Expression for Uncertainty of Arbitrary Operator

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DrPapper
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Hello all, as far as I can see this question is not posted already, my apologies if it is and please provide a link. But I'm watching this video on youtube: And at 22:38 there's an expression given for the uncertainty of an arbitrary operator Q, however I'm concerned the expression is incorrect. Could anyone please look at the expression and tell me if it's true or a mistake?

The expression given is σ= < \hatℚ2 - <ℚ>2 >

Sorry, I can't seem to get LaTeX to work right, though I've managed it in the past.

If so could you please explain why it deviates from eqn 1.12 of Griffiths into QM which is:

σ=√(<j2>-<j>2)
 
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DrPapper said:
And at 22:38 there's an expression given for the uncertainty of an arbitrary operator Q, however I'm concerned the expression is incorrect. Could anyone please look at the expression and tell me if it's true or a mistake?

The expression given is σ= < \hatℚ2 - <ℚ>2 >
I actually I couldn't find that equation in the minute you mentioned above. Anyway, the difference between
DrPapper said:
σ= < \hatℚ2 - <ℚ>2 >
and
DrPapper said:
σ=√(<j2>-<j>2)
is just the presence of the square root. Had the square root been removed from the second equation, they will become equivalent because ##\langle C \rangle = C## if ##C## is just number while at the same time ##\langle Q \rangle^2## is also a number.
 
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Let ##\bar A=\langle A\rangle##. Then ##\sigma_A^2:=\langle(A-\bar A)^2\rangle =\langle A^2\rangle -2\langle A\bar A\rangle+\langle\bar A^2\rangle =\langle A^2\rangle -2\bar A^2+\bar A^2 =\langle A^2\rangle -\bar A^2 =\langle A^2 -\bar A^2 \rangle##.

For the interpretation, see also this post and the discussion here.
 
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A. Neumaier said:
Let ##\bar A=\langle A\rangle##. Then ##\sigma_A^2:=\langle(A-\bar A)^2\rangle =\langle A^2\rangle -2\langle A\bar A\rangle+\langle\bar A^2\rangle =\langle A^2\rangle -2\bar A^2+\bar A^2 =\langle A^2\rangle -\bar A^2 =\langle A^2 -\bar A^2 \rangle##.

For the interpretation, see also this post and the discussion here.
Thank you I'll check those out. :D