Análisis AMOFHIT

We now want to find the update rules for the state (V, w) of a prime dimensional system when we perform a measurement Π described by the subspace VΠ, with generators Σ0_i, and by the

representative ontic vectors rj associated to the different outcomes σj0. These vectors rj can be

obtained from each other by a shift vector (see figure2.3), therefore we will consider the generic measurement element represented by (VΠ, r) for stating the measurement update rules. The

subspace of known variables V can be written in terms of the sets generated by the generators Poisson-commuting with all the Σ0

j, Vcommute, and non-commuting ones, Vother. According to

this definition Vcommute will always be a subspace. We cannot state the same for Vother, since

the null vector does not belong to it. For this reason we augment Vother with the null vector in

order to create a subspace. This implies that we can decompose V as

V = Vcommute⊕ Vother. (2.36)

We will now provide the update rules both for V and w in two steps: first considering the state and measurement to commute, and then the general (non-commuting) case.

Theorem 1. Commuting case. Given the epistemic state (V, w) and the measurement Π that commutes with it, i.e. the generators of V and VΠ all Poisson commute, the epistemic state

(V0, w0) after the outcome σ0 associated to the measurement element (VΠ, r) has occurred, is

described by

V0⊥ = (V⊥+ w− w0)∩ (VΠ⊥+ r− w 0

where w0 _{is given by equation}

w0= w +X

i

Σ0T_i (r_{− w)γ}i, (2.38)

whereΣ0_i are the generators of the measurementΠ and the vectors γj are such thatΣ0Ti γj = δi,j.

Proof. When the state and measurement commute we have to add the generators of the mea- surement to the set of generators of V, as we have seen in the previous subsection2.2.1(learning stage). Therefore the update rule for the subspace V is (equation (2.32))

V → V0= V ⊕ span{Σ00, Σ 0 1, . . . Σ

0

i, . . .} = V ⊕ VΠ. (2.39)

In terms of perpendicular subspaces this implies that V0⊥ = V⊥_{∩ VΠ}⊥.

Let us initially assume the measurement to consist only of one generator Σ0_{. Let us recall}

that the outcome associated with Σ0 is σ0. We assume w is not compatible with this outcome, i.e. Σ0T_{w = σ}0_{+ x, for some shift x∈ Z}

d, and we want to find w0 such that

Σ0Tw0 = σ0. (2.40)

The identity (2.35) we used in the previous section does the job. More precisely,

w0 = w− xγ,

where the vector γ is such that Σ0Tγ= 1. The above expression can be also written as

w0= w_{− k}−1xΣ0,

where k = Σ0T_Σ0_{. The inverse of k always exists because we are in the prime dimensional case.}

Without referring to x we can restate the update rule for the representative ontic vector as

w→ w + k−1(σ0− Σ0Tw)Σ0 = w + k−1Σ0T(r− w)Σ0. (2.41)

over all those generators in the second term. This immediately follows from considering the whole measurement Π as a sequence of measurements given by each generator Σ0_i and apply every time the rule (2.41). In addition, we need also to require that the γj’s are such that

Σ0T_i γj = δi,j, so that Σ0Ti w0 gives σ0 without unwanted additional terms. We state again that

the above formula always holds for prime dimensional systems. We cannot claim the same in non-prime dimensions. The correct update rule for the subspace V0⊥ is found by combining the update rules for V and w as in (2.37). This correction simply sets the subspaces to the same origin in order to correctly compute their intersection, as schematically shown in figure2.8. At the end we obtain for the epistemic state (V0, w0) that V0⊥+ w0 = (V⊥+ w)_{∩ (VΠ}⊥+ r). We recall that the probability associated to each ontic state consistent with the epistemic state is uniform, i.e. given by 1

|V0⊥_+w0_| = _|V10⊥_| = _|(V⊥_+w)∩(V1 ⊥

Π+r)|

, where _{| · | indicates the size of the} subspace.

Figure 2.7shows a basic example of theorem1.

Theorem 2. Non-commuting case. Given the epistemic state (V, w) and the measurement Π that does not commute with it, i.e. some of the generators of VΠ do not Poisson commute

with the generators of V , the epistemic state (V0, w0) after the outcome σ0 associated to the measurement element (VΠ, r) has occurred, is described by

V0⊥ = (V_commute⊥ + w_{− w}0)_{∩ (VΠ}⊥+ r_{− w}0), (2.42)

where V_commute⊥ is given by

V_commute⊥ = V⊥_{⊕ V}other. (2.43)

The representative ontic vector w0 _{is given by}

w0= w +X

i

Σ0T_i (r_{− w)γ}i, (2.44)

where Σ0

i are the generators (even the non-commuting ones) of the measurement Π and the

Nothing known State P=0 P=0 Measurement (V, w) (V0, w0) State after measurement (V⇧, r)

Figure 2.7: Update rules in the prime commuting case. The figure above shows a simple one-trit example of theorem 1 regarding the update rule to predict the state after a sharp measurement that commutes with the original state. The state after measurement is given by V0⊥ + w0 = (V⊥ + w)∩ (V⊥

Π + r). In the above case the shift vectors are all (0, 0), the

perpendicular subspaces are V⊥ _{= Ω, V}⊥

Π = span{(1, 0)}, and V0⊥ = VΠ⊥. Note that with

“measurement” we are here representing one element of the measurement. The other elements can be obtained by simply shifting r as seen in figure 2.3. The final state is associated to each element of the measurement, each one with a corresponding probability of happening. The same reasoning holds for figures 2.9and 2.12.

Proof. Let us assume that Σ0_j, for j ∈ {0, . . . , m − 1}, do not commute with the generators of V. In addition to the learning stage of the previous commuting case, we also have a removal stage of the disturbing part of the measurement. We have already seen that we can split the subspace V in V = Vcommute⊕Vother. Therefore we can reduce to the commuting case if we only

consider Vcommute instead of the whole V. The update rule for the subspace V then becomes

V _{→ V}0= Vcommute⊕ span{Σ00, Σ

0 1, . . . Σ

0

i, . . .} = Vcommute⊕ VΠ.

In terms of the perpendicular subspaces note that we can both write

⌦

V

Figure 2.8: Update rules via Venn diagrams. The figure above schematically shows the subspaces V⊥_{, V}⊥

Π, V0⊥ and the shifted ones (after applying the corresponding representative

ontic vectors w, r, w0). In particular this picture explains the expression V0⊥ = (V⊥+ w−w0_)∩

(V_Π⊥+ r_{− w}0) as a result of combining the update rules for the epistemic subspaces and the representative ontic vectors. It is important to notice that to obtain the correct intersection we have to shift the subspaces V⊥+ w and VΠ+ r back to the same origin (this is the role of w0).

Indeed note that V⊥_{∩ VΠ}⊥ is different from (V⊥+ w)_{∩ (VΠ}⊥+ r). and

V0⊥= V_commute⊥ _{∩ VΠ}⊥,

from the usual property that the perpendicular of a direct sum is the intersection of the perpen- dicular subspaces. The update rule for the representative ontic vector is the same as in the previ- ous case (equation (2.38)). The correct update rule for the subspace V0⊥ is found by combining the update rules for V and w as in the previous case (2.37), where V⊥ _{is replaced by V}⊥

commute.

At the end we obtain for the epistemic state (V0, w0) that V0⊥+w0 = (V_commute⊥ +w)_∩(VΠ⊥+r).

P=0 State X-P=0 Measurement (V, w) (V0, w0) State after measurement X-P=0 (V⇧, r)

Figure 2.9: Update rules in the prime non-commuting case. The figure above shows a simple one-trit example of theorem2regarding the update rule to predict the state after a sharp measurement that does not commute with the original state. The state after measurement is given by V0⊥+ w0 = (V_commute⊥ + w)∩ (V⊥

Π + r). In the above case the shift vectors are all (0, 0),

the perpendicular subspaces are V⊥

commute= Ω, VΠ⊥= span{(1, 1)}, and V0⊥= VΠ⊥.