Dynamical Decomposition of Bilinear Control Systems Subject to Symmetries

Abstract

We describe a method to analyze and decompose the dynamics of a bilinear control system subject to symmetries. The method is based on the concept of generalized Young symmetrizers of representation theory. It naturally applies to the situation where the system evolves on a tensor product space and there exists a finite group of symmetries for the dynamics which interchanges the various factors. This is the case for quantum mechanical multipartite systems, such as spin networks, where each factor of the tensor product represents the state of one of the component systems. We present several examples of application.

Appendix:: Proofs of Formula (38) and of Formula (51)

Consider a Lie algebra ${\mathcal R}$ which has a basis ${\mathcal B}:=\{E_{j}\}$ which is invariant, as a set, under the action of the group G, i.e., if $E \in {\mathcal B}$, $gEg^{-1} \in {\mathcal B}$, ∀g ∈ G. Then, we can derive a basis for ${\mathcal R}^{G}$: Let ${\mathcal O}$ the set of orbits of G in ${\mathcal B}$ under the above action:

Proposition 5.2

The set of elements:

$$ \{ {\sum}_{E_j \in O} E_j | O \in {\mathcal O}\}, $$

(58)

is a basis of ${\mathcal R}^{G}$. In particular, the dimension of ${\mathcal R}^{G}$ is equal to the number of orbits in ${\mathcal B}$ under the above described action of G on ${\mathcal B}$.

Proof

Since the $E_{j} \in {\mathcal B}$ form a basis and the orbits are disjoint, then elements ${\sum }_{E_{j} \in O} E_{j}$ in (58) for different orbits O are linearly independent. Moreover, write $F\in {\mathcal R}^{G}$ as $F= {\sum }_{O \in {\mathcal O}} F_{O}$ where ${\mathcal O}$ is the set of orbits and F_O is a linear combination of elements in ${\mathcal B}$ in the orbit O. Since gFg^− 1 = F for each g ∈ G, and each orbit is invariant, we have:

$$ gFg^{-1}= \sum\limits_{O \in {\mathcal O}} gF_{O}g^{-1}= F= \sum\limits_{O \in {\mathcal O}} F_{O}, $$

which implies that, for every orbit O, and every g ∈ G, gF_Og^− 1 = F_O. Write $F_{O}={\sum }_{j} \alpha _{j} E_{j}$ where E_j are the elements in the basis ${\mathcal B}$ which also belong to the orbit O, and for some coefficients α_j. Fix j and k and a g ∈ G so that g maps E_j to E_k. Such a g always exists because, by definition, the action of G is transitive on its orbits. By imposing gF_Og^− 1 = F_O, using the fact that the map associated with g is a bijection from the orbit to itself, we find that α_j = α_k. Since, this is valid for arbitrary j and k, we find that F_O must be proportional to the elements ${\sum }_{E_{j} \in O} E_{j}$ in the set (58). □

According to the proposition, the dimension of ${\mathcal R}^{G}$ can be calculated using the Burnside’s orbit counting theorem (see, e.g., [24]):

$$ \# \texttt{orbits}=\frac{1}{|G|} \sum\limits_{g \in G}|\texttt{Fix}^g|, $$

(59)

where Fix^g denotes the set of elements fixed by g, in ${\mathcal B}$.

1.1 Proof of Formula (38)

Proof

By Proposition 5.2, we have:

$$ \dim u^{C_{n}}(2^{n}) = \#\text{orbits} $$

(60)

where # orbits is the number of orbits with respect to the action of C_n on the set of all words of length n in the four symbols 1,σ_x,σ_y,σ_z.

Recall Burnside counting theorem (59) which applied to our case gives:

$$ \#\text{orbits} = \frac{1}{|C_{n}|} \sum\limits_{g\in C_{n}} |\text{Fix}^{g}| . $$

(61)

The cyclic group⁶C_n = 〈Z〉 has a unique subgroup H_m of order m for every positive divisor m of n, namely H_m = 〈Z^n/m〉. Since every element g of C_n generates some subgroup, we can partition C_n into subsets corresponding to which subgroup they generate. Then, we get:

$$ \#\text{orbits} =\frac{1}{n}\sum\limits_{m|n}\underset{\langle g\rangle = H_{m}}{\underset{g\in C_{n}}{\sum}} |\text{Fix}^{g}|, $$

(62)

where ${\sum }_{m|n}$ means we sum over all positive integers m which divide n. Next we use the fact that a word is fixed by g if and only if it is fixed by the cyclic subgroup 〈g〉. Thus we get from (62)

$$ \#\text{orbits}=\frac{1}{n}\sum\limits_{m|n} \underset{\langle g\rangle = H_{m}}{\underset{g\in C_{n}}{\sum}} |\text{Fix}^{H_{m}}| . $$

(63)

Now recall that any cyclic group has many possible generators. In particular, if g generates a group G of order m, g^a generates G if and only if $\gcd (a,m)=1$. Applying this to G = H_m, which is cyclic of order m, $(Z^{\frac {n}{m}})^{a}$ generates H_m if and only if $\gcd (a,m)=1$. Euler’s totient function ϕ(m) counts the number of positive integers a less than or equal to m having greatest common divisor 1 with m. Therefore, H_m has ϕ(m) generators. This means that we can rewrite Eq. 63 as follows:

$$ \#\text{orbits}=\frac{1}{n}\sum\limits_{m|n} |\text{Fix}^{H_{m}}|\cdot \phi(m) $$

(64)

If m is a positive integer that divides n, then the number of words of length n in 4 letters that are fixed by H_m (equivalently, by Z^n/m) is 4^n/m because such words are uniquely determined by the first n/m positions, which can be arbitrarily chosen. This gives us the formula we wanted to show:

$$ \#\text{orbits}=\dim u^{C_{n}}(2^{n}) = \frac{1}{n}\sum\limits_{m|n} 4^{n/m} \phi(m). $$

□

1.2 Proof of Formula (51)

With the same steps as in the previous proof applied to X_k rather than the whole set of words, we arrive at (cf., formula (64)):

$$ |X_{k}/C_{n}|=\frac{1}{n}\sum\limits_{m|n} |\text{Fix}^{H_{m}}|\cdot \phi(m), $$

(65)

where now the set fixed by H_m, $\text {Fix}^{H_{m}}$ is considered in X_k rather than in the space of all 2ⁿ binary words. Recall that H_m is the subgroup generated by Z^n/m. A word $\underline {a}$ in X_k is fixed by H_m if and only if $Z^{n/m}(\underline {a})=\underline {a}$. This in turn holds if and only n/m is a multiple of the period T of $\underline {a}$. Therefore, the words in $\text {Fix}^{H_{m}}$ have period T such that n/T divides k and m divides n/T. Their number by definition is w(n,m,k). Moreover, in the sum (65), m has to divide n/T and therefore n, and n/T has to divide k, so that m also has to divide k. Therefore, the nonzero terms are obtained for m at most equal to the greatest common divisor of n and k, i.e., $\gcd (n,k)$ which gives formula (51).