Dalam pengambilan sampel boson , jika kita mulai dengan 1 foton di masing-masing mode pertama $M$dari interferometer, probabilitas mendeteksi 1 foton dalam setiap mode keluaran adalah: $|\textrm{Perm}(A)|^2$ , di mana kolom dan baris $A$ adalah kolom pertama $M$dari matriks kesatuan interferometer $U$, dan semua barisnya.

Ini membuatnya terlihat seperti untuk kesatuan $U$ , kita dapat membangun interferometer yang sesuai, membangun matriks $A$ , dan menghitung nilai absolut dari permanen $A$ dengan mengambil akar kuadrat dari kemungkinan mendeteksi satu foton dalam setiap mode (yang kami dapatkan dari percobaan boson sampling). Apakah ini benar, atau ada tangkapan? Orang-orang mengatakan kepada saya bahwa Anda sebenarnya tidak bisa mendapatkan informasi tentang sampel permanen dari boson.

Juga, apa yang terjadi pada sisa kolom $U$ : Bagaimana tepatnya hasil eksperimen hanya bergantung pada kolom pertama $M$dari $U$ dan semua barisnya, tetapi tidak sama sekali pada kolom $U$ ? Kolom-kolom $U$ itu tidak memengaruhi hasil percobaan dalam mode pertama $M$sama sekali?

Tampaknya benar, sampai batas tertentu. Ketika saya membaca Scott Aaronson ini kertas , ia mengatakan bahwa jika Anda mulai dengan 1 foton di setiap pertama mode interferometer, dan menemukan probabilitas P S yang satu set s i foton adalah output di setiap mode i ∈ { 1 , ... , N } di mana ∑ i s i = M , adalah P s = | Per (A) | $M$$P_S$$s_i$$i\in\{1,\ldots, N\}$$\sum_is_i=M$ Jadi, memang, jika Anda mengambil contoh tertentudimanasi=0atau 1 untuk setiap output yang mungkin, maka, ya probabilitasnya sama dengan permanenA, di manaAadalahkolomMpertamadariUdan subset spesifik dariMbaris yang ditentukan oleh lokasisi=1. Jadi, ini tidak cukup seperti yang ditentukan dalam pertanyaan: tidak semua baris, tetapi hanya beberapa bagian, sehingga

Ini harus cukup jelas. Katakanlah saya memiliki beberapa matriks V . Jika saya mulai dalam keadaan dasar | 0 ⟩ dan menemukan produk, V | 0 ⟩ , kemudian mengetahui bahwa memberitahu saya sedikit tentang output V | 1 ⟩ dan V | 2 $3\times 3$$V$$|0\rangle$$V|0\rangle$$V|1\rangle$$V|2\rangle$ , selain dari apa yang dapat dikatakan dari pengetahuan bahwa adalah kesatuan, dan karenanya kolom dan baris yang ortonormal.$V$

Masalah yang harus diperhatikan adalah keakuratannya: Anda menjalankan ini sekali dan semua yang Anda dapatkan adalah sampel tunggal sesuai dengan distribusi probabilitas . Anda menjalankan ini beberapa kali, dan Anda mulai membangun informasi tentang berbagai probabilitas. Anda menjalankan ini cukup banyak, dan Anda bisa mendapatkan jawaban yang akurat secara sewenang-wenang, tetapi berapa banyak yang cukup? Ada dua cara berbeda yang bisa Anda lakukan untuk mengukur kesalahan dalam estimasi nilai p . Anda dapat meminta kesalahan aditif p ± ϵ atau kesalahan multiplikasi, p ( 1 ± ϵ ) . Karena kami berharap bahwa probabilitas tipikal akan secara eksponensial kecil dalam n + $P_s$$p$$p\pm\epsilon$$p(1\pm\epsilon)$$n+m$, kesalahan multiplikasi menuntut akurasi yang jauh lebih besar, yang tidak dapat dicapai secara efisien melalui pengambilan sampel. Di sisi lain, perkiraan kesalahan aditif dapat dicapai.

Sementara kesalahan multiplikatif adalah apa yang biasanya orang ingin hitung, kesalahan aditif juga bisa menjadi entitas yang menarik. Misalnya, dalam evaluasi polinomial Jones .

Aaronson menunjukkan kita lebih jauh ke masa ketika hubungan antara sampling Boson dan Permanen pertama kali dibuat:

Telah dikenal sejak bekerja dengan Caianiello pada tahun 1953 (jika tidak sebelumnya) bahwa amplitudo untuk proses -boson dapat ditulis sebagai permanents dari n × n matriks.$n$$n\times n$

Sebaliknya, kontribusi utama mereka

adalah untuk membuktikan koneksi antara kemampuan komputer klasik untuk memecahkan masalah BosonSampling perkiraan dan kemampuan mereka untuk memperkirakan permanen

yaitu untuk memahami masalah perkiraan yang terkait dengan, misalnya pengambilan sampel terbatas, dan untuk menggambarkan konsekuensi kompleksitas komputasi yang terkait: bahwa kami percaya hal seperti itu sulit untuk dievaluasi secara klasik.

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.

Memang, masalah yang diselesaikan boson sampler adalah pengambilan sampel dari distribusi probabilitas tertentu. Lebih khusus, pengambilan sampel dari distribusi probabilitas dari kemungkinan hasil (banyak boson) menyatakan.

$(1,1,0,0)\equiv|1,1,0,0\rangle$$\binom{4}{2}=6$ possible output two-photon states: $(1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1)$ 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:

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

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/