Passing on the norm approach

Here we would like to look at a different way of exploring our framework of Sec-tion19.1. This will enable us to find an alternative proof of Conjecture2for dimension 3. This proof will have the advantage to partially generalize to dimension 4. This approach is based on the following observation.

Since we are only interested in reaching marginals ˜pthat are thermal at inverse temperatureβ⁰ < β_R, what we really are interested in is to reach marginals of the following form One way to transformpinto ˜pis to transform eachr_i into its correspondingb_i. However, since doubly stochastic transformations conserve the norm and since in generalkr_ik 6= kb_ik_{, where}k·kdenotes the one norm, i.e.,

kvk=

d−1 i

∑

|v_i|, (19.65)

one cannot perform this transformation directly withM_i. What one can do with M_i instead is to transform the normalizedr_i into the normalizedb_i, as indeed the following holds.

Lemma 26. For any dimension d and Hamiltonian r_i

kr_ik ^bi

kb_ik^, ∀i=0, . . .k. (19.66) Proof. The proof is quite simple. We have

(r_i)_j = p_j(j+i)= ^e

Chapter 19. Symmetric Mixed State Framework 105 What Lemma26tells us is that we are able to transform ther_i’s into theb_i’s, just not in the right amount. Whenkr_ik>kb_ik, we will end up with too muchb_iif one fully transformsr_i intob_i as prescribed by Lemma26. And so, one can instead use the excessiver_ito try and create someb_jfor which Lemma26does not allow to create enough of. One succeeds if the norm can adequately be passed across the different subspaces as such.

Oner_ithat always has excessive norm isr0as indeed

Lemma 27. For all Hamiltonians, inverse temperatureβand dimension d

kr₀k ≥ kb₀k, ∀β⁰ ≤β. (19.71) Proof idea. The proof consists in verifying that∂_βkr₀k ≥0. See Appendix VII of [112]

for more details.

In dimension 3, this excessive norm can also adequately be passed to the otherb_i as the following holds.

Lemma 28. For d=3, any Hamiltonian and inverse background temperatureβ.

r₀

kr0k (₁+_Π)b₁

2kb₁k ^, ∀β⁰ < β. (19.72) Proof idea. The proof consists of two steps. First, one proves that

r₀

kr₀k (₁+_Π)r₁

2kr₁k ^{. (19.73)}

One then proceeds to prove that

(₁+_Π)r₁

2kr₁k (₁+_Π)b₁

2kb₁k ^{. (19.74)}

Using the transitivity of majorization, we get the desired result. The detailed proof of each inequality is found in Appendix IX of [112]. Note in particular that 19.74does not merely follow from

r₁

kr₁k ^b1

kb₁k^{. (19.75)}

Lemma26, Lemma27and Lemma28together prove Conjecture2for dimension 3.

Indeed, according to Lemma26and Lemma28, there exist doubly stochastic matrices M_r0→b0,M_r1→b1, andM_r0→b1 such that

M_r0→b0r₀ = kr₀k

kb₀k^{b0, (19.76)}

M_r1→b1r₁ = kr₁k

kb₁k^{b1, (19.77)}

M_r0→b1r₀ = kr₀k

2kb1k(₁+_Π)b₁. (19.78)

Chapter 19. Symmetric Mixed State Framework 106

as desired. We have therefore proven the following.

Theorem 20. For dimension 3 the “passing on the norm” approach allows to reach any marginalp˜ thermal atβ⁰ <β_Rand therefore proves Conjecture2for d=3.

There are various ways that this approach can unfold to higher dimensions. The fact that Lemma26as well as Lemma27hold for any dimension already gives a good basis to start from. However, the result of Lemma28proves more difficult to generalize. To this end, note that we choose to shift the excessive norm ofr₀ to (₁+_Π)b₁and not tob₁andΠb1separately. This is simply because shifting the norm separately is not always possible to do. Indeed, already in dimension 3

r₀

kr₀k ^b1

kb₁k ^(19.85)

does not always hold. For example forE₂ →_∞,E₀= E₁ =0, the greatest component ofr0/kr0kis ¹₂ while that of b₁/kb₁kis 1. This tells us that shifting the norm to a group of terms is advantageous. The fact that for any dimension

r₀

holds, see the proof below, suggests that a fruitful generalization of Lemma28for higher dimension might be to group as many terms as possible together, namely all ther_i 6=r₀. However, shifting the norm fromr₀to all the otherr_iuniformly oversees the fact that differentb_ineed to have their norm compensated by a different amount.

With this given, there are 3 potential avenues we see to generalize the approach to higher dimensions.

1. Investigate the possibility of shifting some norm between the differentr_i,i6=0 as well as from somer_i,i6=0 tor₀such that a subsequent uniform shift from r₀to all ther_i’s delivers the desired state. In doing so, one has to keep in mind that(₁+_Πⁱ)mixesr_i before its norm can be shifted, which limits the shifting possibilities.

Chapter 19. Symmetric Mixed State Framework 107 2. Shift norm fromr₀to all ther_ibut not uniformly, such that the shifting results

in the desired state. This amounts to checking whether r₀ b₀+

∑

(b²ⁱ_dc+1)⁻¹

1− kr_ik kb_ik

(₁+_Πⁱ)b_i, (19.87) holds.

3. Shift norm fromr₀to the simplest non-trivial grouping ofr_i, namely

(₁+_Πⁱ)r_i. (19.88)

We found option 3 the most tractable alternative and therefore opted for that avenue. Doing so, we were able to prove the following.

Lemma 29. In dimension 4, for any Hamiltonian H = _∑³_i=0E_i|ii hi|such thatδ_i+1 ≤ δ_i, whereδ_i =E_i1−E_i, the following holds for i=1, 2.

r₀

kr₀k (₁+_Πⁱ)b_i

2kb_ik ^{, (19.89)}

kr_ik ≤ kb_ik. (19.90) Proof idea. The idea of the proof is the same as that of Lemma28and Lemma27. We prove the majorization relation in two steps, namely

r₀

kr₀k (₁+_Πⁱ)r_i

2kr_ik ^(19.91)

and

(₁+_Πⁱ)r_i

2kr_ik (₁+_Πⁱ)r_i

2kr_ik ^{. (19.92)}

The inequality is proven by showing that∂_βkr_ik ≤0. For the details of the proof we refer to Appendix IX of [112].

Lemma 29fails to be true in the regime δ_i+1 < δ_i as one may find a counter example to the relation

r₀

kr₀k (₁+_Π)r₁

2kr₁k ^(19.93)

in that regime. See Appendix IX of [112] for more details. With the result of Lemma29 we are able to prove the following for dimension 4.

Theorem 21. For dimension 4 and Hamiltonians H =_∑³_i=0E_i|ii hi|such thatδ_i+1 ≤ δ_i, the “passing on the norm” approach allows to reach any marginalp˜ thermal atβ⁰ <β_Rand therefore proves Conjecture2in the restricted case ofδ_i+1 ≤δ_i for d=4.

Chapter 19. Symmetric Mixed State Framework 108 Proof. The results of Lemma26and Lemma29ensure the existence of doubly stochas-tic matricesM_r0→b0,M₁, M2,M_r0→b1 andM_r0→b2 such that

M_r0→b0r0= kr0k

kb₀k^{b0, (19.94)}

M₁r₁= kr₁k

kb₁k^{b1, (19.95)}

M₂r₂= kr₂k

kb₂k^{b2, (19.96)}

M_r0→b1r₀= kr₀k

2kb₁k(₁+_Π)b₁, (19.97) M_r0→b2r0= kr0k

2kb₂k(₁+_Π²)b2. (19.98) Now let

M₀ =α₁M_r0→b0+α₂M_r0→b1+α₃M_r0→b2, (19.99) with

α₁= kb₀k

kr₀k^{, (19.100)}

α₂=2kb₁k − kr₁k

kr0k ^{, (19.101)}

α₃= kb₂k − kr₂k

kr₀k ^{. (19.102)}

Lemma29ensures thatα₂≥0 andα₃ ≥0. Furthermore, using thatkb₀k+2kb₁k+ kb₂k=kr₀k+2kr₁k+kr₂k=1 we have

α₁+α2+α3=1. (19.103)

SoM₀is doubly stochastic as required and

˜

p= M₀r₀+ (₁+_Π)M₁r₁+¹

2(₁+_Π²)M₂r₂ (19.104)

=b₀+ ^α2kr₀k+2kr₁k

2kb₁k (₁+_Π)b₁+^α3kr₀k+kr₂k kb₂k

1+_Π²

2 b₂ (19.105)

=b₀+ (₁+_Π)b₁+¹+_Π²

2 b₂, (19.106)

as desired.

For completeness, as well as for the sake of potential future investigations, before moving on to the next approach we prove the result claimed in Eq.19.86, namely that Lemma 30.

r₀

kr0k ^s

ksk^{, where s}=

d−1

∑

r_i, (19.107)

Chapter 19. Symmetric Mixed State Framework 109

The equality condition is ensured since bothxandyare normalised. Alternatively, it is also easily directly verifiable. To verify the inequality conditions, note that since the p_i’s form a probability distribution, i.e. 0 ≤ p_i ≤ 1 and∑^d_i=⁻0¹p_i = 1, and that

Chapter 19. Symmetric Mixed State Framework 110