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We were asked to calculate $\Delta x \Delta p$ for the $\psi_0,\psi_1$ of the harmonic oscillator.And so we calculated the answers and verified that $$\langle T \rangle +\langle V\rangle = (n+1/2)\hbar\omega\tag{1}$$ and that indeed they do follow the uncertainty limit.

But why can't we chose what to measure more precisely, position or momentum? Why is it that if we ever try to measure a particle resembling a harmonic oscillator at very low temperatures (at ground state) $\Delta x$ has to be $\sqrt{\hbar/2m\omega}$ and $\Delta p$ has to be $\sqrt{\hbar m \omega/2}$?.

Shouldn't it depend on the observer's wish, what he/she chooses to measure more precisely keeping in mind that the uncertainties follow HU?

Harmonic Oscillator is just an example.

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  • $\begingroup$ BTW, if you like the answer by @Vadim, please accept it. IMHO, it is worth accepting, but that's up to you. $\endgroup$ – garyp Jul 11 at 13:27
  • $\begingroup$ @garyp Remember that comments should just be used for requesting clarification from the OP or suggesting improvements to the post. $\endgroup$ – BioPhysicist Jul 11 at 14:15
  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Jul 11 at 23:20
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The $\Delta x,\Delta p$ in the Heisenberg uncertainty principle refer to the shape of the states, not to measurements. Although this shape tells you what the outcomes of the measurements can be and how likely each outcome is. Let's say you make an experiment where you repeatedly put a particle in the ground state of a harmonic oscillator and then you measure its position. The particle will be in the state $\psi_0(x)$ before measurement. What will the average position be? It will be $\langle \hat x \rangle$. This is a property of this state, not of a particular measurement.

What will the spread be of this state? Will the measurements deviate a lot from the average of will they be close. One way to measure this is by calculating $\sigma_x^2=\langle(\hat x-\mu_x)^2\rangle$ with $\mu_x=\langle\hat x\rangle$. It calculates, on average, the squared distance to the mean. Then $\Delta x$ is usually defined as $\Delta x=\sqrt{\sigma_x^2}$. Just like the average of position this is something you calculate for a state. Not for a measurement.

Experimentally you do have some control over $\Delta x,\Delta p$. For example squeezed states are states with minimal uncertainty that have lower uncertainty in either $x$ or $p$ than that of the ground state. But this is different from your example because in that case you are changing the state. You are trying to get different $\Delta x,\Delta p$ by changing the way you measure while you can only do this by changing the wavefunction.

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    $\begingroup$ Looks good. Thanks. Good answer :) $\endgroup$ – BioPhysicist Jul 11 at 14:22
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Quantum mechanical measurements are carried not on a single object, but an ensemble of objects prepared in the same state. Measurement on a single object produces a specific value, e.g., of the position - $x_i$. Measurements on the ensemble of $N$ objects produce $$\{x_i | i=1...N\},$$ which allow calculating sample mean and variance: $$\bar{x} = \frac{1}{N}\sum_{i=1}^Nx_i,\\ (\Delta x)^2 = \frac{1}{N-1}\sum_{i=1}^N(x_i - \bar{x})^2. $$ For very big $N$ these should ultimately converge to their values estimated from the probability distribution $w(x) = |\psi(x)|^2$: $$\langle x\rangle = \int dx xw(x),\\ \sigma_x^2 = \langle (x-\langle x\rangle)^2\rangle - \int dx (x-\langle x\rangle)^2w(x).$$

This is really about probability, statistics, and measurement theory, not necessarily in the context of quantum mechanics.

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