How Does Quantum Uncertainty Define the Position of a Positron?

[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?

 

[Orodruin]

Where are your units and does your result make sense dimensionally

 

[LCSphysicist]

Where are your units and does your result make sense dimensionally

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

 

[haruspex]

Does that give the full volume or only one eighth?

 

[LCSphysicist]

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.

 

[haruspex]

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 $\Delta\vec x.\Delta\vec p\geq 3\frac\hbar 2$

 

[BvU]

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 ?

 

[Orodruin]

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.