# Minimum volume uncertainty

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?

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Delta2

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

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Does that give the full volume or only one eighth?

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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.

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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##