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Jan 22, 2013 at 0:54 answer added Qmechanic timeline score: 6
Jan 18, 2013 at 23:05 answer added Mauricio Matera timeline score: 0
Jan 18, 2013 at 21:26 history edited Qmechanic CC BY-SA 3.0
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Jan 13, 2013 at 14:22 comment added user17581 I'm afraid that twistor59 is right, you need the set of kets to be a basis of the abstract space.
Jan 13, 2013 at 8:31 comment added twistor59 You said that the $|x_l\rangle$ are not necessarily orthonormal, but do they span the (presumably finite dimensional) space? If not, then you can't construct an arbitrary density matrix in this way.
Jan 12, 2013 at 19:41 history edited N. Virgo CC BY-SA 3.0
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Jan 12, 2013 at 15:46 comment added physics_xyz I don't think so, because here $\left\{|x_{\ell}>\right\}_{\ell = 1}^{N}$ don't form a basis for space they are just an ensemble.
Jan 12, 2013 at 15:17 comment added Luboš Motl It's just the diagonalization of the density matrix, a Hermitian matrix, isn't it? $N$ must be chosen to be nothing else than the dimension of the matrix for generic ones, otherwise the $x$-vectors wouldn't be orthogonal to each other.
Jan 12, 2013 at 14:58 review First posts
Jan 12, 2013 at 15:41
Jan 12, 2013 at 14:56 comment added physics_xyz yeah, that's right but the coefficients of $|x_{\ell}>$ here are all the same and we have factorized them out as $\frac{1}{N}$, besides $|x_{\ell}>$ are normal states! how can it be possible ?
Jan 12, 2013 at 14:48 comment added user17581 I'm not answering right away because right now I can't think of a way to mathematically prove it (and I'm a lil' bit busy ATM), but I'm sure that to construct a density matrix you don't need the states to be orthogonal to each other.
Jan 12, 2013 at 14:39 history asked physics_xyz CC BY-SA 3.0