2022-06-14

A measurement device which can be represented by a 1D quantum system (with canonical observables $X$ and $P$) 'is prepared in a Gaussian state with spread $s$'
$|\psi ⟩=\frac{1}{\left({\pi }^{2}{s}^{2}{\right)}^{1/4}}\int \mathrm{exp}\left[-\frac{{x}^{2}}{2{s}^{2}}\right]\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x|x⟩$
Can somebody tell me what it means that its canonical variables are $X$ and $P$?

pheniankang

To have canonical observables $\stackrel{^}{x}$ and $\stackrel{^}{p}$ means that the eigenvalues of these operators are what you measure (denotes $x,p$), and the operators satisfy the "canonical commutation relation"
$\left[\stackrel{^}{x},\stackrel{^}{p}\right]\equiv \stackrel{^}{x}\stackrel{^}{p}-\stackrel{^}{p}\stackrel{^}{x}=i\hslash$
To prepare a system in an initial state $|\psi ⟩$ means smily that this is the state of the system at $t=0$; it is usually denoted as $|\psi \left(0\right)⟩$
Now, your initial state is given by

Aside: A probability distribution given by $f\left(x\right)=\left({\pi }^{2}{s}^{2}{\right)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{\sigma }^{2}}}$ is said to be a Gaussian with spread $\sigma$, where $\sigma$ is the usual standard deviation from the center of the bell curve.
End Aside
Therefore, the meaning of

is that your system at $t=0$ is in a superposition of "position eigenstates" $|x⟩$ (i.e. eigenvectors of $\stackrel{^}{x}$), weighted by a gaussian distribution. That is, your system isn't just in any old superposition of position eigenstates, but that the ones near $x=0$ are most likely and as you move away from the origin the probability of the system being in that state decreases like ${e}^{-\frac{{x}^{2}}{2{s}^{2}}}$
If you are familiar with wave functions, recall that the definition of $⟨x|\mathrm{\Psi }⟩\equiv \mathrm{\Psi }\left(x\right)$. Then you can get something a bit more useful. Namely, that

where I changed the variable of integration to ${x}^{\prime }$ to make things clearer.
At any rate, what this means is that your initial wave function is given by
$\mathrm{\Psi }\left(x,t=0\right)=\left({\pi }^{2}{s}^{2}{\right)}^{-1/4}{e}^{-\frac{{x}^{2}}{2{s}^{2}}}.$
Where to go from here with whatever you're doing should be familiar at this point (i.e. after finding the initial wave function).

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