Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basis ambiguity

  1. Jun 24, 2007 #1
    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" [Broken]), 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?

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 24, 2007 #2


    User Avatar
    Homework Helper

    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.
    Last edited: Jun 24, 2007
  4. Jun 24, 2007 #3
    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.
  5. Jun 24, 2007 #4


    User Avatar
    Homework Helper

    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.
  6. Jun 24, 2007 #5
    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])
    Last edited: Jun 24, 2007
  7. Jun 24, 2007 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  8. Jun 24, 2007 #7
    It seems you are right, and I misrepresented the claim:

    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]
    Last edited: Jun 24, 2007
  9. Jul 16, 2008 #8
    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"?
  10. Feb 28, 2010 #9
    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook