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Commutators on a discrete QM lattice = ?

  1. Sep 5, 2011 #1
    Commutators on a discrete QM lattice = ???

    Please let me know if any of the following is unclear:

    I was thinking about how you could go about doing QM not in a continuous space but instead on a lattice, take 1D for simplicity. Let's use a finite (not countably infinite) number of positions say N. Now X and P are finite dimensional matrices. Now it turns out that [X,P]=i*I (where I is the NxN identity matrix) cannot be satisfied, ever. Since:

    XP - PX = i*I
    tr( XP - PX ) = tr( i*I ) // tr is the trace of a matrix, which is of course a truth
    // preserving operation
    tr( XP ) - tr( PX ) = i*N // tr is a linear operator
    tr( XP ) - tr( XP ) = i*N // cyclic invariance
    0 = i*N

    or see: http://en.wikipedia.org/wiki/Trace_(linear_algebra)#Lie_algebra

    Needless to say this last equation has a truth value of false. I.e. assuming [X,P]=i*I leads to a contradiction (for finite dimensional vector spaces). Please explain why an infinite dimensional matrix can do this but an arbitrarily large finite dimensional one can't. And what does that mean in the language of physics?
    _______________________________________________________________________

    My "attempts":
    1) Since the 'derivative' operator isn't hermitian for all square finite dimensional approximations, so we could use a non-square matrix but then the matrix dimensions won't come out right. (check for yourself) (failure)
    2) Another possibility is to use XP-PX = i*diag(I,-N+1), so there are i's on the diagonal except for the last entry which is there to balance out the other positive terms, (so the trace is still zero). Such matrices come straight from a truncation of the QSHO X and P matrices in the energy basis (seems the most reasonable approach to me so far).

    Attempting to make it more concrete I typed the following into MATLAB:
    n=15;
    X=diag(1:n);
    P=fft(eye(length(X)))*X*ifft(eye(length(X))); // ie position psi is ifft of momentum psi
    X*P-P*X

    Somehow X*P-P*X manages not to have a single value on the diagonal. Odd when I'd like it to be a diagonal matrix.

    Ok, or just answer this question in continuous QM if you know how (seriously I'd really like to know):
    In continuous ordinary QM what operator distinguishes positive values of x from the negative values ie like sign(X). How do I construct it, what is it's conjugate variable, what does and doesn't it commute with, and so on...
     
  2. jcsd
  3. Sep 6, 2011 #2
    Re: Commutators on a discrete QM lattice = ???

    Nothing stops you from defining an operator whose eigenstates are the position eigenstates |x>, with eigenvalue 1 if x is positive, -1 if it's negative, and maybe 0 if x = 0. This completely specifies the operator. You can think of this operator as acting on a wave function psi(x) by changing psi(x) to -psi(x) for x < 0.
     
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