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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 $s$'

$$\vert \psi \rangle = \frac{1}{(\pi^2s^2)^{1/4}} \int \exp\left[-\frac{x^2}{2s^2}\right]\mathop{}\!\mathrm dx\vert x\rangle$$

Can somebody tell me what it means that its canonical variables are $X$ and $P$?

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  • $\begingroup$ Possible duplicate: link. $\endgroup$ – secavara Dec 24 '18 at 0:13
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To have canonical observables $\hat{x}$ and $\hat{p}$ means that the eigenvalues of these operators are what you measure (denotes $x, p$), and the operators satisfy the "canonical commutation relation"

$$[\hat{x},\hat{p}] \equiv \hat{x}\hat{p} - \hat{p}\hat{x} = i\hbar $$

To prepare a system in an initial state $|\psi \rangle$ means smily that this is the state of the system at $t=0$; it is usually denoted as $|\psi(0)\rangle$.

Now, your initial state is given by

$$\mid \psi(0) \rangle = (\pi^2s^2)^{-1/4} \int dx \ e^{-\frac{x^2}{2s^2}}\mid x\rangle. $$

Aside: A probability distribution given by $f(x) = (\pi^2s^2)^{-1/4} e^{-\frac{x^2}{2\sigma^2}}$ 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

$$\mid \psi(0) \rangle = (\pi^2s^2)^{-1/4} \int dx \ e^{-\frac{x^2}{2s^2}}\mid x\rangle. $$

is that your system at $t=0$ is in a superposition of "position eigenstates" $| x\rangle$ (i.e. eigenvectors of $\hat{x}$), 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}{2s^2}}$.

If you are familiar with wave functions, recall that the definition of $\langle x | \Psi \rangle \equiv \Psi(x)$. Then you can get something a bit more useful. Namely, that

$$ \langle x |\psi(0) \rangle = (\pi^2s^2)^{-1/4} \int dx' \ e^{-\frac{x'^2}{2s^2}} \langle x|x' \rangle = (\pi^2s^2)^{-1/4} \int dx' \ e^{-\frac{x'^2}{2s^2}} \delta(x-x') = (\pi^2s^2)^{-1/4} e^{-\frac{x^2}{2s^2}} $$

where I changed the variable of integration to $x'$ to make things clearer.

At any rate, what this means is that your initial wave function is given by

$$ \Psi(x, t=0)= (\pi^2s^2)^{-1/4} e^{-\frac{x^2}{2s^2}}. $$

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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  • $\begingroup$ Thank you very much ! One question. How can a state $\mid x>$ be near or far from x=0 What does it mean for a state to be at x=0 ? etc. $\endgroup$ – Benjamin Jabl Dec 25 '18 at 15:49
  • $\begingroup$ It means that observing the eigenvalue given by $\hat{x}|x\rangle = x |x\rangle$ has a higher probability the closer $x$ is to zero. That is, the particle being in the state $|x= 0.001\rangle $ is more likely than the particle being in $|x = 10 \rangle$ $\endgroup$ – InertialObserver Dec 26 '18 at 0:22

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