Uncertainty in Frequency of a Photon

[Quelsita]

The question states:
consider a wave packet in the form of a pulse lasting 0.001sand length 3E5m. Since an individual photon of this wave can have an uncertainty of 3E5m (meanng it could be anywhere within the pulse), what is the uncertinty in momentum of the photon? in frequency?


OK, so using the equation $$\Delta$$p=h/$$\Delta$$x, I found the momentum to be 2.2E-39 kgm/s

I also have the equation $$\Delta$$v=$$\Delta$$E/h=1/$$\Delta$$t
but the answer is incorrect, how do you find the $$\Delta$$v for a single photon?

thanks!

 

[jbsteele]

The frequency uncertainty of the photon can be calculated using the equation \Delta\nu = 1/(2\pi\Delta t), where \Delta t is the pulse duration (in this case 0.001 s). This gives an uncertainty of 1.59E7 Hz.

 

[thomate1]

I would like to clarify that the uncertainty in momentum and frequency of a photon are related to the uncertainty in position and time, respectively. The equations you have used are correct, but the uncertainty in frequency can be calculated using the equation \Deltav=\DeltaE/h, where \DeltaE is the uncertainty in energy and h is the Planck's constant. In this case, the energy of a photon is given by E=hf, where f is the frequency. Therefore, the uncertainty in frequency can be calculated as \Deltav=\DeltaE/h=\Deltahf/h=\Deltaf, where \Deltah is the uncertainty in Planck's constant. This means that the uncertainty in frequency is equal to the uncertainty in the Planck's constant divided by the Planck's constant itself, which is a very small value. Therefore, the uncertainty in frequency for a single photon would also be very small.