# I Uncertainty between position and angular frequency - is this correct?

#### Negatratoron

Summary
Derivation of: $\Delta x \Delta \omega \ge \frac{c}{2}$
Motivation: In my thermodynamics + statistical physics class, we derived the equipartition theorem for ideal gasses using Boltzmann factors, dividing the phase space of a gas particle in position+momentum space into units of size x*p=h based on the quantum nature of the space of states that are in principle distinguishable from one another. In so doing, we divided the integral for the partition function by $h^3$ (the integral ended up being equal to $\int e^{-\beta \epsilon} / h^3 dx dy dz dp_x dp_y dp_z$ integrating over all positions and momenta with the density of states being equal to $dxdydzdp_xdp_ydp_z / h^3$). This began to raise a question in my mind. Position and momentum are conjugate variables in quantum mechanics, and obey an uncertainty relation in which their units multiply to angular momentum, i.e. the units of $h$. Similarly, time and energy - and angular position and angular momentum - obey uncertainty relations in which their units again multiply to the units of $h$. I wonder if there is something fundamental to that, or if there could be a relation that has different units, and so I went looking for an uncertainty relation with different units. Here is what I found:

Take Heisenberg's uncertainty principle for the position and momentum of a photon:

\begin{align} \Delta x \Delta p \ge \frac{\hbar}{2} = \frac{h}{4 \pi}\end{align}

Here the uncertainty is standard deviation, so $$\Delta x$$ is the standard deviation of the position of the photon.

Now take the relation

$$E = \frac{hc}{\lambda}$$

Doing a bit of algebra

$$\frac{E \lambda}{h^2} = \frac{c}{h}$$

Applying this to the original equation

\begin{align} \frac{\Delta x \Delta p E \lambda}{h^2} \ge \frac{c}{4 \pi}\end{align}

Photons obey the relation $E = p c$. Standard deviation is linear across coefficients, i.e. the standard deviation of the multiple $a$ of a random variable is equal to the multiple $a$ of the standard deviation of the random variable, i.e. $\Delta (a x) = a \Delta x$, so the photon obeys $\Delta E = \Delta P c$. (I am using $\Delta x$ instead of $\sigma_x$ for the sake of readability). From these two equations, we have $E \Delta p c = \Delta E p c$ therefore $\Delta E p = E \Delta p$. Applying this to (2)

\begin{align} \frac{\Delta x \Delta E p \lambda}{h^2} \ge \frac{c}{4 \pi}\end{align}

Photons obey $p \lambda = h$ so we can simplify to

\begin{align} \frac{\Delta x \Delta E}{h} \ge \frac{c}{4 \pi}\end{align}

Photons also obey $E = h v$ so taking standard deviations, $\Delta E = h \Delta v$. Applying this to the above

\begin{align} \Delta x \Delta v \ge \frac{c}{4 \pi}\end{align}

We can now use $\omega = 2 \pi v$ to convert from frequency to angular frequency

\begin{align} \Delta x \Delta \omega \ge \frac{c}{2}\end{align}

This appears to be an uncertainty relation comparing uncertainty in position and uncertainty in angular frequency with the speed of light, one in which $h$ does not appear at all!

On a philosophical tangent, it is my current state of mind that universal constants should not be thought of as quantities with units, but as assertions that certain units are identical in some situations, giving the conversion factor as data. For instance, the universal constant of the speed of light, if used correctly in dimensional analysis, converts meters to seconds or seconds to meters. So in very abstract terms you can think of the speed of light as an assertion that meters are equal to seconds, with the conversion factor $3 \cdot 10^8$ as evidence, and similar for the other universal constants.

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#### vanhees71

Gold Member
This is a bit dangerous since you seem to use position operators for a photon commuting in the naive way with its momenta as the position operators for massive particles. This is not easily justified since there is no such position operator for a photon (in fact for any massless particle with spin $\geq 1$).

• Negatratoron

#### Negatratoron

Thank you for telling me so. I finally take quantum next semester, so I'll be able to crawl out of the quagmire of self-taught sorcery and onto the dry land of firm, practical foundations.

• vanhees71 and PeroK