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I had the following question in my Quantum Mechanics test.

A particle of mass m is subjected to a potential V(x) = β|x|, where β is a positive constant. Using the uncertainty relations (principle) estimate the ground state (lowest) energy the particle can have.

I had no clue how to approach this question because I didn't know how to find the expected kinetic energy. However, a few people approached this problem by computing the energy using the uncertainty values in position and momentum, and then solving for Δx and Δp by differentiating the energy expression and setting it to 0. These values were substituted back into the energy expression to obtain the least energy. This doesn't make much sense to me because we're supposed to use the expected values and not the uncertainty values.

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Just some observations here:

By symmetry $\langle x\rangle=0$ for your problem, and so does $\langle p\rangle=0$. Thus, for instance, \begin{align} (\Delta p)^2=\langle p^2\rangle -\langle p\rangle^2=\langle p^2\rangle \propto \langle T\rangle \end{align} in this case.

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  • $\begingroup$ Just to be clear, the symmetry in x here arises due to the symmetry in the potential function, right? and how does symmetry in x imply symmetry in p? $\endgroup$
    – Sathvik Swaminathan
    Commented Sep 17, 2020 at 12:58
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    $\begingroup$ yes the parity symmetry gives $\langle x\rangle=0$ and the nature of the stationary state gives $\langle p\rangle =0$. $\endgroup$
    – ZeroTheHero
    Commented Sep 17, 2020 at 13:15
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The question however says explicitly to use the uncertainty relation. The approach taken by those who used the uncertainty relation between momentum and position is correct. The missing link here is the relation between the uncertainty and the expectation value, which should become clear, if one looks up the derivation of the uncertainty relations - they are just expectations of the wave function width in the corresponding space.

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  • $\begingroup$ The uncertainty is the square root of the difference of the expected value of the square of observable and the square of expected value of the observable, so it's basically the standard deviation. How is the standard deviation equivalent to the expected value? $\endgroup$
    – Sathvik Swaminathan
    Commented Sep 17, 2020 at 12:33
  • $\begingroup$ Uncertainty is the (square root of the) expectation value of $(x-\langle x\rangle)^2$, as you have just pointed out. You seem to imply that it should be an expectation value of something else? $\endgroup$
    – Roger V.
    Commented Sep 17, 2020 at 12:37
  • $\begingroup$ No, I don't understand why one would substitute the uncertainty value instead of the expected value in spite of the question explicitly stating to use the uncertainty principle. $\endgroup$
    – Sathvik Swaminathan
    Commented Sep 17, 2020 at 12:39
  • $\begingroup$ There is not such thing as expected value, there is expectation value of something - which expectation value are you talking about? $\endgroup$
    – Roger V.
    Commented Sep 17, 2020 at 12:45
  • $\begingroup$ I'm talking about the expectation value of position and momentum using which I can compute the value of energy. I don't understand how I can use the values of uncertainty in position and momentum instead of their expected values to compute the value of energy $\endgroup$
    – Sathvik Swaminathan
    Commented Sep 17, 2020 at 12:47

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