# Uncertainty in the energy of a laser beam

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## Homework Statement

The duration of a laser pulse is 10^(-8) s. The uncertainty in its energy will be?

## Homework Equations

## \delta x \delta p=\frac{h}{2\pi}##

## The Attempt at a Solution

speed of laser beam=c, duration is given so distance can be calculated exactly-shouldn't that mean delta x=0, therefore delta p=infinite and so uncertainty in energy also infinite.
I'd appreciate some help, thank you

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The Heisenberg Uncertainty Principle comes in several forms. One form is ## \Delta E \, \Delta t \gtrsim \frac{\hbar}{2} ##.

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The Heisenberg Uncertainty Principle comes in several forms. One form is ## \Delta E \, \Delta t > \frac{\hbar}{2} ##.
is the derivation of this expression from the standard one difficult?
We have a large syllabus- I want to memorize the necessities and derive the rest

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With these, I think you need to know of the existence of a couple different forms, and even memorize them if possible. I'll try to find a good "link".

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With these, I think you need to know of the existence of a couple different forms, and even memorize them if possible. I'll try to find a good "link".
Thank you very much, that'd be immensely helpful :D

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The problem is not solved though, delta t still equals zero as the exact duration is given

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which means delta E should be infinite, which is the incorrect answer as per my book

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I should also mention that since ## \Delta p \, \Delta x \gtrsim \frac{\hbar}{2} ##, and ## \Delta E \, \Delta t \gtrsim \frac{\hbar}{2} ##, that also implies, since ## p=\hbar k ## and ## E=\hbar \omega ##, that ## \Delta k \, \Delta x \gtrsim \frac{1}{2} ## and ## \Delta \omega \, \Delta t \gtrsim \frac{1}{2} ##. Sometimes these latest expressions are written with a 1.

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The problem is not solved though, delta t still equals zero as the exact duration is given
You have a valid point. A finite time duration, as opposed to infinite time, means there is a spread in frequencies, and an uncertainty in the photon energy is the result. Perhaps that is a better interpretation of what they are looking for.

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I don't think I'll need that till I get into college, since the uncertainty principle itself isn't supposed to be mandatory syllabus (I'm just studying extra concepts). My main concern is that ## \Delta E## is coming out to be infinite instead of a proper answer required by my book

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You have a valid point. A finite time duration, as opposed to infinite time, means there is a spread in frequencies, and an uncertainty in the photon energy is the result. Perhaps that is a better interpretation off what they are looking for.
The answer is given as 6.6 x 10^(-26) J

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They clearly used ## \Delta E \, \Delta t \gtrsim h ##, which I think is rather inaccurate. Usually, a ## \hbar ## is used, and oftentimes, it is divided by 2.

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They clearly used ## \Delta E \, \Delta t \gtrsim h ##, which I think is rather inaccurate.
Not just that, the wording of the question itself is misleading. Note that they wrote "duration of laser beam" instead of "uncertainty in duration of laser beam".
It might be a mistake but I just want to make sure I'm not missing on any important concept here if by a small chance the question and its answer is accurate

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Not just that, the wording of the question itself is misleading. Note that they wrote "duration of laser beam" instead of "uncertainty in duration of laser beam".
It might be a mistake but I just want to make sure I'm not missing on any important concept here if by a small chance the question and its answer is accurate
See my post 11. I think you have a pretty good understanding of it. I would have to say, yes, here the book did not present the problem in a complete and accurate manner, and even their answer looks to be inaccurate, regardless of the precise question.

Krushnaraj Pandya
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Thank you very much, it is often the misprints and inaccurate questions in my book that end up giving me the most knowledge. I consider it solved on my part :D

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Just an additional comment: These uncertainty relationships come about as the result of the wave nature of what is being measured. Meanwhile, when they say the pulse is of duration ## T=1.0 \cdot 10^{-8} ## seconds, that does not mean that ## \Delta T=0 ##. A typical ## \Delta T ## on this pulse width measurement might be ## \Delta T=\pm 1.0 \cdot 10^{-9} ## seconds. Again though, their statement of the problem is rather incomplete. ## \\ ## Edit=I just deleted something, because I'm not sure that it is correct. ## \\ ## One very relevant result that follows because of a pulsed laser is the photon/laser frequency ## \omega ##. There will necessarily be an uncertainty in ## \omega ## of ## \Delta \omega ## that satisfies ## \Delta \omega \, T \gtrsim 1 ##, or thereabouts, because of the Heisenberg Uncertainty Principle. That would make much more sense than to try to establish an uncertainty in total pulse energy that is measured. Usually, the measurement equipment that measures the energy of a single pulse has a relatively large uncertainty=much, much larger than anything Heisenberg would cause.

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Just an additional comment: These uncertainty relationships come about as the result of the wave nature of what is being measured. Meanwhile, when they say the pulse is of duration ## T=1.0 \cdot 10^{-8} ## seconds, that does not mean that ## \Delta T=0 ##. A typical ## \Delta T ## on this pulse width measurement might be ## \Delta T=\pm 1.0 \cdot 10^{-9} ## seconds. Again though, their statement of the problem is rather incomplete. ## \\ ## Edit=I just deleted something, because I'm not sure that it is correct. ## \\ ## One very relevant result that follows because of a pulsed laser is the photon/laser frequency ## \omega ##. There will necessarily be an uncertainty in ## \omega ## of ## \Delta \omega ## that satisfies ## \Delta \omega \, T \gtrsim 1 ##, or thereabouts, because of the Heisenberg Uncertainty Principle. That would make much more sense than to try to establish an uncertainty in total pulse energy that is measured. Usually, the measurement equipment that measures the energy of a single pulse has a relatively large uncertainty=much, much larger than anything Heisenberg would cause.
Interesting...and enlightening as well. Thank you for the clarification :D