Writing a squared observable in Dirac notation

[Zero1010]

Homework Statement

I need help writing in Dirac notation

Relevant Equations

$<A>=\int_{-\infty}^\infty \Psi^*(x) \hat A \Psi (x) dx$
$<A> = <\Psi|\hat A|\Psi>$
$<A^2>=\int_{-\infty}^\infty |\Psi^*(x)|^2 \hat A^2 dx$

Edited after post below:

Hi,

I need to show that the square of the expectation value of an observable takes a certain form in Dirac notation.

I know in wave notation that the expectation value is a sandwich integral which looks like this:

$<A>=\int_{-\infty}^\infty \Psi^*(x) \hat A \Psi (x) dx$

Which is written in Dirac notation as:

$<A> = <\Psi|\hat A|\Psi>$

And the expectation value of a squared observable is written as:

$<A^2>=\int_{-\infty}^\infty \Psi^*(x) \hat A^2 \Psi(x) dx$

But I am not sure how to write this in Dirac notation.

Thanks for any help and hopefully this makes sense.

 

[PeroK]

And the expectation value of a squared observable is written as:

$<A^2>=\int_{-\infty}^\infty |\Psi^*(x)|^2 \hat A^2 dx$

This can't be right. You have a number on the LHS and an operator on the RHS. It should be:

$<A^2>=\int_{-\infty}^\infty \Psi^*(x) \hat A^2 \Psi (x) dx$

Note also that this is the expectation value of the square of the observable/operator and not to be confused with $\langle A \rangle^2$, which is the square of the expectation value.

 

[Zero1010]

Thanks for the heads up.

I was looking at the expectation value from my textbook (see image) which is obviously different. Thanks

 

[PeroK]

Thanks for the heads up.

I was looking at the expectation value from my textbook (see image) which is obviously different. Thanks

If you have the position operator, or the position operator squared, then we have:

$\Psi^*(x) \hat x^2 \Psi(x) = \Psi^*(x) x^2 \Psi(x) = |\Psi(x)|^2x^2$

But, this does not hold for a general operator $\hat A$. For example, if we have the differential operator:

$\Psi^*(x) \hat D \Psi(x) = \Psi^*(x) \frac{d\Psi}{dx}(x) \ne |\Psi(x)|^2 \frac{d}{dx}$

 

[Zero1010]

Ok that makes sense.

The operator I am dealing with is Hermitian which is important later in the question I'm working on.

 

[Zero1010]

So in Dirac notation is it just written as:

$<\Psi(x)|\hat A^2|\Psi(x)>$

 

[PeroK]

So in Dirac notation is it just written as:

$<\Psi(x)|\hat A^2|\Psi(x)>$

Yes, that's the expectation value for the operator $\hat A^2$. Look at it this way: you could write $\hat B = \hat A^2$, then:

$\langle A^2 \rangle = \langle B \rangle = \langle \Psi | \hat B |\Psi \rangle = \langle \Psi | \hat A^2 |\Psi \rangle$

 

[Zero1010]

Ok. Thanks for your help so far its been great.

One last thing, does this make sense (hopefully):

$<A^2>=<\psi|\hat A^2|\Psi>$

$<A^2>=<\psi|\hat A \hat A|\Psi>$

The operate on the right acts on Psi so:

$<A^2>=<\psi|\hat A|\hat A\Psi>$

Since $\hat A$ is Hermitian (therefore - $<f|\hat A g> = <\hat A f| g>$):

$<A^2>=<\hat A \psi|\hat A\Psi>$

 

[PeroK]

Ok. Thanks for your help so far its been great.

One last thing, does this make sense (hopefully):

$<A^2>=<\psi|\hat A^2|\Psi>$

$<A^2>=<\psi|\hat A \hat A|\Psi>$

The operate on the right acts on Psi so:

$<A^2>=<\psi|\hat A|\hat A\Psi>$

Since $\hat A$ is Hermitian (therefore - $<f|\hat A g> = <\hat A f| g>$):

$<A^2>=<\hat A \psi|\hat A\Psi>$

Yes. In general:

$\langle \Psi|\hat A \hat B|\Psi \rangle = \langle (\hat A^{\dagger} \Psi)|(\hat B\Psi) \rangle$

 

[Zero1010]

Cool.

Thanks again for your help.