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Basis ambiguity? 
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#1
Jun2107, 12:32 AM

P: 32

I'm reading a seminal paper by Zurek on decoherence (preprint here), and am afraid I don't grasp one of the claims he makes. Briefly, consider an entagled state of two qubits:
[tex]\psi{\rangle} = \sum_i x_i A_i{\rangle}B_i{\rangle}[/tex] He claims that one can choose a different basis for the first qubit, and get a different representation for psi: [tex]\psi{\rangle} = \sum_i y_i A'_i{\rangle}B'_i{\rangle}[/tex] However, if psi becomes entangled with a _third_ qubit: [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 don't follow either claim. Can anyone elucidate? Thanks! [edited to use tex] 


#2
Jun2107, 01:39 AM

P: 32

Okay, so I've partially convinced myself of the first part, although I can't yet prove it in general. I started with a singlet state:
I) [tex]\frac{\sqrt{2}}{2} (0{\rangle}0{\rangle} + 1{\rangle}1{\rangle})[/tex] And then considered it in the "Hadamard Basis" (did I just make that term up?):  /+ > = (0> /+ 1>) / 2 And discovered (to my dismay) that (I) can indeed be rewritten as: II) [tex]\frac{\sqrt{2}}{2} ({\rangle}{\rangle} + +{\rangle}+{\rangle})[/tex] Should this have been obvious? Is there a nice way of thinking about the tensor product H2 @ H2 with rotated basis sets? 


#3
Jun2107, 02:22 AM

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PF Gold
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Any pair of bases for the vector spaces V and W naturally yield (in the obvious way) a basis for their tensor product V @W.
I think what he's saying about the 3particle system is not merely that it gets entangled  but we also focus our attention on just two of the particles (by taking a partial trace). Then, even if we started with a pure state, we (usually) now have a mixed state. 


#4
Jun2107, 10:26 AM

P: 32

Basis ambiguity?
Thanks Hurkyl. I'm still not sure I follow:
The curious thing is that psi can (apparently) be written as a sum of tensor products with two different pairs of basis sets ({Ai}, B{i}) and (A'{i}, B'{i}) each in H2. Is that more clear? I'm still inclined to believe the second claim made in the paper, but maybe I'll give it some deeper mathematical thought or take it to the lin alg section of this forum. Thanks! 


#5
Jun2107, 03:50 PM

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By (one) definition, the tensor product [itex]V \otimes W[/itex] of two vector spaces V and W is the set of things that can be written as the finite sum
[tex]\sum v_i \otimes w_i[/tex] for any v_{i} in V and a_{i} in W. If you choose a basis for V and a basis for W, then you can take any of those sums and replace v_{i} and w_{i} as a linear combination of basis vectors. When you distribute, you are again left with a sum [tex]\sum a_j \left( v'_j \otimes w'_j \right)[/tex] but with each of the v_{j}' and w_{j} basis vectors, and the a_{j} scalars. Maybe I just don't understand what question you're asking? 


#6
Jun2107, 05:39 PM

P: 32

Hmm... perhaps I'm just being dense, but I still don't see the light. I agree that I can rewrite this:
[tex]\sum_i v_i \otimes w_i[/tex] as this: [tex]\sum_{i,j} a_{ij} \left( v'_i \otimes w'_j \right) [/tex] but you're implying that [tex]a_{ij} = \delta_{ij}[/tex] when suggesting it can be written as so: [tex]\sum a_j \left( v'_j \otimes w'_j \right)[/tex] Or is there something I'm missing? 


#7
Jun2107, 05:54 PM

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It is true that if the sequences v_{i}' and w_{i}' are each a basis, then every element in [itex]V \otimes W[/itex] can be written uniquely as
[tex]\sum_i \sum_j c_{ij} (v_i' \otimes w_j')[/tex] But in my previous post I was simply stating something weaker. I was not assuming that v_{i}' and w_{i}' were bases; just they were just sequences of basis vectors. In particular, some basis vectors may be repeated, and others might not appear at all. 


#8
Jun2107, 06:00 PM

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Oh, maybe I understand what you're asking  you're wondering specifically about the elements that can be written as
[tex]\sum_i a_i \left( v_i' \otimes w_i' \right)[/tex] where the v_{i}' and w_{i}' are bases, correct? Try making the change of basis 0> = (3/5)L>  (4/5)R> 1> = (4/5)L> + (3/5)R> to both components of 00>. What do you get? 


#9
Jun2107, 06:40 PM

P: 32

[tex]0> = (sin \theta)L>  (cos \theta)R>[/tex] [tex]1> = (cos \theta)L> + (sin \theta)R>[/tex] Should work for any [tex]\theta[/tex], no? But I was hoping there was a simpler explanation. I still have no idea about the second claim, although I've got a rough intuitive idea of what constraints would need to be satisfied for a basis change in all three qubits. P.S. I wrote above: 


#10
Jun2107, 07:01 PM

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#11
Jun2107, 08:30 PM

P: 32

That's what I was getting at with my "Hadamard Basis" nonsense, but it shouldn't have taken this "flash of insight" (or any calculation) to see why the statement should be true. EDIT: This only shows an explicit change of basis that works when all the constants xi are the same in my original state. I haven't figured out how to think about it in general. 


#12
Jun2207, 07:25 AM

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But I wanted you to see what happens to
00> not to (00> + 11>) 


#13
Jun2207, 08:29 AM

P: 32

Okay, I'll bite:
[tex]c_000> = \frac{c_0}{25} (9LL>  12LR>  12RL> + 16RR>)[/tex] 


#14
Jun2307, 09:12 AM

P: 8

I am not an expert in this field, but how about, that the entangled state is a measured state, so according to quantum mechanics: the state must have a "direction" (I couldn't find off hand a better description, what I mean: if you measure the spin of a particle, than you give it a direction).
greetings Kay 


#15
Jun2307, 09:55 AM

P: 451

Regards, Dany. 


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