Can Basis Ambiguity be Preserved for Entangled Particles?

[Talisman]

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:

|\psi{\rangle} = \sum_i x_i |A_i{\rangle}|B_i{\rangle}

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:

|\psi{\rangle} = \sum_i y_i |A'_i{\rangle}|B'_i{\rangle}

However, in the case of three particles:

|\psi{\rangle} = \sum_i x_i |A_i{\rangle}|B_i{\rangle}|C_i{\rangle}

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!

 

[StatusX]

I don't think it's possible in general. The simplest case is when both vector spaces are two dimensional. For example, say the first, V, has basis e_1,e_2 and the second, W, has basis d_1,d_2. Then define the diagonal tensor T = e_1 d_1.

Now we take a new basis for V such that e_1=e_1'+e_2', e_2=e_1'-e_2'. An arbitrary new basis for W will have:

d_1 = a d_1' + b d_2'

d_2 = c d_1' + d d_2'

for some a,b,c,d with ad-bc non-zero. Then in this new system T becomes:

V = e_1 d_1 = (e_1'+e_2')(a d_1' + b d_2' ) = a e_1' d_1' + a e_2'd_1' + b e_1' d_2' + b e_2' d_2'

for this to be diagonal, we must have a=b=0, which is impossible.

 

[Talisman]

Sorry if it wasn't clear, but: the claim wasn't that one can pick an arbitrary new basis for V and find a corresponding one for W, but that such a basis exists.

 

[StatusX]

Maybe you should explain what this is for. I mean, if that's what you're asking, why not just take the original bases, or slightly less trivially, a permutation or scalar multiple of them.

 

[Talisman]

Maybe you should explain what this is for.

I guess I just want to follow that paper in depth, and to do that, I want to get a better intuitive understanding of some of the material.

In any case, you inspired me to prove that it's impossible in general, assuming we're sticking to orthonormal bases:

e_1 = sin \alpha e_1' + cos \alpha e_2'
e_2 = cos \alpha e_1' - sin \alpha e_2'

d_1 = sin \beta d_1' + cos \beta d_2'
d_2 = cos \beta d_1' - sin \beta d_2'

c_1 e_1 d_1 + c2 e_2 d_2 = c_1(sin \alpha sin \beta e_1' d_1' + sin \alpha cos \beta e_1' e_2' + cos \alpha sin \beta e_2' d_1' + cos \alpha cos \beta c_2' d_2') +
c_2(cos \alpha cos \beta e_1' d_1' - cos \alpha sin \beta e_1' e_2' - sin \alpha cos \beta c_2' d_1' + sin \alpha sin \beta c_2' d_2')

The coefficients of e_1' d_2' and e_2' d_1' are
c_1 sin \alpha cos \beta - c_2 cos \alpha sin \beta and
c_1 cos \alpha sin \beta - c_2 sin \alpha cos \beta

respectively. Both must be zero, yielding c_1 = c_2, which is of course not true in general (or alternatively the trivial \alpha = \beta = \frac{\pi}{2})

 

[Hurkyl]

Ah, now I see what you're asking.

Suppose that you have a state that's 'diagonal' with respect to a particular pair of bases for A and B.

The claim is that for every basis of A, there exists a basis for B such that your state is also diagonal with respect to those bases.

 

[Talisman]

It seems you are right, and I misrepresented the claim:

The basis ambiguity – the ability to re-write |\phi\rangle, Eq.
(4.2), in any basis of, say, the system, with the superposition
principle guaranteeing existence of the corresponding
pure states of the apparatus – disappears when an
additional system, E, performs a premeasurement on A

Where |\phi \rangle = \alpha |a0\rangle |b0\rangle + \beta |a1\rangle |b1\rangle

But doesn't my previous post show that this is false?

To be clear, he introduces this 'basis ambiguity' with the following:

|\Psi_t\rangle = \sum_i a_i |s_i\rangle |A_i\rangle = \sum_i b_i |r_i\rangle |b_i\rangle

 

[Talisman]

The claim is that for every basis of A, there exists a basis for B such that your state is also diagonal with respect to those bases.

I think the claim is that there exist some new bases A and B such that the state is diagonal wrt those bases. (Yes, I realize this thread is a year old ;))

Actually, StatusX's idea makes short work of it, I think:

Let T = e_1d_1 as he does and
e_1 = a e_1' + b e_2'
d_1 = c d_1' + d d_2'

Then

T = (a e_1&#039; + b e_2&#039;)(c d_1&#039; + d d_2&#039;) <br /> = ac e_1&#039;d_1&#039; + ad e_1&#039;d_2&#039; + bc e_2&#039;d_1&#039; + bd e_2&#039;d_2&#039;

Then the diagonal constraint gives:

ad = bc = 0

Which leaves us with... scaling the original bases? What's the author really saying? Where is the "basis ambiguity"?

 

[Talisman]

Yes, I know this thread is way old :)

I just stumbled upon something which partially resolves my question. I haven't worked out the details of when the rearrangement is possible, but an easy example is:

|\psi{\rangle} = |x+{\rangle}|x+{\rangle} + |x-{\rangle}|x-{\rangle}
= |y+{\rangle}|y+{\rangle} + |y-{\rangle}|y-{\rangle}
= |z+{\rangle}|z+{\rangle} + |z-{\rangle}|z-{\rangle}