## Uncertainty Principle with time and frequency

ΔxΔp ≥ $\frac{h}{4\pi}$

Since Δx=ct for a photon and Δp=(mv$_{f}$-mv$_{i}$)

Then ct(mv$_{f}$-mv$_{i}$) ≥ $\frac{h}{4\pi}$

Since mv=$\frac{h}{\lambda}$

You have ct($\Delta$$\lambda$)$^{-1}$h ≥ $\frac{h}{4\pi}$

Planck's constant cancels, move the c over $\lambda$, $\frac{c}{\lambda}$=f

This leaves you with t$\Delta$f ≥ $\frac{1}{4\pi}$

Dimensional analysis checks. Is this correct and is there any use to this equation?

t ≥ (4$\pi$Δf)$^{-1}$

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 Recognitions: Homework Help It's usually put in the form: $\Delta E \Delta t = \frac{\hbar}{2}$ (note that E = hf for a photon - giving your relation.) It is used for the lifetime of excited states in solids and the range of fundamental interactions. I'd like to point out that you don't get to do p=mv for a photon though.
 Recognitions: Gold Member Science Advisor Also, the "time-frequency uncertainty relation" is just a natural consequence of Fourier analysis, it is a mathematical results which does not rely on anything from physics. Is is sometimes known as "the mathematical uncertainty principle" See e.g. http://www.ams.org/samplings/feature...rc-uncertainty

## Uncertainty Principle with time and frequency

 Quote by Lord_Sidious ΔxΔp ≥ $\frac{h}{4\pi}$ Since Δx=ct for a photon
What is meant by t in this equation? The meaning of Δx is the uncertainty in the position of the particle. But, photons do not have a defined position.

 Quote by Lord_Sidious ... and Δp=(mv$_{f}$-mv$_{i}$) Then ct(mv$_{f}$-mv$_{i}$) ≥ $\frac{h}{4\pi}$ Since mv=$\frac{h}{\lambda}$
Again, Δp is the uncertainty in momentum, not the change of momentum equal to final - initial momentum. Furthermore, the momentum of the photon is not calculated by the non-relativistic formula $p = m v$. However, your last formula (in this quotation) is correct, if you ignore the intermediate result $m v$, that you never use after that.

 Quote by Lord_Sidious You have ct($\Delta$$\lambda$)$^{-1}$h ≥ $\frac{h}{4\pi}$
You have made a mistake here. If momentum is calculated by the De Broglie relation $p = h/\lambda$, then, an uncertainty in wavelength Δλ implies, by the error propagation formula:
$$\Delta p = \left\vert \frac{d p}{d \lambda} \right\vert \, \Delta \lambda = h \, \frac{\Delta \lambda}{\lambda^2}$$

 Quote by Lord_Sidious Planck's constant cancels, move the c over $\lambda$, $\frac{c}{\lambda}$=f
Planck's constant does cancel. However, because your previous formula was incorrect, so is this one. The corrected version is obtained by going from wavelength to frequency via the error propagation formula:
$$c \, t \, \frac{\Delta \lambda}{\lambda^2} \ge \frac{1}{4 \pi}$$
$$\lambda = \frac{c}{f}, \ \Delta \lambda = \left \vert \frac{d \lambda}{d f} \right\vert \, \Delta f = \frac{c \, \Delta f}{f^2}$$
$$c \, t \, \frac{\frac{c \, \Delta f}{f^2}}{\frac{c^2}{f^2}} \ge \frac{1}{4 \pi}$$
$$t \, \Delta f \ge \frac{1}{4 \pi}$$

 Quote by Lord_Sidious This leaves you with t$\Delta$f ≥ $\frac{1}{4\pi}$ Dimensional analysis checks. Is this correct and is there any use to this equation? t ≥ (4$\pi$Δf)$^{-1}$
but the method by which you derived it is incorrect.

 Quote by Dickfore Planck's constant does cancel. However, because your previous formula was incorrect, so is this one. The corrected version is obtained by going from wavelength to frequency via the error propagation formula: $$c \, t \, \frac{\Delta \lambda}{\lambda^2} \ge \frac{1}{4 \pi}$$ $$\lambda = \frac{c}{f}, \ \Delta \lambda = \left \vert \frac{d \lambda}{d f} \right\vert \, \Delta f = \frac{c \, \Delta f}{f^2}$$ $$c \, t \, \frac{\frac{c \, \Delta f}{f^2}}{\frac{c^2}{f^2}} \ge \frac{1}{4 \pi}$$ $$t \, \Delta f \ge \frac{1}{4 \pi}$$ This is the same as your final answer: but the method by which you derived it is incorrect.
Thanks for clarifying it. What exactly can be done with this now? The "t" is time, but I don't understand the time of what? The time of the change in frequency? I have done some calculations and for a change in frequency from say, UV to violet light, the equation gives:

t ≥ 8.6 attoseconds

But if you reverse it, then the change in frequency from violet light to UV, the equation gives:

t ≥ -8.6 attoseconds

 Again, $\Delta f$ is not the change in frequency. It is the uncertainty with which you know the frequency. Then, t is the minimum time you need to measure the wave-train to achieve the given precision.
 Recognitions: Homework Help The "delta" is not a "change in" - it is the uncertainty in some measurement - a statement of how imprecise something is known. So $\Delta f$ would be the uncertainty in frequency. You calculation shows that if you were making measurements of frequency which lead you to be uncertain to the extent that it could be violet or UV or something in between, then the smallest you could be uncertain about a related time measurement must be 8.6 attoseconds. Now consider what this means in terms of the wave-nature of a particle like the virtual particle that mediates forces? Or energy-width of an electron orbital and it's stability?
 Thanks everyone, this helps.
 The uncertainty relationship between time and frequency has interesting musical applications. The lower the frequency of a pitch, the longer you must hear it in order to ascertain its frequency. That's why a piccolo can play short notes but a tuba can't. A piccolo requires only a tiny fraction of a second to complete several cycles and clearly define its frequency. Notes at the bottom of a piano are only around 30 Hz. If played for a duration of, say, 1/40 of a second, you don't even get a chance to hear a full wavelength, leading to uncertainty about the pitch. Likewise in musical sampling. When digitizing a waveform, it must be sampled at least twice per cycle in order to discern a particular frequency. For frequencies that are high in the range of human hearing, like 4,000 Hz, the waveform must be recorded at least at a rate of 8,000 Hz. Any lower sampling rate will cut off the high frequencies and result in a damped sound. These examples are, to me, some of the clearest ways to illustrate what the uncertainty principle is "really saying." Heisenberg used a very similar example to demonstrate why his location / momentum principle was sensible, though in his case it involved EM waves, not sound. Other aspects of the HUP baffle me! I'll be posting my own question soon, haha.