# Blog

## Ultimate speed limits to the growth of operator complexity

11:31 14 agosto in Artículos por Website

### Vanishing of the anticommutator contribution in the Robertson uncertainty relation for $${{{{{{{\mathcal{K}}}}}}}}$$ and $${{{{{{{\mathcal{L}}}}}}}}$$

We establish a universal feature of Krylov complexity, valid for any physical system: namely, that its anticommutator with the Liouvillian $${{{{{{{\mathcal{L}}}}}}}}$$ has vanishing expectation value over the evolved operator $$|{{{{{{{\mathcal{O}}}}}}}}(t)\left.\right)$$. The relevance of this result relies on the fact that this quantity enters the Schrödinger uncertainty principle for the two operators $${{{{{{{\mathcal{K}}}}}}}}$$ and $${{{{{{{\mathcal{L}}}}}}}}$$

$$4{(\Delta {{{{{{{\mathcal{K}}}}}}}}\Delta {{{{{{{\mathcal{L}}}}}}}})}^{2} \ge {\left|\left({{{{{{{\mathcal{O}}}}}}}}(t)\right|[{{{{{{{\mathcal{K}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}]\left|{{{{{{{\mathcal{O}}}}}}}}(t)\right)\right|}^{2}\\ +{\left|\left({{{{{{{\mathcal{O}}}}}}}}(t)\right|\{{{{{{{{\mathcal{K}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}\}\left|{{{{{{{\mathcal{O}}}}}}}}(t)\right)\right|}^{2},$$

(7)

from which one can bound the complexity rate ∂tK. We have that

$$\big({{{{{{{\mathcal{O}}}}}}}}(t)\big|[{{{{{{{\mathcal{K}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}]\big|{{{{{{{\mathcal{O}}}}}}}}(t)\big)=2i\,{{{{{{\mathrm{Im}}}}}}}\,\left({{{{{{{\mathcal{O}}}}}}}}(t)\big|{{{{{{{\mathcal{K}}}}}}}}{{{{{{{\mathcal{L}}}}}}}}\big|{{{{{{{\mathcal{O}}}}}}}}(t)\right)$$

(8)

and

$$\big({{{{{{{\mathcal{O}}}}}}}}(t)\big|\{{{{{{{{\mathcal{K}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}\}\big|{{{{{{{\mathcal{O}}}}}}}}(t)\big)=2{{{{{{\mathrm{Re}}}}}}}\,\left({{{{{{{\mathcal{O}}}}}}}}(t)\big|{{{{{{{\mathcal{K}}}}}}}}{{{{{{{\mathcal{L}}}}}}}}\big|{{{{{{{\mathcal{O}}}}}}}}(t)\right),$$

(9)

where

$${{{{{{{\mathcal{K}}}}}}}}{{{{{{{\mathcal{L}}}}}}}}=\mathop{\sum }\limits_{n=0}^{D-1}{b}_{n+1}\left[n\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right|+(n+1)\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n}\right|\right].$$

(10)

Let us now demonstrate that the anticomutator term in Eq. (9) is identically zero. By expanding $$|{{{{{{{\mathcal{O}}}}}}}}(t)\left.\right)$$ over the Krylov basis, we obtain

$$\big({{{{{{{\mathcal{O}}}}}}}}(t)\big|{{{{{{{\mathcal{K}}}}}}}}{{{{{{{\mathcal{L}}}}}}}}\big|{{{{{{{\mathcal{O}}}}}}}}(t)\big)= \mathop{\sum }\limits_{m,n,k=0}^{D-1}{(-i)}^{m}{i}^{k}{\varphi }_{m}{\varphi }_{k}{b}_{n+1}\big(n{\delta }_{mn}{\delta }_{n+1,k}\\ +(n+1){\delta }_{m,n+1}{\delta }_{nk}\big),$$

(11)

which, by performing the sums over k and n, yields

$$\big({{{{{{{\mathcal{O}}}}}}}}(t)\big|{{{{{{{\mathcal{K}}}}}}}}{{{{{{{\mathcal{L}}}}}}}}\big|{{{{{{{\mathcal{O}}}}}}}}(t)\big) = i\mathop{\sum }\limits_{m=0}^{D-1}m{\varphi }_{m}({\varphi }_{m+1}{b}_{m+1}-{\varphi }_{m-1}{b}_{m}) \\ =-i\mathop{\sum }\limits_{m=0}^{D-1}m{\varphi }_{m}{\partial }_{t}{\varphi }_{m}.$$

(12)

Since the amplitudes φn and the coefficients bn are real quantities, comparing Eqs. (9) and (12) we immediately conclude that

$$\left({{{{{{{\mathcal{O}}}}}}}}(t)\right|\{{{{{{{{\mathcal{K}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}\}\left|{{{{{{{\mathcal{O}}}}}}}}(t)\right)=0\quad \forall t.$$

(13)

Let us note that the key condition to obtain this result is the fact that the Liouvillian connects only states that are nearest neighbors on the Krylov lattice so that we are left with a purely imaginary phase (−i)m(i)m±1 = ±i. It is this peculiar property that allows the Liouvillian to be interpreted as a sum of generalized ladder operators $${{{{{{{{\mathcal{L}}}}}}}}}_{\pm }$$42. However, let us point out that here we are not making any assumption regarding the commutation rules between these operators: we are considering the structure of Krylov space in full generality.

Moreover, from Eq. (8), we immediately obtain the relation between the anticommutator $$[{{{{{{{\mathcal{K}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}]$$ and the complexity rate ∂tK:

$$\big({{{{{{{\mathcal{O}}}}}}}}(t)\big|[{{{{{{{\mathcal{K}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}]\big|{{{{{{{\mathcal{O}}}}}}}}(t)\big)=-2i\mathop{\sum }\limits_{m=0}^{D-1}m{\varphi }_{m}{\partial }_{t}{\varphi }_{m}=-i{\partial }_{t}K.$$

(14)

Therefore, the Schrödinger uncertainty relation (7) can be recast as the dispersion bound (2) on the growth of Krylov complexity:

$$| {\partial }_{t}K| \le 2{b}_{1}\Delta {{{{{{{\mathcal{K}}}}}}}}.$$

(15)

### Geometrical interpretation of the saturation of the bound

For the geometrical interpretation of the saturation of the bound, we assume the Krylov space to be of finite dimension. However, the results could potentially be extended to infinite-dimensional Krylov spaces as well.

The Krylov space is isomorphic to a 2D-dimensional real vector space, and we can therefore consider the Euclidean metric g, given by the real part of the inner product. The evolution curve of $${{{{{{{\mathcal{O}}}}}}}}$$ will then be restricted to the unit sphere of the Krylov space. This unit sphere forms a Riemannian manifold and we can consider the Krylov complexity as a function on this manifold defined by $$K({{{{{{{\mathcal{A}}}}}}}})=({{{{{{{\mathcal{A}}}}}}}}| {{{{{{{\mathcal{K}}}}}}}}{{{{{{{\mathcal{A}}}}}}}}),$$ for any element $$\left|{{{{{{{\mathcal{A}}}}}}}}\right)$$ in the Krylov space with a unit norm. In this sense, when we write K(t) we simply mean $$K({{{{{{{\mathcal{O}}}}}}}}(t))$$ which is consistent with how we defined complexity for the evolution. The differential of Krylov complexity will be denoted by dK and its action on any tangent vector $${\dot{{{\mathcal{A}}}}}$$ at $${{{{{{{\mathcal{A}}}}}}}}$$ is given by $$dK({{{{{\dot{{{\mathcal{A}}}}}}}}})= ({{{{{\dot{{{\mathcal{A}}}}}}}}}| {{{{{{{\mathcal{A}}}}}}}})+ ({{{{{{{\mathcal{A}}}}}}}}| {{{{{\dot{{{\mathcal{A}}}}}}}}})$$. This differential together with the metric can be used to define the gradient of Krylov complexity. It follows from the theory of differential geometry that the gradient of Krylov complexity at $${{{{{{{\mathcal{A}}}}}}}}$$, denoted by $$\nabla K({{{{{{{\mathcal{A}}}}}}}})$$, is the unique vector satisfying the expression $$g(\nabla K({{{{{{{\mathcal{A}}}}}}}}),{{{{{\dot{{{\mathcal{A}}}}}}}}})=dK({{{{{\dot{{{\mathcal{A}}}}}}}}})$$ for all tangent vectors $${{{{{\dot{{{\mathcal{A}}}}}}}}}$$ at $${{{{{{{\mathcal{A}}}}}}}}$$48. It can be checked that the gradient must then be given by $$\nabla K({{{{{{{\mathcal{A}}}}}}}})=2({{{{{{{\mathcal{K}}}}}}}}-\left\langle {{{{{{{\mathcal{K}}}}}}}}\right\rangle ){{{{{{{\mathcal{A}}}}}}}}$$, which indeed is tangent to the unit sphere at $${{{{{{{\mathcal{A}}}}}}}}$$. The change of Krylov complexity along the curve $${{{{{{{\mathcal{O}}}}}}}}(t)$$, generated by the Liouvillian, is given by $${\partial }_{t}K(t)=g(\nabla K(t),{\partial }_{t}{{{{{{{\mathcal{O}}}}}}}}(t))$$, where K(t) is the gradient at $${{{{{{{\mathcal{O}}}}}}}}(t)$$. Applying the Cauchy-Schwarz inequality on the right-hand side gives us the inequality

$$\left|{\partial }_{t}K(t)\right|\le \parallel \!\!\nabla K(t)\!\!\parallel \parallel \!\!{\partial }_{t}{{{{{{{\mathcal{O}}}}}}}}(t)\!\!\parallel .$$

(16)

The right-hand side of this inequality is exactly $$2{b}_{1}\Delta {{{{{{{\mathcal{K}}}}}}}}$$ and we note that it is saturated if and only if the tangent vector of $${{{{{{{\mathcal{O}}}}}}}}(t)$$ is parallel to the gradient of Krylov complexity. We also note that the gradient is the zero vector at time zero and so the dispersion bound is always initially saturated.

The unitary orbit of $${{{{{{{\mathcal{O}}}}}}}}$$ is the set of all points $${U}^{{{{\dagger}}} }{{{{{{{\mathcal{O}}}}}}}}U$$, where U is a unitary operator. We emphasize that this is a proper subset of the unit sphere in Krylov space which, in contrast, is the set of all points $${{{{{{{\mathcal{U}}}}}}}}{{{{{{{\mathcal{O}}}}}}}}$$, where $${{{{{{{\mathcal{U}}}}}}}}$$ is a unitary superoperator. The gradient we have considered is with respect to the unit sphere, and it is therefore not obvious that this gradient will ever be tangential to the unitary orbit of $${{{{{{{\mathcal{O}}}}}}}}$$. However, the gradient is indeed tangential to the unitary orbit at time zero and at all times, provided the simplicity algebra is fulfilled.

### On the closure of the complexity algebra

Here we show the proof that the only possible closure of the complexity algebra introduced by42 is given by Eq. (3). The (anti-Hermitian) operator $${{{{{{{\mathcal{B}}}}}}}}={{{{{{{{\mathcal{L}}}}}}}}}_{+}-{{{{{{{{\mathcal{L}}}}}}}}}_{-}$$ “conjugated” to the Liouvillian can be expanded in Krylov space as

$${{{{{{{\mathcal{B}}}}}}}}=\mathop{\sum }\limits_{n=0}^{D-1}{b}_{n+1}\left[\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n}\right|-\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right|\right].$$

(17)

We note that one can establish a formal analogy with the harmonic oscillator: $${{{{{{{\mathcal{L}}}}}}}}$$ plays the role of the position of the harmonic oscillator, while $$i{{{{{{{\mathcal{B}}}}}}}}$$ corresponds to its momentum. However, in general, the commutator between $${{{{{{{\mathcal{L}}}}}}}}$$ and $${{{{{{{\mathcal{B}}}}}}}}$$ is not proportional to the identity, indeed:

$$\tilde{{{{{{{{\mathcal{K}}}}}}}}}=2[{{{{{{{{\mathcal{L}}}}}}}}}_{+},{{{{{{{{\mathcal{L}}}}}}}}}_{-}]=2\mathop{\sum }\limits_{n=0}^{D-1}\big({b}_{n+1}^{2}-{b}_{n}^{2}\big)\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n}\right|,$$

(18)

where it is understood that b0 has to be replaced with 0. Let us now investigate the conditions under which $${{{{{{{\mathcal{L}}}}}}}}$$, $${{{{{{{\mathcal{B}}}}}}}}$$ and $$\tilde{{{{{{{{\mathcal{K}}}}}}}}}$$ form a closed algebra with respect to the operation [,]: the so-called complexity algebra42. This happens if and only if the commutators $$[{{{{{{{\mathcal{L}}}}}}}},\tilde{{{{{{{{\mathcal{K}}}}}}}}}]$$ and $$[{{{{{{{\mathcal{B}}}}}}}},\tilde{{{{{{{{\mathcal{K}}}}}}}}}]$$ can be written as linear combinations of the operators $${{{{{{{\mathcal{L}}}}}}}}$$, $${{{{{{{\mathcal{B}}}}}}}}$$ and $$\tilde{{{{{{{{\mathcal{K}}}}}}}}}$$ themselves. These commutators can be expanded over the Krylov basis as follows:

$$[{{{{{{{\mathcal{L}}}}}}}},\tilde{{{{{{{{\mathcal{K}}}}}}}}}]=2\mathop{\sum }\limits_{n=0}^{D-1}f(n){b}_{n+1}\left[\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n}\right|-\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right|\right],$$

(19)

$$[{{{{{{{\mathcal{B}}}}}}}},\tilde{{{{{{{{\mathcal{K}}}}}}}}}]=2\mathop{\sum }\limits_{n=0}^{D-1}f(n){b}_{n+1}\left[\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n}\right|+\left|{{{{{{{{\mathcal{O}}}}}}}}}_{n}\right)\left({{{{{{{{\mathcal{O}}}}}}}}}_{n+1}\right|\right],$$

(20)

where we have defined

$$f(n)={b}_{n+1}^{2}-{b}_{n}^{2}-\big({b}_{n+2}^{2}-{b}_{n+1}^{2}\big)=\frac{{\tilde{{{{{{{{\mathcal{K}}}}}}}}}}_{nn}-{\tilde{{{{{{{{\mathcal{K}}}}}}}}}}_{n+1,n+1}}{2}.$$

(21)

Now, it is clear that the commutator (19) between $${{{{{{{\mathcal{L}}}}}}}}$$ and $$\tilde{{{{{{{{\mathcal{K}}}}}}}}}$$ cannot contain any element of the complexity algebra other than $${{{{{{{\mathcal{B}}}}}}}}=\mathop{\sum }\nolimits_{n = 0}^{D-1}{b}_{n+1}[|{{{{{{{{\mathcal{O}}}}}}}}}_{n+1})({{{{{{{{\mathcal{O}}}}}}}}}_{n}|-|{{{{{{{{\mathcal{O}}}}}}}}}_{n})({{{{{{{{\mathcal{O}}}}}}}}}_{n+1}|]$$, while the commutator (20) can only contain $${{{{{{{\mathcal{L}}}}}}}}=\mathop{\sum }\nolimits_{n = 0}^{D-1}{b}_{n+1}[|{{{{{{{{\mathcal{O}}}}}}}}}_{n+1})({{{{{{{{\mathcal{O}}}}}}}}}_{n}|+|{{{{{{{{\mathcal{O}}}}}}}}}_{n})({{{{{{{{\mathcal{O}}}}}}}}}_{n+1}|]$$. Moreover, the only possibility for the algebra to be closed is that the discrete function f(n) is a constant. By looking at Eq. (21), we conclude that f(n) is constant if and only if

$$2\left({b}_{n+1}^{2}-{b}_{n}^{2}\right)=\alpha n+2\gamma ,$$

(22)

for some constants α and γ (the factors 2 are included for convenience). Again, b0 has to be replaced with 0, so that Eq. (22) holds for n ≥ 1, while $$2{b}_{1}^{2}=\alpha +2\gamma$$. Then, the function f(n) takes the constant value f = −α/2, so that the only possible closure of the complexity algebra is given by:

$$[{{{{{{{\mathcal{L}}}}}}}},{{{{{{{\mathcal{B}}}}}}}}]=\tilde{{{{{{{{\mathcal{K}}}}}}}}},\quad [\tilde{{{{{{{{\mathcal{K}}}}}}}}},{{{{{{{\mathcal{L}}}}}}}}]=\alpha {{{{{{{\mathcal{B}}}}}}}},\quad [\tilde{{{{{{{{\mathcal{K}}}}}}}}},{{{{{{{\mathcal{B}}}}}}}}]=\alpha {{{{{{{\mathcal{L}}}}}}}}.$$

(23)

Moreover, from Eq. (22), we immediately conclude that

$$\tilde{{{{{{{{\mathcal{K}}}}}}}}}=\alpha {{{{{{{\mathcal{K}}}}}}}}+\gamma .$$

(24)

Therefore, if α ≠ 0, the Krylov complexity is related to $$\tilde{{{{{{{{\mathcal{K}}}}}}}}}$$ by a shift. Conversely, if α = 0, there is no simple relation between the Krylov complexity and the operator $$\tilde{{{{{{{{\mathcal{K}}}}}}}}}$$. In this case, $$\tilde{{{{{{{{\mathcal{K}}}}}}}}}$$ is proportional to the identity and the complexity algebra reduces to the Heisenberg–Weyl algebra43, being $$[{{{{{{{{\mathcal{L}}}}}}}}}_{+},{{{{{{{{\mathcal{L}}}}}}}}}_{-}]=\gamma {\mathbb{1}}$$.

### Possible scenarios under the closure of the complexity algebra

As already discussed, if $${{{{{{{\mathcal{L}}}}}}}}$$, $${{{{{{{\mathcal{B}}}}}}}}$$ and their commutator $$\tilde{K}$$ closes an algebra, then the only possible commutation relations are given by (23). This complexity algebra is then reduced to the Heisenberg–Weyl algebra whenever α = 0. We next show that for the cases α < 0 and α > 0, the complexity algebra reduces to the SU(2) algebra and the $${{{{{{{\rm{SL}}}}}}}}(2,{\mathbb{R}})$$ algebra, respectively. Let us introduce the operators J+ and J, which are defined by $$\nu {J}_{+}={{{{{{{{\mathcal{L}}}}}}}}}_{+}$$ and $$\nu {J}_{-}={{{{{{{{\mathcal{L}}}}}}}}}_{-}$$, where ν is a strictly positive scaling parameter. We can then write $${{{{{{{\mathcal{L}}}}}}}}=\nu ({J}_{+}+{J}_{-})$$ and $${{{{{{{\mathcal{B}}}}}}}}=\nu ({J}_{+}-{J}_{-})$$. Let us also introduce the operator J0 defined by $${J}_{0}=-\frac{1}{2{\nu }^{2}}\tilde{K}$$. By substituting these operators into (23), one can rewrite the commutation relations as

$$[{J}_{+},{J}_{-}]={J}_{0},\quad [{J}_{0},{J}_{\pm }]=\mp \frac{\alpha }{2{\nu }^{2}}{J}_{\pm }.$$

(25)

By choosing the scaling parameter such that 2ν2 = α, we find that the algebra (23) is equivalent to

$$[{J}_{+},{J}_{-}]={J}_{0},\quad [{J}_{0},{J}_{\pm }]=\pm {J}_{\pm }\quad \alpha \, < \,0\quad {{{{{{{\rm{SU}}}}}}}}(2),$$

(26)

$$[{J}_{+},{J}_{-}]={J}_{0},\quad [{J}_{0},{J}_{\pm }]=\mp {J}_{\pm }\quad \alpha \, > \,0\quad {{{{{{{\rm{SL}}}}}}}}(2,{\mathbb{R}}).$$

(27)

What we have shown is that, whenever the simplicity hypothesis holds, then the algebra generated by $${{{{{{{\mathcal{L}}}}}}}}$$, $${{{{{{{\mathcal{B}}}}}}}}$$ and their commutator can always be reduced to either SU(2), $${{{{{{{\rm{SL}}}}}}}}(2,{\mathbb{R}})$$ or the Heisenberg–Weyl algebra, and for which of these it reduces to depends on the value of α.