在玻色子采样中，如果我们从干涉仪的前$M$模式的每一个中以1个光子开始，则在每种输出模式中检测到1个光子的概率为：$|\textrm{Perm}(A)|^2$参照图2，其中的列和行是干涉仪的ary矩阵$A$的前$M$列及其所有行。$U$

这使得它看起来像对于任何单一$U$，我们可以构造适当的干涉仪，构建矩阵$A$，并计算永久的绝对值$A$通过利用在每个模式检测一个光子的概率的平方根（我们从玻色子采样实验中获得）。这是真的，还是有收获？人们告诉我，您实际上无法从玻色子采样中获得有关永久物的信息。

此外，会发生什么情况的其余列$U$：究竟它是如何，实验结果仅取决于第一$M$的列$U$及其所有的行，但不是在所有上的其他列$U$？U的那些列$U$根本不影响前$M$模式下的实验结果吗？

在某种程度上，这似乎是真的。当我读到斯科特·阿伦森的纸，它说，如果你在每个第1个光子启动的干涉模式，并找到概率P 小号一套小号我的光子是在每种模式下输出我∈ { 1 ，... ，N }，其中∑ i s i = M，是 P s = | Per（A）| $M$$P_S$$s_i$$i\in\{1,\ldots, N\}$$\sum_is_i=M$ 因此，的确，如果您针对每个可能的输出采用si=0或1的特定实例，那么，是的，概率等于A的永久性，其中A是U的前M列和M的特定子集位置si=1指定的行。因此，这与问题中所指定的不完全相同：它不是所有行，而只是某些子集，因此

这应该是很明显的。比方说，我有一些矩阵V。如果我以某种基础状态开始| 0 ⟩并找到自己的产品，V | 0 ⟩，然后知道告诉我甚少输出V | 1 ⟩和V | 在图2⟩中，除了可以从V表示单数的知识得知之外，因此列和行是正交的。$3\times 3$$V$$|0\rangle$$V|0\rangle$$V|1\rangle$$V|2\rangle$$V$

必须注意的问题是准确性：您运行一次，得到的只是根据概率分布的一个样本。您运行了几次，然后开始建立有关不同概率的信息。您运行了足够的次数，就可以得到任意准确的答案，但是多少才足够？您可以通过两种不同的方法来估计值p的误差。您可以要求相加误差p ± ϵ或乘法误差p （1 ± ϵ ）。由于我们期望在n + m中典型的概率将呈指数减小$P_s$$p$$p\pm\epsilon$$p(1\pm\epsilon)$$n+m$，乘法误差要求更高的精度，而这不能通过采样有效地实现。另一方面，可以实现加法误差近似。

人们通常要计算乘法误差，而附加误差也可以是一个有趣的实体。例如，在对琼斯多项式的求值中。

亚伦森（Aaronson）向我们指出了玻色子采样与永久采样之间的这种联系最早出现的时间：

自1953年Caianiello的工作（如果不是更早的话）以来，人们已经知道玻色子过程的振幅可以写成n × n矩阵的永久物。$n$$n\times n$

相反，他们的主要贡献

证明经典计算机解决近似BosonSampling问题的能力与它们逼近永久性的能力之间的联系

也就是说，要了解与有限采样相关的近似问题，并描述与计算相关的复杂性所带来的后果：我们认为这样的事情很难经典地评估。

您无法有效地恢复振幅的绝对值，但是如果您允许任意多个样本，则可以以您喜欢的任何精度来估计它们。

更具体地说，如果在前模式的每一个中输入状态都是单个光子，并且一个人愿意从输出中抽取任意数量的样本，则原则上可以将A的永久性估计为任意程度通过计算n个输入光子在前n个不同的输出端口中出射的次数的分数，人们喜欢这样的精度。值得注意的是，这与BosonSampling并没有多大关系，因为硬度结果取决于模式数量远大于光子数量的范围，而采样效率与采样效率有关。$n$$A$$n$$n$

玻色子采样

我将简要介绍什么是玻色子采样，但是应该指出的是，我在这方面做不到比亚伦森本人更好的工作，因此，看看他的相关博客文章可能是一个好主意。 （例如blog /？p = 473和blog /？p = 1177），以及其中的链接。

BosonSampling是一个采样问题。这可能会有些混乱，因为人们通常更习惯于思考具有确定答案的问题。采样问题的不同之处在于，该问题的解决方案是从某种概率分布中提取的一组样本。

确实，玻色子采样器解决的问题是从特定概率分布中采样的问题。更具体地说，抽样从可能结果（玻色子）状态的概率分布中。

考虑作为一个简单的例子，2个光子的情况下在4种模式，并让我们说我们固定输入状态为（即，在每两个第一两种输入方式的单一光子）。忽略每种模式中一个以上光子的输出状态，有（$(1,1,0,0)\equiv|1,1,0,0\rangle$可能的输出双光子状态： （1，1，0，0），（1，0，1，0），（1，0，0，1），（0，1，1，0），（0，1，0，1）和（0$\binom{4}{2}=6$$(1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1)$。让我们表示为了方便与 Ø 我，我= 1 ，。，6的我第一个（因此，例如， Ô 2 = （1 ，0 ，1 ，0 ））。然后，可能的一系列结果可能是BosonSampling的解决方案： o 1，o 4，o 2，o 2，o 5$(0,0,1,1)$$o_i, i=1,.,6$$i$$o_2=(1,0,1,0)$

为了比喻一个可能更熟悉的情况，这就像说我们要从高斯概率分布中采样。这意味着我们想要找到一个数字序列，如果我们将它们抽出足够多并将其放入直方图中，将产生接近高斯的值。

计算永久物

$|\boldsymbol r\rangle$$|\boldsymbol s\rangle$

$\boldsymbol R$${}^{(1)}$与$|\boldsymbol r\rangle$, $\boldsymbol S$ that of $|\boldsymbol s\rangle$, and $U$ is the unitary matrix describing the evolution, then the probability amplitude $\mathcal A(\boldsymbol r\to\boldsymbol s)$ of going from $|\boldsymbol r\rangle$ to $|\boldsymbol s\rangle$ is given by

Thus, considering the fixed input state $|\boldsymbol r_0\rangle$, the probability distribution of the possible outcomes is given by the probabilities

BosonSampling is the problem of drawing "points" according to this distribution.

This is not the same as computing the probabilities $p_s$, or even computing the permanents themselves. Indeed, computing the permanents of complex matrices is hard, and it is not expected even for quantum computers to be able to do it efficiently.

The gist of the matter is that sampling from a probability distribution is in general easier than computing the distribution itself. While a naive way to sample from a distribution is to compute the probabilities (if not already known) and use those to draw the points, there might be smarter ways to do it. A boson sampler is something that is able to draw points according to a specific probability distribution, even though the probabilities making up the distribution itself are not known (or better said, not efficiently computable).

Furthermore, while it may look like the ability to efficiently sample from a distribution should translate into the ability of efficiently estimating the underlying probabilities, this is not the case as soon as there are exponentially many possible outcomes. This is indeed the case of boson sampling with uniformly random unitaries (that is, the original setting of BosonSampling), in which there are $\binom{m}{n}$ possible $n$-boson in $m$-modes output states (again, neglecting states with more than one boson in some mode). For $m\gg n$, this number increases exponentially with $n$. This means that, in practice, you would need to draw an exponential number of samples to even have a decent chance of seeing a single outcome more than once, let alone estimate with any decent accuracy the probabilities themselves (it is important to note that this is not the core reason for the hardness though, as the exponential number of possible outcomes could be overcome with smarter methods).

In some particular cases, it is possible to efficiently estimate the permanent of matrices using a boson sampling set-up. This will only be feasible if one of the submatrices has a large (i.e. not exponentially small) permanent associated with it, so that the input-output pair associated with it will happen frequently enough for an estimate to be feasible in polynomial time. This is a very atypical situation, and will not arise if you draw unitaries at random. For a trivial example, consider matrices that are very close to identity - the event in which all photons come out in the same modes they came in will correspond to a permanent which can be estimated experimentally. Besides only being feasible for some particular matrices, a careful analysis of the statistical error incurred in evaluating permanents in this way shows that this is not more efficient than known classical algorithms for approximating permanents (technically, within a small additive error) ${}^{(2)}$.

Columns involved

Let $U$ be the unitary describing the one-boson evolution. Then, basically by definition, the output amplitudes describing the evolution of a single photon entering in the $k$-th mode are in the $k$-th column of $U$.

The unitary describing the evolution of the many-boson states, however, is not actually $U$, but a bigger unitary, often denoted by $\varphi_n(U)$, whose elements are computed from permanents of matrices built out of $U$.

Informally speaking though, if the input state has photons in, say, the first $n$ modes, then naturally only the first $n$ columns of $U$ must be necessary (and sufficient) to describe the evolution, as the other columns will describe the evolution of photons entering in modes that we are not actually using.

(1) This is just another way to describe a many-boson state. Instead of characterizing the state as the list of occupation numbers for each mode (that is, number of bosons in first mode, number in second, etc.), we characterize the states by naming the mode occupied by each boson. So, for example, the state $(1, 0, 1, 0)$ can be equivalently written as $(1, 3)$, and these are two equivalent ways to say that there is one boson in the first and one boson in the third mode.

(2): S. Aaronson and T. Hance. "Generalizing and Derandomizing Gurvits's Approximation Algorithm for the Permanent". https://eccc.weizmann.ac.il/report/2012/170/