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In this question, I want to to know wether my reasoning on the plausibility of the Heisenberg equation is flawed:

Let's say I want to describe my system in the quantum-mechanics framework: Observables (like $\hat{x}$ are represented by operators acting on vectors, and the decomposition of a vector into eigenvectors of the operator tells me something about the propabilities to measure a specific value. To preserve the norm of the vector (and thus to preserve the propabilitiy for "ANY" value of the observable to be 1), one needs unitary time evolution, which is generated by an hermitian operator (which I now simply will call "$\hat{H}$"). What follows (in the Heisenberg picture) is $$ \dot{\hat{x}} = \frac{i}{\hbar}[\hat{H}, \hat{x}] $$ So far, so good, I could use the same argument to derive the Heisenberg equation for $\hat{p}$ (with a yet to determine operator $\hat{\tilde{H}}$. But who tells me that the generators of those two time evolutions will be the same? If the only argument is to preserve the propabilistic interpretation, then those two observables could be time-evoluted by two different Operators.

My Idea now is the following: $\hat{p}$ isn't just any abitrary operator, it is supposed to always be the generator of translations of $\hat{x}$, which means that the commutation relations should always hold (by definition): I define $\hat{p}(t)$ in a way, so that it always is the generator of infinitesimal translations of $\hat{x}$. Then the commutation relations $[\hat{x}, \hat{p}]=\hbar$ are supposed to hold at anytime, and for that to be true, $\hat{x}$ and $\hat{p}$ need to be time-evoluted by the same unitary transformation.

Is that argument flawed, or can I think of it that way? The way I do it, is the only assumption I make "observables are represented by operators" and "time evolution has to preserve the norm"? Or do I make additional assumptions, and I just don't see them here?

EDIT

I did some calculations: I assume that the time evolution of operator $\hat{x}$ is generated by the operator $\hat{H}$, while the time volution of operator $\hat{p}$ is generated by the operator $\hat{H}+\hat{A}$. If initially \begin{align} \hat{x}(0) = \hat{x}_0 \\ \hat{p}(0) = \hat{p}_0 \end{align}

Then after an infinitesimal amount of time t, the operators will be

\begin{align} \hat{x}(t) = \hat{x}_0 + i\hbar[\hat{x}, \hat{H}] \\ \hat{p}(t) = \hat{p}_0 + i\hbar[\hat{x}, \hat{H} + \hat{A}] = \hat{x}(t) + i \hbar [\hat{x}, \hat{A}] \end{align}

If I still require $[\hat{x}(t), \hat{p}(t)] = i \hbar$ to hold, then $\hat{A}$ only has to satisfy:

\begin{align} [\hat{x}, [\hat{p}, \hat{A}]] = 0 \end{align}

So clearly, demanding the commutation relation to hold for $\hat{p}$ and $\hat{x}$ is not a sufficient condition for them to be governed by the same time evolution. The generator for their unitary transformations can differ by an Operator $\hat{A}$, that satisfied the above written condition. The Answer to my question should be "NO". An additional question would be what the further assumptions are that make sure that position and momentum are governed by the same time evolution.

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  • $\begingroup$ You say that $\hat{p}$ is not an operator but the "generator of the translations". Under my point of view, that is contradictory: when you say the "generator", you imply that the element is inside a Lie group. The representation of the elements of the groups in a vector space are operators. So, if $\hat{p}$ is not an operator, it cannot generate anything. $\endgroup$ – VictorSeven Oct 13 '17 at 8:50
  • $\begingroup$ I edited the text. What I meant was that $\hat{p}$ is not an abitrary (that's what I wrongly translated to "any") operator, but it is always a certain operator, defined in a certain way. $\endgroup$ – Quantumwhisp Oct 13 '17 at 9:27
  • $\begingroup$ By the very definition of the Heisenberg picture, $i\frac{dA}{dt} = [A,H]$ for any operator $A$ isn't it? $\endgroup$ – user154997 Oct 13 '17 at 9:33
  • $\begingroup$ a) It isn't for any operator, for example not for operators that have a "built in time dependency. b) I don#t want to take the Heisenberg picture as given. I seek for a plausibility argument, why the equation you stated should hold (at least for x and p) $\endgroup$ – Quantumwhisp Oct 13 '17 at 10:00

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