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UCL, 23 Feb 2006Entanglement Probability Distribution of Random Stabilizer StatesOscar C.O. DahlstenMartin B. Plenio

- The title is ‘Entanglement Probability Distribution of Random Stabilizer States’
- Entanglement is the amount of quantum correlations, here taken between two parties sharing a pure state.
- By Entanglement Probability Distribution we mean P(E), the likelihood of having entanglement of value E.
- Stabilizer states are an important discrete subset of general states.
- By random stabilizer states, me mean that we are sampling
at random and without bias from states that are restricted to be stabilizer states.

This talk aims to explain the paper:

Exact Entanglement Probability Distribution in

Randomised Bipartite Stabilizer States.

[Dahlsten, Plenio, quant-ph/0511119]

1. Introduction, aim of work

2. Entanglement Probability Distribution

3. Properties of Distribution

4. Summary and Outlook

- Entanglement is a fundamental resource in quantum information tasks.
- We can classify and quantify entanglement between two parties quite well, but there is a plethora of classes for more than two parties.
- Here we consider two simplifications to the problem:
A. Restrict the states to be ‘stabilizer states’, a discrete subset of all possible quantum states.

B. Restrict entanglement types to those that are ‘typical’.

- Stabilizer states are an important discrete subset of all possible states [Gottesman, Caltech PhD thesis].
- Called stabilizer states as the state is defined by listing the Pauli Matrices that ’stabilize it’, i.e.
- They can be parametrised efficiently, yet form a rich variety of states: and etc.
- Bipartite Entanglement in stabilizer states comes in integer values, E=0,1,2,…Emax
[Audenaert, Plenio, quant-ph/0505036 ]

[Fattall et al. quant-ph/0406168]

- Second simplification: Consider only the typical entanglement in a completely randomised system. [Hayden et al., quant-ph/0407049]
- Physical setting: imagine two-level atoms in a gas colliding at random, causing entanglement between energy levels.
- Asymptotically the system is completely randomised.
- Alice and Bob E-Entangled with probability P(E).

Alice

Bob

t=0

t=1

- In general states it is known that the average typical/generic entanglement is near maximal.
(Page’s conjecture).

- Here typical is defined relative to the uniform distribution on states, given by the ’Haar measure’ on unitaries.
- There is a concentration of the distribution around this average with increasing N -’concentration of measure’.
- Is the above still true under the restriction of stabilizer states?

- The first question in this line of enquiry is: what is the typical bipartite entanglement in randomised stabilizer states?
- To answer this we need the probability distribution P(E).
- Entanglement value E is typical if P(E) significant, atypical if P(E) insignificant.
- Hence theobjective is to find and study P(E) for randomised bipartite stabilizer states.

1.(Done) Introduction, aim of work

-Simplify entanglement classification by

restricting classes to those that are typical in

stabilizer states.

-Therefore aim to find P(E) of randomised stabilizer

states, where E isbipartite entanglement .

Next

2. Entanglement Probability Distribution

We derive an expression for P(E).

3. Properties of Distribution

4. Summary and Outlook

- Notation: The N qubits are grouped such that NA belong to Alice(the smaller party) and NB to Bob.
- The total state is pure and N=NA+NB.
- The state is restricted to be a stabilizer state, but any such state is equally likely.
- Then P(E), the probability of E entanglement between Alice and Bob is:

- Take probability distribution on stabilizer states as flat. Then p(state)=1/ntot where ntot is the total number of states for the given N.
- Entanglement E is an integer,
- So P(E)=nE/ntot where nE(N,NA) is the number of possible stabilizer states with entanglement E.
- Simplest example: N=2, NA=1 whereby
Then an explicit count gives ntot=60, n0=36, n1=24.

Thus P(0) =36/60 and

p(1)=24/60

n0

n1

All ntot states

- Finding nE(N, NA) for any N and NA is tricky. Use three lemmas:
- Lemma 1: The total number of states is known to be
[Gottesmann, Aaronson quant-ph/052328]

[Gross, quant-ph/0602001]

- Lemma 2: The number of unentangled states n0is
- Lemma 3: There is an invariant ratio (proof complicated)
- The lemmas together give an iterative expression for nE. This gives P(E) as P(E)=nE/ntot

1.(Done) Introduction, aim of work

2. (Done)Entanglement Probability Distribution

Derived

Next

3. Properties of Distribution

-Distribution is ‘Gaussianish’

-Average is nearly maximal

-Concentration around average

-Similar to general states

4. Summary and Outlook

- An entirely equivalent form of the distribution is
- Where is messy but comparatively small
- Therefore P(E) is roughly the side of a Gaussian curve, centred on N/2

- An example of P(E), for N=12, NA=5.

- Recall maximal entanglement possible is NA, the number of qubits in the smallest of the two groups.
- By the main P(E) theorem, one sees the average entanglement, , is nearly maximal for large N.
- Therefore if we pick stabilizer states at random we expect to get near maximal entanglement on average.

- Distribution squeezes up around the average with increasing N.
- Typical entanglement for large N is thus nearly maximal.
- Animation to the right shows P(E) with fixed NA but N increasing.

- The average entanglement in general states is also near maximal [‘Page’s conjecture’]. The figure below compares the averages for N=10 and varying NA.
- There isconcentration around the average for general states too [Hayden et al., quant-ph/0407049].

- We give the Probability Distribution of Entanglement in randomised stabilizer states.
- It shows the typical entanglement is near maximal.
- Surprisingly this is very similar to the case for general states.
- Note: [Smith&Leung, quant-ph/0510232] also interesting.
Outlook

- Is there a stabilizer-general state similarity for other quantities than entanglement?
- What about multipartite entanglement?
- What happens during the process of randomisation?