A particle is in a harmonic oscillator potential, not in the ground state. The position of the particle is known with an rms spread of $1\mathring{\text{A}}$. Which of the following are possible spreads for a simultaneous measurement of the momentum?

  1. $2.63 \times 10^{−25} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  2. $5.27 \times 10^{−25} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  3. $1.05 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  4. $2.10 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$

I think the answer is $5.27 \times 10^{−25} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$ from which I used the minimum uncertainty relation $$\Delta x \Delta p= \frac{\hbar}{2}$$ With $\Delta x=10^{-10}$ so $$\Delta p=\frac{\hbar}{2\Delta x}\approx\frac{1.055\times 10 ^{-34}}{2\times 10^{-10}}\approx 5.275\times 10^{-25}\mathrm{\,kg\,m}\, \mathrm{s}^{−1}$$

The correct answers are

  1. $1.05 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$
  2. $2.10 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}$

I note that my (wrong) answer is precisely double another answer on that list $$\frac{5.275\times 10^{-25}}{2}=2.6375\mathrm{\,kg\,m}\, \mathrm{s}^{−1}$$

I am very curious as to why there must be two values that are connected somehow. I managed to use the uncertainty relation to get one answer, but how on Earth do you get the other answer?

More importantly for now, could anyone please explain or give hints so that I can understand why the correct answers are

$$1.05 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1} \qquad \text{&} \qquad 2.10 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}\,?$$


closed as off-topic by Emilio Pisanty, AccidentalFourierTransform, John Rennie, Kyle Kanos, BLAZE Aug 13 '18 at 10:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, AccidentalFourierTransform, John Rennie, Kyle Kanos, BLAZE
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @Aaron Okay then, the saturated uncertainty bound was wrong for me to assume. But it was the only way I could think of a way to try to solve the problem. So I used the HUP since I am only given one piece of information in the question. I can't think of another way to approach this problem. $\endgroup$ – BLAZE Aug 12 '18 at 21:29
  • $\begingroup$ Look at my answer. You found what the spreads must be larger than. This is why the correct answers are the two that are larger than the number you calculate $\endgroup$ – Aaron Stevens Aug 12 '18 at 21:32
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    $\begingroup$ @BLAZE: showing effort isn't a magic fix that makes off-topic questions on topic. you can always ask a classmate or instructor for such a thing. See this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Aug 13 '18 at 10:22
  • $\begingroup$ @Kyle I'm well aware of the rules, I've been a member of this site for many years so your 'macro message' isn't going to tell me anything I don't know. But this is one case where the question really shouldn't have been closed. "Off-topic" is not even well defined, it's just a matter of opinion and in this case the opinions were not in favour of keeping the question open. I have now voted to close the question as well since it seems the majority is in such a hurry to see this one get closed. All I'm saying is; please use more caution before you so readily choose to close a question, thanks. $\endgroup$ – BLAZE Aug 13 '18 at 11:04
  • $\begingroup$ @BLAZE as the guy with more close reviews than anyone else, I'm pretty well aware of what constitutes on- or off-topic. You certainly can rephrase the question to make it on-topic, but the question you've asked ("How is this calculation done to get value X?") is a very clear off-topic question, as per the HW policy. There is also no sense of hurry, closures happen in their own time. $\endgroup$ – Kyle Kanos Aug 13 '18 at 11:08

It says it's not in the ground state. You can only get the lowest bound in the Heisenberg Uncertainty Principal when the harmonic oscillator is in the ground state. Since we are not in the ground state, we know the spread must be larger than the number you calculate.

In other words, since it says the system is not in the ground state, you cannot assume minimum uncertainty. The correct answers are therefore all answers larger than $5.275 \times 10^{-25} kg\cdot m \cdot s^{-1}$ (Answers 3&4).

There is not any more information to actually calculate answers 3&4. Any choice the question would have given larger than the minimum spread would be correct. There are no calculations to be done to get those two choices. You did everything right in the math, just not in the interpretation of the math given the assumptions stated in the problem.

There are an infinite amount of possible correct answers. You just have to pick the two that the maker of the question gave that are larger than the spread from the minimum uncertainty.

  • $\begingroup$ @BLAZE I changed it to an answer because it does answer the question. You found what the spread must be larger than. Notice how 3&4 are the numbers larger than what you calculate. $\endgroup$ – Aaron Stevens Aug 12 '18 at 21:34
  • $\begingroup$ Does it say in your "answer" how to calculate $1.05 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1} \qquad \text{&} \qquad 2.10 \times 10^{−24} \mathrm{\,kg\,m}\, \mathrm{s}^{−1}\tag{?}$ $\endgroup$ – BLAZE Aug 12 '18 at 21:35
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    $\begingroup$ No, because you already did the calculation correctly. You didn't do anything wrong in your calculation. You found the spread for the minimum uncertainty. Since the system is not in the ground state, you know that the spread must be larger than this. This is why the answers are 3&4. These are the two that are larger than what you found. There is nothing to do to actually calculate 3&4. Any answer choice larger than the number you calculate would be considered correct. $\endgroup$ – Aaron Stevens Aug 12 '18 at 21:37
  • $\begingroup$ So you're using process of elimination. Your answer and comments do still not explain why their are two correct answers. Right now you are telling me what all my mistakes and not explaining how why those answers come in pairs $\endgroup$ – BLAZE Aug 12 '18 at 21:44
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    $\begingroup$ It's a pair of answers because it gives two choices larger than the one you calculate. There is nothing special about the numbers other than they are larger than what you calculate. They could have only given one number larger than your number and there would only be one correct answer. It's just how whoever made the question decided to make the answers what they are. $\endgroup$ – Aaron Stevens Aug 12 '18 at 21:47

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