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In Heisenberg Picture, for a free particle, $[x_i(t),x_i(0)]=\frac{-i\hbar t}{m}$.

This relation implies that even if the particle is well localized at t=0, its position becomes more and more uncertain with time. (Sakurai)

Applying the same approach to Schrodinger picture,

Now, the position operators do not depend upon time ($x_i(t)=x_i(0)$, for any time t) and $[x_i(t),x_i(0)]=0$. This means if the particle is well localized at t=0, its position remains localized with time.

But this is the violation of Heisenberg's Principle as we're providing some amount of certainty to the location of the particle. Even if we measure the location at t=0, leaving the particle free for some time changes its location because the state kets change with time in Schrodinger's Picture and hence the location changes.

Where am I wrong?

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    $\begingroup$ Observables are independent of the picture. They can be always expressed in terms of $\langle\psi_1|\hat{O}|\psi_2\rangle$, where $|\psi_1\rangle$ and $|\psi_2\rangle$ may or may not coincide. Where are you wrong? You only considered operators, not their expectation values. Once you sandwich the operators between states, you'll see that the difference cancels out. $\endgroup$ – user178876 Oct 10 '18 at 6:01
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    $\begingroup$ @marmot this should have been an answer $\endgroup$ – Aaron Stevens Oct 10 '18 at 10:01
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    $\begingroup$ The mistake you are making is that you are trying to use the uncertainty principle intuitively without thinking what exactly it says. The uncertainty relation is a precise mathematical inequality that constrains the variances of two different operators (observables) in the same state. Now in the Heisenberg picture you can indeed take $x(0)$ and $x(t)$ for the two operators. In the Schroedinger picture, you however want to relate the variances of the same operator (coordinate) in two different states. You simply cannot apply the uncertainty relation to such a situation. $\endgroup$ – Tomáš Brauner Aug 6 at 15:33
  • $\begingroup$ @TomášBrauner thanks. I understood it wrong. $\endgroup$ – Asit Srivastava Aug 7 at 11:31
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Observables are independent of the picture. They can be always expressed in terms of $\langle\psi_1|\hat{O}|\psi_2\rangle$, where $|\psi_2\rangle$ and $|\psi_2\rangle$ may or may not coincide. Where are you wrong? You only considered operators, not their expectation values. Once you sandwich the operators between states, you'll see that the difference cancels out.

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One has to be precise about what "well-localized" means. Naively, one would maybe want to ask that, if $\psi_{0}$ is the initial wave-function, supposed to be localized at $x_0$, and we denote the position operator with a capital $X$

$ X \psi_{0} = x_0 \psi_{0}$

An equation which is "solved by Dirac's $\delta$-function $\delta(x-x_0)$". You may have heard it, but this has the problem that the Dirac's $\delta$-function is not in the Hilbert space $\mathcal{H}$ of the theory. In fact, the position operator has no eigenvectors! It has, however, approximate eigenvectors for $x_0$, that is, a sequence of vectors $\{\delta_{n}\}_{n\in \mathbb{N}} \subset \mathcal{H}$ where all elements have unit norm, $\|\delta_{ n}\| = 1$ and

$ \| X \delta_{n} - x_0 \delta_{ n} \| \stackrel{n\rightarrow \infty}{\longrightarrow} 0 $ .

One explicit sequence is

$\delta_n(x) = \begin{cases} \sqrt{n} && |x-x_0| < \frac{1}{2n} \\ 0 && \text{else} \end{cases} \ .$

Note that

$ \int_\mathbb{R} x |\delta_n(x)|^2 d x = x_0 \ \ \ \ \text{ and } \ \ \int_\mathbb{R} (x-x_0)^2 |\delta_n(x)|^2 d x = \frac{1}{12 n^2} \ .$

Now we can ask for a particle to be initially well-localized. We should say, that a particle described by a wave-function $\psi_0$ is well localized in an interval $I_\epsilon = (x_0 - \epsilon/2,x_0+\epsilon/2)$ if initially,

$\langle \psi_0,X\psi_0 \rangle = x_0 $

and

$ \langle \psi_0,(X-x_0)^2\psi_0 \rangle < \frac{\epsilon}{2} $

Then you may compute the time evolution and see that it will spread.

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