Your question is not completely clear to me, since you can define the momentum operator with the opposite sign in wave mechanics, and then also the energy operator. The current convention has been chosen about one hundred years ago. Or do you want to keep the usual convention and multiply X and P with i or something like that, rendering the operators non-hermitian? What exactly do you want to do? Are you considering the uncertainty relation - there, of course, you also can not measure the position and the *negative* momentum at the same time...

Dear Won Sang Chung,

The important point is that the operators, corresponding to the vectors X and P, are conjugate, so if you choose one sign convention for e.g. X, the sign for P is given. This yields the given forms of the commutation relations. If you want to "play" with this be careful that you know what you do. 

Dear Won Sang Chung

please make me clear about your question! and What exactly do you want to do? 

Thanks for answers

I am now finding the deformed Heisenberg algebra with both minimal lenth and maximal lemgth. If we are to deform the HA with maximal lemgth, we can use the HA on anti-de Sitter background:

[ X_i, P_j] = i ( 1 - a R^2 )\delta_{ij}, ( a>0)   R^2 = X_k X_k

In this case from the invariant hyper volume in D dim ( which should be positive), we have

R^2 \le  1/a

If we consider

[ X_i, P_j] = i ( 1 - a R^2 ) ( bR^2 -1) \delta_{ij}, ( a>0, b>0 a 0

we have

[ X_i, P_j] = -i \delta_{ij},

??????

Is this unphysical?

Dear Dear Won Sang Chung,

if you refer to the derivation of the uncertainty relation you may find in any good QM textbook, you'll realize that the relation between Deltax and Deltap comes from an inequality for the squares of these quentities. As such,  it is irrelevant the sign you have for your commutator.

Thank  Giorgio Pastore for helpful comment

That [X,P] = ih(cut) will be clear once we go for a quantum-classical correspondence along the path of Ehrenfest. On the average, Newton's laws should be obeyed. You should arrive at

m(d/dt) =

.

The commutator with the opposite sign will lead to

m(d/dt) =-

.

Of course, one needs to keep the Schrodinger equation intact. Hope this will be useful.

 To Kamal Bhattacharyya

The possion bracket {x,p} =1 is replaced with Dirac's bracket

[x,p]=i\har if we assume

{f,g} -> \frac{1}{i\hbar} [x,p]

But, if we assume that

{f,g} -> \frac{1}{-i\hbar} [x,p] in quantum case,

we have

[x,p]=-i\hbar from {x,p}=1

The remaning problem for the thus case is that we should regard

e^{ i ( k x + w t )} as a right moving wave.....

Dear Giorgio,

The sign of the commutator between conjugate operators cannot be arbitrary as I said in my initial post. This follows from the fundamental structure of QM.

Best

erkki

If we assume {f,g}_{PB} -> \frac{1}{-i\hbar} [f,g] in quantum case, we get [x,p]=-i

from now on we set hbar =1

For free Hamiltonian H = p^2/2m, the eq of motion is classically,

dx/dt = { x, H} = p/m

If we demand {f,g}_{PB} -> \frac{1}{-i} [f,g] ( hence  [x,p]=-i) in quantum case, we get

dx_{op}/dt =  \frac{1}{-i} [x,H]  = p_{op}/m

which indicates that Ehrenfest theorem still holds in this setting.

A general recommendation: to receive more answers and comments, why not use emoj and symbols to write direct readable equations?!

Best

erkki

Dear Erkki,

the original question does not contain enough information for a unique answer. I concentrared my answer to the uncertainty principle derivable from the propose commutator.

Your observation is correct but has to do with the problem if X and P such that [X,P]=-i hbar can or cannot represent conjugate operators for building the QM evolution.

In that respect, I observe that the result obtained by the OP could be interpreted as an inversion of role between position and momentum. But I am not able to judge how plausible this hypothesis could be.

My point was this: If one employs the standard equation of quantum dynamics, or the equation of motion, (by the statement 'in QM' that appeared in the question, I assume this is so), then there is no further choice available. One has to stick to what I said.

The choice exists only before the formulation of QM.