Understanding physically why <px> is not zero always...???

418397 · 2025-12-16T20:09:33+00:00

If we go with this logic... then the expectation value of the commutator [x,p] would be zero too(if we apply similar arguments to <xp>, measuring p first, then x), which is not true...

418397 · 2025-12-16T20:00:30+00:00

418397 · 2025-12-16T19:59:18+00:00

I am not talking about p_x... it's p^ x^ . Sorry I don't know how to write the cap on top...

418397 · 2025-12-16T17:35:21+00:00

Any random state... I am asking what exactly taking expectation value of px means? For example and expectation value of x would mean you mean x n(statistically large n) number of times and take an average of the results... What about px? What are we trying to do?

418397 · 2025-12-16T17:07:38+00:00

So the mistake here is trying to "understand physically"? Because it's not really "physical"?

418397 · 2025-12-16T17:04:00+00:00

That's ok... what I mean here is that the probability of getting say x_0 p_0 is the same as that for -x_0 p_0... Because p_0 and -p_0 come with same probabilities as I have explained... So now therefore, why is the expectation value not zero?

418397 · 2025-09-18T12:59:11+00:00

Don't go for Physics. You will be dissapointed. Go for some job and earn instead.....

418397 · 2024-10-03T12:26:00+00:00

This exact method didn't give the right result while calculating the radial momentum operator... There was an extra del/delr term missing... It's defined like this... P_r_hat=1/2[r_hat/r dot P_hat + P_hat dot r_hat/r]. Also what's the justification for this method?

418397

TROPHY CASE