# Does Heisenberg Indeterminism have a lower bond in $\Delta \chi$?

Take the Heisenberg's indeterminism law:

$$\Delta \chi \cdot \Delta \rho \geq h/ 2$$

Does the momentum pose a limit so that we cannot measure the position with a precision greater than:

$$h / (2\cdot \Delta \rho)$$

where $\Delta \rho$ is at maximum $mc$?

• $m$ is mass of particle
• $c$ is speed of light
• $\rho$ is the momentum
• $\chi$ is the position
• mass * speed of light – CoffeDeveloper Nov 3 '16 at 16:26
• Someone edited the question in a wrong way. m is the mass of the particle, and we assume we can somehow make it fast (almost) as light, then we have a lower bond for position precision measurement? – CoffeDeveloper Nov 3 '16 at 16:41
• Have youo learned about the Compton wavelength? – Cosmas Zachos Nov 3 '16 at 16:43
• Just learned now: it is the same quantity (apart the 2 constant), but I think it has not the same meaning of what my question ask – CoffeDeveloper Nov 3 '16 at 17:23
• Minor comment to the notation (v8): Why $\chi$ and $\rho$ rather than $x$ and $p$? – Qmechanic Nov 5 '16 at 15:24

Your question is a very good candidate for considerations in the formalism of relativistic quantum mechanics. What you are pointing to is the uncertainty principle in relativistic quantum mechanics.

Consider the uncertainty relation:

$$\Delta x\Delta p\sim \hbar$$

If $v'$ and $v$ represent the velocities before and after measurement made in a time interval $\Delta t$, we can write

$$\left(v'-v\right)\Delta t\Delta p\sim \hbar$$

If the measurement is made such that both $\Delta t$ and $\Delta p$ both small simultaneously, then $\left(v'-v\right)$ will be very much large. Yes, your measurement of both time interval and momentum simultaneously with accuracy lead to a very large change in the velocity of the particle, a direct consequence of such a measurement. In fact, if the momentum eigen states are not energy eigen states, you can measure both time and momentum simultaneously with accuracy. This will give an infinite increase in the velocity of the particle.

However in the light of special relativity, there exists a speed limit $c$. This restriction on the speed limit of the particle will give you the limit in the accuracy of your measurement.

$$\Delta p\Delta t\sim \frac{\hbar}{c}$$

This is the highest accuracy theoretically attainable when the momentum is measured by a process occupying a given time $\Delta t$. Hence, in the relativistic theory, it is , in principle, impossible to make an arbitrary accurate and rapid measurement of the momentum. An exact measurement of momentum ($\Delta p\longrightarrow 0$) happen only in the limit the duration of measurement $\Delta t\longrightarrow\infty$.

In the rest frame of the particle, the least possible error in the measurement of its coordinates is

$$\Delta q\sim \frac{\hbar}{mc}$$

This value implies a momentum uncertainty of $mc$, while in a frame of reference in which the particle is moving with an energy $\varepsilon$, the momentum uncertainty becomes $\varepsilon/c$. In the ultra-relativistic limit

$$\Delta q\sim \frac{\hbar}{p}$$

i.e, the error in the measurement of coordinates is the same as the de Broglie wavelength of the particle. In the case of photons, it is always valid. So, it means that talking about the "coordinates of a photon" is meaningful only when the characteristic dimensions of the problem are large in comparison with its wavelength. This is the classical limit of geometrical optics, after which light tends to behave like wavess (diffraction). However in the quantum scale, coordinates of a photon has no meaning at all as the wavelength cannot be regarded as small in the quantum scale.

Not directly. One lower bound on $\Delta x$ is imposed by the fact that as $\Delta x$ goes down, the kinetic energy grows like $(\Delta p)^2$. So it takes a lot of energy to localize a particle. In principle, if you had infinite energy to spare (and higher order theories didn't become relevant), you'd could have $\Delta x \rightarrow 0$.

A lower bound on $\Delta p$ comes from the limit of our ability to setup an arbitrarily large undisturbed space for a long enough time for the particle to spread out and not decohere/be knocked out of a pure momentum state by interactions.

As for questions about $mc$, they aren't a problem because the marriage of special relativity and quantum mechanics was handled by quantum field theory (QFT). The only particularly special thing that happens at $p=mc$ is it is the changeover point from the low momentum approximation of energy: $$E \approx mc^2 + \frac{p^2}{2m} + \ldots$$ to the high momentum approximation: $$E \approx pc + \frac{p^3c}{2m^2c} + \ldots$$ of the exact formula (dispersion relation): $$E = \sqrt{[mc^2]^2 + [pc]^2}.$$

This is good question, because it combines the notions of quantum mechanics and special relativity to arrive at a particular conclusion. The thing is that basic quantum mechanics is non-relativistic. The best way to include special relativity into quantum mechanics is to use quantum field theory.

In the context of quantum field theory, one can now consider the question. The basic issue is whether one can limit the uncertainty in the momentum by the maximum momentum allowed according to special relativity. If the particle is on-shell (obeys the dipersion relation) then the answer would be yes. However, in quantum field theory one can also have virtual partiles, which play an important role in interaction. There particles don't need to be on-shell and therefore their uncertainty can exceed the maximum allowed by special relativity.

So in general one would then say that: no, Heisenberg's uncertainty does not set a lower bound.