- #1

peguerosdc

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- TL;DR Summary
- Why we can approximate the uncertainty (std. dev) as 1) the difference between two measurements, 2) the value of one measurement?

Hi!

I am checking Zettili's explanation on the uncertainty principle and I have this confusion on what the "uncertainty" really means which arises from the following statements:

When introducing the uncertainty principle, for the case of position and momentum it states that:

$$

\Delta x \Delta p \geq \hbar / 2

$$

Similarly for the energy and time:

$$

\Delta E \Delta t \geq \hbar / 2

$$

But the two given examples doesn't seem to fit with that definition.

The energy example says that:

Now, this doesn't makes sense to me when the more formal statement of the uncertainty principle is given in terms of the standard deviation ##\sigma##:

$$

\sigma_A \sigma_B \geq \frac {|\langle[A,B]\rangle|} 2

$$

Then, the next example calculates the uncertainty of the position of a 50kg person moving at 2m/s:

$$

\Delta x \geq \frac \hbar {2\Delta p} \approx \frac \hbar {2mv} = \frac \hbar {2 \times 50kg \times 2 ms^-1}

$$

This doesn't feel consistent neither with the definition in terms of the standard deviation nor with the first example.

Thanks!

I am checking Zettili's explanation on the uncertainty principle and I have this confusion on what the "uncertainty" really means which arises from the following statements:

When introducing the uncertainty principle, for the case of position and momentum it states that:

*if the x-component of the momentum of a particle is measured with an uncertainty ##\Delta p##, then its x-position cannot, at the same time, be measured more accurately than ##\Delta x = \hbar / (2\Delta p)##*:$$

\Delta x \Delta p \geq \hbar / 2

$$

Similarly for the energy and time:

$$

\Delta E \Delta t \geq \hbar / 2

$$

But the two given examples doesn't seem to fit with that definition.

The energy example says that:

*if we make two measurements of the energy of a system and if these measurements are separated by a time interval ##\Delta t##, the measured energies will differ by an amount ##\Delta E## which can in no way be smaller than ##\hbar / \Delta t##*.Now, this doesn't makes sense to me when the more formal statement of the uncertainty principle is given in terms of the standard deviation ##\sigma##:

$$

\sigma_A \sigma_B \geq \frac {|\langle[A,B]\rangle|} 2

$$

How is the difference between two measurements equivalent to the standard deviation?How is the difference between two measurements equivalent to the standard deviation?

Then, the next example calculates the uncertainty of the position of a 50kg person moving at 2m/s:

$$

\Delta x \geq \frac \hbar {2\Delta p} \approx \frac \hbar {2mv} = \frac \hbar {2 \times 50kg \times 2 ms^-1}

$$

This doesn't feel consistent neither with the definition in terms of the standard deviation nor with the first example.

- In this case we only have one measurement for the momentum, so when comparing with the previous example,
**Why is the "uncertainty" of p approximated just to the value of p (instead of taking the difference between two measurements)?** - When comparing with the definition of the uncertainty principle,
**Why we are now approximating the standard deviation of p to the value of p?**

Thanks!