# Does DQN really learn?
Exploring adversarial training schemes in Pong

Bowen He
Department of Computer Science
Brown University
Providence, RI 02912
<EMAIL_ADDRESS>Sreehari Rammohan
Department of Computer Science
Brown University
Providence, RI 02912
<EMAIL_ADDRESS>Jessica Forde
Department of Computer Science
Brown University
Providence, RI 02912
<EMAIL_ADDRESS>Michael L. Littman
Department of Computer Science
Brown University
Providence, RI 02912
<EMAIL_ADDRESS>
###### Abstract

In this work, we study two self-play training schemes, _Chainer_ and _Pool_ ,
and show they lead to improved agent performance in Atari Pong compared to a
standard DQN agent—trained against the built-in Atari opponent. To measure
agent performance, we define a robustness metric that captures how difficult
it is to learn a strategy that beats the agent’s learned policy.Through
playing past versions of themselves, Chainer and Pool are able to target
weaknesses in their policies and improve their resistance to attack. Agents
trained using these methods score well on our robustness metric and can easily
defeat the standard DQN agent. We conclude by using linear probing to
illuminate what internal structures the different agents develop to play the
game. We show that training agents with Chainer or Pool leads to richer
network activations with greater predictive power to estimate critical game-
state features compared to the standard DQN agent.

Keywords:

Adversarial Training, DQN, Pong, Deep Reinforcement Learning, Robustness,
Chainer, Pool

##  1 Introduction

Since the introduction in 2013, DQN (Mnih et al., 2013) has been accepted as
outperforming human-level play in a range of Atari games. However, a
fundamental question about what these agents truly learn remains: Are they
picking up on gameplay techniques and fundamental scoring concepts, or are
they merely exploiting idiosyncrasies in the game dynamics? As an example, a
small change to the opponent policy can totally confuse a standard DQN-trained
agent because it overfits to the opponent gameplay weaknesses instead of
learning the essential skills of the game (Witty et al., ). Furthermore, are
there ways of training Deep RL agents that can facilitate their learning of
the essential rules of the game, leading to more robust policies? We aim to
answer these questions in the game of Pong, showing that adversarial training
schemes can extract more context-relevant information leading to stronger
policies.

##  2 Background

The environment is modeled as a a time-homogeneous Markov Decision Process
$(X,A,R,P,\gamma)$. As usual, $X$ and $A$ represent state and action spaces,
respectively, while the transition kernel $P(\cdot|x,a)$ defines the dynamics
of the environment. Reward function $R(x,a)$ provides a signal to direct
learning. Deep Q Networks, or DQN, is an approach to learning to optimize
behavior in such environments. It applies several strategies to stabilize
learning compared to naive learning approaches: replay buffer sampling, a
target network, and frame pre-processing tricks.

Recent years have seen an uptick in the attention researchers have paid to
questions of generalization of reinforcement-learning models. Packer et al.
(2018) found that the vanilla deep RL algorithms have better generalization
performance than specialized schemes that were proposed specifically to tackle
generalization. In addition, Kansky et al. (2017) argued that generalizing
from limited data and learning causal relationships are essential abilities on
the path toward general intelligent systems.

Probe techniques used by Tenney et al. (2019) were useful to determine where
different kinds of linguistic information is encoded in a large language model
by correlating network activations with known properties of inputs. We apply
these methods to our work, using linear probing to investigate the ability of
the learned networks to support a heuristic task that well-trained agents
should be able to perform well on.

Up until now, DQN in Pong has been trained non-adversarially, where a single
agent competes against a static opponent modelled by the computer. The thesis
of this paper is that adversarially trained agents are better than those
trained non-adversarially in that they should develop policies prepared to
handle a broader array of scenarios.

We observe in the training of a _standard DQN agent_ (used to refer a learner
trained non-adversarial) in Pong that the agent learns to exploit weaknesses
in the built-in Atari opponent—after a few million training steps, it perfects
a “kill shot” aimed always at the upper or lower quadrants of the board that
the built-in Atari agent cannot return. In our work, we change the dynamics of
the opponent, and we demonstrate that doing so pushes the learner to focus on
learning a more generalizable understanding of the game itself, robustly
improving its skills in the process.

##  3 Methodology

Our adversarial training schemes pit the current agent against a previous
version of itself. The key difference between the two algorithms we propose is
which previous agent the current agent faces. We used the Gym Retro Atari
implementation of Pong, which allows us to control both the left and right
paddles independently. We chose Pong as a test bed because it is challenging,
but small scale, allowing many training runs in a short amount of time.

In addition, Pong is well suited to adversarial training because the board is
vertically symmetric, allowing us to easily create self-playing agents using a
single policy. Specifically, if an agent is trained on one side, we can
symmetrically reflect the game board as input to the agent to apply the same
policy to the opposite side. To avoid overfitting the agent to a single side
of the board, we randomly assign the agent a game side to play periodically.
Under this uniform setting, we empirically tested two methods of self-play,
_Pool_ and _Chainer_ , described in the following sections. We use the
standard DQN agent as a baseline and our empirical results show that it is
much weaker than the pool and chainer agents.

###  3.1 Pool

Pool, described in Algorithm 1, plays against a randomly selected previous
agent from a queue of fixed size. Periodically, the current version of the
agent is frozen and appended to the queue, creating a set of most up-to-date
adversaries to play against. The opponent is updated periodically by sampling
from this queue. In practice, we set the queue size to $5$ and add new agents
to the pool every $250,000$ steps.

###  3.2 Chainer

Chainer, described in Algorithm 2, plays the current version of the agent,
$A^{n}$, solely against its immediate predecessor, $A^{n-1}$. When $A_{n}$
achieves a fixed evaluation threshold, defined here as getting above an
average score of $15$ over $10$ matches, the current agent $A^{n}$ replaces
the opponent’s policy and a successor $A^{n+1}$ will continue to be trained
from the policy of $A^{n}$. We initialize the first member of the chain,
$A^{0}$, with a standard DQN agent trained for $45$ million steps. Agent
$A^{1}$ is trained from scratch (tabla rasa) to play against agent $A^{0}$.
This training approach allows the agent to continually improve its skill
without overfitting to a specific opponent (which would stagnate the skill
learning of the agent).

Algorithm 1 Pool Self-Play Algorithm

$P\leftarrow p$, pool update frequency

$S\leftarrow s$, step counter

Initialize _Agent_

Initialize _Env_

Initialize _Pool_ of fixed size

Initialize _Opponent_

while True do

 _interact_(_Agent_ , _Opponent_ , _Env_)

 _update_(_Agent_)

if $S\%P==0$ then

 _Pool_ $\leftarrow$ _freeze_(_Agent_)

 _Opponent_ $\leftarrow$ _sample_(_Pool_)

end if

$S+=1$

end while

Algorithm 2 Chainer Self-Play Algorithm

$S\leftarrow 0$, the number of steps

$C\leftarrow c$, the evaluation frequency

$i\leftarrow 1$, current length of chain

$t\leftarrow 15$, the iteration threshold

Initialize $A^{i}$, the current agent.

Initialize Opponent $A^{0}$, the standard DQN agent

while True do

 _interact_(_Agent_ : $A^{i}$, _Opponent_ : $A^{i-1}$, _Env_)

 _update_(_Agent_ : $A^{i}$)

if $S\%C==0$ then

average_score = evaluate($A^{i}$, $A^{i-1}$)

if average_score $>$ t then

 _Opponent_ $\leftarrow A^{i}$

$i+=1$

end if

end if

$S+=1$

end while

###  3.3 Linear Probing

Linear Probing is a technique used to determine if the learned representation
in layers of a neural network contains information relevant for solving a
sample task. This entails using simple machine-learning models like linear
regressors to map layer activations to target values (Tenney et al., 2019).
Designing the heuristic task is entirely problem-dependent; the task often
includes targets that a human expects is important toward achieving the end
goal.

In the setting of Pong, we assume that a good agent should have the ability to
predict, with high accuracy, where the ball is going to land on its side of
the game board. Therefore, we attempted to linearly map the agent’s final
layer activations (a $512$-length vector) to the landing $y$ position of the
ball. A ball landing at the top of the board on the agent’s side has a landing
value of $0$, and a ball landing on the bottom has a landing value of $82$,
the height of board in pixels. We collect a training and test set of $10,000$
frames each; frames in this set are taken $\leq 30$ steps before the ball
collides with the agent’s side of board. We report the MSE and $R^{2}$
statistic using $A^{7}$ from Chainer, Pool trained with $25$ million steps,
and the standard DQN agent trained for $45$ million steps.

##  4 Experiments

In this section, we show empirical results for Chainer and Pool. We present
learning curves, a new robustness metric, and the linear probing results.

###  4.1 Chainer and Pool Results

(a) Chainer  (b) Pooling

Figure 1: Learning Curves for Chainer and Pool

The Chainer learning curve shown in Figure 1 (a) shows the game reward
achieved throughout training for different iterations of the opponent. Each
vertical red dashed line represents a changing of the opponent (for example,
the first dashed line represents the point in time when the opponent becomes
$A^{2}$, the frozen agent at the second iteration of the chain). We move on to
the next agent in the chain after the current agent has converged, which we
define as reaching an average score of $15$ over $10$ matches against the
opponent ($21$-point game). Notice how, as training progresses, the number of
steps it takes for the agent to converge against its adversary
increases—indicating the strength of opponents is increasing in the chain.

The Pooling learning curve in Figure 1 (b) shows the game reward achieved
throughout training for different samplings of opponents from the queue. In
this graph, the vertical dashed lines represent points in time where we sample
a new opponent from the pool (every $250,000$ steps). Notice that, as time
goes on, the overall reward curve dips downward. That is because the agents in
the pool are getting better, making it more difficult for the current agent to
perform well. Also note, in comparison to Figure 1 (a), this curve is not
training agents to convergence. As a result, the agent finds it more difficult
to adapt to the changing opponent. We show in Table 2 how, even though the
agent is challenged by these new, more powerful opponents, it retains the
ability to defeat the standard DQN agent.

###  4.2 Robustness

A truly robust agent should be hard to defeat by an opponent policy because
its play is strong across the board. Thus, we define the _robustness_ of an
agent as the difficulty for an opponent to defeat it through learning. More
robust agents will have flatter adversary score curves, indicating less reward
achieved by the opponent over the course of learning. We picked the scores
achieved by the opponent after $6e6$ steps, after which point the learning
curve was stable with no large fluctuations.

In our robustness experiments, we tested $3$ Pooling agents
($A_{\mathrm{0.5M}},A_{\mathrm{4M}},A_{\mathrm{25M}}$) and $3$ Chainer agents
($A^{1},A^{4},A^{7}$). Later agents are consistently more robust than earlier
ones, as shown by their lower lines in the graphs. In comparison, the standard
DQN agent is easily defeated by an opponent, both standard DQN agent curves
reach $21$ within about $1$ million steps.

(a) Chainer Robustness Score (b) Pooling Robustness

Figure 2: Robustness Tests (lower is more robust)

###  4.3 Linear Probing

Training Scheme | MSE | $R^{2}$
---|---|---
Standard DQN | $276.97$ | $0.41$
Pool | $211.84$ | $0.55$
Chainer | $202.41$ | $0.57$
Table 1: Linear Probing Results

The results for Linear Probing are shown in Table 1. Probes for Chainer and
Pool have significantly lower MSE and higher $R^{2}$ statistics than the probe
for standard DQN agent. A regression model that randomly guesses the landing
$y$ position of the ball would have MSE $\approx(82-41)^{2}\approx 1600$. It
is important to note that while the true model regressing network activations
onto the landing $y$ position of the ball is likely non-linear, even using the
crudest form of estimation with a linear model shows that the activations from
adversarially trained agents have greater predictive power on this task,
suggesting these agents more thoroughly capture the dynamics of the game.

###  4.4 Adversarial Agents vs. Standard DQN

An obvious question that the learning curves in Figure 1 raises is whether the
adversarial agents can consistently defeat the standard DQN. For example, do
Chainer agents in link $A^{n+1}$ remember how to defeat the standard DQN agent
($A^{0}$, trained for $45$ million steps) even after being trained to defeat
more difficult opponents? To answer this type of question, we play Pooling
agents ($A_{\mathrm{0.5M}},A_{\mathrm{4M}},A_{\mathrm{25M}}$) and Chainer
agents ($A^{1},A^{4},A^{7}$) against the standard DQN agent and report the
10-match average score achieved. The general trend is clear, as agents get
further along in the adversarial training, they are able to still easily
defeat the standard DQN agent (indicating that these gameplay skills are not
forgotten). In fact, their relative strength appears to be steadily
increasing. We reiterate that it is not simply a matter of the adversarial
agents being trained longer—the longest training agents have just over half
the experience used to train the standard DQN agent. The difference is that
the adversarial agents are taught to handle a wider variety of situations,
resulting in stronger overall play.

Chainer Agents | Scores Achieved
---|---
| Against Standard DQN
$A^{1}$ | $16.7$
$A^{4}$ | $19.5$
$A^{7}$ | $20$

Pooling Agents | Scores Achieved
---|---
| Against Standard DQN
$0.5M$ | $18.2$
$4M$ | $15.1$
$10M$ | $20.2$
$25M$ | $20.4$

Table 2: Adversarial Agents vs. Standard DQN Agent (trained for $45$ million
steps)

##  5 Conclusion

These results validate adversarial training schemes as a way to produce robust
agents capable of opponent generalization in Pong. Agents trained using either
Chainer or Pool are capable of handily defeating a standard DQN agent. Lastly,
Chainer and Pool agents are more robust (more difficult to defeat by a
learning opponent) and have better informed internal activations when compared
to standard DQN agents.

## References

  * Mnih et al. [2013] Volodymyr Mnih, Koray Kavukcuoglu, David Silver, Alex Graves, Ioannis Antonoglou, Daan Wierstra, and Martin Riedmiller.  Playing Atari with Deep Reinforcement Learning.  _arXiv e-prints_ , art. arXiv:1312.5602, December 2013.
  * [2] Sam Witty, Jun Ki Lee, Emma Tosch, Akanksha Atrey, Michael L. Littman, and David D. Jensen.  Measuring and characterizing generalization in deep reinforcement learning.  _Applied AI Letters_.  doi: http://doi.org/10.1002/ail2.45.
  * Packer et al. [2018] Charles Packer, Katelyn Gao, Jernej Kos, Philipp Krähenbühl, Vladlen Koltun, and Dawn Song.  Assessing Generalization in Deep Reinforcement Learning.  _arXiv e-prints_ , art. arXiv:1810.12282, October 2018.
  * Kansky et al. [2017] Ken Kansky, Tom Silver, David A. Mély, Mohamed Eldawy, Miguel Lázaro-Gredilla, Xinghua Lou, Nimrod Dorfman, Szymon Sidor, Scott Phoenix, and Dileep George.  Schema Networks: Zero-shot Transfer with a Generative Causal Model of Intuitive Physics.  _arXiv e-prints_ , art. arXiv:1706.04317, June 2017.
  * Tenney et al. [2019] Ian Tenney, Dipanjan Das, and Ellie Pavlick.  BERT Rediscovers the Classical NLP Pipeline.  _arXiv e-prints_ , art. arXiv:1905.05950, May 2019.


# The Quantum Steering Ellipsoid

Sania Jevtic  Matthew Pusey  David Jennings  Terry Rudolph  Controlled Quantum
Dynamics Theory, Department of Physics, Imperial College London, London SW7
2AZ

###### Abstract

The quantum ellipsoid of a two qubit state is the set of Bloch vectors that
Bob can collapse Alice’s qubit to when he implements all possible measurements
on his qubit. We provide an elementary construction of the ellipsoid for
arbitrary states, and explain how this geometric representation can be made
faithful. The representation provides a range of new results, and uncovers new
features, such as the existence of ‘incomplete steering’ in separable states.
We show that entanglement can be analysed in terms of three geometric features
of the ellipsoid, and prove that a state is separable if and only if it obeys
a ‘nested tetrahedron’ condition. We provide a volume formula for the
ellipsoid in terms of the state density matrix, which identifies exactly when
steering is fully three dimensional, and identify this as a new feature of
correlations, intermediate between discord and entanglement.

###### pacs:

03.65.Ta, 03.67.Mn

The Bloch sphere provides a remarkably simple representation for the state
space of the most primitive quantum unit - the single qubit system - and
results in geometric intuitions that are invaluable in countless fundamental
information-processing scenarios. Unfortunately, such a direct three-
dimensional depiction of the state space is impossible for any quantum system
of larger dimension.

While the qubit is the primitive unit of quantum information, the two qubit
system constitutes the primitive unit for the theory of bipartite quantum
correlations. However, the two qubit state space is already 15-dimensional and
possesses a surprising amount of structure and complexity. As such, it is a
highly non-trivial challenge to both faithfully represent its states and to
acquire natural intuitions for their properties Avron et al. (2007); Horodecki
and Horodecki (1996a); Bengtsson and Życzkowski (2006).

The phenomenon of steering in quantum states was first uncovered by
Schrödinger Schrödinger (1935) (and subsequently rediscovered by others Gisin
(1984); Hughston et al. (1993); Spekkens and Rudolph (2002)), who realised
that measurements on B on the pure entangled state $|\psi\rangle_{AB}$ could
be used to “steer” the state at $A$ into all different convex decompositions
of the reduced state $\rho_{A}$; in particular for rank-one POVM measurements
at $B$ we have that $A$ is steered into ensembles of pure states. More
significantly, given any ensemble decomposition of $\rho_{A}$ into either pure
or mixed states there always exists a POVM measurement at $B$ that generates
that ensemble. In this regard we say that the steering at $A$ for the pure
state $|\psi\rangle_{AB}$ is complete within the Bloch sphere of $A$. More
generally, for a 2-qubit mixed state $\rho_{AB}$ it is known Verstraete (2002)
that the convex set of states that $A$ can be steered to is instead an
ellipsoid ${\cal E}_{A}$ that contains $\rho_{A}$, as in figure 1.

In this Letter we show that all correlation features of a two qubit state are
encoded in its steering ellipsoid and local Bloch vectors, and argue that this
ellipsoid representation for two qubits provides the natural generalization of
the Bloch sphere picture, in that it gives a simple geometric representation
of an arbitrary two qubit state $\rho$ in three dimensions, which makes the
key properties of the state manifest in simple geometric terms.

By adopting the ellipsoid representation we are lead to a range of novel
results and fresh insights into quantum correlations for both entangled and
separable states. Firstly, we provide an alternative construction to the
steering ellipsoid ${\cal E}_{A}$ to that in Verstraete et al. (2001), which
applies even when ${\cal E}_{A}$ is degenerate. We then provide a
reconstruction of a state $\rho_{AB}$ from its geometric data to explore the
faithfulness of the representation. This analysis reveals a new phenomenon of
incomplete steering for certain separable quantum states, in which some
decompositions of $\rho_{A}$ within of the steering ellipsoid ${\cal E}_{A}$
are inaccessible.

Figure 1: Ellipsoid representation of a two-qubit state. For any two-qubit
state $\rho_{AB}$, the set of states to which Bob can steer Alice forms an
ellipsoid ${\cal E}_{A}$ in Alice’s Bloch sphere, containing her Bloch vector
$\mathbf{a}$. The inclusion of Bob’s Bloch vector $\mathbf{b}$ determines
$\rho_{AB}$ up to a choice of basis for Bob, which can be fixed by indicating
the orientation of ${\cal E}_{A}$.

The representation then allows us to decompose entanglement into simple
geometric features, and show how it depends only on (a) the spatial
orientation of the ellipsoid, (b) its distance from the origin and (c) its
size. The representation also leads us to establish the surprising nested
tetrahedron condition: a state $\rho_{AB}$ is separable if and only if its
ellipsoid fits inside a tetrahedron that itself fits inside the Bloch sphere.
We then study the minimal number of product states in the ensemble of a
separable state, which we show is determined solely by the dimension of ${\cal
E}_{A}$. We then develop a useful volume formula that identifies exactly when
steering is fully three-dimensional. Finally the work suggests a new feature
of quantum correlations, called obesity, which is neither discord nor
separability nor entanglement. We observe that quantum discord arises from a
combination of both the obesity of the state and the orientation of its
ellipsoid.

Beyond these new results, we also feel that the method of compactly depicting
any two qubit state $\rho_{AB}$ in three dimensions, via a single ellipsoid
and two local Bloch vectors, should be of interest to a range of researchers
in both the theoretical and experimental quantum sciences.

The Pauli basis. Let $\sigma_{0}$ denote the $2\times 2$ identity matrix and
$\sigma_{i}$, $i=1,2,3$ be the usual Pauli matrices. Any single-qubit
Hermitian operator $E$ can be written
$E=\frac{1}{2}\sum_{\mu=0}^{3}X_{\mu}\sigma_{\mu}$, where the
$X_{\mu}=\operatorname{tr}(E\sigma_{\mu})$ are real. $E$ is positive iff
$X_{0}\geq 0$ and $X_{0}^{2}\geq\sum_{i=0}^{3}X_{i}^{2}$. With $X$ viewed as a
4-vector, this identifies the set of positive operators with the usual forward
light cone in Minkowski space. In the case of a qubit state, $X_{0}=1$ and the
remaining components form the Bloch vector.

Similarly, we can write a two-qubit state as
$\rho=\frac{1}{4}\sum_{\mu,\nu=0}^{3}\Theta_{\mu\nu}\sigma_{\mu}\otimes\sigma_{\nu}$.
The components of the 4 $\times$ 4 real matrix $\Theta$ are
$\Theta_{\mu\nu}=\operatorname{tr}(\rho\sigma_{\mu}\otimes\sigma_{\nu})$. As a
block matrix we have $\Theta=\begin{pmatrix}1&\bm{b}^{T}\\\
\bm{a}&T\end{pmatrix}$, where $\bm{a},\bm{b}$ are the Bloch vectors of the
reduced states $\rho_{A}$ and $\rho_{B}$ of $\rho$ and $T$ is a 3 $\times$ 3
matrix encoding correlations Horodecki and Horodecki (1996a). Now if Bob does
a measurement and obtains a POVM outcome $E$, then Bob steers Alice to the
state proportional to $\operatorname{tr}_{B}(\rho(\sigma_{0}\otimes E))$, with
the latter being given by $\frac{1}{2}\Theta X$ in the Pauli basis.

The matrix $\Theta$ transforms, up to a normalization, under SLOCC operations
$\rho^{\prime}=S_{A}\otimes S_{B}\rho(S_{A}\otimes S_{B})^{\dagger}$, where
$S_{A},S_{B}$ are invertible complex matrices, as
$\Theta^{\prime}=\Lambda_{A}\Theta\Lambda^{T}_{B}$ Verstraete et al. (2001)
where $\Lambda_{A},\Lambda_{B}$ are proper orthochronous Lorentz
transformations (see the appendices).

In the case of a SLOCC operation affecting only Bob
($\Theta^{\prime}=\Theta\Lambda_{B}$), the set of states Bob can steer Alice
to is unaffected: $X$ is in the forward light cone if and only if
$X^{\prime}=\Lambda_{B}X$ is, and $\Theta^{\prime}X=\Theta X^{\prime}$.

Construction of the quantum steering ellipsoid for a given state $\rho$.
Steering ellipsoids have been studied for the representation of SLOCC
transformations Verstraete (2002), and more recently in relation to quantum
discord Shi et al. (2011).

The steering ellipsoid is easiest to understand for states with
$\bm{b}=\bm{0}$. Suppose that Bob projects onto some pure state
$X=\begin{pmatrix}1\\\ \bm{v}\end{pmatrix}$ with $v=1$. Then

$Y=\frac{1}{2}\Theta X=\frac{1}{2}\begin{pmatrix}1&\bm{0}^{T}\\\
\bm{a}&T\end{pmatrix}\begin{pmatrix}1\\\
\bm{v}\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1\\\
\bm{a}+T\bm{v}\end{pmatrix},$ (1)

showing that Bob will obtain that outcome with probability $1/2$ and Alice’s
Bloch vector will then be $\bm{a}+T\bm{v}$. The set of all states Alice can
end up with is simply the unit sphere of possible $\bm{v}$, shrunk and rotated
by $T$ and translated by $\bm{a}$, i.e. an ellipsoid centred at $\bm{a}$
with orientation and semiaxes given by the eigenvectors and eigenvalues of
$TT^{T}$. The dimension of the ellipsoid is
$\mbox{rank}(T)=\mbox{rank}(\Theta)-1$. Points inside the ellipsoid can be
reached via convex combinations of projective measurements, and conversely any
POVM element for Bob can be decomposed into a mixture of projections, thus
giving a point within the ellipsoid.

Now consider a general state with $\bm{b}\neq\bm{0}$. If $b=1$ then $\rho$ is
a product state, and so the steering ellipsoid is a single point. Hence assume
that $b<1$. Then the SLOCC operator
$\mathbbm{1}\otimes\rho_{B}^{-\frac{1}{2}}$ that corresponds to a Lorentz
boost $\Lambda_{B}$ by a ‘velocity’ $\bm{b}$, transforms $\rho_{B}$ to the
maximally mixed state. We refer to this filtered state $\widetilde{\rho}$ as
the canonical state on the SLOCC orbit $\mathcal{S}(\rho)$ (see appendix). As
already noted, SLOCC operations on Bob do not affect Alice’s steering
ellipsoid. Therefore we can find the parameters of an arbitrary state’s
steering ellipsoid by boosting $\Theta$ by $\Lambda_{B}$ and then reading off
the ellipsoid parameters. This gives (see appendix) a steering ellipsoid
centred at $\bm{c}_{A}=\frac{\bm{a}-T\bm{b}}{1-b^{2}}$, and orientation and
semiaxes lengths $s_{i}=\sqrt{q_{i}}$ given by the eigenvectors and
eigenvalues $q_{i}$ of the ellipsoid matrix

$Q_{A}=\frac{1}{1-b^{2}}\left(T-\bm{a}\bm{b}^{T}\right)\left(\mathbbm{1}+\frac{\bm{b}\bm{b}^{T}}{1-b^{2}}\right)\left(T^{T}-\bm{b}\bm{a}^{T}\right).$
(2)

This together with $\bm{c}_{A}$, specify the ellipsoid ${\cal E}_{A}$.

To obtain the ellipsoid at B, $\mathcal{E}_{B}$, the roles of A and B are
reversed, which amounts to a transposition of the matrix $\Theta$. In terms of
its components, this reads as
$\bm{b}\rightarrow\bm{a},\bm{a}\rightarrow\bm{b},T\rightarrow T^{T}$, and thus
${\cal E}_{A}$ and ${\cal E}_{B}$ always have the same dimensionality,
$\mbox{rank}(\Theta)-1$. This completes the construction of the geometric data
$({\cal E}_{A},\bm{a},\bm{b})$ for a given state $\rho$. We next describe the
reverse direction, obtaining $\rho_{AB}$ from an ellipsoid ${\cal E}_{A}$ and
the ectors $\bm{a}$ and $\bm{b}$ Verstraete (2002).

Faithfulness of the representation: the reconstruction of $\rho$ from
geometric data. Since we are given $\bm{a}$ and $\bm{b}$, all that remains is
to obtain the correlation matrix $T$ as a function of
$(Q_{A},\bm{c}_{A},\bm{a},\bm{b})$. We find that the components of the matrix
are given by

$\displaystyle
T_{ij}=(c_{A})_{i}b_{j}+\sum_{k=1}^{3}(\sqrt{Q_{A}}M)_{ik}\operatorname{tr}(\sqrt{\rho_{B}}\sigma_{k}\sqrt{\rho_{B}}\sigma_{j})$
(3)

where $M$ is an orthogonal matrix, obtained from
$M\bm{b}=(\sqrt{Q_{A}})^{-1}(\bm{a}-\bm{c}_{A})$. This specifies $M$ up to an
orthogonal matrix $O_{B}\bm{b}=\bm{b},$ with $O_{B}\in\mathrm{O}(3)$, which
corresponds to the choice of bases at $B$. This choice of basis can be simply
encoded, for example, by providing two contour lines on ${\cal E}_{A}$. The
derivation is provided in the appendices.

A theorem on “complete” and “incomplete” steering. The steering ellipsoid
specifies for which states there is at least one measurement outcome for Bob
that steers Alice to it. A more subtle question is for which decompositions of
Alice’s reduced state $\rho_{A}$ is there a measurement for Bob that steers to
the decomposition. Clearly a necessary condition is that all of the states in
the decomposition must be in $\mathcal{E}_{A}$, surprisingly however it turns
out that this is not sufficient. The situation is captured by the following
result.

Consider some non-product two-qubit state $\Theta$ with ellipsoids
$\mathcal{E}_{A}$ and $\mathcal{E}_{B}$. The following are equivalent:

  1. 1.

(_Complete steering_) For any convex decomposition of $\bm{a}$ into states in
$\mathcal{E}_{A}$ or on its surface, there exists a POVM for Bob that steers
Alice to it.

  2. 2.

Alice’s Bloch vector $\bm{a}$ lies on the surface of $\mathcal{E}_{A}$ scaled
down by $b$.

  3. 3.

The affine span of $\mathcal{E}_{B}$ contains the maximally mixed state.

The proof is in the appendices. In particular, these conditions hold for all
non-degenerate ellipsoids (which includes all entangled states) as well as all
states where $\bm{b}=\bm{0}$. An example of a state where the above conditions
fail is the state $\rho=\frac{1}{2}\left(|00\rangle\langle
00|+|1+\rangle\langle 1+|\right)$.

The three geometric contributions to entanglement. The Peres-Horodecki
criterion Peres (1996); Horodecki et al. (1996) asserts that a two-qubit state
$\rho_{e}$ is entangled if and only if $\rho_{e}^{T_{B}}$ has a negative
eigenvalue. It was shown in Verstraete et al. (2001) that at most one
eigenvalue of $\rho_{e}^{T_{B}}$ can be negative, furthermore in Sanpera et
al. (1998) it is demonstrated that $\rho_{e}^{T_{B}}$ is full rank for all
entangled states. It follows that $\det\rho_{e}^{T_{B}}<0$ is a necessary and
sufficient condition for entanglement.

Dim. |  Type |  Alice’s ellipsoid |  Bob’s ellipsoid
---|---|---|---
3  (Obese) |  Entangled |  |
3  (Obese) |  Separable |  |
2  (Pancake) |  Incomplete |  |
2  (Pancake) |  Complete |  |
1  (Needle) |  Incomplete |  |
Table 1: The classes of steering ellipsoids. The green dot is the reduced
state. States with ${\cal E}_{A}$ being 3-dimensional have obesity and can be
either entangled or separable. The state $\rho$ is separable if and only if
${\cal E}_{A}$ fits inside a tetrahedron inside the Bloch sphere of $A$. For
separable states, the set ${\cal E}_{A}$ can also be 2-dimensional (a steering
pancake), or 1-dimensional (a steering needle), or trivially 0-dimensional
(not shown). For these cases, steering is either “complete” if all ensemble
decompositions of $\bm{a}$ in ${\cal E}_{A}$ are attainable (when the span of
${\cal E}_{B}$ contains $\frac{1}{2}\mathbbm{1}$), otherwise the steering is
“incomplete”. Zero discord occurs only for radial steering needles.

Suppose $\rho$ is entangled, then any state in its SLOCC orbit
$\mathcal{S}(\rho)$ is also entangled Verstraete et al. (2001) and in
particular, the canonical state $\tilde{\rho}\in\mathcal{S}(\rho)$. Therefore,
we can use $\det(\rho^{T_{B}})<0\Leftrightarrow\det(\tilde{\rho}^{T_{B}})<0$
to show that a quantum steering ellipsoid corresponds to an entangled state if
and only if

$c^{4}+c^{2}(1-\operatorname{tr}(Q)+\hat{\bm{n}}^{T}Q\hat{\bm{n}})+h(Q)<0$ (4)

where $\bm{c}=c\hat{\bm{n}}$ and
$h(Q)=1-8\det\sqrt{Q}+2\operatorname{tr}(Q^{2})-(\operatorname{tr}(Q))^{2}-2\operatorname{tr}(Q)$.
We have dropped the $A,B$ labels for the centre $\bm{c}$ and $Q$ because
entanglement is a “symmetric” relation. This equation is manifestly invariant
under global rotations of the ellipsoid, corresponding to local unitaries on
the quantum state. It elucidates that entanglement correlations between the
qubits manifest themselves in three geometric properties: (1) the distance of
${\cal E}_{A}$ from the origin, (2) The dimensions of ${\cal E}_{A}$, given by
singular values of $Q$ and (3) the “skew” of ${\cal E}_{A}$, captured by the
term $\hat{\bm{n}}^{T}Q\hat{\bm{n}}$, which depends on the alignment of the
ellipsoid relative to the radial direction.

The nested tetrahedron condition. The condition for entanglement given by
equation (4), provides a compact algebraic condition for non-separability and
uncovers contributions from different geometric aspects. However, the steering
setting allows us to go further, and capture the distinction between separable
and non-separable states in a remarkably elegant and intuitive way. The result
is an single, geometric criterion for separability:

> A two qubit state $\rho$ is separable if and only if its steering ellipsoid
> ${\cal E}_{A}$ fits inside a tetrahedron that fits inside the Bloch sphere.

To prove necessity, suppose Alice and Bob share a separable state
$\rho=\sum_{i=1}^{n}p_{i}\alpha_{i}\otimes\beta_{i}$. Since we can always take
$n\leq 4$ Wootters (1998), the Bloch vectors of the $\alpha_{i}$ define a
(possibly degenerate) tetrahedron $\mathcal{T}$ within Alice’s Bloch sphere.
If Bob measures an outcome $E$ then Alice is collapsed to
$\sum_{i=1}^{n}\frac{p_{i}\operatorname{tr}(E\alpha_{i})}{\operatorname{tr}(E\rho_{B})}\alpha_{i}$.
Hence her new Bloch vector will be a convex combination of the Bloch vectors
for the $\alpha_{i}$ — in other words her steering ellipsoid is contained in
$\mathcal{T}$.

We prove in the appendix that the non-trivial converse holds: any ellipsoid
that fits inside a tetrahedron that itself fits inside the Bloch sphere must
arise from a separable state, and thus the nested tetrahedron condition is
both necessary and sufficient for separability of the state.

Building further on this geometric understanding, we can prove the following
result. Let $\rho$ be separable and $n$ minimal such that
$\rho=\sum_{i=1}^{n}p_{i}\alpha_{i}\otimes\beta_{i}$. Then
$n=\mbox{rank}(\Theta)$. Proof: since the $\Theta$ matrix for a product state
has rank 1, it is clear that $n\geq\mbox{rank}(\Theta)$. Since any separable
state can be written using 4 (pure) product states Wootters (1998) we have
$n\leq 4$. If $\mbox{rank}(\Theta)=1$ then $\rho$ is a product state and so
$n=1$. If $\mbox{rank}(\Theta)=2$ then ${\cal E}_{A}$ is a line segment and we
can form a decomposition of $\rho$ using the endpoints of its steering
ellipsoid, giving $n=2$. The case $\mbox{rank}(\Theta)=3$ (where ${\cal
E}_{A}$ is an ellipse) is non-trivial, but is solved in the appendix by using
the fact that any ellipse inside a tetrahedron inside the unit sphere also
fits inside a triangle inside the unit sphere Vidrighin (2012).

Volume of the ellipsoid. We can now arrive at a compact and useful expression
for the volume of the steering ellipsoids, which allows a simple inference as
to the type of steering a given state $\rho$ provides. The volume of any
ellipsoid is proportional to the product of its semiaxes
$V=\frac{4\pi}{3}s_{1}s_{2}s_{3}$. Therefore the $\mathcal{E}_{A}$ has volume
$V_{A}=\frac{4\pi}{3}|\sqrt{\det Q_{A}}|$, which may be rewritten as
$V_{A}=\frac{4\pi}{3}\frac{\left|\mathrm{det}\Theta\right|}{(1-b^{2})^{2}}$.
However, it turns out (see appendix) that
$|\mathrm{det}\Theta|=16|\mathrm{det}\rho-\mathrm{det}\rho^{T_{B}}|$, and so
we obtain that

$V_{A}=\frac{64\pi}{3}\frac{\left|\det\rho-\det\rho^{T_{B}}\right|}{\left(1-b^{2}\right)^{2}}.$
(5)

The $\mathcal{E}_{B}$ volume can be calculated from $V_{A}$ via the simple
relation $V_{B}=\frac{(1-b^{2})^{2}}{(1-a^{2})^{2}}V_{A}$, and thus if one
side has non-zero volume then so too must the other. This volume is a witness
of entanglement, in the sense any state with $V_{A}>V_{\star}=4\pi/81$ must be
entangled. This can be seen either from the nested tetrahedron theorem in the
earlier section or from equation (4). With the former one simply constructs
the largest tetrahedron possible in the Bloch sphere, while for the latter one
observes that $c$ must vanish, and then maximizes subject to $h(Q)=0$. In both
cases one obtains $V_{\star}$, namely the volume of the Werner state on the
separable-entangled boundary.

Quantum discord and the ellipsoid. Quantum discord has received much attention
as a measure of the quantumness of correlations (see Ollivier and Zurek (2001)
for details) in which zero discord for one party roughly corresponds to them
possessing a non-disturbing projective measurement. In the appendix, we show
that $\rho$ has zero discord for $A$ if and only if ${\cal E}_{A}$ is a
segment of a diameter, while $\rho$ has zero discord for $B$ if and only if
${\cal E}_{A}$ is one-dimensional and $b=\frac{2|\bm{c}_{A}-\bm{a}|}{l_{A}}$,
where $l_{A}$ is the length of ${\cal E}_{A}$.

A new perspective on quantum correlations: obesity. Consider a game in which
Bob must convince Alice that the qubits they each hold possess more-than-
classical correlations: he succeeds in his task if he can steer Alice to 3
states with linearly independent Bloch vectors. A state that is a resource for
this game is called “obese”: it has an ellipsoid with finite volume. Obesity
is neither the same as entanglement nor separability nor discord. All
entangled states have $\det\rho^{T_{B}}<0$ and so it follows from (65) that
all entangled states have non-zero volume and hence are obese. Equation (65)
also implies that the set of separable states divides into obese and non-obese
(skinny) states. All zero discord states are skinny, however one can have
${\cal E}_{A}$ being one-dimensional (steering needles) but the state having
positive discord due to ${\cal E}_{A}$ not being radially aligned. In this
sense, quantum discord can be viewed as arising from a combination of an
obesity component and a skew contribution.

It turns out that $V_{A}\propto|\mathrm{det}(T-\bm{a}\bm{b}^{T})|$, and so an
obese state must have $T-\bm{a}\bm{b}^{T}$ being full rank. Since
$\bm{a}\bm{b}^{T}$ is rank-one we must have $\mathrm{rank}(T)\geq 2$, however
we cannot simply reduce the condition of non-zero obesity to full rank $T$,
since there exist states with full rank $T$ that have degenerate steering
ellipsoids (ellipses) and so have zero volume, for example
$\bm{a}=\bm{b}=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)^{T}$ and
$T=\mbox{diag}\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. However, if
we replace this $T$ with $T=\mbox{diag}\left(\frac{1}{3},\frac{1}{3},0\right)$
we obtain an obese state with $\mathrm{rank}(T)=2$.

To contrast entanglement, discord and obesity, we can consider a family of
canonical states
$\rho(\theta)=\frac{1}{4}\left(\mathbbm{1}+\frac{1}{2}\sigma_{z}\otimes\mathbbm{1}+\sum_{ij}T_{ij}(\theta)\sigma_{i}\otimes\sigma_{j}\right)$
in which the skew of the ellipsoid smoothly varies via the elements of
$T(\theta)=R_{y}(\theta)TR^{T}_{y}(\theta)$, with $R_{y}(\theta)$ being a
rotation about the $y$ axis by $\theta\in[0,\pi)$ and
$T=\mbox{diag}(-\frac{9}{20},-\frac{3}{10},-\frac{3}{10})$. The ellipsoids
have equal volumes and are centred at $\bm{c}_{A}=(0,0,\frac{1}{2})^{T}$. This
family of states illustrate an opposing behavior of the discord and
concurrence as a function of skew, see Fig. 2. The entanglement favors an
orientation in which the longest ellipsoid semiaxis is aligned radially, while
discord is maximised when this axis is orthogonal to the radial direction
111Discord can be calculated analytically when $\theta=0,\frac{\pi}{2}$, and
at these points $\rho$ is an X-state Mingjun Shi and Du (2011) and this agrees
with our findings..

Conclusion. The quantum steering ellipsoid provides a faithful representation
of any two qubit state and a natural geometric classification of states (as in
Table 1). It provides clear and intuitive understanding into the usual key
aspects of two qubit states, uncovers surprising new features (such as the
nested tetrahedron condition, skew & obesity, and incomplete steering) while
prompting novel questions, such as: can we use (4) to define a class of
“least-classical” separable states for fixed $(\bm{a},\bm{b},\bm{c})$? Can we
use the nested tetrahedron condition to provide a simple construction for the
best separable approximation Lewenstein and Sanpera (1998) for a state $\rho$?
What is the characterisation of steering ellipsoids for real states? This
would be particularly useful experimentally, acting as a new type of
tomography for two qubits.

Figure 2: Discord (solid) and concurrence (dotted) of the states
$\rho(\theta)$ as a function of the orientation of the ellipsoid. Entanglement
is maximized when the major axis is radial.

###### Acknowledgements.

We wish to acknowledge Zuzana Gavorova and Peter Lewis for their early
contributions to this topic. This work was supported by the EPSRC. DJ is
funded by the Royal Commission for the Exhibition of 1851.

This following appendices contain the details for some of the statements and
theorems that are presented in the main article.

##  Appendix A The steering ellipsoid for a general two qubit state

Previous approaches to representing two qubit states have included
partitioning the set of all two qubit states into SLOCC (Stochastic Local
Operations and Classical Communication) equivalence classes Avron et al.
(2007), which results in a three dimensional representation of a state $\rho$
through its SLOCC orbit, defined as

$\mathcal{S}(\rho):=\\{\frac{S_{A}\otimes S_{B}\rho(S_{A}\otimes
S_{B})^{{\dagger}}}{\operatorname{tr}(S_{A}\otimes S_{B}\rho(S_{A}\otimes
S_{B})^{{\dagger}})}:\,S_{A},S_{B}\in\mathrm{GL}(2,\mathbb{C})\\}$

with $\mathrm{GL}(2,\mathbb{C})$ the group of invertible, complex 2 $\times$ 2
matrices. However replacing $\rho$ with its SLOCC orbit is far from faithful,
and amounts to a highly coarse-grained representation of the state that erases
much of its detail. Another approach is to start with a Pauli basis expansion
of $\rho$, which can be converted via a state-dependent choice of local
unitaries on both qubits, to a representation involving three spatial vectors
Horodecki and Horodecki (1996b). Again, this is still not a faithful
representation, and more importantly it is extremely difficult to develop any
intuition for what the vector representing correlations actually means.

In this section we provide the details for constructing the steering ellipsoid
representation of an arbitrary two qubit quantum state.

Consider a two-qubit state

$\Theta=\begin{pmatrix}1&\mathbf{b}^{T}\\\ \mathbf{a}&T\end{pmatrix}.$ (6)

The matrix $\Theta$ transforms, up to a normalization, under SLOCC operations
$\rho^{\prime}=S_{A}\otimes S_{B}\rho(S_{A}\otimes S_{B})^{\dagger}$ as
$\Theta^{\prime}=\Lambda_{A}\Theta\Lambda^{T}_{B}$ Verstraete et al. (2001)
where $\Lambda_{A},\Lambda_{B}$ are proper orthochronous Lorentz
transformations given by $\Lambda_{W}=\Upsilon S_{W}\otimes
S_{W}^{*}\Upsilon^{\dagger}/|\det S_{W}|,$ $W\in\\{A,B\\}$ and

$\Upsilon=\frac{1}{\sqrt{2}}\begin{pmatrix}1&0&0&1\\\ 0&1&1&0\\\ 0&i&-i&0\\\
1&0&0&-1\end{pmatrix}.$ (7)

In particular, local unitaries rotate the Pauli basis: $\bm{a}\rightarrow
O_{A}\bm{a}$, $\bm{b}\rightarrow O_{B}\bm{b}$, $T\rightarrow O_{A}TO_{B}^{T}$,
where $O_{A},O_{B}\in\mathrm{SO}(3)$.

If $b=1$ then $\Theta$ must be a product state and no steering can occur, so
assume otherwise. Define $\gamma=1/\sqrt{1-b^{2}}$, and a new state

$\Theta^{\prime}=\gamma\Theta L_{\mathbf{b}},$ (8)

where $L_{\mathbf{b}}$ is a Lorentz boost by $\mathbf{b}$ Frosch (2006):

$L_{\mathbf{b}}=\begin{pmatrix}\gamma&-\gamma\mathbf{b}^{T}\\\
-\gamma\mathbf{b}&\mathbbm{1}+\frac{\gamma-1}{b^{2}}\mathbf{b}\mathbf{b}^{T}\end{pmatrix}.$
(9)

The $\gamma$ in (8) ensures that $\Theta^{\prime}$ is normalized: the top-left
element is

$\gamma\begin{pmatrix}1&\mathbf{b}^{T}\end{pmatrix}\begin{pmatrix}\gamma\\\
-\gamma\mathbf{b}\end{pmatrix}=\gamma^{2}(1-b^{2})=1.$ (10)

The boost means Bob’s reduced state is maximally mixed: the top-right block of
$\Theta^{\prime}$ is

$\gamma\begin{pmatrix}1&\mathbf{b}^{T}\end{pmatrix}\begin{pmatrix}-\gamma\mathbf{b}^{T}\\\
\mathbbm{1}+\frac{\gamma-1}{b^{2}}\mathbf{b}\mathbf{b}^{T}\end{pmatrix}=\\\
\gamma(-\gamma\mathbf{b}^{T}+\mathbf{b}^{T}+(\gamma-1)\mathbf{b}^{T})=\mathbf{0}^{T}.$
(11)

By the above two observations we can write

$\Theta^{\prime}=\begin{pmatrix}1&\mathbf{0}^{T}\\\
\mathbf{a}^{\prime}&T^{\prime}\end{pmatrix}.$ (12)

The set of states Bob can steer Alice to will be exactly the same for $\Theta$
and $\Theta^{\prime}$: $Y=\frac{1}{2}\Theta X\iff
Y=\frac{1}{2}\Theta^{\prime}X^{\prime}$ where
$X^{\prime}=L_{-\mathbf{b}}X/\gamma$ and $X$ corresponds to a positive
operator iff $X^{\prime}$ does because $L_{-\mathbf{b}}/\gamma$ preserves the
forward light cone.

What is that set of states? Writing $X=\begin{pmatrix}t\\\
\mathbf{x}\end{pmatrix}$ we see that without loss of generality we can take
$t=1$ since the effect of multiplying $X$ by a positive number is undone when
normalizing $Y$. Hence the positivity condition becomes $x\leq 1$. If $x<1$
then we could write $X$ as a convex combination of ones with $x=1$, so we
restrict attention to the latter case. So any state $\mathbf{y}$ that Alice
can steer Bob to is a convex combination of states of the form
$\mathbf{y}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}$ with $x=1$. But this
defines a linear image of the unit sphere displaced by $\mathbf{a}^{\prime}$,
or in other words an ellipsoid centred at $\mathbf{a}^{\prime}$.

Consider the singular value decomposition $T^{\prime}=O_{1}DO_{2}$. $O_{2}$
simply rotates the unit sphere and so can be ignored. $D$ stretches the sphere
and thus gives the lengths of the semi-axes of the resulting ellipsoid.
$O_{1}$ rotates the semi-axes. We have $T^{\prime}T^{\prime
T}=O_{1}D^{2}O_{1}^{T}$ and so the lengths of the semi-axes can also be found
by square rooting the eigenvalues of $T^{\prime}T^{\prime T}$ whilst their
directions can be found from its eigenvectors.

Combining (8) with (12) we find

$\mathbf{a}^{\prime}=\gamma\begin{pmatrix}\mathbf{a}&T\end{pmatrix}\begin{pmatrix}\gamma\\\
-\gamma\mathbf{b}\end{pmatrix}=\gamma^{2}(\mathbf{a}-T\mathbf{b}),$ (13)
$\displaystyle T^{\prime}$ $\displaystyle=$
$\displaystyle\gamma\begin{pmatrix}\mathbf{a}&T\end{pmatrix}\begin{pmatrix}-\gamma\mathbf{b}^{T}\mathbbm{1}+\frac{\gamma-1}{b^{2}}\mathbf{b}\mathbf{b}^{T}\end{pmatrix}$
(14) $\displaystyle=$
$\displaystyle\gamma\left(-\gamma\mathbf{a}\mathbf{b}^{T}+T+\frac{\gamma-1}{b^{2}}T\mathbf{b}\mathbf{b}^{T}\right).$
(15)

And so, after some algebra,

$\displaystyle T^{\prime}T^{\prime
T}=\gamma^{2}(TT^{T}-\mathbf{a}\mathbf{a}^{T})+\mathbf{a}^{\prime}\mathbf{a}^{\prime
T}.$ (16)

Using
$-\mathbf{a}\mathbf{b}^{T}\left(\mathbbm{1}+\frac{\gamma-1}{b^{2}}\mathbf{b}\mathbf{b}^{T}\right)=-\gamma\mathbf{a}\mathbf{b}^{T}$
we can also write

$T^{\prime}=\gamma\left(T-\mathbf{a}\mathbf{b}^{T}\right)\left(\mathbbm{1}+\frac{\gamma-1}{b^{2}}\mathbf{b}\mathbf{b}^{T}\right).$
(17)

leading to the form in the main text:

$T^{\prime}T^{\prime
T}=\gamma^{2}(T-\mathbf{a}\mathbf{b}^{T})(\mathbbm{1}+\gamma^{2}\mathbf{b}\mathbf{b}^{T})(T^{T}-\mathbf{b}\mathbf{a}^{T})=:Q_{A}$
(18)

Since $\gamma L_{\mathbf{b}}$ is invertible we have
$\mbox{rank}(\Theta^{\prime})=\mbox{rank}(\Theta)$. By counting linearly
independent columns in (12) we have
$\mbox{rank}(\Theta^{\prime})=\mbox{rank}(T^{\prime})+1$ thus proving that the
dimension of the ellipsoid is $\mbox{rank}(\Theta)-1$.

##  Appendix B Proof of the complete steering theorem

We present an extended formulation of the theorem:

Theorem Consider some non-product two-qubit state
$\Theta=\begin{pmatrix}1&\mathbf{b}^{T}\\\ \mathbf{a}&T\end{pmatrix}$ with
ellipsoids $\mathcal{E}_{A}$ and $\mathcal{E}_{B}$. The following are
equivalent:

  1. 1.

Complete steering: for all convex decompositions of $\mathbf{a}$ into states
in $\mathcal{E}_{A}$, there exists a POVM for Bob that steers Alice to it.

  2. 2.

Any surface steering: there exists a convex decomposition of $\mathbf{a}$ into
states on the surface of $\mathcal{E}_{A}$ with a POVM for Bob that steers
Alice to it.

  3. 3.

Alice’s Bloch vector lies on the surface of her ellispoid scaled down by $b$.

  4. 4.

The affine span of $\mathcal{E}_{B}$ contains the maximally mixed state.

  5. 5.

$\begin{pmatrix}1\\\ \mathbf{0}\end{pmatrix}\in\mathrm{range}(\Theta^{T})$.

  6. 6.

$\begin{pmatrix}1\\\ \mathbf{0}\end{pmatrix}\in\ker(\Theta)^{\perp}$.

###### Proof.

Preliminaries: Let $\gamma=1/\sqrt{1-b^{2}}$ and $\Theta^{\prime}=\gamma\Theta
L_{\mathbf{b}}$. Then $\Theta=\Theta^{\prime}\frac{L_{-\mathbf{b}}}{\gamma}$
and so 6 is equivalent to $\begin{pmatrix}1\\\
\mathbf{b}\end{pmatrix}\in\ker(\Theta^{\prime})^{\perp}$. If we write
$\Theta^{\prime}=\begin{pmatrix}1&\mathbf{0}^{T}\\\
\mathbf{a}^{\prime}&T^{\prime}\end{pmatrix}$ then we see that any vector in
$\ker(\Theta^{\prime})$ is of the form $\begin{pmatrix}0\\\
\mathbf{x}\end{pmatrix}$ with $T^{\prime}\mathbf{x}=\mathbf{0}$ and so 6 is
equivalent to $\mathbf{b}\in\ker(T^{\prime})^{\perp}$.

$\mathcal{E}_{A}$ are the points that can be written
$\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}^{\prime}$ where $x^{\prime}\leq 1$.
Therefore the surface of $\mathcal{E}_{A}$ are the points that can be written
$\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}^{\prime}$ where $x^{\prime}=1$ and
$\mathbf{x}^{\prime}\in\ker(T^{\prime})^{\perp}$. Hence the scaled down
surface is $\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}^{\prime}$ where
$x^{\prime}=b$ and $\mathbf{x}^{\prime}\in\ker(T^{\prime})^{\perp}$.

From $\Theta=\Theta^{\prime}\frac{L_{-\mathbf{b}}}{\gamma}$ we calculate that
$\mathbf{a}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{b}$.

1 $\implies$ 2: Trivial.

2 $\implies$ 6: Let $\mathbf{y}_{i}$ on the surface of $\mathcal{E}_{A}$ form
a convex decomposition $\sum_{i}p_{i}\mathbf{y}_{i}=\mathbf{a}$. Since they
are on the surface, we have
$\mathbf{y}_{i}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}_{i}^{\prime}$ where
$x_{i}^{\prime}=1$ and $\mathbf{x}_{i}^{\prime}\in\ker(T^{\prime})^{\perp}$.
Suppose we also have
$\mathbf{y}_{i}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}_{i}^{\prime\prime}$
with $x_{i}^{\prime\prime}\leq 1$. Then
$\mathbf{x}_{i}^{\prime}-\mathbf{x}_{i}^{\prime\prime}\in\ker(T^{\prime})$,
and the only way that the difference between two vectors can be perpendicular
to the longer one is if they are equal. Therefore $2p_{i}\begin{pmatrix}1\\\
\mathbf{x}^{\prime}_{i}\end{pmatrix}$ is the unique element of the forward
light cone that $\frac{1}{2}\Theta^{\prime}$ maps to $p_{i}\begin{pmatrix}1\\\
\mathbf{y}_{i}\end{pmatrix}$, and therefore $\gamma L_{\mathbf{b}}$ times
these form the only possible POVM elements for Bob. But to be a valid POVM,
they must sum to the identity $\begin{pmatrix}2\\\ \mathbf{0}\end{pmatrix}$,
i.e. $\sum_{i}2p_{i}\begin{pmatrix}1\\\
\mathbf{x}^{\prime}_{i}\end{pmatrix}=\frac{L_{-\mathbf{b}}}{\gamma}\begin{pmatrix}2\\\
\mathbf{0}\end{pmatrix}=2\begin{pmatrix}1\\\ \mathbf{b}\end{pmatrix}$. Since
the $\mathbf{x}_{i}^{\prime}\in\ker(T^{\prime})^{\perp}$, this implies
$\mathbf{b}\in\ker(T^{\prime})^{\perp}$ which is equivalent to 6.

6 $\implies$ 1: Let $\mathbf{y}_{i}\in\mathcal{E}_{A}$ form a convex
decomposition $\sum_{i}p_{i}\mathbf{y}_{i}=\mathbf{a}$. Since
$\mathbf{y}_{i}\in\mathcal{E}_{A}$ we have
$\mathbf{y}_{i}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}_{i}^{\prime}$ where
$x_{i}^{\prime}\leq 1$. Write
$\mathbf{x}_{i}^{\prime}=\mathbf{k}_{i}+\mathbf{c}_{i}$ where
$\mathbf{k}_{i}\in\ker(T^{\prime})$ and
$\mathbf{c}_{i}\in\ker(T^{\prime})^{\perp}$. This implies $c_{i}\leq
x_{i}^{\prime}\leq 1$ and
$\mathbf{y}_{i}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{c}_{i}$. So
$2p_{i}\begin{pmatrix}1\\\ \mathbf{c}_{i}\end{pmatrix}$ are in the forward
light cone and map to $p_{i}\begin{pmatrix}1\\\ \mathbf{y}_{i}\end{pmatrix}$
under $\frac{1}{2}\Theta^{\prime}$. Hence $\gamma L_{\mathbf{b}}$ times these
are in the forward line cone and map to $p_{i}\begin{pmatrix}1\\\
\mathbf{y}_{i}\end{pmatrix}$ under $\frac{1}{2}\Theta$. Since
$\sum_{i}p_{i}\mathbf{y}_{i}=\mathbf{a}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{b}$
we have $T^{\prime}\sum_{i}p_{i}\mathbf{c}_{i}=T^{\prime}\mathbf{b}$. By
construction $\mathbf{c}_{i}\in\ker(T^{\prime})^{\perp}$ and by assumption
$\mathbf{b}\in\ker(T^{\prime})^{\perp}$, and so this implies
$\sum_{i}p_{i}\mathbf{c}_{i}=\mathbf{b}$. Then $\sum_{i}\gamma
L_{\mathbf{b}}2p_{i}\begin{pmatrix}1\\\ \mathbf{c}_{i}\end{pmatrix}=2\gamma
L_{\mathbf{b}}\begin{pmatrix}1\\\ \mathbf{b}\end{pmatrix}=\begin{pmatrix}2\\\
\mathbf{0}\end{pmatrix}$ so we have a valid POVM.

6 $\implies$ 3: Immediate from form of scaled down $\mathcal{E}_{A}$ and
$\mathbf{a}$ in preliminaries.

3 $\implies$ 6: If $\mathbf{a}$ is on the scaled down surface then
$\mathbf{a}^{\prime}+T^{\prime}\mathbf{x}^{\prime}=\mathbf{a}^{\prime}+T^{\prime}\mathbf{b}$
where $x^{\prime}=b$ and $\mathbf{x}^{\prime}\in\ker(T^{\prime})^{\perp}$.
Hence $\mathbf{x}^{\prime}-\mathbf{b}\in\ker(T^{\prime})$. The only way the
difference between two vectors of the same length can be perpendicular to one
of them is if they are the same, and so 6 follows.

4 $\implies$ 5: Suppose $\sum q_{i}\mathbf{x}_{i}=0$, $\sum q_{i}=1$ with
$\mathbf{x}_{i}\in\mathcal{E}_{B}$. Recalling that swapping parties sends
$\Theta\to\Theta^{T}$ we see that there exists $Y_{i}$ in the forward light-
cone with $\begin{pmatrix}1\\\
\mathbf{x}_{i}\end{pmatrix}=\frac{1}{2}\Theta^{T}Y_{i}$. But then
$\Theta^{T}\frac{1}{2}\sum_{i}q_{i}Y_{i}=\sum_{i}q_{i}\frac{1}{2}\Theta^{T}Y_{i}=\sum_{i}q_{i}\begin{pmatrix}1\\\
\mathbf{x}_{i}\end{pmatrix}=\begin{pmatrix}1\\\ \mathbf{0}\end{pmatrix}$.

5 $\implies$ 4: Suppose there exists $Y$ with
$\frac{1}{2}\Theta^{T}Y=\begin{pmatrix}1\\\ \mathbf{0}\end{pmatrix}$. If
$Y=\begin{pmatrix}t\\\ \mathbf{y}\end{pmatrix}$ is in the forward light cone
(i.e. $t\geq y$) then $\mathcal{E}_{B}$ itself contains the maximally mixed
state and we are done. Otherwise, notice that $Y_{1}=\begin{pmatrix}y-t\\\
0\end{pmatrix}$ and $Y_{2}=\begin{pmatrix}y\\\ \mathbf{y}\end{pmatrix}$ are in
the forward light cone. Writing $\frac{1}{2}\Theta^{T}Y_{i}$ as
$q_{i}\begin{pmatrix}1\\\ \mathbf{x}_{i}\end{pmatrix}$ we have
$\mathbf{x}_{i}\in\mathcal{E}_{B}$. Noting that $Y=\sum_{i}Y_{i}$ we have
$\sum_{i}\frac{1}{2}\Theta^{T}Y_{i}=\begin{pmatrix}1\\\
\mathbf{0}\end{pmatrix}$, in other words $\sum_{i}q_{i}=1$ and
$\sum_{i}q_{i}\mathbf{x}_{i}=\mathbf{0}$.

5 $\iff$ 6: $\mbox{range}(A^{T})=\ker(A)^{\perp}$ is a theorem of linear
algebra, which follows straightforwardly from the singular value
decomposition. ∎

##  Appendix C Reconstructing the state from its steering ellipsoid and the
Bloch vectors

Alice’s steering ellipsoid $\mathcal{E}_{A}$ is described by matrix $Q_{A}$
and centre $\mathbf{c}_{A}$. One quantum state that corresponds to it is the
“canonical” state $\tilde{\rho}$ with Bloch vectors
$\tilde{\mathbf{b}}=\mathbf{0}$, $\tilde{\mathbf{a}}=\mathbf{c}_{A}$ and
T-matrix $\tilde{T}=\sqrt{Q_{A}}O$, with $O\in\mathrm{O}(3)$ (since
$Q_{A}=\tilde{T}\tilde{T}^{T}$). Written explicitly it is

$\displaystyle\tilde{\rho}=\frac{1}{4}(\mathbbm{1}+\bm{c}_{A}\cdot\sigma\otimes\mathbbm{1}+\sum_{i,j=1}^{3}(\sqrt{Q_{A}}O)_{ij}\sigma_{i}\otimes\sigma_{j})$
(19)

The orthogonal matrix $O$ is not fixed.

If in addition to $Q_{A},\bm{c}_{A}$ we are also given the Bloch vectors
$\bm{a},\bm{b}$, a corresponding state is

$\displaystyle\rho=\mathbbm{1}\otimes\sqrt{2\rho_{B}}\,\tilde{\rho}\,\mathbbm{1}\otimes\sqrt{2\rho_{B}}$
(20)

where

$\rho_{B}=\frac{1}{2}(\mathbbm{1}+\bm{b}\cdot\bm{\sigma})$ (21)

This is because $\sqrt{2\rho_{B}}$ is a SLOCC operator for
$\mathbf{b}\neq\mathbf{0}$ and Alice’s ellipsoid is invariant under such local
operations on Bob’s qubit.

Note that

$\operatorname{tr}(\rho\mathbbm{1}\otimes\bm{\sigma})=\bm{b},$ (22)

where $\bm{\sigma}$ is a vector of Paulis, as required. However

$\operatorname{tr}(\rho\bm{\sigma}\otimes\mathbbm{1})=\bm{c}_{A}+\sqrt{Q_{A}}O\bm{b}$
(23)

and the right hand side of this should be $\bm{a}$. This constrains the
orthogonal matrix $O$ as a solution to

$O\bm{b}=(\sqrt{Q_{A}})^{-1}(\bm{a}-\bm{c}_{A})$ (24)

We can split up the orthogonal matrix into two orthogonal rotations: $O=MB$
where $M\in\mathrm{SO}(3)$ (properly) rotates $\bm{b}$ to
$(\sqrt{Q_{A}})^{-1}(\bm{b}-\bm{c}_{A})$ about a unit vector perpendicular to
the plane they lie in, and $B\in\mathrm{O}(3)$ rotates about $\bm{b}$, that is
$B\bm{b}=\bm{b}$. The equation (24) only determines $M$.

Define the unitary matrix $U_{B}$ as the one corresponding to $B$, it has the
property $[U_{B},\rho_{B}]=0$. Then we may write

$\tilde{\rho}=\mathbbm{1}\otimes U_{B}\chi\mathbbm{1}\otimes U^{\dagger}_{B}$
(25)

where

$\displaystyle\chi$ $\displaystyle=$ $\displaystyle\mathbbm{1}\otimes
U^{\dagger}_{B}\tilde{\rho}\mathbbm{1}\otimes U_{B}$ (26) $\displaystyle=$
$\displaystyle\frac{1}{4}(\mathbbm{1}+\bm{c}_{A}\cdot\sigma\otimes\mathbbm{1}+\sum_{i,j=1}^{3}(\sqrt{Q_{A}}M)_{ij}\sigma_{i}\otimes\sigma_{j})$
(27)

Thus

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle\mathbbm{1}\otimes\sqrt{2\rho_{B}}U_{B}\,\chi\,\mathbbm{1}\otimes
U^{\dagger}_{B}\sqrt{2\rho_{B}}$ (28) $\displaystyle=$
$\displaystyle\mathbbm{1}\otimes
U_{B}\sqrt{2\rho_{B}}\,\chi\,\mathbbm{1}\otimes\sqrt{2\rho_{B}}U^{\dagger}_{B}$
(29)

since if $[U_{B},\rho_{B}]=0$ then $[U_{B},\sqrt{\rho_{B}}]=0$.

Therefore the state $\rho$ compatible with $Q_{A},\bm{c}_{A},\bm{a},\bm{b}$ is

$\displaystyle\rho=\frac{1}{4}\left(\mathbbm{1}+\bm{a}.\bm{\sigma}\otimes\mathbbm{1}+\mathbbm{1}\otimes\bm{b}.\bm{\sigma}+\sum_{i,j=1}^{3}T_{ij}\sigma_{i}\otimes\sigma_{j}\right)$
(30)

where $T=RB$ and

$\displaystyle R_{ij}$ $\displaystyle=$
$\displaystyle\operatorname{tr}(\mathbbm{1}\otimes\sqrt{2\rho_{B}}\,\chi\,\mathbbm{1}\otimes\sqrt{2\rho_{B}}\sigma_{i}\otimes\sigma_{j})$
(31) $\displaystyle=$
$\displaystyle(c_{A})_{i}b_{j}+\sum_{k=1}^{3}(\sqrt{Q_{A}}M)_{ik}\operatorname{tr}(\sqrt{\rho_{B}}\sigma_{k}\sqrt{\rho_{B}}\sigma_{j})$
(32)

and so we can reconstruct the state up to a local unitary $B$ on Bob’s qubit
that leaves it invariant.

##  Appendix D Steering ellipsoids in a tetrahedron in the Bloch sphere
correspond to separable states

We prove that for any $\mathcal{E}$ inside a tetrahedron inside the Bloch
sphere, there is a separable state with $\rho_{B}=\frac{1}{2}\mathbb{I}$ and
$\mathcal{E}$ as its steering ellipse. We present the proofs for each possible
dimension of $\mathcal{E}$ separately, although each one is basically a
slightly more involved version of the previous one. Note that in the 0 and 1
dimensional cases the requirement to fit inside a tetrahedron is trivially
satisfied by any $\mathcal{E}$ inside the Bloch sphere.

This result suffices to show that any state with an ellipse that fits inside a
tetrahedron is separable by the following argument. Suppose $\rho_{AB}$ has an
ellipse that fits inside the tetrahedron. If $b=1$ then $\rho_{AB}$ is a
product state and we are done. Otherwise, apply a SLOCC operator to Bob and
obtain $\tilde{\rho}_{AB}$ with $\mathbf{b}=0$, recalling that SLOCC operators
cannot change a state from being entangled to separable. This will leave
Alice’s ellipse unchanged whilst moving her reduced state to the centre of her
ellipse. Since by the above statement there exists a separable state with the
correct ellipse and reduced states, $\tilde{\rho}_{AB}$ must equal the
separable state up to a choice of basis for Bob, and hence must itself be
separable.

In fact the separable states constructed below use a number of product states
equal to the dimension of the ellipse plus one. Since the SLOCC operator and
choice of basis for Bob do not affect the number of product states in a
decomposition, we furthermore have that $\rho_{AB}$ can be built using that
number of product states.

###  D.1 0-dimensional

If the steering ellipsoid is a single point $\mathbf{r}$ then simply take
$\rho$ with Bloch vector $\mathbf{r}$ and let
$\rho_{AB}=\rho\otimes\frac{1}{2}\mathbb{I}$.

###  D.2 1-dimensional

Suppose $\mathcal{E}$ is a line segment from $\mathbf{r}_{0}$ to
$\mathbf{r}_{1}$. Take $\rho_{i}$ with Bloch vectors $\mathbf{r}_{i}$ and let
$\rho_{AB}=\frac{1}{2}\sum_{i}\rho_{i}\otimes|i\rangle\langle i|$.

###  D.3 2-dimensional

If an ellipse fits inside a tetrahedron in the Bloch sphere, it also fits
inside a triangle in the Bloch sphere Vidrighin (2012). Therefore, suppose an
ellipse $\mathcal{E}$ fits within a triangle in the Bloch sphere whose
vertices are $\\{\mathbf{r}_{0},\mathbf{r}_{1},\mathbf{r}_{2}\\}$. Without
loss of generality we can take the ellipse to be tangent to each edge of the
triangle, at points $\\{\mathbf{s}_{i}\\}$ where $\mathbf{s}_{i}$ is on the
face opposite to $\mathbf{r}_{i}$. Denote the centre of the ellipse by
$\mathbf{c}$. Clearly there exists unique $p_{i}\geq 0$ such that
$\sum_{i}p_{i}\mathbf{r}_{i}=\mathbf{c}$ and $\sum_{i}p_{i}=1$.

By the definition of an ellipse, there is an invertible affine transformation
${\cal A}$ that maps $\mathcal{E}$ to the unit circle in the $(x,z)$-plane,
centred at the origin. Let $\rho_{i}$ have Bloch vectors $\mathbf{r}_{i}$
and$|\psi_{i}\rangle$ be such that the Bloch vector of
$|\psi_{i}\rangle\langle\psi_{i}|$ is $-{\cal A}(\mathbf{s}_{i})$. We claim
that the (manifestly separable) state

$\rho_{AB}=\sum_{i}p_{i}\rho_{i}\otimes|\psi_{i}\rangle\langle\psi_{i}|$ (33)

has $\rho_{B}=\frac{1}{2}\mathbbm{1}$ and that Alice’s steering ellipsoid for
this state is $\mathcal{E}$. To prove the first part, notice that the Bloch
vector of $\rho_{B}$ is $-\sum_{i}p_{i}{\cal A}(\mathbf{s}_{i})$. Since ${\cal
A}$ is affine, the unit circle will be tangent to the triangle with vertices
$\\{{\cal A}(\mathbf{r}_{i})\\}$ at the points $\\{{\cal
A}(\mathbf{s}_{i})\\}$, and $\sum_{i}p_{i}{\cal A}(\mathbf{r}_{i})={\cal
A}(\mathbf{c})=\mathbf{0}$. Hence it suffices to prove

######  Lemma 1.

Suppose the triangle with vertices $\\{\mathbf{v}_{i}\\}$ contains the unit
circle centered at the origin, and the circle is tangent to each edge of the
triangle at the points $\\{\mathbf{t}_{i}\\}$ (where $\mathbf{t}_{i}$ is on
the edge opposite $\mathbf{v}_{i}$). Fix $p_{i}$ by the requirements that
$\sum_{i}p_{i}\mathbf{v}_{i}=\mathbf{0}$ and $\sum_{i}p_{i}=1$. Then
$\sum_{i}p_{i}\mathbf{t}_{i}=\mathbf{0}$.

###### Proof.

We use $\mathbf{x}$ to represent points on or within the tetrahedron using
normalized barycentric co-ordinates $(x_{0},x_{1},x_{2})$ where
$\sum_{i}x_{i}=1$ and $\vec{x}=\sum_{i}x_{i}\mathbf{v}_{i}$. Let $A_{0}$ be
the volume of the triangle with vertices
$\\{\mathbf{x},\mathbf{v}_{1},\mathbf{v}_{2}\\}$, $A_{1}$ be the area of the
triangle with vertices $\\{\mathbf{v}_{0},\mathbf{x},\mathbf{v}_{2}\\}$ and
similarly for $A_{2}$. Let $A$ be the area of the original triangle (notice
$A=\sum_{i}A_{i}$). Then $x_{i}=A_{i}/A$. By definition the barycentric co-
ordinates of the origin are $(p_{0},p_{1},p_{2})$.

Let $L_{i}$ be the length of the edge opposite $\mathbf{v}_{i}$, and let
$L=\sum_{i}L_{i}$. By using that the area of a triangle = $\frac{1}{2}$ (base)
$\times$ (perpendicular height) and noting that by the tangency assumption the
relevant triangles have a perpendicular height of 1, we obtain that
$p_{i}=L_{i}/L$.

Figure 3: The various quantities used in proving Lemma 1. The dashed lines
form the three triangles used to show $p_{i}=L_{i}/L$, the dotted lines
indicate their perpendicular heights (which are equal to the radius of the
circle: 1).

Let $M_{0}^{(1)}=\left\lvert\mathbf{v}_{0}-\mathbf{t}_{1}\right\rvert$,
$M_{0}^{(2)}=\left\lvert\mathbf{v}_{0}-\mathbf{t}_{2}\right\rvert$. In fact
$M_{0}^{(1)}=M_{0}^{(2)}$ because they are both the unique length defined by
the requirement of being from a fixed point to a point on the circle such that
the line between them is tangent to the sphere, so we can write this length
simply as $M_{0}$. Define the other two $M_{i}$ by a similar argument. All
this is illustrated in Figure 3. Notice that

$\displaystyle L_{0}$ $\displaystyle=M_{1}+M_{2},$ (34) $\displaystyle L_{1}$
$\displaystyle=M_{0}+M_{2},$ (35) $\displaystyle L_{2}$
$\displaystyle=M_{0}+M_{1},$ (36)

The barycentric co-ordinates of $\mathbf{t}_{0},\mathbf{t}_{1}$ and
$\mathbf{t}_{2}$ can now be calculated as

$(0,M_{2},M_{1})/L_{0},$ (38) $(M_{2},0,M_{0})/L_{1},$ (39)

and

$(M_{1},M_{0},0)/L_{2},$ (40)

respectively. Using $p_{i}=L_{i}/L$ and the fact that barycentric coordinates
respect convex combinations the required result is now immediate. ∎

Suppose Bob projects qubit onto $|\psi\rangle$ and the orthogonal state. Since
$\rho_{B}=\frac{1}{2}\mathbbm{1}$ he will obtain each outcome with probability
$\frac{1}{2}$. Therefore if he obtains the $|\psi\rangle$ outcome then Alice’s
state will be

$\displaystyle\rho_{A}(|\psi\rangle)$ $\displaystyle:=$
$\displaystyle\frac{\operatorname{tr}_{B}(\rho_{AB}(I\otimes|\psi\rangle\langle\psi|))}{\operatorname{tr}(\rho_{B}|\psi\rangle\langle\psi|)}$
(41) $\displaystyle=$
$\displaystyle\frac{\sum_{i}p_{i}\rho_{i}\left\lvert\langle\psi_{i}|\psi|\right\rangle\rvert^{2}}{\frac{1}{2}}$
(42) $\displaystyle=$ $\displaystyle
2\sum_{i}p_{i}\rho_{i}\left\lvert\langle\psi_{i}|\psi|\right\rangle\rvert^{2}$
(43)

Recalling that the Bloch vector of $|\psi_{i}\rangle\langle\psi_{i}|$ is
$-{\cal A}(\mathbf{s}_{i})$, then if $|\psi\rangle\langle\psi|$ has Bloch
vector $\mathbf{r}$ then the Bloch vector of $\rho_{A}(|\psi\rangle)$ will be

$f(\mathbf{r}):=2\sum_{i}p_{i}\mathbf{r}_{i}\frac{1-\mathbf{r}\cdot{\cal
A}(\mathbf{s}_{i})}{2}=\sum_{i}p_{i}\mathbf{r}_{i}\left(1-\mathbf{r}\cdot{\cal
A}(\mathbf{s}_{i})\right).$ (44)

Let us extend this expression to all $\mathbf{r}$ to define an affine function
$f$. The statement that Alice’s steering ellipsoid is $\mathcal{E}$ is
equivalent to the statement that $\mathcal{E}$ is the image of the unit sphere
under $f$. Since all the ${\cal A}(\mathbf{s}_{i})$ are in the $(x,z)$-plane,
we have $f\left((0,1,0)\right)=f(\mathbf{0})$, i.e. we can think of $f$ as
first projecting onto the $(x,z)$-plane and then applying some affine
transformation. The image of the unit sphere under that projection is the unit
disc, and so it suffices to check that ${\cal E}$ is the image of the unit
circle under $f$. Define $g(\mathbf{r})={\cal A}(f(\mathbf{r}))$. Since ${\cal
A}$ is invertible and maps $\mathcal{E}$ to unit circle it suffices to prove
that $g$ is the identity on the $(x,z)$-plane. Since $g$ is the composition of
two affine functions it is also affine. By the definition of the $p_{i}$,
$g(\mathbf{0})={\cal A}(\mathbf{c})=\mathbf{0}$ so $g$ is in fact linear.
Hence it suffices to check that $g(\mathbf{u}_{j})=\mathbf{u}_{j}$ for some
spanning set of vectors $\\{\mathbf{u}_{j}\\}$. Since the triangle cannot be
degenerate, its vertex set $\\{\mathbf{r}_{j}\\}$ span some plane. Since
${\cal A}$ is invertible, $\\{{\cal A}(\mathbf{r}_{j})\\}$ must span the
$(x,z)$ plane. For $i\neq j$, $\\{\mathbf{0},{\cal A}(\mathbf{s}_{i}),{\cal
A}(\mathbf{r}_{j})\\}$ form a right-angle triangle, and $\left\lvert{\cal
A}(\mathbf{s}_{i})\right\rvert=1$. Therefore ${\cal
A}(\mathbf{r}_{j})\cdot{\cal A}(\mathbf{s}_{i})=1$ whenever $i\neq j$. But
$\sum_{i}p_{i}(1-\mathbf{r}\cdot{\cal
A}(\mathbf{s}_{i}))=\sum_{i}p_{i}-\mathbf{r}\cdot\left(\sum_{i}p_{i}{\cal
A}(\mathbf{s}_{i})\right)=1-\mathbf{r}\cdot\mathbf{0}=1$ (the penultimate
equality is from Lemma 1). Hence $p_{i}(1-{\cal A}(\mathbf{r}_{j})\cdot{\cal
A}(\mathbf{s}_{i}))=\delta_{ij}$ and we are done.

###  D.4 3-dimensional

Suppose an ellipsoid $\mathcal{E}$ fits within a tetrahedron in the Bloch
sphere whose vertices are
$\\{\mathbf{r}_{0},\mathbf{r}_{1},\mathbf{r}_{2},\mathbf{r}_{3}\\}$. Without
loss of generality we can take the ellipsoid to be tangent to each face of the
tetrahedron, at points $\\{\mathbf{s}_{i}\\}$ where $\mathbf{s}_{i}$ is on the
face opposite to $\mathbf{r}_{i}$. Denote the centre of the ellipsoid by
$\mathbf{c}$. Clearly there exists unique $p_{i}\geq 0$ such that
$\sum_{i}p_{i}\mathbf{r}_{i}=\mathbf{c}$ and $\sum_{i}p_{i}=1$.

By the definition of an ellipsoid, there is an invertible affine
transformation ${\cal A}$ that maps $\mathcal{E}$ to the unit sphere centred
at the origin. Let $\rho_{i}$ have Bloch vectors $\mathbf{r}_{i}$ and
$|\psi_{i}\rangle$ be such that the Bloch vector of
$|\psi_{i}\rangle\langle\psi_{i}|$ is $-{\cal A}(\mathbf{s}_{i})$. We claim
that the (manifestly separable) state

$\rho_{AB}=\sum_{i}p_{i}\rho_{i}\otimes|\psi_{i}\rangle\langle\psi_{i}|$ (45)

has $\rho_{B}=\frac{1}{2}\mathbbm{1}$ and that Alice’s steering ellipsoid for
this state is $\mathcal{E}$. To prove the first part, notice that the Bloch
vector of $\rho_{B}$ is $-\sum_{i}p_{i}{\cal A}(\mathbf{s}_{i})$. Since ${\cal
A}$ is affine, the unit sphere will be tangent to the tetrahedron with
vertices $\\{{\cal A}(\mathbf{r}_{i})\\}$ at the points $\\{{\cal
A}(\mathbf{s}_{i})\\}$, and $\sum_{i}p_{i}{\cal A}(\mathbf{r}_{i})={\cal
A}(\mathbf{c})=\mathbf{0}$. Hence it suffices to prove the following
3-dimensional analogue to Lemma 1:

######  Lemma 2.

Suppose the tetrahedron with vertices $\\{\mathbf{v}_{i}\\}$ contains the unit
sphere centered at the origin, and the sphere is tangent each face of the
tetrahedron at the points $\\{\mathbf{t}_{i}\\}$ (where $\mathbf{t}_{i}$ is on
the face opposite $\mathbf{v}_{i}$). Fix $p_{i}$ by the requirements that
$\sum_{i}p_{i}\mathbf{v}_{i}=\mathbf{0}$ and $\sum_{i}p_{i}=1$. Then
$\sum_{i}p_{i}\mathbf{t}_{i}=\mathbf{0}$.

###### Proof.

We use $\mathbf{x}$ to represent points on or within the tetrahedron using
normalized barycentric co-ordinates $(x_{0},x_{1},x_{2},x_{3})$ where
$\sum_{i}x_{i}=1$ and $\vec{x}=\sum_{i}x_{i}\mathbf{v}_{i}$. Let $V_{0}$ be
the volume of the tetrahedron with vertices
$\\{\mathbf{x},\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}\\}$, $V_{1}$ be
the volume of the tetrahedron with vertices
$\\{\mathbf{v}_{0},\mathbf{x},\mathbf{v}_{2},\mathbf{v}_{3}\\}$ and so on. Let
$V$ be the volume of the original tetrahedron (notice $V=\sum_{i}V_{i}$). Then
$x_{i}=V_{i}/V$. By definition the barycentric co-ordinates of the origin are
$(p_{0},p_{1},p_{2},p_{3})$.

Let $A_{i}$ be the area of the face opposite $\mathbf{v}_{i}$, and let
$A=\sum_{i}A_{i}$. By using that the volume of a tetrahedron = $\frac{1}{3}$
(area of base) $\times$ (perpendicular height) and noting that by the tangency
assumption the relevant tetrahedra have a perpendicular height of 1, we obtain
that $p_{i}=A_{i}/A$.

Let $A_{23}^{(0)}$ be the area of the triangle with vertices
$\\{\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{t}_{0}\\}$. Let $A_{23}^{(1)}$ be
the area of the triangle with vertices
$\\{\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{t}_{1}\\}$. Now we have that
$\left\lvert\mathbf{v}_{2}-\mathbf{t}_{0}\right\rvert=\left\lvert\mathbf{v}_{2}-\mathbf{t}_{1}\right\rvert$
because they are both the unique length defined by the requirement of being
from a fixed point to a point on the sphere such that the line between them is
tangent to the sphere. Similarly
$\left\lvert\mathbf{v}_{3}-\mathbf{t}_{0}\right\rvert=\left\lvert\mathbf{v}_{3}-\mathbf{t}_{1}\right\rvert$.
Hence the two triangles are congruent and we can simply write their areas as
$A_{23}$. Define the other five $A_{ij}$ by a similar argument. Notice that

$\displaystyle A_{0}$ $\displaystyle=A_{12}+A_{13}+A_{23},$ (46)
$\displaystyle A_{1}$ $\displaystyle=A_{02}+A_{03}+A_{23},$ (47)
$\displaystyle A_{2}$ $\displaystyle=A_{01}+A_{03}+A_{13},$ (48)
$\displaystyle A_{3}$ $\displaystyle=A_{01}+A_{02}+A_{12}.$ (49)

The barycentric co-ordinates of $\mathbf{t}_{0},\mathbf{t}_{1},\mathbf{t}_{2}$
and $\mathbf{t}_{3}$ can now be calculated as

$(0,A_{23},A_{13},A_{12})/A_{0},$ (50) $(A_{23},0,A_{03},A_{02})/A_{1},$ (51)
$(A_{13},A_{03},0,A_{01})/A_{2},$ (52)

and

$(A_{12},A_{02},A_{01},0)/A_{3}$ (53)

respectively. Using $p_{i}=A_{i}/A$ and the fact that barycentric coordinates
respect convex combinations the required result is now immediate. ∎

As in the 2-dimensional case we find that if Bob projects onto the state with
Bloch vector $\mathbf{r}$ than Alice’s Bloch vector is

$f(\mathbf{r}):=\sum_{i}p_{i}\mathbf{r}_{i}\left(1-\mathbf{r}\cdot{\cal
A}(\mathbf{s}_{i})\right).$ (54)

Let us extend this expression to all $\mathbf{r}$ to define an affine function
$f$. The statement that Alice’s steering ellipsoid is $\mathcal{E}$ is
equivalent to the statement that $\mathcal{E}$ is the image of the unit sphere
under $f$. Define $g(\mathbf{r})={\cal A}(f(\mathbf{r}))$. Since ${\cal A}$ is
invertible and maps $\mathcal{E}$ to unit sphere it suffices to prove that $g$
is the identity. Since $g$ is the composition of two affine functions it is
also affine. By the definition of the $p_{i}$, $g(\mathbf{0})={\cal
A}(\mathbf{c})=\mathbf{0}$ so $g$ is in fact linear. Hence it suffices to
check that $g(\mathbf{u}_{j})=\mathbf{u}_{j}$ for some spanning set of vectors
$\\{\mathbf{u}_{j}\\}$. Since the tetrahedron cannot be degenerate, its vertex
set $\\{\mathbf{r}_{j}\\}$ must be spanning. Since ${\cal A}$ is invertible,
$\\{{\cal A}(\mathbf{r}_{j})\\}$ is also spanning. As in the 2-dimensional
case, for $i\neq j$, $\\{\mathbf{0},{\cal A}(\mathbf{s}_{i}),{\cal
A}(\mathbf{r}_{j})\\}$ form a right-angle triangle, and $\left\lvert{\cal
A}(\mathbf{s}_{i})\right\rvert=1$. Therefore ${\cal
A}(\mathbf{r}_{j})\cdot{\cal A}(\mathbf{s}_{i})=1$ whenever $i\neq j$. But
$\sum_{i}p_{i}(1-\mathbf{r}\cdot{\cal
A}(\mathbf{s}_{i}))=\sum_{i}p_{i}-\mathbf{r}\cdot\left(\sum_{i}p_{i}{\cal
A}(\mathbf{s}_{i})\right)=1-\mathbf{r}\cdot\mathbf{0}=1$ (the penultimate
equality is from Lemma 2). Hence $p_{i}(1-{\cal A}(\mathbf{r}_{j})\cdot{\cal
A}(\mathbf{s}_{i}))=\delta_{ij}$ and we are done.

##  Appendix E Discord and steering ellipsoids

Below we outline the condition for zero discord for Alice from either her or
Bob’s ellipsoid.

> A state has zero discord for Alice iff her ellipsoid is a segment of a
> diameter.

The “only if” part: A general zero discord state for Alice
$\rho=p|e\rangle\langle
e|\otimes\rho_{0}+(1-p)|\bar{e}\rangle\langle\bar{e}|\otimes\rho_{1}$ has
$\langle e|\bar{e}\rangle=0$ and

$\displaystyle\mathbf{a}$ $\displaystyle=$ $\displaystyle t\mathbf{e}$ (55)
$\displaystyle\mathbf{b}$ $\displaystyle=$ $\displaystyle\mathbf{x}$ (56)
$\displaystyle T$ $\displaystyle=$ $\displaystyle\mathbf{ey}^{T}$ (57)

where $t=2p-1$, $\mathbf{e}=\langle e|\bm{\sigma}|e\rangle$ and
$\mathbf{x}=\operatorname{tr}\left[\left(p\rho_{0}+(1-p)\rho_{1}\right)\bm{\sigma}\right]$,
$\mathbf{y}=\operatorname{tr}\left[\left(p\rho_{0}-(1-p)\rho_{1}\right)\mathbf{\sigma}\right]$
Dakić et al. (2010).

Alice’s steering ellipsoid $\mathcal{E}_{A}$ has centre
$\mathbf{c}_{A}=\left(\frac{t-\mathbf{x}\cdot\mathbf{y}}{1-x^{2}}\right)\mathbf{e}$
and matrix $Q_{A}=s_{A}^{2}\mathbf{e}\mathbf{e}^{T}$ with

$s_{A}^{2}=\frac{1}{1-x^{2}}\left[\left(\mathbf{y}-t\mathbf{x}\right)^{T}\left(\mathbbm{1}+\frac{\mathbf{x}\mathbf{x}^{T}}{1-x^{2}}\right)\left(\mathbf{y}-t\mathbf{x}\right)\right]$
(58)

So $\mathcal{E}_{A}$ is a segment of the diameter.

The “if” part: suppose we are given $\mathcal{E}_{A}$, a segment of the
diameter. Denote the states endpoints of the ellipsoid $\rho_{0}$ and
$\rho_{1}$. Alice’s state can always be decomposed as
$\rho_{A}=q\rho_{0}+(1-q)\rho_{1}$. However since all the Bloch vectors of
$\rho_{A},\rho_{0},\rho_{1}$ are collinear they will eigendecompose into the
same pair of orthogonal states, call them $|\psi\rangle,|\bar{\psi}\rangle$.
Writing
$\rho_{i}=p_{i}|\psi\rangle\langle\psi|+(1-p_{i})|\bar{\psi}\rangle\langle\bar{\psi}|$,
for $i=0,1$ then
$\rho_{A}=p|\psi\rangle\langle\psi|+(1-p)|\bar{\psi}\rangle\langle\bar{\psi}|$
with $p=qp_{0}+(1-q)p_{1}$. Then the joint state

$\displaystyle\rho=p|\psi\rangle\langle\psi|\otimes\beta_{0}+(1-p)|\bar{\psi}\rangle\langle\bar{\psi}|\otimes\beta_{1}$
(59)

is a zero discord state for Alice with the correct $\mathcal{E}_{A}$ and
$\rho_{A}$ for any mixed states on Bob’s side $\beta_{0},\beta_{1}$.

> There is zero discord for Alice iff Bob’s ellipsoid is a line segment and
> the length of Alice’s Bloch vector is equal to the distance from the centre
> of Bob’s ellipsoid to his Bloch vector divided by the radius of his
> ellipsoid.

The “only if” part can easily be checked: it requires
$a=\frac{|\bm{c}_{B}-\bm{b}|}{s_{B}}$. Since, after some algebra,

$\displaystyle\mathbf{c}_{B}$ $\displaystyle=$
$\displaystyle\frac{\mathbf{x}-t\mathbf{y}}{1-t^{2}}$ (60) $\displaystyle
Q_{B}$ $\displaystyle=$
$\displaystyle\frac{1}{(1-t^{2})^{2}}(\mathbf{y}-t\mathbf{x})(\mathbf{y}-t\mathbf{x})^{T}$
(61)

then $|\bm{c}_{B}-\bm{b}|=\frac{t|\mathbf{y}-t\mathbf{x}|}{1-t^{2}}=as_{B}$.

For the “if” part, let $\rho_{0}$ and $\rho_{1}$ be the endpoints of Bob’s
ellipsoid, and let the state corresponding to Alice’s Bloch vector have eigen-
decomposition
$\rho_{A}=p_{0}|\psi_{0}\rangle\langle\psi_{0}|+p_{1}|\psi_{1}\rangle\langle\psi_{1}|$.
Then the joint state

$\rho=p_{0}|\psi_{0}\rangle\langle\psi_{0}|\otimes\rho_{0}+p_{1}|\psi_{1}\rangle\langle\psi_{1}|\otimes\rho_{1}$
(62)

has zero discord for Alice, the correct Bloch vector for Alice and the correct
ellipsoid for Bob. If necessary swapping $\rho_{0}$ and $\rho_{1}$, it also
has the right Bloch vector for Bob. Bob’s ellipsoid $\mathcal{E}_{B}$ is
invariant under local unitaries on Alice’s qubit, so Alice and Bob’s actual
state is therefore equivalent to $\rho$ up to this transformation, which
preserves discord.

##  Appendix F Volume formula for the steering ellipsoid

The volume of any ellipsoid is proportional to the product of its eigenvalues
$V=\frac{4\pi}{3}s_{1}s_{2}s_{3}$. Therefore the $\mathcal{E}_{A}$ has volume
$V_{A}=\frac{4\pi}{3}|\sqrt{\det Q_{A}}|$, which may be rewritten as

$V_{A}=\frac{4\pi}{3}\frac{\left|\det(T-\bm{a}\bm{b}^{T})\right|}{\left(1-b^{2}\right)^{2}}=\frac{4\pi}{3}\frac{\left|\mathrm{det}\Theta\right|}{(1-b^{2})^{2}}.$
(63)

To express this in terms of the density matrix $\rho$, we use the equation
$\Theta=2\Upsilon\rho^{R}\Upsilon^{T}$ Verstraete et al. (2001), where the
unitary matrix $\Upsilon$ is given in equation (7) and $R$ denotes a
reshuffling operation: if $\rho=\sum_{i,j=0}^{1}\rho_{ij;kl}|ij\rangle\langle
kl|$ then $\rho^{R}=\sum_{i,j=0}^{1}\rho_{ik;jl}|ij\rangle\langle kl|$. We
also require a curious relation that holds for any 4$\times$4 or 9$\times$9
matrix $M$. For any such $M$ we have that

$\det M=\det M^{T_{B}}-\det(M^{T_{B}})^{R}.$ (64)

Applying this relation to (63), together with the reshuffled form of $\Theta$,
we obtain

$V_{A}=\frac{64\pi}{3}\frac{\left|\det\rho-\det\rho^{T_{B}}\right|}{\left(1-b^{2}\right)^{2}}.$
(65)

##  Appendix G The steering ellipsoid zoo

In this section we illustrate the main types of ellipsoid.

###  G.1 Entangled states

For every pure entangled state the ellipsoid coincides with the Bloch sphere.
When the state is mixed and entangled, the ellipsoid does not satisfy the
tetrahedral condition because, loosely speaking, the ellipsoid is either _too
big_ (with volume $V\geq V_{\star}>0$ ) or _too near_ (large $c$) to the
surface of the Bloch sphere, see figure 4. Every entangled state is completely
steerable.

Figure 4: A generic entangled state $\rho_{AB}$: both ellipsoids ${\cal
E}_{A}$ and ${\cal E}_{B}$ are always full rank and neither can be inscribed
within a tetrahedron within the Bloch sphere.

###  G.2 Separable states with full-dimensional ellipsoids

Separable states admit a convex decomposition in terms of product states, and
have “more classical” correlations. Steering is still possible, however the
steering ellipsoids necessarily obey the tetrahedral condition, as in figure
5.

If the state has a three dimensional ${\cal E}_{A}$ then it has non-zero
obesity and non-zero discord, and furthermore, it can be written as a mixture
of just four product states. Such states are also completely steerable.

Figure 5: A generic separable state $\rho_{AB}$, where both ellipsoids ${\cal
E}_{A}$ and ${\cal E}_{B}$ are full rank fit inside a tetrahedron. Figure 6: A
Bell-diagonal state: the ellipsoid is centred at the origin and its semiaxes
are given by the three singular values of $T$. The vector of these singular
values $\bm{t}=(t_{1},t_{2},t_{3})$ lives in a tetrahedron with vertices at
(1,1,-1),(1,-1,1),(-1,1,1),(-1,-1,-1), and when it is inside an octahedron
inside of this tetrahedron, then the state is necessarily separable. This
defines the set of ellipsoids that fit inside the nested tetrahedron (these
are not the same tetrahedra). Figure 7: A Werner state. The steering ellipsoid
${\cal E}_{A}$ is a sphere centered at the origin. The ellipsoid fits inside a
tetrahedron when its radius is less than $\frac{1}{3}$ and thus the state is
separable.

###  G.3 Steering pancakes

The set of states that Bob can steer Alice to may become degenerate, and form
a _two-dimensional_ set. This “steering pancake” will not only fit inside a
tetrahedron, but will fit within a _triangle_ that is inscribed within the
Bloch sphere as shown in 8.

Figure 8: A generic separable state $\rho_{AB}$, where both ellipsoids ${\cal
E}_{A}$ and ${\cal E}_{B}$ are steering pancakes.

Recall that we have a novel feature for some steering pancakes (and steering
needles) of _incomplete_ steering. For steering pancakes we have complete
steering of qubit A if and only if the affine span of ${\cal E}_{B}$ contains
the origin of the Bloch sphere.

###  G.4 Steering needles

The steering can become even more degenerate, and the steering set collapse to
a one-dimensional line segment, or “steering needle”. These states include
perfectly classical (doubly zero-discord states) with needles being radial
(figure 9), but also includes non-zero discord states for which either one or
both of the steering needles ${\cal E}_{A}$ and ${\cal E}_{B}$ is not radial
(figure 10). Being radial is indicated by dashed lines on the figures, which
depict diameters.

Figure 9: A doubly zero discord state, where both ${\cal E}_{A}$ and ${\cal
E}_{B}$ are radial line segments.

Figure 10: A state with zero discord at $B$, but _non-zero_ discord at $A$. We
have ${\cal E}_{B}$ being a radial line segment, but ${\cal E}_{A}$ is not a
radial line segment.

## References

  * Avron et al. (2007) J. E. Avron, G. Bisker, and O. Kenneth, J. Math. Phys. 48, 102107 (2007).
  * Horodecki and Horodecki (1996a) R. Horodecki and M. Horodecki, Phys. Rev. A 54, 1838 (1996a).
  * Bengtsson and Życzkowski (2006) I. Bengtsson and K. Życzkowski, _Geometry of Quantum States: An Introduction to Quantum Entanglement_ (Cambridge University Press, 2006).
  * Schrödinger (1935) E. Schrödinger, Proc. Cam. Phil. Soc. 31, 555 (1935).
  * Gisin (1984) N. Gisin, Helvetica Physica Acta 62, 363 (1984).
  * Hughston et al. (1993) L. P. Hughston, R. Jozsa, and W. K. Wootters, Phys. Lett. A 183, 14 (1993).
  * Spekkens and Rudolph (2002) R. W. Spekkens and T. Rudolph, Quantum Inform. Compu. 2, 66 (2002).
  * Verstraete (2002) F. Verstraete, PhD Thesis (2002).
  * Verstraete et al. (2001) F. Verstraete, J. Dehaene, and B. DeMoor, Phys. Rev. A 64, 010101 (2001).
  * Shi et al. (2011) M. Shi, W. Yang, F. Jiang, and J. Du, J. Phys. A: Math. Theor. 44, 415304 (2011).
  * Peres (1996) A. Peres, Phys. Rev. Lett. 77, 1413 (1996).
  * Horodecki et al. (1996) M. Horodecki, P. Horodecki, and R. Horodecki, Phys. Lett. A 223, 1 (1996).
  * Sanpera et al. (1998) A. Sanpera, R. Tarrach, and G. Vidal, Phys. Rev. A 58, 826 (1998).
  * Wootters (1998) W. K. Wootters, Phys. Rev. Lett. 80, 2245 (1998).
  * Vidrighin (2012) M. Vidrighin (2012), URL http://mathoverflow.net/questions/109153.
  * Ollivier and Zurek (2001) H. Ollivier and W. H. Zurek, Phys. Rev. Lett. 88, 017901 (2001).
  * Lewenstein and Sanpera (1998) M. Lewenstein and A. Sanpera, Phys. Rev. Lett. 80, 2261 (1998).
  * Horodecki and Horodecki (1996b) R. Horodecki and M. Horodecki, Phys. Rev. A 54, 1838 (1996b).
  * Frosch (2006) R. Frosch, _Four-tensors, the Mother Tongue of Classical Physics_ (vdf Hochschulverlag AG, 2006).
  * Dakić et al. (2010) B. Dakić, V. Vedral, and Č. Brukner, Phys. Rev. Lett. 105, 190502 (2010).
  * Mingjun Shi and Du (2011) C. S. Mingjun Shi, Fengjian Jiang and J. Du, N. J. Phys. 13, 073016 (2011).