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 Talisman Jun24-07 02:56 PM

Basis ambiguity

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 here), 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 Jun24-07 04:50 PM

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 Jun24-07 05:11 PM

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 Jun24-07 05:18 PM

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 Jun24-07 05:45 PM

Quote:
 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 Jun24-07 05:50 PM

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 Jun24-07 06:25 PM

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

Quote:
 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 Jul16-08 12:42 AM

Re: Basis ambiguity

Quote:
 Quote by Hurkyl (Post 1363847) 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

$$T = (a e_1' + b e_2')(c d_1' + d d_2') = ac e_1'd_1' + ad e_1'd_2' + bc e_2'd_1' + bd e_2'd_2'$$

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 Feb28-10 07:48 PM

Re: Basis ambiguity

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:

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

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