Commutators on a discrete QM lattice = ?

In summary: It's not hermitian, but you could multiply it by i to make it hermitian. In summary, the conversation discusses the possibility of doing quantum mechanics on a discrete lattice instead of a continuous space. However, it is shown that the commutator [X,P]=i*I cannot be satisfied in finite dimensional vector spaces, leading to a contradiction. The conversation also explores the idea of using a non-square matrix or a truncation of QSHO matrices to solve this problem. The question of what operator distinguishes positive and negative values of x in continuous QM is also discussed.
  • #1
cheeseits
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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...
 
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  • #2


cheeseits said:
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...

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.
 

Related to Commutators on a discrete QM lattice = ?

1. What is a commutator on a discrete QM lattice?

A commutator on a discrete QM lattice is a mathematical operation that represents the non-commutativity of operators in quantum mechanics. It is used to determine the uncertainty in the measurement of two observables.

2. How is a commutator calculated on a discrete QM lattice?

A commutator on a discrete QM lattice can be calculated by taking the difference between the product of two operators and the product of the same operators in reverse order.

3. What is the significance of commutators in quantum mechanics?

Commutators play a crucial role in quantum mechanics as they determine the uncertainty in the measurement of two observables. They also help to define the fundamental principles of quantum mechanics, such as the Heisenberg uncertainty principle and the concept of non-commutativity.

4. Can commutators be generalized to continuous QM systems?

Yes, commutators can be generalized to continuous QM systems. In fact, the commutator on a discrete QM lattice is a special case of the commutator in a continuous system.

5. How are commutators related to the wave function in quantum mechanics?

Commutators are related to the wave function in quantum mechanics through the Schrödinger equation, which describes the time evolution of the wave function. Commutators are also used to define the momentum and position operators in the wave function.

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