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Is it possible to have an infinite potential well of such width, that for ground state $\sigma_x\sigma_p=\frac{\hbar}{2}$?

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    $\begingroup$ Have you tried writing down a potential well with the width as a variable (say, $a$), and computing the ground state uncertainties for that, then solving your equation for $a$? If yes, what problems did you face? If no, why not? $\endgroup$ – ACuriousMind Apr 15 at 21:02
  • $\begingroup$ @ACuriousMind Yes I did. This is what I've done: $$<x> = \frac{a}{2} \\<x^2> = a^2[\frac{1}{3}-\frac{1}{2(n\pi)^2}]\\<p>=0\\<p^2>=(\frac{n\pi\hbar}{a})^2$$ Then: $$\sigma_x^2\sigma_p^2 = \frac{\pi^2}{12}-1$$ which worries me, becuase it's not dependent on a. Have I done something wrong? $\endgroup$ – NotStudent Apr 15 at 21:12
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    $\begingroup$ You should edit that into your question! However, you should also explain why this worries you - why do you expect $\sigma_x \sigma_p$ to depend on $a$ (also, from your first four expressions, it's not clear how $h$ vanished from the final expression)? $\endgroup$ – ACuriousMind Apr 15 at 21:18
  • $\begingroup$ I am confused. Do you mean $\sigma_p$ instead of $\sigma_y$? $\endgroup$ – Aaron Stevens Apr 15 at 22:21

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