- #1
Feynman.12
- 14
- 0
Homework Statement
Show that ## \Delta\lambda\Delta\ x>lamdba^2/4*pi##
The Attempt at a Solution
When I substitute de Broglie's p=h/lambda I get the equation of
##\frac {\Delta\x}{\Delta\lambda} > 1/(4*pi )##
Heisenberg Uncertainty in wavelength and position
- Thread starter Feynman.12
- Start date
In summary, when substituting de Broglie's p=h/lambda into equation of motion, you get that \Delta\x > 1/(4*pi). However, intuitively this does not imply that \Delta\p = h/\Delta\lambda. You need to provide more evidence that your solution is correct before we can accept it.
Show that ## \Delta\lambda\Delta\ x>lamdba^2/4*pi##
When I substitute de Broglie's p=h/lambda I get the equation of
##\frac {\Delta\x}{\Delta\lambda} > 1/(4*pi )##
- #2
- 21,509
- 12,670
You need to show more of your work. How are we supposed to tell where you went wrong if you only provide us with your final result? (I anyway have a pretty good idea of where you have gone wrong, but I want to see exactly what you did first.)
- #3
- #4
- 21,509
- 12,670
The first row is not correct. ##p = h/\lambda## does not imply ##\Delta p = h/\Delta \lambda##. What is ##d\lambda/dp##?
- #5
- #6
- 21,509
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Feynman.12 said:
Because it is a matter of how a small change in ##\lambda## changes ##p##. If you used your formula, a small change in ##\lambda## would give a huge change in ##p##.
- #7
BvU
Science Advisor
Homework Helper
- 15,978
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If ##\ \ y = {1\over x}\ \ ## then surely ##\ \ {dy\over dx} = -{1\over x^2}\ \ \Rightarrow \ \ dy = -{dx\over x^2}\ ## . Change d to ##\Delta## and voila !Feynman.12 said:
(Sorry for barging in, Oro...)
1. What is Heisenberg Uncertainty?
Heisenberg Uncertainty is a principle in quantum mechanics that states that it is impossible to know both the exact position and exact momentum of a particle at the same time. This means that the more accurately we measure one variable, the less accurately we can measure the other.
2. How does Heisenberg Uncertainty apply to wavelength and position?
In quantum mechanics, the wavelength of a particle is related to its momentum, and the position of a particle is related to its location. Heisenberg's Uncertainty Principle states that the more precisely we measure the wavelength of a particle, the less accurately we can know its position, and vice versa.
3. Why is Heisenberg Uncertainty important in science?
Heisenberg's Uncertainty Principle is important because it fundamentally changes how we understand the behavior of particles at a microscopic level. It shows that there are inherent limits to our ability to measure and predict the behavior of particles, and this has major implications for fields such as quantum mechanics, chemistry, and even biology.
4. How does Heisenberg Uncertainty impact technology?
Heisenberg's Uncertainty Principle has led to the development of technologies such as electron microscopes and atomic force microscopes, which allow us to observe particles at a much smaller scale than was previously possible. However, the principle also places limits on the precision of these technologies, as there will always be a degree of uncertainty in the measurements.
5. Can Heisenberg Uncertainty be overcome?
No, Heisenberg's Uncertainty Principle is a fundamental part of quantum mechanics and cannot be overcome. However, scientists have developed techniques to minimize the effects of uncertainty, such as using multiple measurements and statistical analysis to improve the accuracy of their results.
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