Timeline for Weak simulation of Clifford circuits
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2022 at 8:30 | history | edited | Markus Heinrich | CC BY-SA 4.0 |
added 93 characters in body
|
| Jan 26, 2022 at 8:20 | comment | added | Markus Heinrich | @DorianoBrogioli Unfortunately, the "Feynman-style" sampling is not fixable unless you bring the circuit into a certain form. I think that is what the van den Nest paper is doing. However, in the worst case, that should be as costly as computing the generators of the output state, and the support of $U|0\rangle$, so you can also do that ... Sorry about that, but I'm usually thinking in terms of global unitaries. I'll update the answer. | |
| Jan 25, 2022 at 22:11 | comment | added | Doriano Brogioli | I had a look at the thesis. Actually, it's quite technical. But I saw that it mentions the interesting topic of magic states. Inspired by it, I posted a further question, here: quantumcomputing.stackexchange.com/questions/23815/… . If you have time, please have a look at my question, I'm sure you can help. | |
| Jan 25, 2022 at 21:58 | vote | accept | Doriano Brogioli | ||
| Jan 25, 2022 at 15:42 | comment | added | Markus Heinrich | @DorianoBrogioli mmh you're right. I'll think about the proper formulation and update my answer soon. | |
| Jan 25, 2022 at 12:59 | comment | added | Doriano Brogioli | I'm not convinced of 2), maybe I did not understand. For what you say, I start from a classical bit 0 and I apply H, I get 0 or 1 with equal probability. Then if I apply H again, I get again 0 or 1 with same probability (two consecutive branchings). But if I apply H twice to |0>, I get |0>, not a superposition of |0> and |1>. But probably I misunderstood your proposed method. | |
| Jan 25, 2022 at 11:28 | history | edited | Markus Heinrich | CC BY-SA 4.0 |
link to thesis added
|
| Jan 25, 2022 at 11:25 | comment | added | Markus Heinrich | @DorianoBrogioli 2) if you have elementary Clifford gates, i.e. H, S, CNOT, then you can see their action as changing extending the support of the $|0\rangle$ state. Of course, they also change the amplitudes, but for sampling that's irrelevant. While $S$ and other diagonal gates simply add phases, $CNOT$, $X$ (and more) do not extend the support, but merely do a deterministic mapping on the bitstrings $x$ in the support (a linear reversible transformation). $H$ does "branchings": with probability 1/2 you flip the bit, with 1/2 you don't. | |
| Jan 25, 2022 at 11:21 | comment | added | Markus Heinrich | @DorianoBrogioli 1) Yes, you can compute this representation from the stabilizer "tableau". Having brought this into a standard form, you can read of a basis for the linear part of $K = V + u$. An affine shift $u$ can be constructed from the character/phases of the stabilizer group. Getting the amplitudes is a bit more complicated, see the above paper or Eq. (4.47) ff in my thesis (warning: technical!). | |
| Jan 24, 2022 at 21:53 | comment | added | Doriano Brogioli | 1) You wrote the "representation of U|0⟩ in the computational basis". Is it possible to calculate the state, in this representation, from the "tableau" (the representation of the $n$ Pauli operators $N_j$)? 2) Could you please expand a bit the sentence with "you could simply propagate your samples"? There is something similar in arxiv 0811.0898, but it does not directly refer to generic circuits made of Clifford gates, but rather to a particular form they can be rewritten, called HT. | |
| Jan 24, 2022 at 11:08 | history | edited | Markus Heinrich | CC BY-SA 4.0 |
language
|
| Jan 24, 2022 at 9:20 | history | edited | Markus Heinrich | CC BY-SA 4.0 |
Added reference
|
| Jan 24, 2022 at 9:12 | history | answered | Markus Heinrich | CC BY-SA 4.0 |