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Basis ambiguity

by Talisman
Tags: ambiguity, basis
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Jun24-07, 02:56 PM
P: 32
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:

[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?

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Jun24-07, 04:50 PM
HW Helper
P: 2,567
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 [itex]e_1,e_2[/itex] and the second, W, has basis [itex]d_1,d_2[/itex]. Then define the diagonal tensor [itex]T = e_1 d_1[/itex].

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

[tex] d_1 = a d_1' + b d_2' [/tex]

[tex] d_2 = c d_1' + d d_2' [/tex]

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

[tex] 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' [/tex]

for this to be diagonal, we must have a=b=0, which is impossible.
Jun24-07, 05:11 PM
P: 32
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.

Jun24-07, 05:18 PM
HW Helper
P: 2,567
Basis ambiguity

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.
Jun24-07, 05:45 PM
P: 32
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:

[tex]e_1 = sin \alpha e_1' + cos \alpha e_2'[/tex]
[tex]e_2 = cos \alpha e_1' - sin \alpha e_2'[/tex]

[tex]d_1 = sin \beta d_1' + cos \beta d_2'[/tex]
[tex]d_2 = cos \beta d_1' - sin \beta d_2'[/tex]

[tex]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') + [/tex]
[tex]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')[/tex]

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

respectively. Both must be zero, yielding [tex]c_1 = c_2[/tex], which is of course not true in general (or alternatively the trivial [tex]\alpha = \beta = \frac{\pi}{2}[/tex])
Jun24-07, 05:50 PM
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,091
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.
Jun24-07, 06:25 PM
P: 32
It seems you are right, and I misrepresented the claim:

The basis ambiguity – the ability to re-write [tex]|\phi\rangle[/tex], 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 [tex]|\phi \rangle = \alpha |a0\rangle |b0\rangle + \beta |a1\rangle |b1\rangle[/tex]

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

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

[tex]|\Psi_t\rangle = \sum_i a_i |s_i\rangle |A_i\rangle = \sum_i b_i |r_i\rangle |b_i\rangle[/tex]
Jul16-08, 12:42 AM
P: 32
Quote Quote by Hurkyl View Post
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 [tex]T = e_1d_1[/tex] as he does and
[tex] e_1 = a e_1' + b e_2' [/tex]
[tex] d_1 = c d_1' + d d_2' [/tex]


[tex]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'[/tex]

Then the diagonal constraint gives:

[tex]ad = bc = 0[/tex]

Which leaves us with... scaling the original bases? What's the author really saying? Where is the "basis ambiguity"?
Feb28-10, 07:48 PM
P: 32
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:

[tex]|\psi{\rangle} = |x+{\rangle}|x+{\rangle} + |x-{\rangle}|x-{\rangle}[/tex]
[tex] = |y+{\rangle}|y+{\rangle} + |y-{\rangle}|y-{\rangle}[/tex]
[tex] = |z+{\rangle}|z+{\rangle} + |z-{\rangle}|z-{\rangle}[/tex]

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