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(I updated the question, the new last passage is the important one)

If I assume the position of a particle to be represented by an operator $\hat{x}$, and the time evolution to be carried out by the a unitary transformation generated by an operator $\hat{H}$ (called the "Hamiltonian", then I don't have any notion that there has to be something like a "momentum operator".

Now I am trying to come up with a plausibility argument that there must be another, just as fundamental operator, which we call momentum, and which will contribute to any observable that we can define as function of operators, especially $\hat{H}$, in the same sense that in classical mechanics, we naturally see that the state of a system is not described by an element of the configurationspace $q$, but instead by an element of the phase space $(q, p)$.

The explanation I came up with is the following, and I want to know whether it's a good plausibility argument, or a bad one: The Hamiltonian $\hat{H}$ has to be an abitrary operator acting on the Hilbert space that consists of all the non-degenerate eigenstates of the position operator. My idea is that this most general operator $\hat{H}$ could only be represented by a polynomial containing $\hat{X}$ AND $\hat{P}$ with $[\hat{X}, \hat{P}]= i \hbar$.

If I look at the position representation of the Hilbert space, then the hilbert space becomes the space of all square-integrable functions over the domain of the real numbers, and the most general linear map acting on those functions $f(x)$ would be a combination of derivations $\frac{d}{dx}$ ( that would be the momentum operator) and multiplications with $x$ (that would be the position operator).

Does this reasoning hold in general? If the canonical commutation $[\hat{X}, \hat{P}]= i \hbar$ relation holds, does that mean I can represent any hermitian operator by a polynomial of $\hat{X}$ and $\hat{P}$?

If not, is there an other reason there has to be an observable like momentum called momentum with the properties we are used to momentum to have, and that contributes to the generator of time translations $\hat{H}$ in the sense that $\hat{H} = h(\hat{X}, \hat{P})$? I'm aware that we need momentum to reproduce the classical limit of the theory, but that is not the path of reasoning I want to follow. I want to argument from "the other side".

** ** I think I can reformulate the question in the following way, that makes it clearer what I want to ask: Why do we need a conjugate variable for every observable that we look at? In classical mechanics we are originally left with a generalized coordinate $q$. Later on, we add a canonical momentum $p$, because then the movement can be described in the phase space, which is a symplectic manifold, in a very elegant way, with every observable possible being a function on said phase space.

In quantum mechanics I don't see a similar reasoning. We have the postulates of a hilbert space, of operators and of unitary time development, but momentum as an operator and the canonical commutation relation is just introduced to mimic classical mechanics. There is no argument given about the structure of the space of observables, like it is the case with classical mechanics. Is there a reasoning for the plausibility of a conjugate variable in quantum mechanics, that is similar to the reasoning (symplectic manifold, elegant treatment of dynamics) in classical mechanics?

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    $\begingroup$ What it is true is that every self-adjoint operator, also unbounded, defined in a dense domain in $L^2(\mathbb R^3)$ is the strong limit of polynomials of the spectral measures of the three position and the three momentum operators. Some of these operators are in fact (the closures of) polynomials of these operators directly. $\endgroup$ – Valter Moretti Dec 31 '17 at 9:34
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    $\begingroup$ However the existence of the momentum operator arises from another way. For evident physical reasons, it is assumed that a non-relativistic particle supports a (strongly continuous unitary) representation of the (central extension of) Galileian group acting on the space of states. In particular the subgroup of spatial translations acts on states through unitary operators. These kinds of representations are generated by (are the exponetial of) observables in view of general theorems. Spatial translations are generated by the momentum operators by definition. $\endgroup$ – Valter Moretti Dec 31 '17 at 9:43
  • $\begingroup$ @ValterMoretti to maybe rephrase my question, I see that I haven't expressed myself well: The question is not why there should be a momentum operator, it instead is why it should appear in the hamiltonian, and thus the time evolution of $\hat{x}$ automatically always be paired with the time evolution of $\hat{P}$. $\endgroup$ – Quantumwhisp Dec 31 '17 at 10:38
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    $\begingroup$ I am not sure to understand well your question, and I thino the title of your question is misleading. However once we have introduced the momentum operator as I wrote, from the theory of Galilean group representations, it turns out that $H-P^2/2m$ is a function of $X$ only. Where $H$ (actually the statement would need further details) is the generator of time displacements. It is natural to classify quanum systems by means of that difference.... $\endgroup$ – Valter Moretti Dec 31 '17 at 11:28
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    $\begingroup$ @Quantumwhisp: You need to edit your question to reflect the actual question as you rephrased it in your comment. Otherwise someone that gives an answer may not address the question that your are really meant to ask. $\endgroup$ – flippiefanus Jan 3 '18 at 4:09
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This may not be what you are looking for, but this is how I would proceed :

As you said, time evolution is carried by a unitary operator $\hat{U} (\Delta t)$ whose generator is the Hamiltonian $\hat{H}$, i.e. $\hat{U}(\Delta t) = e^{-i\hat{H}\Delta t/\hbar}$, and $\hat{U}(\Delta t) \vert \psi (t) \rangle = \vert \psi (t+\Delta t) \rangle$. The Hamiltonian is thus defined as the operator whose action on the representation $\vert \psi \rangle$ of a particle is infinitesimal time translation $\hat{H} \vert \psi \rangle = i \hbar \frac{d}{dt} \vert \psi \rangle$.

Similarly to time evolution, I can define a unitary operator $\hat{U}(\Delta x)$ whose action is to displace a particle by the distance $\Delta x$ in the $x$-direction. Momentum in the $x$-direction is thus defined as the generator of this unitary operator, and must be hermitian : $\hat{U} (\Delta x) = e^{i \hat{p}\Delta x / \hbar }$. Since \hat{p} is hermitian, it has an eigenbasis $\hat{p} \vert p \rangle = p \vert p \rangle$ , $p \in \mathbb{R}$. From the superposition principle, I can create another state from the momentum eigenstates as : $\vert \Delta x \rangle := \int \frac{dp}{2 \pi \hbar} \hat{U}(\Delta x)\vert p \rangle = \int \frac{dp}{2 \pi \hbar} e^{ip\Delta x}\vert p \rangle$. This allows me to construct a position operator as $\hat{x} := \int d \Delta x \, \Delta x \vert \Delta x \rangle \langle \Delta x \vert$.

By these definitions we retrieve the commutation relations $[\hat{x},\hat{p}] = i \hbar \mathbb{1}$. Therefore any quantum theory with unitary representations of spatial translations will have canonically conjugated position and momentum operators.

This definition of momentum as the operator of infinitesimal spatial displacement seems the most natural to me, both in quantum and classical physics : in the Lagrangian formalism, momentum $p = \frac{\partial L}{\partial \dot{x}}$ is the conserved charge associated with the symmetry $x \rightarrow x + \Delta x$ of the action. In the Hamiltonian formalism, momentum defines an operator whose action on a phase space function $F(x,p)$ is given by the Poisson bracket $\{ F, p \} = \frac{\partial F}{\partial x}$, that is the effect on $F$ of infinitesimal displacement in $x$.

This brings me to your second question : "Why do we need a conjugate variable for every observable that we look at?" In general, conjugated variables are related in the same way as time and energy : one variable defines a transformation of the system (often a symmetry), and the other defines the generator of that transformation (and the conserved charge in the case of a symmetry). In this sense : energy generates time translation, momentum generates space translation, the number operator (or charge operator in relativistic QFT) generates phase rotation $\hat{\Psi} \rightarrow e^{i \alpha} \hat{\Psi}$ ($\hat{\Psi}$ here denotes a quantum field operator in second quantization or relativistic QFT), and so on $^*$. Furthermore, some of the conjugate variables that define a transformation have no corresponding observables/hermitian operators$^{**}$, like time and phase, so it isn't always helpful to consider conjugate variables as canonically commuting observables.

To resume : for conjugate variables, the counterpart of momentum generates an infinitesimal change in its conjugate "coordinate". The unitarity of QM implies that finite changes in the "coordinate" variable must be represented by unitary operators, whose hermitian generator is the "momentum" variable. This defines the "momentum" variable eigenstates, and the action of the unitary "coordinate" changes on the latter leads to the uncertainty relations. In particular, if the "coordinate" has a corresponding observable & eigenbasis, the two observables are related by canonical commutation relations.

$^*$ Since the phase is multivalued ($\phi \equiv \phi + 2 \pi n$), it is difficult to provide a general & rigorous charge-phase or number of particles-phase uncertainty relation going beyond the naive inequality $\Delta N \Delta \phi \geq 1/2$. Nevertheless, this concept is widely used in the study of coherent states of many-body systems (condensates), which are the closest thing to "phase eigenstates", as discussed here in the context of quantum optics.

$^{**}$ This thread discusses the possibility of a time operator, and this review article covers the attempts at defining a phase operator.

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