Sunday, February 20, 2011

Mastering Physics: The Uncertainty Principle: Virtual Particles

The uncertainty principle can be expressed as a relation between the uncertainty \Delta E in the energy state of a system and the time interval \Delta t during which the system remains in that state. In symbols,
\Delta E \Delta t \ge \hbar,
where \hbar = \frac{h}{2\pi}, h is Planck's constant.
The energy-time uncertainty principle says that the longer a system remains in the same energy state, the higher the accuracy (or the smaller the uncertainty) a measurement of that energy can be. Another implication is that physical processes can violate the law of energy conservation as long as the violation occurs for only a short time, determined by the uncertainty principle. This idea is at the base of the theory of virtual particles.

Part A
Consider two electrons that interact with each other. Classically, their interaction would be described in terms of the electrostatic force. In quantum mechanics, their interaction is interpreted in terms of emission and absorption of photons: One of the two electrons emits a photon with energy DeltaE, which is then absorbed by the other electron after a short period of time.
How long can the photon survive before it is absorbed without violating the uncertainty principle?
Express your answer in terms of DeltaE, and h or hbar.

  Deltat = \frac{{\hbar}}{{\Delta}E}
 The total energy of the two electrons after the absorption of the photon is the same as that before the emission of that photon, so the energy of the system is conserved at every time except during the intermediate state during which the photon is created. Such a short-lived photon is called a virtual photon, and it represents the mediator of the electromagnetic force. By exchanging a virtual photon, two charged particles interact with each other, and the observable effect of their interaction is either repulsion, when the particles have the same charge, or attraction, when the particles have opposing charges.

Part B
The same concepts used to describe the emission and absorption of virtual photons in electromagnetic interactions can be used to describe nuclear interactions. According to this theory, when two protons of a nucleus interact with each other, a virtual particle is created. In order to effectively mediate the nuclear force, the virtual particle must exist long enough to travel a distance comparable to the size of the nucleus: a distance on the order of 1.5 \times 10^{-15}\;\rm m.
Assuming that the virtual particle moves at a speed close to c, what is its minimum mass?
To simplify the calculation, assume that the kinetic energy of the particle is negligible compared to its rest energy. This assumption is not really valid for particles moving at close to the speed of light, but it will lead to an answer of a reasonable order of magnitude.

Express your answer numerically in kilograms.

  Deltam = 2.34×10−28
 \rm kg
If we convert the answer into units more widely used for measuring the masses of fundamental particles, the result above is about 131 {\rm MeV}/c^2.
The theory that postulates a virtual particle as the mediator of the nuclear force was put forward by the Japanese physicist H. Yukawa in 1935. He called this particle a meson, but no experimental evidence of its existence was found until 1947, when a series of three particles, called pions, were discovered. Pions have masses (about 140 {\rm MeV}/c^2) comparable to that predicted by Yukawa using a more sophisticated theory than what we have used in this problem.


  1. You forgot to divide Hbar by 2, so A and B are wrong.

    A is (Hbar/2)/deltaE and thus
    B is 1.17*10^-28 kg

  2. why would you divide hbar by 2?