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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• Possible duplicate: link. – secavara Dec 24 '18 at 0:13

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).

• 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. – Benjamin Jabl Dec 25 '18 at 15:49
• 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 $|x = 10 \rangle$ – InertialObserver Dec 26 '18 at 0:22