Unclear about the Heisenberg uncertainty principle

[Nerro]

Hi everybody,

I have a slightly noobish question and I've searched for the answer unsuccesfully so I'm posting it here.

I'm was reading up on the uncertainty principle when I ran into two phrases that I don't quite understand. They are <A²> and <A>². Am I too apply this as \int \psi^* A^2 \psi d\tau and \int(\psi^*A\psi)^2 d\tau?

Without understanding this the Heisenberg uncertainty principle makes not much sense to me...

 

[Meir Achuz]

Your <A^2> is correct, but &lt;A&gt;^2=[\int\psi^* A\psi d\tau]^2.

 

[Nerro]

So \Delta A = \sqrt{\int{\psi^* A^2 \psi d\tau} - (\int{\psi^* A \psi d\tau})^2}?

 

[jtbell]

Yes. If you're wondering where that formula comes from, one starts by defining \Delta A as the square root of the expectation value of the square of the deviation of A from <A>, i.e. as the standard deviation of A:

\Delta A = \sqrt { &lt; A - &lt;A&gt; &gt;^2}

By multiplying out the square and collecting terms one arrives at

\Delta A = \sqrt { &lt;A^2&gt; - &lt;A&gt;^2 }

which is usually easier to calculate.

 

[Anonym]

If you're wondering where that formula comes from, one starts by defining \Delta A as the square root of the expectation value of the square of the deviation of A from <A>. i.e. as the standard deviation of A.

I start by defining delta (x) as the size of the single QM object (eigenvalue=1 of the s-aj number operator, which commute with the Hamiltonian). And I am wondering how it happens that it is closely related (but not identical) to the standard deviation of X which is the macroscopic quantity. Then I ask what the standard deviation of X describes in the classical statistical mechanics: the size of the statistical sub ensemble or the size of the volume (ball) occupied by that sub ensemble. Please explain all that.

Regards, Dany.