Spekken’s Toy Theory: Elementary System

27May12

Robert Spekkens will be one of our invited speakers at APCWQIS 2012. I thought it would be nice to review some simple aspects of of his highly cited papers namely “Evidence for the epistemic view of quantum states: A toy theory”, Phys. Rev. A 75 (2007) 032110. (arXiv version: quant-ph/0401052). As the title explains, it is an attempt to consider quantum theory (as a probabilistic theory) to be simply a reflection of our knowledge about quantum systems. The word epistemic refers to this nature of referring to the state of knowledge. The other relevant notion is ontic (as in ontology) which refers to the nature of the systems themselves independent of their experimental access. The two notions are indeed intertwined (please refer to Atmanspacher and Primas’ article “Epistemic and Ontic Quantum Realities” for deeper discussions). In fact, as Spekkens explains, the distinction between an ontic state and an epistemic state disappears when epistemic states describe complete knowledge. Thus, his hope for the epistemic view of the quantum state describing an incomplete knowledge.

He begins by proposing a foundantional principle namely the knowledge balance principle which posits a balance between knowledge and ignorance in a state of maximal knowledge. He demonstrates this in a simple set-up of toy theory where there are four ontic states labelled by numbers 1, 2, 3, 4 in a system. To investigate the system, one questions (experiments) yes-no questions e.g.

  • Is it 1?
  • Is it 2?
  • Is it 3?
  • Is it 4?

This is the maximal set of four questions but it is an ineffective one. One can do better by combining two ontic states in one single question and the grouping will be considered an epistemic state. For example, the table

\begin{tabular} {l | c | r}  Is it in {1,2}? &Is it in {1,3}? &Ontic state \\  \hline  Yes & Yes & 1 \\  Yes & No & 2 \\  No & Yes & 3 \\  No & No & 4  \end{tabular}

shows that two questions are enough to find out what the ontic state is and  is in fact the minimum number. The amount of ignorance in such a system is defined as

Ignorance = (Max. no. of yes-no queries) – (Min. no of yes-no queries) = 4 – 2 = 2.

This equals the (maximal) number of questions for which the ontic state answer is known (the knowledge). Thus the knowledge balance principle. The two questions above form what is called a canonical set but it is not unique since we can have another that can equally well do the job such as the table below:

\begin{tabular} {l | c | r}  Is it in {1,2}? &Is it in {2,3}? &Ontic state \\  \hline  Yes & Yes & 2 \\  Yes & No & 3 \\  No & Yes & 4 \\  No & No & 1  \end{tabular}

The epistemic states (set of two ontic states) can now be denoted by e.g. 1 \vee 2 or {1, 2, x, x}, 1 \vee 3 or {1, x, 3, x}, and are states of maximal knowledge. Here, we avoid the boxes shown in Spekkens’ paper simply for brevity.

The reader may ask why with the four ontic states? Well, let us consider the other small possibilities. For the case of one ontic state, there is nothing to ask and represents a completely trivial system. If there are two ontic states, we can have the following table:

\begin{tabular} {l | c | r}  Is it in {1}? &Is it in {2}? &Ontic state \\  \hline  Yes & No & 1 \\  No & Yes & 2  \end{tabular}

The maximum number of questions is two and the minimum is one and hence the measure of ignorance is one. It seems we are able to obey the knowledge balance principle here. The trouble is that the epistemic state here is simply the ontic state. So we do not consider this case. It can be shown that for the case of three ontic states, the knowledge balance principle is not satisfied.

Thus, we are left with four epistemic states being the smallest interesting one and with two in the canonical set. This is called an elementary system and it consists of six epistemic states:

1\vee 2 \equiv \{1, 2, x, x\}
3\vee 4 \equiv \{x, x, 3, 4\}
1\vee 3 \equiv \{1, x, 3, x\}
2\vee 4 \equiv \{x, 2, x, 4\}
2\vee 3 \equiv \{x, 2, 3, x\}
1\vee 4 \equiv \{1, x, x, 4\}

The less than maximal state is when there is no questions answered, namely the state

1\vee 2\vee 3\vee 4 \equiv \{1, 2, 3, 4\}.

The logical jump: Spekkens made the identication of these six epistemic states with qubit states (and this is only a convention):

1\vee 2\quad\Leftrightarrow\quad\lvert 0\rangle
3\vee 4\quad\Leftrightarrow\quad\lvert 1\rangle
1\vee 3\quad\Leftrightarrow\quad\lvert +\rangle = \frac{1}{\sqrt{2}}\left(\lvert 0 \rangle + \lvert 1 \rangle\right)
2\vee 4\quad\Leftrightarrow\quad\lvert -\rangle = \frac{1}{\sqrt{2}}\left(\lvert 0 \rangle - \lvert 1 \rangle\right)
2\vee 3\quad\Leftrightarrow\quad\lvert +i\rangle = \frac{1}{\sqrt{2}}\left(\lvert 0 \rangle + i\lvert 1 \rangle\right)
1\vee 4\quad\Leftrightarrow\quad\lvert -i\rangle = \frac{1}{\sqrt{2}}\left(\lvert 0 \rangle - i\lvert 1 \rangle\right)

Finally the non-maximal epistemic state is identified with the completely mixed state:

1\vee 2\vee 3\vee 3\vee 4\quad\Leftrightarrow\quad I/2.

One thing we know about mixed states is that it does not have a unique decomposition with respect to a basis. Spekkens indeed define an operation +_{\textrm{cv}} between epistemic states. The condition is that the two epistemic states to be combined together must have intersection of their ontic support to be empty:

1\vee 2\vee 3\vee 4
\qquad\qquad\quad= (1\vee 2) +_{\textrm{cv}} (3\vee 4)
\qquad\qquad\quad= (1\vee 3) +_{\textrm{cv}} (2\vee 4)
\qquad\qquad\quad= (2\vee 3) +_{\textrm{cv}} (1\vee 4)

This is analogous to

\frac{I}{2}  = \frac{1}{2} \lvert 0\rangle\langle 0 \rvert + \frac{1}{2} \lvert 1\rangle\langle 1\rvert
\qquad\qquad = \frac{1}{2} \lvert +\rangle\langle + \rvert + \frac{1}{2} \lvert -\rangle\langle -\rvert
\qquad\qquad = \frac{1}{2} \lvert +i\rangle\langle +i \rvert + \frac{1}{2} \lvert -i\rangle\langle -i\rvert

So this came out nicely. There is another important combination that the toy model needs to mimick namely coherent superposition. This binary operation has to take a pair of epistemic states to another pure epistemic state. This operation has to be different since it cannot simply take the union of ontic support but results in different ontic support. There are in fact four possibilities, namely:

(1\vee 2) +_1 (3\vee 4) = 1\vee 3
(1\vee 2) +_2 (3\vee 4) = 2\vee 4
(1\vee 2) +_3 (3\vee 4) = 2\vee 3
(1\vee 2) +_4 (3\vee 4) = 1\vee 4

which are analogous to

\frac{1}{\sqrt{2}} (\vert 0\rangle + \lvert 1\rangle = \lvert +\rangle
\frac{1}{\sqrt{2}} (\vert 0\rangle - \lvert 1\rangle = \lvert -\rangle
\frac{1}{\sqrt{2}} (\vert 0\rangle + i\lvert 1\rangle = \lvert +i\rangle
\frac{1}{\sqrt{2}} (\vert 0\rangle - i\lvert 1\rangle = \lvert -i\rangle

It is not as “straightforward” as the convex combination. For example

(1\vee 3) +_3 (2\vee 4) = 2\vee 3

has the analogue

\frac{1}{\sqrt{2}}(\lvert +\rangle + i\lvert -\rangle)
\quad = \frac{(1+i)}{2}\,\lvert 0 \rangle + \frac{(1-i)}{2}\,\lvert 1 \rangle
\quad = \frac{e^{i\pi/4}}{\sqrt{2}}\,\lvert 0 \rangle + \frac{e^{-i\pi/4}}{\sqrt{2}}\,\lvert 1 \rangle
\quad = e^{i\pi/4}\,\frac{1}{\sqrt{2}}\,\left(\lvert 0 \rangle + e^{-i\pi/2}\,\lvert 1 \rangle\right)
\quad = e^{i\pi/4}\,\frac{1}{\sqrt{2}}\, (\lvert 0 \rangle - i\vert 1 \rangle) = e^{i\pi/4}\,\lvert -i \rangle.

Note the extra phase. There are other such examples. The intent is however not to set up a precise correspondence but just a mere reflection that some quantum properties can be mirrored in this toy model. In fact for qubit superposition has a continuum of possibilities, unlike the epistemic states. The theory is in fact closer to the quantum mechanics of stabiliser states (see Matthew Pusey arXiv: 1193.5037).

What are indeed some of these quantum properties? We focus on measurement which requires an initial (prepared) state, the outcome and the post measurement state. The post measurement state must obey the knowledge balance principle and the fewest ontic state associated to a single outcome is two. Thus we need to partition the four ontic states into sets of of two ontic states which are disjoint:

\{ 1\vee 2, 3\vee 4\} \textrm{ comparable to }\lbrace \lvert 0 \rangle, \lvert 1 \rangle\rbrace
\{ 1\vee 3, 2\vee 4\} \textrm{ comparable to }\lbrace \lvert + \rangle, \lvert - \rangle\rbrace
\{ 1\vee 4, 2\vee 3\} \textrm{ comparable to }\lbrace \lvert +i \rangle, \lvert -i \rangle\rbrace

Schematically this can be depicted the slot templates

{ I, I, II, II }
{ I, II, I , II }
{ I, II, II, I }

Given state \{ 1, 2, x, x \} \equiv 1\vee 2, measuring { I, I, II, II } will give outcome 1\vee 2 with certainty. If on the other hand, measurement { I, II, I, II } is made instead, then two outcomes will be possible namely 1\vee 3, 2\vee 4 with equal frequencies. An immediate consequence is that measurements are now noncommutative since the following two sets of consecutive measurements

{ I, I, II, II } then { I, II, I, II }

{ I, II, I, II } then { I, I, II, II }

completely give different outcomes.

One hallmark of quantum physical results is the interference pattern arising from constructive and destructive interference. Spekkens considered the following measurements for qubit states \{\lvert + \rangle , \lvert - \rangle\} on the following prepared states:

(a) \lvert 0 \rangle;
(b) \lvert 1 \rangle;
(c) \frac{1}{\sqrt{2}} (\lvert 0 \rangle + \lvert 1 \rangle ).

From the following probability amplitudes,

\langle + \vert 0 \rangle = \frac{1}{\sqrt{2}}
\langle - \vert 0 \rangle = \frac{1}{\sqrt{2}}
\langle + \vert 1 \rangle = \frac{1}{\sqrt{2}}
\langle - \vert 1 \rangle = -\frac{1}{\sqrt{2}}

we have the 50-50 outcomes of the first two experiments i.e

P(\lvert \pm \rangle) = \lvert\langle \pm\vert 0 \rangle\rvert^2 = \frac{1}{2} = \lvert\langle \pm\vert 1 \rangle\rvert^2.

For the third experiment, if one recognises the input state is really \lvert +\rangle then the probability outcome is (1,0). But it is better to consider the result given the first two experiments. Suppose we write instead \lvert\psi\rangle \equiv \{\lvert +\rangle , \lvert - \rangle \}, the probability outcome is calculated as

\lvert \frac{\langle\psi\vert 0\rangle + \langle\psi\vert 1 \rangle}{\sqrt{2}} \rvert^2
= \frac{1}{2} (\lvert\langle\psi\vert 0 \rangle\rvert^2 + \lvert\langle\psi\vert 1 \rangle\rvert^2
\qquad + \langle\psi\vert 0\rangle^* \langle\psi\vert 1 \rangle + \langle\psi\vert 1\rangle^* \langle\psi\vert 0 \rangle )

Note that the last two terms (on the third line) are the interference terms. Putting \lvert\psi\rangle = \lvert +\rangle shows the constructive interference effect:

P(\lvert +\rangle ) = \frac{1}{2}(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2})=1

Putting \lvert\psi\rangle = \lvert -\rangle shows the destructive interference effect:

P(\lvert -\rangle ) = \frac{1}{2}(\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2})=0

The analogous toy model measurements are:

(a) Prepare {1, 2, x, x} then measure {I, II, I, II}.
The probability outcomes are \{ \frac{1}{2},\frac{1}{2}\}

(b) Prepare {x, x, 3, 4} then measure {I, II, I, II}.
The probability outcomes are again \{ \frac{1}{2},\frac{1}{2}\}

(c) Prepare {1, x, 3, x} then measure {I, II, I, II}.
The probability outcomes are \{ 1,0\} which reflects the earlier interference signature.

These similarities do not stop here but can even be further extended form of the elementary systems which will be discussed in a future post.

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