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Basis ambiguity 
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#1
Jun2407, 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? Thanks! 


#2
Jun2407, 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 adbc nonzero. 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. 


#3
Jun2407, 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.



#4
Jun2407, 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.



#5
Jun2407, 05:45 PM

P: 32

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]) 


#6
Jun2407, 05:50 PM

Emeritus
Sci Advisor
PF Gold
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. 


#7
Jun2407, 06:25 PM

P: 32

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


#8
Jul1608, 12:42 AM

P: 32

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] Then [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"? 


#9
Feb2810, 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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