orthogonal state. Let bold E be any eigenstate of that arbitrary Hamiltonian H. Assume without loss of generality that the eigenvalue is zero. So, we want to show that those assumptions alone imply that the distance between that eigenstate and the code state can never get small. Begin by writing down the norm of H acting on the vector difference between psi and E. This inequality is just a simple norm inequality. Also, we can simplify the expression on the left, by using the fact that H acting on E gives the zero vector, since E has a zero eigenvalue. Immediately you can see that we’ve got the distance between psi and E as one of the factors, which is good. Rearranging, the norm of psi minus E is greater than or equal to this ratio of norms. What can we say about that fraction? Well, lets substitute the expansion for H into the numerator. Remember that the Bn acting on psi give a set on orthonormal vectors, so the norm of that is simply the Euclidean (or L2) norm of the vector of coefficients c. Lets look at the denominator. We chuck in the expansion for H again. This time we use this norm inequality. But the Bn matrices are chosen to have norm one, so that just reduces to the sum of the absolute values of coefficients. That is the Manhattan, or L1 norm of the vector of coefficients C. In fact, no matter what that vector c actually is, you can prove that that ratio of L2 to L1 norms can never get smaller than one on the square-root of the number of elements in the vector. That number of elements is just the number of two-body Pauli operators on N qubits, which is a quadratic in N which I wont bother giving you the exact form of.

12
**Nondegenerate QECCs Radius of the holes is**

So we return to this picture of the set of states that could possibly be the ground state of a physically plausible Hamiltonian. We’ve explored this set a little bit. Enough to know that there it’s a bit like Swiss cheese, with these holes centred about NGECC. Nondegenerate QECCs Radius of the holes is

13
**Part 2: When far-apart objects don’t interact**

In the ground state, how much entanglement is there between the ●’s? We find that the entanglement is bounded by a function of the energy gap between ground and first exited states So lets move on to the second part of my talk, which looks at a different assumption about the interactions in a many-body quantum system. It is simply that there are two parts of the system, denoted by the red circles here, which don’t directly interact. But they’re allowed to indirectly interact via the rest of the system. We don’t make any assumptions about the internal structure or size of those three objects, so this picture can actually apply to a wide range of situations. For example, we could be looking at two sites in a lattice of spins, where each spin only interacts with its nearest neighbours. In many physical situations, the fact that the two objects don’t interact is going to be more of a good approximation rather than an exact truth, but this type assumption is still used quite successfully in lots of different models. The question is, “in the ground state, how much entanglement, and other types of correlations, is there between the two red objects?”. We’ll show that the answer to the question depends on the size of the energy gap between the ground state and first excited state. The larger the energy gap the smaller the amount of entanglement that is allowed. Don’t worry if it’s not obvious that there should even be a connection between the entanglement and the energy gap, because it is not obvious. First of let me try and convince you that such a relationship would be interesting if it existed.

14
**Energy gap E1-E0: Entanglement:**

Physical quantity: how much energy is needed to excite to higher eigenstate Needs to be nonzero in order for zero-temperature state to be pure Adiabatic QC: you must slow down the computation when the energy gap becomes small Entanglement: Uniquely quantum property A resource in several Quantum Information Processing tasks Is required at intermediate steps of a quantum computation, in order for the computation to be powerful So, I’ve just listed a few ways in which both of these quantities is interesting in itself, from which it would follow that a connection between the two is also interesting. First of all, the energy gap between the ground state and first excited state. It’s a physical quantity that can often be measured: it’s just the amount of energy you need to excite the system from the ground to the first excited state. It has implications for whether the T=0 state is pure or not: the gap needs to be nonzero in order for the zero temperature state to be pure. At which point I should make it clear that in my definition of energy gap, whenever the ground state is degenerate, that means you have a zero gap. That’s as opposed to if you were to take the gap from the degenerate level to the next highest distinct level in that situation. Also, in adiabatic quantum computation: as the gap size becomes small, you must slow the computation down to prevent the system being accidentally exited. So the gap is relevant to the power of adiabatic QC. What about entanglement? It is a uniquely quantum property. It is an important resource in several Quantum Information Processing tasks. And, you can prove that unless you have large amounts of entanglement at intermediate steps of a quantum computation, your quantum computer will be no more powerful than a classical computation.

15
Some related results Theory of quantum phase transitions. At a QPT, one sees both a vanishing energy gap, and long-range correlations in the ground state. Theory usually applies to infinite quantum systems. Non-relativistic Goldstone Theorem. Diverging correlations imply vanishing energy gap. Applies to infinite systems, and typically requires additional symmetry assumptions The idea that there is a connection between the gap and entanglement or correlations is not a particularly new one. It crops up quite a bit in condensed matter theory. Here are a couple of examples. First, the theory of quantum phase transitions. A quantum phase transition is a phenomenon where you see two types of behaviour at once: a vanishing energy gap, and the appearance of long-range correlations in the ground state. So, you see a connection between the gap and correlations. However, the theory usually only applies to infinite quantum systems. Also, there’s the non-relativistic Goldstone Theorem. This result proves that diverging correlations in the ground state imply a vanishing energy gap. But the theorem relies on having an infinite system, and you typically need to use additional symmetry assumptions. I’m still in the process of trying to understand some of the related results in Condensed Matter Theory, but so far it seems like our results unique in giving a relationship that is applicable for any quantum system where two parts aren’t directly coupled.

16
**Extreme case: maximum entanglement**

B C Assume the ground state has maximum entanglement between A and C or A C B I’ll try and convince you that there should be a relationship between the amount of entanglement between A and C in the ground state to the size of the energy gap, by considering the extreme case first. That is, to assume that A and C are maximally entangled in the ground state. For simplicity, we’ll assume that A and C are qubits. Up to a relabelling of the basis vectors for the systems A and C, any state on that system that has the most possible entanglement can be written like this. So we assume the ground state has this form. The important feature is the fact that it is a product of any state on system B, with this equal superposition of 00 and 11 on the subsystem AC. The fact that A and C don’t directly couple means that the Hamiltonian can be written as a term acting on the subsystem AB, plus a term acting on the subsystem BC. This means that the energy of a state can be written as this function of the reduced state on subsystem AB, plus this function of the reduced state on subsystem BC. So given this form for the ground state, what are the reduced states on AB and BC? Well, to find the state of the subsystem AB, you need to take the partial trace over the object C. That means we’re taking a partial trace over half of a maximally entangled pair AC. It happens that whenever you take the partial trace over half of a maximally entangled pair, you’re left with the maximally mixed state on the other half, which for a qubit is the identity matrix divided by two. The state on system B is unaffected by the partial trace on C. Similarly, when you find the reduced state on subsystem BC, you get the maximally mixed state on C, and an unchanged state on B. The crucial step is when we introduce a slightly different definition of E0, call it E0 prime. We’ve just turned this plus sign in to a minus. E0-prime is certainly a different state than E0. In fact, E0 and E0-prime are orthogonal. But they have the same reduced states on AB and BC. That’s because they have the same state on B, that’s just phi, and they both have maximal entanglement between A and C, which gives you the maximally mixed state here and here. And because the reduced states are the same, the energy of the two states are the same. So, we have two different states both having the ground state energy. This can only happen if the ground state is doubly-degenerate, that is if the energy gap is zero. So, to recap this slide, whenever you have a state on a system like this, that has maximal entanglement between A and C, you can always find a different state that has the same energy.

17
**That is, whenever you have couplings of the form**

B C it is impossible to have a unique ground state that maximally entangles A and C. So, a maximally entangled ground state implies a zero energy gap Same argument extends to any maximally correlated ground state That means it is impossible to have a unique ground state for this system, such that A and C are maximally entangled. So a maximal entanglement between parts that don’t directly interact, implies a zero energy gap. In fact, the same argument generalises somewhat to show that having A and C maximally correlated in the ground state implies a zero energy gap. Up to relabelling of basis states on system A and C, any state that has maximum correlation between qubits A and C can be written like this. We say that there is maximum correlation simply because whenever you measure a 0 on qubit, you end up with a zero state on qubit C, and whenever you measure a one on qubit A, you end up with a one qubit C. You get the maximally entangled state that we saw before if you set p=1/2 and phi equal to psi. But, for example, if phi is orthogonal to psi, you’ll still have maximum correlation, but you’ll have zero entanglement. Anyway, like before, you can introduce a different state E0’, which can be shown to have the same reduced states on subsystems AB and BC as does E0, so has the same energy, and so the same result holds: you can have it as the unique ground state.

18
**Can we get any entanglement between A and C in a unique ground state?**

Yes. For example (A, B, C are spin-1/2): 0.1X X 0.1X 1.4000 1.0392 1.0000 0.6485 = 0.1 (XX + YY + ZZ) … has a unique ground state having an entanglement of formation of 0.96 So, you might ask, if you can’t get maximum entanglement between A and C in a unique ground state, can you get any entanglement at all? Yes, you can. Lets look at a simple numerical example. Let A, B and C be qubits. Let the Hamiltonian be such that there are these one-body terms, which you can think of as a magnetic field in the X direction, of strength 0.1, 1 and .1. Also, let the blue lines indicate the interaction XX + YY + ZZ, with strength 0.1, which is just the Heisenberg interaction. When you diagonalise that, you find that it does indeed have a unique ground state, and the entanglement of formation, which is one of the measures of entanglement, between A and C is 0.96, compared with a maximum of 1. But you also notice that the energy gap is very small. Here is the energy spectrum for that system. The difference between the lowest two energy levels is really small compared to the overall energy scale. So can we prove a general trade-off between the ground-state entanglement, or more generally the ground-state correlation, and the energy gap? Can we prove a general trade-off between ground-state entanglement and the gap?

19
General result A B C Have a “target state” |i that we want “close” to being the ground state |E0i --- measure of closeness of target to ground --- measure of correlation between A and C Yes, and I’ll give you one of our main results along those lines. So, the situation is we have some target state psi that we want to be close to being the ground state E0 of the system. By close, it means that we have in mind some value F representing the inner product between psi and the ground state. We express our target state like this. It’s actually a completely general way of expressing a state of the system. The j states are an orthonormal basis for system A, and the k states are an orthonormal basis for system C. The ejk states are any arbitrary normalised states on system B, they don’t have to be orthogonal to each other. The way to interpret a term in this sum, is that with probability pjk, you measure a “j” outcome on system A, and a “k” outcome on system C, and when that happens you’re left with the state ejk on system B. This sum is just taking a superposition over those different outcomes. Now we want to introduce some notion of the amount correlation between A and C in that state, so we define C to be the probability that you the measurement on system A will have the same outcome as the measurement on system C. Then we have the following result for a bound on the energy gap. Actually, it’s not just a bound on the gap between ground and exited state, but is also a bound on the gap to higher exited states, depending on the number of possible measurement outcomes you’re summing over here. You see that the closer psi gets to being the ground state, that is the closer F gets to 1, and the more correlated psi becomes, that is the closer C gets to 1, the closer the right hand side gets to zero. Note that Etot is the total energy scale of the system, which means the maximum energy eigenvalue minus the minimum energy eigenvalue. The closer the RHS gets to zero, the more these low-lying states will get sandwiched together. So, we do indeed have the explicit trade-off between energy gap and correlation that we were looking for, but as a bonus we have a bound on gaps to higher energy levels as well.

20
The future… At the moment, our bound on the energy gap becomes very weak when you make the system very large. Can we improve this? The question of whether a state can be a unique ground state is closely related to the question of when a state is uniquely determined by its reduced density matrices. Explore this question further: what are the conditions for this “unique extended state”?

21
Conclusions Simple yet widely-applicable assumptions on the interactions in a many-body quantum system, lead to interesting and powerful results regarding the ground states of those systems Assuming two-locality affects the error-correcting abilities Assuming that two parts don’t directly interact, introduces a correlation-gap trade-off.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google