4
$\begingroup$

enter image description here

I am working through my quantum optics textbook by Grynberg, Aspect and Fabre, and this concept has tripped me up a little.

(13) is an inseparable state, whereas (15) IS separable - but they are the same state, just written using different bases. I can't get my head around how changing our basis has somehow changed what I thought to be inherent to the state.

If this is a stupid question and I need to go back in the textbook, let me know! :)

$\endgroup$
3
  • $\begingroup$ The edition of Grynberg Aspect & Fabre Introduction to Quantum Optics that I am looking at has 700 pages and equation numbers of the form (V.W.XY) none of the form (XY). Without further background from what exactly you are reading (edition/page number) it is hard to handle this question. $\endgroup$
    – Kurt G.
    Jan 17, 2022 at 14:13
  • $\begingroup$ Sorry, I should have clarified, the picture is not from the textbook, it's a supplementary set of notes. I mentioned the textbook to give an idea of what level I am at and what information I'm using most. $\endgroup$
    – cpp
    Jan 17, 2022 at 15:13
  • $\begingroup$ A local matrix operation can make it separable if it is irreversible, like a measurement process, for example the matrix whose colums are all the corresponding endstate. $\endgroup$
    – Cretin2
    May 29, 2022 at 9:34

1 Answer 1

3
$\begingroup$

Any state, no matter how entangled in the original basis, can be written as a product state in another basis, and vice versa. The catch is that the basis transformation has to be global, not local. Entanglement between subsystems $A$ and $B$ cannot be changed by changes of basis which affect $A$ and $B$ alone ($U=U_A\otimes U_B$), but they can be changed by global changes of basis ($U=U_{AB}$).

A change of basis can be any unitary operator and for any given initial state $\lvert\psi\rangle$ and any final state $\lvert\phi\rangle$, there is a unitary which maps $\lvert\psi\rangle$ to $\lvert\phi\rangle$, so any state can be fully disentangled by some change of basis.

The original entangled degrees of freedom are still entangled after the basis change, but the new basis is written in terms of new (perhaps complicated nonlocal) unentangled degrees of freedom. Those degrees of freedom were still unentangled in the original basis. The basis change doesn't change anything physical: it just changes which degrees of freedom we explicitly keep track of.

$\endgroup$
2
  • 1
    $\begingroup$ This makes total sense. Thank you! $\endgroup$
    – cpp
    Jan 18, 2022 at 16:15
  • $\begingroup$ What I wonder is what is the meaning of non separability ? Is it that a hidden variable dictates the choice of both together ? $\endgroup$
    – Cretin2
    Jun 25, 2022 at 6:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.