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I was always under the impression that the Heisenberg uncertainty principle tells us to what precision we can make a momentum measurement given a position one.

However, when doing quantum mechanics problems, or even solid state, I noticed quite often we were asked to use the Heisenberg uncertainty principle to estimate the momentum of an object given its size (or vice versa) and then quite often we would use that momentum to determine the energy of the object. Examples of a question that asks us to do this are as follows:

  1. Applying the uncertainty principle on the smallest scales, estimate the characteristic energy of the three relativistic quarks confined within the neutron. In the light of your answer comment on the mass of the neutron.
  2. Applying the uncertainty principle on the largest scale of the visible Universe ($\sim 8.8 \times 10^{26}$ m), find the lowest allowable energy of a photon in the Universe. Find the wavelength of such a photon and comment on this value.

Seeing as though the uncertainty principle tells us to what precision we can expect to make a momentum measurement given we have made a position one how is it that so often Physicists use it to instead determine the momentum of an object given its position.

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  • $\begingroup$ The uncertainty principle gives you an inequality which states that the product of the error in your measurement of of position and momentum should be greater than a particular value. Assuming the case when you had maximum precision, the above would turn into an equality. This gives you a way to roughly estimate the spread in momentum if the spread in position (length, diameter, size) is known. Is this useful or did I misinterpret your question ? $\endgroup$ – M111 Jan 17 at 23:31
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    $\begingroup$ @M111 calling $\sigma_x$ and $\sigma_p$ "error" is highly misleading. $\endgroup$ – DanielSank Jan 18 at 0:34
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Your impression of the Heisenberg uncertainty principle is correct, but the uncertainty principle also has a more fundamental interpretation in terms of constraints it puts on the wavefunction itself, independent of any measurements.

The Heisenberg uncertainty principle states that a quantum mechanical wavefunction $\psi(x)$ must satisfy the following relation:

$$\sigma_x \sigma_p \geq \frac{\hbar}{2}$$

where

$$\sigma_x = \sqrt{\int x^2 |\psi(x)|^2 dx - \left(\int x |\psi(x)|^2 dx\right)^2}$$

is the standard deviation of the position and

$$\sigma_p = \sqrt{-\hbar^2\int \psi^*(x)\frac{\mathrm{d}^2\psi(x)}{\mathrm{dx}^2} dx - \left(-i\hbar\int \psi^*(x)\frac{\mathrm{d}\psi(x)}{\mathrm{dx}} \mathrm{dx}\right)^2}$$

is the standard deviation of the momentum.

So, independent of any actual measurements, we can say that the wavefunction must obey this inequality.

Roughly what this means is that the product of the "extent" of the wavefunction in space times the "extent" of the wavefunction in momentum must be greater than some fundamental value.

Your interpretation does follow from this since if you assume you have made a measurement of the position with some error $\sigma_x$ then you know that shortly after the measurement the wavefunction will be in a state whose spatial "extent" is roughly $\sigma_x$ and therefore there will be a minimum spread in the momentum of the wavefunction of $\frac{1}{\sigma_x}\frac{\hbar}{2}$.

To use this information to answer these questions you have to rely on a few additional assumptions. Let's take a simple example. Suppose you know that the electron in a hydrogen atom orbits roughly at the Bohr radius of $5\cdot 10^{-11}$ m. Can we use this to estimate the kinetic energy of the electron?

Well given that the electron orbits "roughly" at a distance of the Bohr radius we can make a rough estimate that the spatial extent of the wavefunction is approximately equal to the Bohr radius too. Then, using the Heisenberg uncertainty principle we can say that the spread in the electron momentum must be greater than $\frac{1}{\sigma_x}\frac{\hbar}{2}$:

$$\sigma_{p_x} \geq \frac{1}{\sigma_x}\frac{\hbar}{2} $$

The spread in the electron's kinetic energy must be greater than

$$\sigma_E \geq \frac{1}{2m}\left(\sigma_{p_x}^2 + \sigma_{p_y}^2 + \sigma_{p_z}^2\right)$$

What we really want is the mean value of the kinetic energy, but all we have is an inequality on the spread of the kinetic energy. Here, the assumption is usually that the mean value of the energy is approximately equal to the standard deviation. Therefore we can estimate the ground state energy as:

$$\sigma_E \approx \frac{3}{2m}\left(\frac{1}{\sigma_x}\frac{\hbar}{2}\right)^2$$

In this case it gives an answer of 10.2 $\mathrm{eV}$ which is pretty close to the correct answer.

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  • $\begingroup$ I really like that you explained the assumptions required to use the uncertainty principle to estimate the energy. Thank you. I would say that the integrals near the beginning aren't really necessary and are kind of distracting, but that's probably a matter of taste. $\endgroup$ – DanielSank Jan 18 at 0:34
  • $\begingroup$ Thanks! Yeah I agree. I wanted to convey directly that these quantities can be calculated from the wavefunction itself and don't rely on measurements, but the term standard deviation usually seems to imply measurements. Do you know of any way I could get across that these quantities can be calculated directly from the wavefunction without measurement without writing out the definition? $\endgroup$ – user545424 Jan 18 at 0:37
  • $\begingroup$ I see your point. It might be worth saying what you just wrote in your comment explicitly in the post, although it's really already there... even in italics. $\endgroup$ – DanielSank Jan 18 at 0:57

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