Expressing a quantum state as a polynomial

Question

Let us consider the state $\left|\psi\right>$ obtained by applying $m$ 1- and 2-qubit gates to $n$ qubits, starting from the state $\left|0,0,\dots\right>$. Let us express it as: $$ \left|\psi\right> = \sum_{b_1, b_2, \dots} f(b_1, b_2, \dots) \left| b_1, b_2, \dots \right>$$ where the $b_i$ take the values 0 or 1.

It is trivial to see that there is a polynomial $p(b_1, b_2, \dots)$ (something like $b_2 b_4 + 3 b_3^2 b_6 +\dots$) that takes the same values of $f(b_1, b_2, \dots)$ when $b_i$ take the values 0 or 1. This is actually true for whatever $f$, it depends on the discrete values of the $b_i$. Please let me know if I am wrong.

The polynomial obtained with the trivial method is very large but there could be smaller representations. Here, I am asking if there is a know scaling between the the size of the polynomial $p$ (with whatever representation) and the numbers $n$ and $m$. For example, a lower limit, or some method to get a small $p$.

ADDED: the size of the polynomial can be defined in various ways. For example, as the length of the string encoding the polynomial (with a reasonable encoding). Another example: the sum of the degrees of the monomials. The number of monomials or the degree could also be considered.

The question is not about the time needed to calculate $p$, but only about its size.

Answer 1

I assume $f$ is real because, in your example, the polynomial you gave appears to be real.

The dependence on the number of qubits $n$ is pretty straightforward to see. Once you see that, you realize that $m$ doesn't contribute meaningfully to a polynomial degree.

Briefly, given any function $f:\{0,1\}^n \rightarrow \mathbb{R}$, there exists a polynomial $p(x_1, \ldots, x_n)$ of a degree at most $n$ such that $f(x_1, \ldots, x_n)=p(x_1, \ldots, x_n)$ for all $x_i \in \{0,1\}$. This result follows directly from Lagrange interpolation polynomials. So, the scaling of the degree of a polynomial is linear in the number of qubits, i.e. $\deg{p} = O(n)$.

We can even find a very compact expression of the polynomial by following the Lagrange polynomials' construction in the link above. The polynomial is: $$p(x_1, x_2,\ldots, x_n) = \sum_{b \in \{0,1\}^n}f(b)L_b(x_1, \ldots, x_n), \tag{1}$$ where $L_b(x_1, \ldots, x_n)$ is a multivariate Lagrange polynomial of degree at most $n$ such that for $b \in \{0,1\}^n$ we have: $$L_b(x_1, \ldots, x_n)=\begin{cases} 1, \textrm{ if } x_i = b_i, \textrm{ for all } i, \\ 0, \textrm{ otherwise}. \end{cases} $$ The basis polynomial $L_b(x_1, \ldots, x_n)$ is a product of $n$ univariate linear polynomials: $$L_b(x_1, \ldots, x_n)=\prod^n_{i=1}( b_i x_i +(1-b_i)(1-x_i)).$$ If it is unclear how this is derived, I suggest following the construction of Lagrange polynomials in the link above and considering a special case where $x$ is a binary variable. I hope that, on the intuitive level, (1) makes sense. Basically, we have a sum over exponentially many $n$-degree polynomials $L_b$ where each polynomial acts similarly to the indicator or Kronecker delta function.

Note that we have the explicit equation of a polynomial, which only requires the knowledge of the values of $f(b)$ for $b \in \{0,1\}^n$, but it doesn't require knowledge of $m$. So, you can have as many gates as you want, but the scaling of the degree will still be linear in $n$, i.e. $O(n)$.

Analyzing (1), you can always find the lowest degree. For example, assume $|\psi\rangle$ is a uniform superposition of all states. Then (1) simplifies to: $$p(x_1, \ldots, x_n) = \frac{1}{\sqrt{2^n}}\sum_{b \in \{0,1\}^n} L_b(x_1, \ldots, x_n) = \frac{1}{\sqrt{2^n}}.$$ Recall that $L_b$ add up to 1 because only one term evaluates to 1, and the rest are 0. We conclude that the lowest degree we can get is 0 for the choice of $|\psi\rangle$ above.