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visits member for 3 years, 7 months
seen May 5 '12 at 20:03

Scientist


May
4
comment Information conservation during quantum measurement in $\psi$-epistemic interpretations
May be interesting: mathoverflow.net/questions/95537/…
Apr
4
comment Negative probabilities in quantum physics
Feynman wrote in this essay: "Trying to think of negative probabilities gave me a cultural shock at first, but when I finally got easy with the concept I wrote myself a note so I wouldn't forget my thoughts."
Mar
26
comment Multiqubit state tomography by performing measurement in the same basis
@Piotr Migdal: Using other words I am simply showing that the set of that nonorthogonal measurements is not "informationally complete".
Mar
26
comment Multiqubit state tomography by performing measurement in the same basis
Even measurement in single basis produces $2^n-1$ parameters. My preliminary estimation give equation $\sum_k C^n_k C^{k+2}_k$, but seems Chris' curve corresponds a bit different values.
Mar
25
comment Multiqubit state tomography by performing measurement in the same basis
Indeed, I see - $\rho_1 = (1 + \sigma_z \otimes \sigma_x)/4$, $\rho_2 = (1 + \sigma_x \otimes \sigma_z)/4$.
Mar
25
comment Multiqubit state tomography by performing measurement in the same basis
@Peter Shor: I already wrote, I do not require that - we may distinuish the swap of all components due to assymetry between $\sigma_0 \otimes \sigma_k$ and $\sigma_k \otimes \sigma_0$.
Mar
25
comment Multiqubit state tomography by performing measurement in the same basis
@Peter Shor: they are equal only for $k,j \neq 0$ but it is not so for terms with $\sigma_0 =1$ (so my note about "a swap operator" in earlier answer was misleading).
Mar
9
comment Analyticity and Causality in Relativity
I doubt an answer may be short. From experience of discussion about that problem (also confirmed by answers and comments here) I learn, that even statement of the problem is not very simple. One my colleague even had idea to use it as PhD theme ...
Feb
20
comment Constructing a Hamiltonian (as a polynomial of $q_i$ and $p_i$) from its spectrum
I think, the Newton interpolation formula may be used instead of Gaussian elimination in this case
Jan
21
comment Time reversal symmetry and T^2 = -1
I could better understand a question $Pin(3,1)$ vs $Pin(1,3)$. How we could answer what we are using in Euclidean case? Even in Lorentzian case we sometimes have to talk about experimental data (e.g. discovery of antiparticles, search for Majorana neutrino, etc.) to clarify such questions.
Dec
3
comment Charged black holes in equilibrium
It is quite common to claim that even single black hole does not result a stationary spacetime lightandmatter.com/html_books/genrel/ch07/ch07.html
Nov
29
comment Is this a simple Lie algebra?
In such a case OP could write $[i\sigma_a,i\sigma_b]=\cdots$. It is not a convention, it is rather a convenient trick to work with Lie algebra of unitary group, because the algebra is anti-Hermitian matrices. If to use the convention without warning we could not distinguish $sl(2,C)$ and $su(2)$
Nov
24
comment Hilbert-Schmidt basis for many qubits - reference
Maybe the discrete Wigner function is a bit other story, because they need to use trace on $GF(2^n)$ and exchange components in Hilbert-Schmidt scalar product.
Nov
24
comment Is this a simple Lie algebra?
It is rather $so(4,{\mathbb C}) \approx sl(2,{\mathbb C}) \oplus sl(2,{\mathbb C})$, because OP defines it as a complex algebra.
Nov
24
comment Is this a simple Lie algebra?
@Luboš Motl: any Pauli matrix meets two last equations and after all ${\rm diag}(+1,-1)$ is also Pauli matrix $\sigma_z$.
Nov
23
comment Hilbert-Schmidt basis for many qubits - reference
I guess, but I never heard about simple expressions for constrains
Nov
23
comment Hilbert-Schmidt basis for many qubits - reference
I supposed the decomposition itself is rather standard consequence of axioms of quantum mechanics and linear algebra, e.g. section 5.3 in my e-print arxiv.org/abs/quant-ph/0104126v1
Nov
22
comment Hilbert-Schmidt basis for many qubits - reference
So the term due to Hilbert-Schmidt inner product for matrices?
Nov
22
comment Hilbert-Schmidt basis for many qubits - reference
It is simply decomposition of $4^n$ dimensional “vector” with an orthogonal basis. Vector space is space of $2^n \times 2^n$ Hermitian matrices with respect to norm $(A,B) =Tr(AB) = Tr(AB^*)$. But I doubt, it could be called Hilbert-Schmidt decomposition because it is defined for any $n$ and for $n=2$ produces up to 16 terms instead of 4.
Nov
22
comment What proof techniques have failed for solving the SIC-POVM problem and what new insights have been gleaned from them?
Just mentioned related question on MO mathoverflow.net/questions/2897/…