Using Mathematica to solve for Jacobi Identity

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The discussion revolves around using Mathematica to solve problems related to the Jacobi Identity in Quantum Mechanics, specifically in calculating commutators. A user seeks guidance on deriving the commutator [p_i, p_j] from known commutators [x_i, x_j] and [x_i, p_j], but it is clarified that the Jacobi Identity does not allow for this derivation due to the absence of a unit element in Lie algebras. Instead, it is noted that one can only obtain expressions involving [x_i, [p_i, p_j]] and similar forms. The user expresses a need for computational assistance given the complexity of their research on deformed quantum mechanics, while another participant mentions having created a program to automate part of the process. Ultimately, the consensus is that without additional assumptions, deriving these relationships remains challenging.
Clarence Liu
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Hi everyone, I'm new to Physics Forums and to Mathematica, as well as Jacobi Identity.

In any case, I was wondering on how I may use Mathematica to solve various Quantum Mechanics related problems through commutators. Like if it's possible to find out what is the form of a particular commutator from the Jacobi Identity.

To be more specific, say I define [x_i, x_j] and [x_i,p_j] to be something, how do I use mathematica to find [p_i, p_j].

As well as the coding to supposedly find the uncertainty relation of 2 operators, suppose I define the commutator to be of a certain form/value.

In general, I'm simply not familiar with mathematica and I would like to make my life easier in doing these relevant calculations. Hope someone might be able to enlighten me on this aspect.

Thanks!
 
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I don't know how this is done in mathematica. I programmed the Jacobi-Identity once myself with what was at hand.
Anyway, you cannot derive ##[p_i,p_j]## from ##[x_i,x_j]## and ##[x_i,p_j]## by the Jacobi-Identity because there is no ##1## in a Lie algebra. You will only get expressions ##[x_i, [p_i,p_j]] = ...## and therefore only a result up to commuting parts. Why want you to do it? Usually you have a certain Lie algebra and its structure constants given.
 
fresh_42 said:
I don't know how this is done in mathematica. I programmed the Jacobi-Identity once myself with what was at hand.
Anyway, you cannot derive ##[p_i,p_j]## from ##[x_i,x_j]## and ##[x_i,p_j]## by the Jacobi-Identity because there is no ##1## in a Lie algebra. You will only get expressions ##[x_i, [p_i,p_j]] = ...## and therefore only a result up to commuting parts. Why want you to do it? Usually you have a certain Lie algebra and its structure constants given.
yes, I do understand that. but essentially, given a particular form of [xi, xj] and [xi, pj] you can in directly proof a certain form of [pi, pj] that obeys the Jacobi Identity. At least as far as I know. I'm actually doing some a research on deformed quantum mechanics from the commutator relationship under a supervisor from my university (I'm an sophomore). So in the research, if we are to suggest a particular deformed [xi, xj] and assuming [pi, pj] to still be 0, I'll be able to get a form of [xi, pj] which obeys the Jacobi Identity. I realized if I were to solve it by hand it's going to be quite a pain in the ass, so I was wondering if I'd be able to do it via Mathematica.
 
Well, it is painful, I know. That's why I've made a program that at least wrote me the equations so I'was left with simply comparing the coefficients. Without additional assumptions I don't see short cuts. Sorry, for I'm no help with mathematica.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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