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Think of a particle known to be trapped in a box of size $\Delta x$ and cooled down to near absolute zero. I know that attempting to measure the momentum of this particle repeatedly will give a random range of answers.

So if I had to write down a formula that gives an estimate of the spread in the possible results of the measurement of momentum of the particle, would it just be the uncertainty principle? $\Delta x \Delta p \geq \frac{\hbar}{2}$?

Then considering an electron, what size $\Delta x$ would result in this spread of results being so large that a positron-electron pair could be formed out of the kinetic energy of the original electron?

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    $\begingroup$ if the original electron had enough energy/momentum to create an electron positron pair it would not be close to absolute zero. $\endgroup$
    – anna v
    Commented May 9, 2014 at 11:18

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A positron-electron pair can be produced if the energy is greater than $3m_ec^2$. Using the uncertainty principle, we can estimate that $\Delta x \sim \hbar/6m_ec = 64~\text{fm}$. We could also ask how small must a 3D box be in order to make the ground state energy large enough to produce a pair of new particles. Assuming a cube, the length of a side is

$$L = \sqrt{\hbar^2\pi^2\over 2 m_e^2c^2} \approx 860~\text{fm}. $$

For comparison, the bohr radius is about $5\times 10^4~\text{fm}$, so this box is much smaller than a hydrogen atom.

I'm not sure how to calculate temperature in this situation, but if the energy is high enough that more states are available, i.e. we could have one high energy particle or three low energy particles, then the temperature cannot be zero.

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So if I had to write down a formula that gives an estimate of the spread in the possible results of the measurement of momentum of the particle, would it just be the uncertainty principle? ΔxΔp≥ℏ2?

Not in general, the spread would be affected by measurement errors too that are dependent on the measuring setup, temperature etc. The Heisenberg inequality states the least possible product of spreads theoretically that is consistent with the rules we use to derive average position and momentum from $\psi$ functions. It may be hard to attain this experimentally, usually the product is many times greater.

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  • $\begingroup$ But why is inequality and not an equality? $\endgroup$ Commented Feb 14, 2020 at 11:23
  • $\begingroup$ The spread (statistical deviation) does not have the same value for all psi functions, for some it is bigger, for some it is smaller. The smallest possible value is $\hbar/2$. $\endgroup$ Commented Feb 15, 2020 at 14:43

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