# Unclear about the Heisenberg uncertainty principle

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 priciple 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...

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Meir Achuz
Homework Helper
Gold Member
Your <A^2> is correct, but $$<A>^2=[\int\psi^* A\psi d\tau]^2.$$

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

jtbell
Mentor
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:

$$\Delta A = \sqrt { < A - <A> >^2}$$

By multiplying out the square and collecting terms one arrives at

$$\Delta A = \sqrt { <A^2> - <A>^2 }$$

which is usually easier to calculate.

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.

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