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Why is the "decision" version of the local Hamiltonian problem promised to have a positive gap?

I think I figured it out. The issue is that you can only use a finite number $m$ of bits to specify your problem instance. Therefore, you can only specify $H$ and $E_0$ to some finite precision $\...
tparker's user avatar
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1 vote
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Stationary state of Lindblad equation

To give an explicit example, consider $H=\sigma_x$, $V=|0\rangle\langle 1|$, and $$\mathcal L:=-i[H,\cdot]+V(\cdot)V^\dagger-\frac12V^\dagger V(\cdot)-(\cdot)\frac12V^\dagger V\,.$$ The (in fact ...
Frederik vom Ende's user avatar
5 votes
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In an $n$-body system, do all bodies accelerate towards center of mass?

Consider a simple 3-body system, star-planet-moon. Its center of mass will generally be inside the body of the star or at least very close, because the star is by far the heaviest. So if your ...
leftaroundabout's user avatar
0 votes
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Two-body operators in second-quantized form

$$ a^\dagger(\textbf{r})a^\dagger(\textbf{r}')a(\textbf{r}')a(\textbf{r})|\textbf{r}_1,\textbf{r}_2,...,\textbf{r}_N\rangle=a^\dagger(\textbf{r})a^\dagger(\textbf{r}')a(\textbf{r}')a(\textbf{r})a^\...
hft's user avatar
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3 votes

General expression for fermionic creation operator subspace

Any operator $\hat{a}^\dagger$ that maps $H_n \rightarrow H_{n+1}$ for all $n$, with $H_{N+1} \equiv 0$, is defined by set of matrix elements $$ B^{(n)}_{i_1,\ldots,i_{n+1};j_1,\ldots,j_n} = \langle 0|...
Gec's user avatar
  • 5,377
2 votes

General expression for fermionic creation operator subspace

Yes, you can. To prove this, let us first define what we mean by a general fermionic creation operator. Let $f_n: \mathcal{H}_n \to \mathcal{H}_{n+1}$ be a linear map from the Hilbert space of $n$ ...
Nandagopal Manoj's user avatar

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