- #1
Talisman
- 95
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I asked this question over in the QM forum, but it fizzled out there. I think it's more appropriate here anyway so I'll post it. If this is against forum rules, I apologize!
I'm reading a paper on decoherence (preprint http://arxiv.org/abs/quant-ph/0105127" ), and am afraid I don't grasp one of the claims the author makes. Briefly, consider an entagled state of two particles:
[tex]|\psi{\rangle} = \sum_i x_i |A_i{\rangle}|B_i{\rangle}[/tex]
He claims that it is always possible to choose a different basis for the first particle, and find a new basis for the second so that the sum still has the same form:
[tex]|\psi{\rangle} = \sum_i y_i |A'_i{\rangle}|B'_i{\rangle}[/tex]
However, in the case of three particles:
[tex]|\psi{\rangle} = \sum_i x_i |A_i{\rangle}|B_i{\rangle}|C_i{\rangle}[/tex]
Then the basis ambiguity is lost: one cannot, in general, pick a different basis for A and expect to get a similar representation with alternate bases for B and C.
Perhaps my lin alg is a bit rusty, but I cannot prove either claim. Can anyone elucidate?
Thanks!
I'm reading a paper on decoherence (preprint http://arxiv.org/abs/quant-ph/0105127" ), and am afraid I don't grasp one of the claims the author makes. Briefly, consider an entagled state of two particles:
[tex]|\psi{\rangle} = \sum_i x_i |A_i{\rangle}|B_i{\rangle}[/tex]
He claims that it is always possible to choose a different basis for the first particle, and find a new basis for the second so that the sum still has the same form:
[tex]|\psi{\rangle} = \sum_i y_i |A'_i{\rangle}|B'_i{\rangle}[/tex]
However, in the case of three particles:
[tex]|\psi{\rangle} = \sum_i x_i |A_i{\rangle}|B_i{\rangle}|C_i{\rangle}[/tex]
Then the basis ambiguity is lost: one cannot, in general, pick a different basis for A and expect to get a similar representation with alternate bases for B and C.
Perhaps my lin alg is a bit rusty, but I cannot prove either claim. Can anyone elucidate?
Thanks!
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