# How can the uncertainty principle be used to deduce range of a force from properties of the force carrier?

Question:

How can the uncertainty principle be used to deduce range of a force from properties of the force carrier?

My thoughts?

Not too sure how this can be done and would like some advice. My understanding of the uncertainty principle is that a particles position and velocity cannot be precisely determined at a given time. I'm not sure how this relates though.

Any help?

• Are you talking about quantum field theory? – Jimmy360 May 3 '15 at 15:26
• Yes. But I'm quite new to the topic so please keep it fairly elementary. – tey yreryt May 3 '15 at 15:28

$$\Delta E \Delta t \sim \hbar$$
This is a version of the Heisenberg Uncertainty Principle. Instead of using momentum and position,however,the above form uses energy and time ($\Delta$ means change in).
The uncertainty relation, $\Delta E \Delta t \sim \hbar$, tells us that if we only observe a system for a time interval $\Delta$$t, there is no way that we can know the energy of the system better than to within an uncertainty \Delta E. So the photon can have energy \Delta E for a time interval \Delta t \sim \hbar /\Delta E, without anybody being able to know if energy conservation is violated. As long as the photon is reabsorbed quickly enough, there is no measurable violation of energy conservation. Since the photon must be reabsorbed and cannot be detected, it is called a virtual photon. A QFT calculation of the effect of this virtual photon exchange gives rise to the usual Coulomb force. Like charges repel and unlike charges attract. virtual photon can exist for a time Δt ~ ћ/ΔE. Since for a photon the wavelength can be arbitrarily large and therefore the energy arbitrarily small, we can have ΔE --> 0, Δt --> infinity. The virtual photon can have an infinite range. We can do this same thing, with slighty different methods, to all force carriers. Good reading on the topic In quantum field theory we describe the interaction between two particles, A and B, as being due to the exchange of a gauge boson - call this X. So A emits a gauge boson X of mass m_X that travels over to particle B and is absorbed. If the lifetime of the gauge boson is t_X then the range will be of order ct_X:$$ d \approx ct_X \tag{1} $$The lifetime t_X can be obtained from the energy-time uncertainty principle:$$ \Delta t \Delta E \approx \frac{\hbar}{2} \tag{2} $$The argument is that in this case \Delta t is the lifetime of the gauge boson and \Delta E is the rest mass of the gauge boson, that is:$$ \Delta E = m_X c^2 $$If we substitute this into equation (2) and rearrange we get:$$ t_X \approx \frac{\hbar}{2m_Xc^2} $$And substituting this into equation (1) gives the range of the force d as:$$ c \approx \frac{\hbar}{2m_Xc} $$So the range is inversely proportional to the mass of the X boson. But note that this is an exceedingly rough and ready calculation and at best can give only an order of magnitude approximation. The particle exchange actually produces a Yukawa potential:$$ V(r) = -g^2 \frac{e^{-m_Xr}}{r}$$where$g\$ is a constant called the coupling constant. Actually the coupling constant isn't constant, but depends on energy. This is why the strength of the fundamental forces changes with energy.