Stabilizer Notation for Spekkens’ Toy Theory

Abstract

Spekkens has introduced a toy theory (Spekkens in Phys. Rev. A 75(3):032110, 2007) in order to argue for an epistemic view of quantum states. I describe a notation for the theory (excluding certain joint measurements) which makes its similarities and differences with the quantum mechanics of stabilizer states clear. Given an application of the qubit stabilizer formalism, it is often entirely straightforward to construct an analogous application of the notation to the toy theory. This assists calculations within the toy theory, for example of the number of possible states and transformations, and enables superpositions to be defined for composite systems.

Appendices

Appendix A: Proofs

Proof of Lemma 1

The second statement is equivalent to “m(g ₁),m(g ₂),…,m(g _l ) commute, have linearly independent check vectors, and each square to I ^⊗n” [5]. By definition check vectors and commuting conditions are preserved by m. Furthermore m(g)²=I ^⊗n since the only elements of P _n that don’t square to I ^⊗n are those with phases α∈{i,−i}, which aren’t in the image of m.

Hence the second statement is equivalent to “g ₁,g ₂,…,g _l commute and have linearly independent check vectors”. The proof is completed by showing that this is equivalent to the first statement using a similar argument to the qubit case [5]. First note that g ₁,g ₂,…,g _l commute if and only if 〈g ₁,g ₂,…,g _l 〉 commute.

Suppose $-{\mathcal {I}}^{\otimes n} \in \langle g_{1}, g_{2}, \ldots, g_{l}\rangle$. Then the check vector of $-{\mathcal {I}}^{\otimes n}$ can be written is a linear combination of the check vectors of g ₁,g ₂,…,g _l . But the check vector of $-{\mathcal {I}}^{\otimes n}$ is 0 and so the check vectors of g ₁,g ₂,…,g _l are linearly dependent. Suppose that the first statement holds but the check vectors of g ₁,g ₂,…,g _l are linearly dependent. Then the check vector of one of them, say of g ₁, can be written as a linear combination of the others. That means that either g ₁ or −g ₁ can be written as a product of the others. The first possibility contradicts the assumption of independent generators, and since $g_{1}(-g_{1}) = -{\mathcal {I}}^{\otimes n}$ the second contradicts the assumption that the subgroup does not contain $-{\mathcal {I}}^{\otimes n}$. □

Lemma 7

An epistemic state is permitted by the toy theory if and only if it corresponds to a toy stabilizer subgroup.

Proof

In order to prove that the toy stabilizer subgroups correspond exactly to the valid epistemic states of Spekkens’ toy theory, we need a precise formulation of the valid states for n elementary systems. Define them inductively as follows. An epistemic state is a subset of the ontic states (which we assign a uniform probability distribution over). For n=1 a valid epistemic state consists of either two or four ontic states. Suppose the valid states are defined for n≤k. Define a valid measurement on n≤k elementary systems as the discovery of which member of a partition of the ontic states into valid epistemic states the ontic state lies in. A valid epistemic state on n=k+1 elementary systems is such that for any non-trivial partition of the elementary systems into two subsystems A and B, any possible outcome of any valid measurement of A (which is assumed not to affect the ontic state of B) results in a valid epistemic state for B.

The definition of the “knowledge balance principle” in [1] is not as mathematical as the one above. Since the valid epistemic states of an elementary system are stated explicitly in [1], the definitions certainly agree for n=1. For n>1 the inductive structure made explicit above is implicit in [1]: examples are given of particular epistemic states of a composite system, such that a measurement on one subsystem leads to an invalid epistemic state for the other. This is always taken as proof that the epistemic state of the composite system was invalid, the validity of the definitions for the smaller subsystems is never questioned.

Also implicit in [1] is that for any epistemic state on n elementary systems, there must exist a canonical set such that the agent is certain about the answers to l of the questions, and completely ignorant of the remaining 2n−l. Hence the number of ontic states compatible with the epistemic state must be 2^(2n−l). This implicit requirement rules out the epistemic state 1∨2∨3. Furthermore, by the knowledge balance principle we must have l≤n. Except for when n=1, this requirement might appear to be missing from the above definition. But since the Lemma can be proven without it, and any state corresponding to a toy stabilizer subgroup certainly satisfies this requirement, it turns out that there is no need to include it in the definition.

We now proceed with the proof using induction on n. The only toy stabilizer subgroups for n=1 are $\langle \rangle,\langle {\mathcal {X}}\rangle, \langle-{\mathcal {X}}\rangle,\langle {\mathcal {Y}}\rangle, \langle-{\mathcal {Y}}\rangle,\langle {\mathcal {Z}}\rangle$ and $\langle-{\mathcal {Z}}\rangle$ which correspond to the 7 valid epistemic states for an elementary system.

Suppose the claim has been proven for all n≤k and consider an epistemic state of an n=k+1 system. For the “if” part let S be a toy stabilizer subgroup. Consider a non-trivial partitioning into subsystems A and B of sizes n _A and n _B . Since n _A ,n _B <k we can assume the lemma holds for each. Therefore some outcome of some valid measurement of A corresponds to a valid epistemic state and hence a toy stabilizer subgroup for A, which we denote N′. Tensor the operators in N′ with an ${\mathcal {I}}^{\otimes n_{B}}$ to obtain the corresponding knowledge about the entire system, denoted N. It is now known that before the measurement the ontic state was in the support of P _N P _S =P _〈N∪S〉.

〈N∪S〉 may not represent a valid epistemic state since we have not yet taken into account the disturbance to A due to the measurement. But since system B is not disturbed the new epistemic state for B alone will be given by the subgroup S _B of 〈N∪S〉 that applies to B alone. It is useful to note that since G _n is abelian, 〈N∪S〉={gh|g∈N,h∈S}.

Since neither N nor S contain $-{\mathcal {I}}^{\otimes n}$, the only way $-{\mathcal {I}}^{\otimes n}$ could be in 〈N∪S〉 is if there is some g∈N with −g∈S. But then the measurement would never have returned the result it did. Hence $-{\mathcal {I}}^{\otimes n} \notin \langle N \cup S \rangle$.

Let g,h∈S _B , and decompose g=ab, h=cd with a,c∈N and b,d∈S. We have that a commutes with c and b commutes with d. Since a,c∈N are results from a measurement of A alone we can write $a = a_{A} \otimes {\mathcal {I}}^{\otimes n_{B}}$ and $c = c_{A} \otimes {\mathcal {I}}^{\otimes n_{B}}$ with $a_{A}, c_{A} \in G_{n_{A}}$. Since g,h∈S _B apply to B alone we must have b=a _A ⊗b _B and d=c _A ⊗d _B with $b_{B}, d_{B} \in G_{n_{B}}$. This demonstrates that a commutes with d and b commutes with c also. Therefore g and h commute. Hence S _B is a toy stabilizer subgroup, which represents a valid epistemic state by the inductive hypothesis. Therefore S represents a valid epistemic state.

For the “only if” part consider some valid epistemic state. First I show that the probability of obtaining an outcome from the measurement of a toy observable is always either 0, $\frac{1}{2}$ or 1. Let g ⁺ be the event of obtaining the +1 outcome from measuring some g∈G _n . Decompose g=a⊗b with a∈G ₁ and b∈G _n−1. If g v=v then, writing v=v _a ⊗v _b we must have a v _a =v _a ,b v _b =v _b or a v _a =−v _a and b v _b =−v _b . Hence

$$P\bigl(g^+\bigr) = P\bigl(a^+ \cap b^+\bigr) + P\bigl(a^- \cap b^-\bigr) = P\bigl(a^+\bigr)P\bigl(b^+|a^+\bigr) + P\bigl(a^-\bigr)P\bigl(b^-|a^-\bigr).$$

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All of the probabilities in the last expression are 0, $\frac{1}{2}$ or 1 by the inductive hypothesis and the fact that measuring a will result in a valid epistemic state for the other subsystem. Hence the only way P(g ⁺) could fail to be 0, $\frac{1}{2}$ or 1 is if $P(a^{+})= P(a^{-}) = \frac{1}{2}$ and $(P(b^{+}|a^{+}), P(b^{-}|a^{-})) \in\{(0, \frac{1}{2}),(\frac{1}{2}, 0), (\frac{1}{2}, 1), (1, \frac{1}{2})\}$. But by the inductive hypothesis $P(b^{+}) \in\{0, \frac{1}{2}, 1\}$, and we have

$$P\bigl(b^+\bigr) = P\bigl(b^+|a^+\bigr)P\bigl(a^+\bigr) + P\bigl(b^+|a^-\bigr)P\bigl(a^-\bigr),$$

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which would give a contradiction in each of those four cases.

We now construct a list of independent generators for the toy stabilizer subgroup corresponding to the valid epistemic state as follows. Identify the first elementary system as subsystem A and the rest as a subsystem B. Consider an ${\mathcal {I}}$ measurement on system A, which is valid by the inductive hypothesis. This is certain to return the outcome 1 and results in a valid epistemic state for B. By the inductive hypothesis this corresponds to a toy stabilizer subgroup S _B , which will have some list of independent generators. Tensor each element of this list with ${\mathcal {I}}$ to obtain the first generators for S.

Next consider an $\langle {\mathcal {X}}\rangle$ versus $\langle-{\mathcal {X}}\rangle$ measurement on A (i.e. a measurement of the ${\mathcal {X}}$ toy observable). If we know that only a single outcome, ±1 is possible then doing the measurement cannot give us any knowledge about system B that we didn’t already have. Therefore we just need to add $\pm {\mathcal {X}}_{A} \otimes {\mathcal {I}}^{\otimes(n-1)}$ to our generators for S (it is clear that this commutes with, and has a linearly independent check vector from, the existing elements of the list).

If either outcome is possible then they must result in valid epistemic states and hence toy stabilizer subgroups $S_{B}^{\pm}$. Suppose there is a $g \in S_{B}^{+}$ with neither g nor −g in $S_{B}^{-}$. Recall that the measurement of A does not disturb B. Consider a g measurement on system B and denote the event of the +1 outcome g ⁺. Denote the event of a ±1 outcome of the ${\mathcal {X}}$ measurement on A by ${\mathcal {X}}^{\pm }$. Then we have

$$P\bigl(g^+\bigr) = P\bigl(g^+|{\mathcal {X}}^+\bigr) \times P\bigl({\mathcal {X}}^+\bigr) + P\bigl(g^+|{\mathcal {X}}^-\bigr) \times P\bigl({\mathcal {X}}^-\bigr) = 1 \times\frac{1}{2} + \frac{1}{2} \times\frac{1}{2} =\frac{3}{4}$$

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contradicting the earlier proof that $P(g^{+}) \in\{0, \frac{1}{2}, 1\}$.

There is a similar contradiction if there is $g \in S_{B}^{-}$ with neither g nor −g in $S_{B}^{+}$. Hence (in the language of Sect. 5) $S_{B}^{-}$ is a rephasing of $S_{B}^{+}$. Therefore either $S_{B}^{+} =S_{B}^{-}$ (in which case the measurement provides no information about B and we do nothing), or we can write a list of independent generators for $S_{B}^{\pm}$ as g ₁,…,g _l−1,±g _l . g ₁,…,g _l−1 hold for either outcome so they must already be in S. Hence we just need to add ${\mathcal {X}}\otimes g_{l}$ to our generators for S. It is clear that the list still commutes and has linearly independent check vectors.

Now consider a ${\mathcal {Y}}$ measurement on A and repeat the process. If there is a new g _l , it must anticommute with the old (if present), since otherwise we could measure two commuting toy observables and determine the value of both ${\mathcal {X}}$ and ${\mathcal {Y}}$ on system A. Hence the new generator ${\mathcal {Y}}\otimes g_{l}$ commutes with the one added in the previous step, and certainly commutes with all the rest and has a linearly independent check vector.

Finally repeat the process for a ${\mathcal {Z}}$ measurement on A, but don’t add a generator with a check vector linearly dependent on the existing ones. Any such element will already be in the subgroup.

Consider the measurement of some toy observable g with neither g nor −g in S. It must either return one outcome with certainty, or either outcome with equal probability. In the former case it will be possible to find some contradiction with the above procedure, i.e. to find a reason why g or −g would have been added to S. The latter case is exactly what is predicted by the toy stabilizer subgroup S.

We have now constructed a toy stabilizer subgroup S whose corresponding epistemic state makes identical predictions about the expectation values of all toy observables. It remains to show that there is only one epistemic state with this property and hence S represents exactly the epistemic state we began with. This is analogous to the fact that knowing the expectation values of every Pauli observable uniquely determines a quantum state, and the proof is similar. We note that the elements $G_{n}^{+}$ of G _n that have α=1 are a basis for the real vector space of real diagonal 4ⁿ×4ⁿ matrices. This is true because there are 4ⁿ of them, and they are orthogonal under the trace inner product and hence linearly independent. Hence any probability distribution over the ontic states, written as a diagonal real 4ⁿ×4ⁿ matrix, is a linear combination of the $G_{n}^{+}$ where the coefficients are proportional to the expectation values of those observables. □

Proof of Lemma 3

For one system this can be seen by inspection.

Consider a four-outcome valid measurement on two systems and represent the outcomes, which are valid epistemic states, by toy stabilizer subgroups S ₁,S ₂,S ₃,S ₄. I claim there are three toy observables g,h ^± with {g,h ⁺}, {g,−h ⁺}, {−g,h ⁻}, {−g,−h ⁻} each contained in one of the S _k . Hence the measurement can be implemented by measuring g, and then, based on the outcome ±1, measuring h ^±. For example if $S_{1} = \langle {\mathcal {X}}_{1}{\mathcal {X}}_{2},{\mathcal {Y}}_{1}{\mathcal {Y}}_{2} \rangle$, $S_{2} = \langle-{\mathcal {X}}_{1}{\mathcal {X}}_{2},-{\mathcal {Y}}_{1}{\mathcal {Y}}_{2} \rangle$, $S_{3} =\langle {\mathcal {Z}}_{1}, -{\mathcal {Z}}_{2} \rangle$ and $S_{4} =\langle-{\mathcal {Z}}_{1}, {\mathcal {Z}}_{2} \rangle$ we can take $g = {\mathcal {Z}}_{1}{\mathcal {Z}}_{2}$, $h^{+} = {\mathcal {X}}_{1}{\mathcal {X}}_{2}$ and $h^{-} = {\mathcal {Z}}_{1}$.

First note that since the S _k are disjoint and cover all the ontic states

$$P_{S_1} + P_{S_2} + P_{S_3} + P_{S_4} ={\mathcal {I}}^{\otimes2}. $$

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Denote $S_{1} = \{ {\mathcal {I}}^{\otimes2}, a, b, ab \}$, where a and b are independent generators of S ₁. By (17) −a, −b and −ab must appear in the other S _k . Suppose without loss of generality that −a appears in S ₂ and denote $S_{2} = \{ {\mathcal {I}}^{\otimes 2}, -a, c, -ac \}$. If c=b then by (17) we have −b∈S ₃,S ₄ and so we can take g=b. Similarly if c=−b take g=ab, if c=ab take g=ab and if c=−ab take g=b. Suppose c is otherwise. Since −b, −c and −ac must be in S ₃ or S ₄, two of them must be in the same subgroup. If −c and −ac are in the same subgroup then a is also in it and so by (17) we can take g=a. If −b and −c are in the same subgroup then they must commute. But then {a,b,c} would be independent generators of a toy stabilizer subgroup, even though the maximum length of such a list is n=2. Similarly if −b and −ac being in the same subgroup creates a contradiction using {a,b,ac}.

Once we have found g, use (17) to verify that g must be in two of the subgroups and −g must be in the other two. Take the two subgroups with g in. To ensure that they represent disjoint states there must be some h ⁺ in one with −h ⁺ in the other. Define h ⁻ similarly.

The proofs for a valid measurement with two or three outcomes are simpler, but use similar observations. □

Proof of Lemma 5

We begin with the “if” part. For any θ∈ℝ and l∈{1,…,n−1} we have g _l |ψ〉=|ψ〉 and g _l |ψ′〉=|ψ′〉, hence

$$g_l \frac{1}{\sqrt{2}}\bigl(\vert \psi \rangle + e^{i\theta}\bigl \vert \psi' \bigr\rangle\bigr) = \frac{1}{\sqrt{2}}\bigl(\vert \psi \rangle + e^{i\theta}\bigl \vert \psi' \bigr\rangle\bigr).$$

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Next calculate the effect of h on |ψ〉. Since T is a stabilizer subgroup h must commute with g ₁,…,g _n−1. But it must anticommute with g _n since otherwise g ₁,…,g _n−1,g _n ,h would be a list of n+1 independent generators of a stabilizer subgroup, contradicting the maximum length of such a list being n. Therefore under conjugation by h, S is mapped to S′. Hence h|ψ〉=e ^iθ|ψ′〉 for some θ∈ℝ. Since h is Hermitian and unitary we also have he ^iθ|ψ′〉=|ψ〉. Therefore $h\frac{1}{\sqrt{2}}(\vert \psi \rangle + e^{i\theta} \vert \psi' \rangle) = \frac{1}{\sqrt{2}}(\vert \psi \rangle +e^{i\theta} \vert \psi' \rangle)$, and we have that T stabilizes this state.

Now for the “only if” part. Suppose S′ is not a rephasing of S. Then there exists some g∈S with ±g∉S′. Denote $\vert \phi(\theta) \rangle = \frac{1}{\sqrt{2}}(\vert \psi \rangle + e^{i\theta} \vert \psi' \rangle$ and calculate

$$\bigl\langle\phi(\theta) \bigr| g \bigl| \phi(\theta) \bigr\rangle = \frac{ \langle\psi|g|\psi \rangle + \langle\psi'|g|\psi' \rangle + e^{i\theta}\langle\psi|g|\psi' \rangle +e^{-i\theta} \langle\psi'|g|\psi \rangle}{2} =\frac{1}{2},$$

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which contradicts the expectation value of any Pauli observable in a stabilizer state being −1,0 or 1.

So S′ is a rephasing of S. Since S′≠S we can therefore write S=〈g ₁,…,g _n−1,g _n 〉 and S=〈g ₁,…,g _n−1,−g _n 〉. It is easy to check that for l∈{1,…,n−1} we have g _l |ϕ(θ)〉=|ϕ(θ)〉 and hence g _l ∈T. Therefore we can take g ₁,…,g _n−1 as the first n−1 independent generators of T. Write the final independent generator as h. Suppose h∈S. Since h∉〈g ₁,…,g _n−1〉, and S′ is a rephasing of S, we must have −h∈S. But then h|ϕ(θ)〉=|ϕ(θ+π)〉 contradicting h being a stabilizer of |ϕ(θ)〉. Similarly if h∈S′. Therefore h∉S∪S′. □

Proof of Lemma 6

We closely follow the proof of Proposition 1 in [10]. The number of distinct stabilizer subgroups of the required form is G/A where G is the number of choices for the final generator h, and A is the number of choices of h giving rise to the same subgroup.

First we calculate G. Ignoring phases there are 4ⁿ choices in G _n . But h must commute with g ₁,…,g _n−1 which gives 4ⁿ/2ⁿ⁻¹ options. Also, it must not be in the subgroup generated by g ₁,…,g _n−1, which gives 4ⁿ/2ⁿ⁻¹−2ⁿ⁻¹. Finally, it must not be in S or S′, i.e. it cannot be g _n multiplied by something generated by g ₁,…,g _n−1. There are two possible overall phases ±1. This gives G=2(4ⁿ/2ⁿ⁻¹−2ⁿ⁻¹−2ⁿ⁻¹)=2(4×2ⁿ⁻¹−2×2ⁿ⁻¹)=2ⁿ⁺¹.

Next we calculate A. Fix a stabilizer subgroup. Then for h we can choose any of the elements of the stabilizer subgroup not generated by g ₁,…,g _n−1. This gives A=2ⁿ−2ⁿ⁻¹=2ⁿ⁻¹.

Finally we calculate G/A=4 as required. □

Appendix B: Relation to Spekkens’ New Formulation

Spekkens’ has previously outlined [8] a new formulation of the toy theory which is very closely related to the stabilizer notation presented here. The new formulation is based around “linear functionals” or “canonical variables”. These can be put in 1-to-1 correspondence with the elements of G _n with α=1. For an elementary system, the canonical variables are 0, X, P, and X+P which may be taken to correspond to ${\mathcal {I}}$, ${\mathcal {X}}$, ${\mathcal {Z}}$ and ${\mathcal {Y}}$ respectively. The correspondence for composite systems can then be built up from this, so that X ₁+X ₃+P ₃ corresponds to ${\mathcal {X}}\otimes {\mathcal {I}}\otimes {\mathcal {Y}}$ and so on. The toy stabilizer notation combines canonical variables with their values, for example having ${\mathcal {X}}\otimes {\mathcal {Z}}$ in a toy stabilizer subgroup represents knowledge that X ₁+P ₂=0 whereas $-{\mathcal {X}}\otimes {\mathcal {Z}}$ represents the knowledge X ₁+P ₂=1.

The new formulation’s “Poisson bracket” condition for jointly-knowable variables is exactly the usual condition on the check vectors of commuting observables.

Measurements in the new formulation determine the value of some set of jointly-knowable variables, or equivalently a set of commuting toy observables. Therefore Lemmas 3 and 4 show that for one or two systems the notion of measurement is identical in both formulations, whereas for three or more systems not all of the measurements in the original formulation are included in the new formulation.

The new formulation can be compared to qubit quantum mechanics by using the discrete Wigner representation [16, 17] of the latter. From this perspective the difference between the two theories comes from the fact the discrete Wigner function is sometimes negative, whereas the toy theory only uses positive probabilities.