ボソンサンプリングを使用してパーマネントの絶対値を「計算」することは可能ですか？

でボソンサンプリング、我々は最初のそれぞれに1個の光子で開始した場合、 $M$ 干渉計のモード、各出力モードで1個の光子を検出する確率は、次のとおりです。 $|\textrm{Perm}(A)|^2$ 、ここで列と行は、干渉計のユニタリ行列の $A$ 最初の $M$ 列とそのすべての行です。 $U$

これにより、ユニタリように見え $U$ 、適切な干渉計を構築し、行列構築し、各モードで1光子を検出する確率の平方根を取ること $A$ でのパーマネントの絶対値を計算でき $A$ ます（これは、ボソンサンプリング実験から取得）。これは本当ですか、それともキャッチがありますか？ボソンのサンプリングからパーマネントに関する情報を実際に取得することはできないと人々は私に言った。

また、何がの列の残りの部分に起こる $U$ ：どのように正確には、実験結果は初回のみに依存していることである $M$ の列 $U$ とそのすべての行ではなく、他の列の上のすべての $U$ ？これらの列は、最初のモードでの実験の結果にまったく影響しませんか？ $U$ $M$

— user1271772
ソース

photonicsを作成したので、タグの抜粋を書くことを検討してください。ここに行きます。ありがとうございました。

— Sanchayan Dutta

回答:

ある時点までは真実のようです。私はスコット・アーロンソンの読みとして紙を、それはあなたが最初のそれぞれにおける1個の光子で始まる場合と言う干渉計のモード、および確率見つけ設定することの光子が各モードで出力されるここで、であり、 $M$ $P_S$ $s_i$ $i\in\{1,\ldots, N\}$ $\sum_is_i=M$ したがって、実際には、可能なすべての出力に対してまたは1特定のインスタンスを使用する場合、はい、確率はのパーマネントに等しくなります。ここで、は最初の列と特定のサブセットです場所指定された行。したがって、これは質問で指定されたとおりではありません。すべての行ではなく、一部のサブセットのみであるため、

P_{s} = \frac{| Per(A) |^{2}}{s_{1}! s_{2}! \dots s_{M}!} .

$P_s=\frac{|\text{Per(A)}|^2}{s_1!s_2!\ldots s_M!}.$

s_{i} = 0

$s_i=0$

A

$A$

A

$A$

M

$M$

U

$U$

M

$M$

s_{i} = 1

$s_i=1$

A

$A$ は、実験が「見る」ビット、つまり入力行と出力行に対応する正方行列です。光子は他に何も入力しないため、ユニタリ行列

他の要素は見えません。

U

$U$

これはかなり明白なはずです。のは、私はいくつか持っていると言う行列。何らかの基本状態で起動する場合と見つけるその製品、、その出力は約非常に少ないと言われます知っと、脇という知識から言うことができるものと一体であるため、列と行が正規直交です。 $3\times 3$ $V$ $|0\rangle$ $V|0\rangle$ $V|1\rangle$ $V|2\rangle$ $V$

注意しなければならない問題は、精度です。これを1回実行すると、確率分布に従って1つのサンプルしか得られません。これを数回実行し、さまざまな確率に関する情報の構築を開始します。これを十分な回数実行すると、任意の正確な答えを得ることができますが、いくつあれば十分ですか？値推定値の誤差を測定できる方法は2つあります。加算誤差または乗算誤差いずれかを要求できます。典型的な確率はで指数関数的に小さくなると予想されるため $P_s$ $p$ $p\pm\epsilon$ $p(1\pm\epsilon)$ $n+m$ 、乗法誤差ははるかに高い精度を要求しますが、これはサンプリングでは効率的に達成できません。一方、加法誤差近似を実現できます。

乗法誤差は通常人々が計算したいものですが、加法誤差も興味深いエンティティです。たとえば、ジョーンズ多項式の評価では。

アーロンソンは、ボソンのサンプリングと恒久的なものとの間のこのつながりが最初に作られた場所について、さらに時間をさかのぼって指摘します。

1953年のCaianielloの研究以来（以前でなければ）、ボソンプロセスの振幅は行列のパーマネントとして記述できることが知られています。 $n$ $n\times n$

代わりに、彼らの主な貢献

近似的なBosonSampling問題を解決する古典的なコンピューターの能力と、恒久的なコンピューターを近似する能力との関係を証明することです。

すなわち、例えば有限サンプリングに関連する近似問題を理解し、関連する計算の複雑さの結果を説明すること：そのようなことを古典的に評価するのは難しいと信じていること。

— ダフトウォリー
ソース

I'm not sure whether this is what you are saying, but it is not true that solving efficiently BosonSampling allows to efficiently estimate the permanents, which would imply that quantum computers are able to solve #P-hard problems. In other words, quantum computers can efficiently simulate the output of a boson sampler, but not efficiently compute its output probability distribution

— glS

@glS No, that's very much what I'm saying. The Aaronson paper is very careful to distinguish that issue, but it makes the computational complexity statement a lot messier, which is why I didn't state it.

— DaftWullie

@DaftWullie sorry, now I'm confused. Do we agree that boson sampling does not allow to efficiently estimate permanents? (see e.g. bottom of left column at pag 6 of arxiv.org/pdf/1406.6767.pdf)

— glS

@gls I agree that you cannot do it if you want an estimate of the permanent with some multiplicative error bound, which, admittedly, is the standard way of defining things (but since I carefully avoided defining anything...). If you’re willing to tolerate an additive error bound, then I believe you can do it.

— DaftWullie

| 0 ⟩

$|0\rangle$

V | 0 ⟩

$V|0\rangle$ , then knowing that tells me very little about the outputs

V | 1 ⟩

$V|1\rangle$ and

V | 2 ⟩

$V|2\rangle$ ", but every single element of

V

$V$ is involved in giving you

V | 0 ⟩

$V|0\rangle$ . But for boson sampling, only the first

M

$M$ columns are involved, isn't that amazing?

— user1271772

You cannot efficiently recover the absolute values of the amplitudes, but if you allow for arbitrary many samples, then you can estimate them to whatever degree of accuracy you like.

More specifically, if the input state is a single photon in each of the first $n$ modes, and one is willing to draw an arbitrary number of samples from the output, then it is in principle possible to estimate the permanent of $A$ to whatever degree of accuracy one likes, by counting the fraction of the times the $n$ input photons come out in the first $n$ different output ports. It is to be noted though that this does not really have much to do with BosonSampling, as the hardness result holds in the regime of the number of modes much larger than the number of photons, and it's about the efficiency of the sampling.

BosonSampling

I'll try a very brief introduction to what boson sampling is, but it should be noted that I cannot possibly do a better job at this than Aaronson himself, so it's probably a good idea to have a look at the related blog posts of his (e.g. blog/?p=473 and blog/?p=1177), and links therein.

BosonSampling is a sampling problem. This can be a little bit confusing in that people are generally more used to think of problems having definite answers. A sampling problem is different in that the solution to the problem is a set of samples drawn from some probability distribution.

Indeed, the problem a boson sampler solves is that of sampling from a specific probability distribution. More specifically, sampling from the probability distribution of the possible outcome (many-boson) states.

Consider as a simple example a case with 2 photons in 4 modes, and let's say we fix the input state to be $(1,1,0,0)\equiv|1,1,0,0\rangle$ (that is, a single photon in each of the two first two input modes). Ignoring the output states with more than one photon in each mode, there are $\binom{4}{2}=6$ $(1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1)$ and $(0,0,1,1)$ . Let us denote for convenience with $o_i, i=1,.,6$ the $i$ -th one (so, for example, $o_2=(1,0,1,0)$ ). Then, a possible solution to BosonSampling could be the series of outcomes:

o_{1}, o_{4}, o_{2}, o_{2}, o_{5} .

$o_1, o_4, o_2, o_2, o_5.$

To make an analogy to a maybe more familiar case, it's like saying that we want to sample from a Gaussian probability distribution. This means that we want to find a sequence of numbers which, if we draw enough of them and put them into a histogram, will produce something close to a Gaussian.

Computing permanents

It turns out that the probability amplitude of a given input state $|\boldsymbol r\rangle$ to a given output state $|\boldsymbol s\rangle$ is (proportional to) the permanent of a suitable matrix built out of the unitary matrix characterizing the (single-boson) evolution.

More specifically, if $\boldsymbol R$ denotes the mode assignment list ${}^{(1)}$ associated to $|\boldsymbol r\rangle$ , $\boldsymbol S$ 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

A (r \to s) = \frac{1}{\sqrt{r! s!}} perm U [R | S],

$\mathcal A(\boldsymbol r\to\boldsymbol s) = \frac{1}{\sqrt{\boldsymbol r!\boldsymbol s!}} \operatorname{perm} U[\boldsymbol R|\boldsymbol S],$ with

U [R | S]

$U[\boldsymbol R|\boldsymbol S]$ denoting the matrix built by taking from

U

$U$ the rows specified by

R

$\boldsymbol R$ and the columns specified by

S

$\boldsymbol S$ .

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

p_{s} = \frac{1}{r_{0}! s!} | perm U [R | S] |^{2} .

$p_{\boldsymbol s} = \frac{1}{\boldsymbol r_0! \boldsymbol s!} \lvert \operatorname{perm}U[\boldsymbol R|\boldsymbol S] \rvert^2.$

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/

— glS
ソース

I started with 1 photon in each input mode, and said we're looking at the probability of having 1 photon in each output mode, so that we could avoid all these more complicated general equations involving the permanent, which you provide. In fact if

M

$M$ is the number of columns in

U

$U$ , we get that the probability of having 1 photon in each output mode is

| Perm (U) |^{2}

$|\textrm{Perm}(U)|^2$ from which we can easily get

| Perm (U) |

$|\textrm{Perm}(U)|$ . If we let the experiment go on for long enough and get enough samples, can we not obtain an estimate for

| Perm (U) |

$|\textrm{Perm}(U)|$ ?

— user1271772

In no part of the question did I mention "efficiency" or "sub-exponentially". I'm just interested to know whether or not it's possible to estimate

| Perm (U) |

$|\textrm{Perm}(U)|$ using boson sampling.

— user1271772

@user1271772 I see. That's the standard way of talking about these things in this context so I might have automatically assumed you meant to talk about efficiency. If you don't care about the number of samples you have to draw then sure, you can compute the output probability distribution, and therefore the absolute values of the permanents, to whatever accuracy you like

— glS

@gIS, Aram Harrow once told me you cannot calculate Permanents using boson sampling, so I thought there was some "catch". The best classical algorithm for simulation of exact boson sampling is:

O (m 2^{n} + m n^{2})

$\mathcal{O}\left(m2^n + mn^2\right)$ , for

n

$n$ photons in

m

$m$ output modes, what is the cost using the interferometer?

— user1271772

@user1271772 I answered more specifically your first point in the edit. I guess I got confused because the setting you are mentioning does not seem to have really much to do with boson sampling, but is more generally about the dynamics of indistinguishable bosons through an interferometer

— glS