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I read in Griffith's quantum mechanics that

In a particular system, the second time measurement of the position (say) would yield the same result (the same collapse or the same spike) given that the measurement is done quickly (since it soon spreads out).

I don't understand how quick this is supposed to be. Could somebody give a quantitative feeling for this quickness?

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  • $\begingroup$ Its actually an axiom in Diracs Principles of Quantum Mechanics as far as I can recall. $\endgroup$ – Mozibur Ullah Dec 4 '17 at 6:41
  • $\begingroup$ can you give the page? I found a pdf of Griffiths but it is an image and cannot be searched $\endgroup$ – anna v Dec 4 '17 at 6:54
  • $\begingroup$ you might be interested in my answer to a similar question here physics.stackexchange.com/questions/89690/… $\endgroup$ – anna v Dec 4 '17 at 7:08
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This answer depends on the timescale of your system. Often time, the problem can be rephrased as follows. If you start with a state a $\vert\psi_0(0)\rangle$, evolving under a Hamiltonian $\hat H$ so that $U(t)$ is the resulting evolution, one can ask for what small time $\Delta t$ will be probability of finding the system in $\vert\psi_0(0)\rangle$ be greater than $1-\epsilon$, where $\epsilon$ is assumed small and will depend on $\Delta t$. In other words, one ask for what $\Delta t$ is $$ \vert \langle \psi_0(\Delta t)\vert \psi_0(0)\rangle \vert^2<1-\epsilon. \tag{1} $$ with $\vert\psi_0(t)\rangle =U(t)\vert\psi_0(0)\rangle$.

This defines "quickly" in the sense that, if you remeasure within the time interval $\Delta t$, there is only a small probability $\epsilon$ that your system will have evolved out of your initial state. Given your tolerance $\epsilon$ you can find the appropriate $\Delta t$ to stay within this tolerance.

Assuming $\hat H$ does not depend explicit on $t$ for simplicity, $U(t)=e^{-i\hat H t/\hbar}$ and for small times one can usually write \begin{align} \vert\psi_0(\Delta t)\rangle &\approx \left(\hat 1-i\frac{\Delta t}{\hbar} \hat H -\frac{(\Delta t)^2}{2\hbar^2}\hat H^2 \right) \vert\psi_0(0)\rangle\\ \langle \psi_0(0)\vert\psi_0(\Delta t)\rangle&= 1 -i\frac{\Delta t}{\hbar} \langle \psi_0(0)\vert \hat H\vert \psi_0(0)\rangle -\frac{(\Delta t)^2}{2\hbar^2} \langle\psi_0(0)\vert H^2\vert\psi_0(0)\rangle \end{align} from where you can complete the calculation of $$ \vert \langle \psi_0(0)\vert\psi_0(\Delta t)\rangle \vert^2\le 1-\epsilon $$ and find $\epsilon$ in terms of $\Delta t$ and the matrix elements of $\hat H$.

Thus for instance (using $\hbar=1$), let's say we take $$ \hat H=\sigma_x+\sigma_z=\left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{array} \right) $$ and suppose $\vert\psi_0(0)\rangle=\left(\begin{array}{c}1\\0\end{array}\right)$. Then $$ U(\Delta t)= \left( \begin{array}{cc} \cos \left(\sqrt{2} \Delta t\right) -\frac{i \sin \left(\sqrt{2} \Delta t\right)}{\sqrt{2}} & -\frac{i \sin \left(\sqrt{2} \Delta t \right)}{\sqrt{2}} \\ -\frac{i \sin \left(\sqrt{2} \Delta t\right)}{\sqrt{2}} & \cos \left(\sqrt{2} \Delta t \right)+\frac{i \sin \left(\sqrt{2} \Delta t\right)}{\sqrt{2}} \\ \end{array} \right) $$ and $$ \langle \psi_0(0)\vert U(\Delta t) \vert\psi_0(0)\rangle =\cos \left(\sqrt{2} \Delta t \right)-\frac{i \sin \left(\sqrt{2} \Delta t\right)}{\sqrt{2}} $$ so that, expanding, one finds $$ \vert\langle\psi_0(0)\vert U(\Delta t)\vert\psi_0(0)\rangle\vert^2 \approx 1-(\Delta t)^2 $$ so that, in this case, "quickly" means $(\Delta t)^2<\epsilon$.

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  • $\begingroup$ as far as I can see the subscript for the wave function is redundant. I guess, for this answer there is no need for 0 (zero) under the wave function to be the hero. $\endgroup$ – physicopath Dec 4 '17 at 14:48
  • $\begingroup$ @physicopath You are right... it's a habit because for some slightly different calculation you need a complete set of states $\vert \psi_m\rangle$ $\endgroup$ – ZeroTheHero Dec 4 '17 at 15:30

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