The Quantum Uncertainty of a Particle

AI Thread Summary
The discussion revolves around understanding the quantum harmonic oscillator problem and the significance of the result 2.34 in relation to the uncertainty of a particle. The participant is confused about the trend observed in the values of ##\sqrt{<x^2>}## as n increases, noting that the results approach 2.34. It is clarified that as n increases, the uncertainty also increases, and the numerical integration may introduce inaccuracies. The exact relationship between ##<x^2>## and the energy levels of the harmonic oscillator is introduced, revealing that ##\sqrt{<x^2>}## correlates with the energy levels. Ultimately, the participant gains clarity on their question regarding the quantum harmonic oscillator.
maguss182
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Summary:: I'm trying to understand the meaning behind the answer of 2.34... but I haven't taken a quantum class yet so I'm utterly lost.

So I took this physics class that's teaching us python and for our final project our teacher assigned random problems to work out, well I got stuck with a quantum harmonic oscillator problem, and while I worked out the code easy enough, the final question I have no idea how to answer it.

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The answer I get is 2.34... but I'm not sure what the trend is? It gets closer and closer to 2.34? Why? Any help would be greatly appreciated, thanks!
 
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They are asking if you can see a trend in the numbers that you get for ##\sqrt{<x^2>}## for ##n = 1, 2, 3, 4, 5##.

This might be hard to recognize. However, try looking at the square of these numbers and see if you recognize a trend.
 
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Well I see that as we get to 5 the answer gets closer to 2.34,

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Is it saying that as the number goes up the uncertainty gets closer and closer to 2.34? So a more precise calculation?
 
What do the squares of these numbers look like?
 
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Ok so they are almost the same thing right?
 
Calculate (1.22474)2, (1.58109)2, etc.
 
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Seems to increase by ~1? Sorry I'm so lost, math is not my strong suit
 
wait uncertainty increases as n increases
 
maguss182 said:
View attachment 293955

Seems to increase by ~1? Sorry I'm so lost, math is not my strong suit
Yes, the numbers in the second column increase by 1. You are doing numerical integration. So, you know it is not perfectly accurate and will tend to get less accurate as n increases (because ##\psi_n(x)## varies more rapidly as n increases). Suppose you round off your second column of numbers to the nearest tenth. Can you guess what the exact values of these numbers would be if there wasn't any error in the numerical integration?

Note that the second column of numbers represents ##<x^2>## for the states ##n = 1, 2, 3, 4, 5##.

Can you guess the exact formula for ##<x^2>## as a function of ##n##?
 
  • #10
maguss182 said:
wait uncertainty increases as n increases
Yes. Could this be the only trend that they wanted you to notice? I don't know. If so, then you can ignore what I was trying to get you to see. :oldsmile:
 
  • #11
I think so lol sorry, thinking on what you said though would it be the quantum harmonic oscillation? that is what the problem is about
 
  • #12
maguss182 said:
I think so lol sorry, thinking on what you said though would it be the quantum harmonic oscillation? that is what the problem is about
You mentioned that you have not taken a quantum mechanics course. So, you might not be familiar with the energy levels, ##E_n##, of the harmonic oscillator: $$E_n = hf (n+1/2) \,\,\,\,\,\,\, n = 1, 2, 3, 4, ...$$ where ##h## is Planck's constant and ##f## is the frequency of the oscillator.

For a specific oscillator, you can choose units such that ##hf = 1##. Then $$E_n = n+1/2$$
You can show that in these units ##<x^2>## for a particular quantum state ##\psi_n(x)## actually equals ##E_n##. So, ##\sqrt{<x^2>} = \sqrt{E_n} = \sqrt{n+1/2}##.

Hope I haven't caused too much confusion. As I said, you can ignore all of this if it's something you are not familiar with.
 
  • #13
Oh ok yeah no I haven't taken a course yet so that's new to me but thanks for all your help, I was able to figure out my question, thank you!
 
  • #14
maguss182 said:
Oh ok yeah no I haven't taken a course yet so that's new to me but thanks for all your help, I was able to figure out my question, thank you!
OK, great.
 
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