# How Does Quantum Uncertainty Define the Position of a Positron?

• LCSphysicist
In summary: No, that would not represent the situation fully as you also have bounds on the direction. The easiest is to work in the three independent directions, noting that the uncertainty relation holds in each of these.
LCSphysicist
Homework Statement
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Relevant Equations
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The velocity of a positron is measured to be: ##v_x = (4.00 \pm 0.18)10^5## m/sec, ##v_y = (0.34 \pm 0.12) 10^5## m/sec, ##v_z = (1.41 \pm 0.08) 10^5 ## m/sec. Within what minimum volume was the positron located at the moment the measurement was carried out?

So, let's assumed ##v_i = (v_i \pm \Delta v_i)##. We can say that, for minimum values, $$\Delta X_i \Delta v_i = \hbar / (2m)$$.

$$\Delta X_i = \frac{\hbar}{2 m \Delta v_i} \implies V_{min} = \Delta X \Delta Y \Delta Z = \frac{\hbar}{2 \Delta v_x m } \frac{\hbar}{2 \Delta v_y m} \frac{\hbar}{2 \Delta v_z m} = (\frac{\hbar}{2m})^3 \frac{1}{ \Delta v_x \Delta v_y \Delta v_z}$$

Where, for example, ##|\Delta v_x| = 0.18*10^5##

Is that right?

Last edited by a moderator:
Delta2

Orodruin said:
Ops, i forgot i should have been used momentum. Wait a minut, will edit it.

Does that give the full volume or only one eighth?

haruspex said:
Does that give the full volume or only one eighth?
I don't know? At least i think it should give the full volume. But i am asking Because i am not sure of my answer also.

Herculi said:
I don't know? At least i think it should give the full volume. But i am asking Because i am not sure of my answer also.
Your ##\Delta X_i## are ##\pm##, so aren't the sides of the cube ##2\Delta X_i##?
As to whether your approach is in itself correct I do not know. I have never been able to find a description of quantum physics that uses vectors, so I could have believed the vector form was ##\Delta\vec x.\Delta\vec p\geq\frac\hbar 2##, or maybe ##\Delta\vec x.\Delta\vec p\geq 3\frac\hbar 2##

Naive ? If the errors in the measurements are independent, don't we get ##|v| = (4.25\pm0.25)\ \times 10^5 ## m/s ? And take it from there ?

BvU said:
Naive ? If the errors in the measurements are independent, don't we get ##|v| = (4.25\pm0.25)\ \times 10^5 ## m/s ? And take it from there ?
No, that would not represent the situation fully as you also have bounds on the direction. The easiest is to work in the three independent directions, noting that the uncertainty relation holds in each of these.

BvU

## What is minimum volume uncertainty?

Minimum volume uncertainty is a concept in measurement science that refers to the smallest possible uncertainty that can be assigned to a measurement result. It is the minimum amount of uncertainty that can exist due to the limitations of the measurement process.

## Why is minimum volume uncertainty important?

Minimum volume uncertainty is important because it allows scientists to accurately assess the reliability and accuracy of their measurements. It also helps to identify the sources of uncertainty in a measurement and determine ways to reduce or eliminate them.

## How is minimum volume uncertainty calculated?

Minimum volume uncertainty is calculated using statistical methods and taking into account various sources of uncertainty, such as instrument precision, calibration errors, and environmental factors. It is typically expressed as a standard deviation or confidence interval.

## What factors can affect minimum volume uncertainty?

Several factors can affect minimum volume uncertainty, including the quality and precision of the measuring instrument, the skill and experience of the person performing the measurement, and the environmental conditions in which the measurement is taken.

## How can minimum volume uncertainty be reduced?

Minimum volume uncertainty can be reduced by using more accurate and precise measuring instruments, calibrating instruments regularly, and controlling environmental factors that can affect the measurement. It can also be reduced by increasing the number of measurements taken and using statistical methods to analyze the data.

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