Uncertainty principle in terms of expectations values in Dirac notatio

[Leb]

Homework Statement

Show that
(\Delta A)^{2} = \langle \psi |A^{2}| \psi \rangle - \langle \psi |A| \psi \rangle ^{2}\\<br /> \phantom{(\Delta A)^{2} }=\langle \psi | (A - \langle A \rangle )^{2} | \psi \rangle ,
where \Delta A is the uncertainty of an operator A and \langle A \rangle is the expectation value of A.

Homework Equations

\langle \psi |A| \psi \rangle = \langle A \rangle
completeness relation: |\psi \rangle = \sum_{n}b_{n}|\psi_{n}\rangle. We can use the completeness relation to expand out any wavefunction and derive an expression for the expectation value in terms of the weights in the expansion.

The Attempt at a Solution

I tried expanding the first line and got up to \sum_{n}\left( b_{n}^{2}a_{n}^{2} - b_{n}^{2}a_{n} \right). To be honest, I am just guessing the left result with the square term.

 

[dextercioby]

You don't need the completeness relation. This is a standard calculation, if you assume everything to be well-defined mathematically. Start with the definitions and show where you get 'lost'.

 

[Leb]

To be honest I do not know the definition of it. It was just introduced as a notation and that is all I know. I've looked up online and it does not seem to me to have a formal definition, just notation.

I checked that it has linearity and associativity properties, but do not see how they help.

I have spent maybe 4 hours on this and am feeling like the idiot I am.

Anyway, I tried expanding the second line to get

\langle \psi|(A-\langle A \rangle )^2|\psi \rangle = \langle \psi|A^2 -2A\langle A \rangle + \langle A \rangle ^2 |\psi \rangle = \langle \psi|A^{2}|\psi \rangle + \langle \psi | \langle A \rangle ^2 |\psi \rangle - \langle \psi | 2A\langle A\rangle | \psi \rangle
Realizing that <A> should be a number (found it somewhere in these forums). How do I now get rid of -2A<A> and get - instead of +?

 

[dextercioby]

For A being a (compact self-adjoint) linear operator acting everywhere on a complex (separable) Hilbert space, the definition of \Delta A for a (pure) state \psi is

\Delta A =: \sqrt{\left\langle \psi \left| \left(A-\langle A\rangle_{\psi}\hat{1}\right)^2 \right| \psi\right\rangle}

 

[dextercioby]

The 3rd term is (-2) times the second one. Can you see why ?

 

[Leb]

Because \langle \psi | 2A\langle A\rangle | \psi \rangle = \langle \psi | 2A\langle\psi | A |\psi \rangle | \psi \rangle = \langle \psi | 2AA\langle\psi |\psi \rangle | \psi \rangle = \langle \psi | 2A^{2} | \psi \rangle = 2A^{2}\langle \psi | \psi \rangle ?
And it does not have to be a_{n}^{2} because we can choose whether we are interested in an eigenfunction or eigenvalue ?I think my problem is that I am not comfortable manipulating them.

 

[dextercioby]

Not really <A> is a number, so

<psi|2A<A>|psi> = 2<A> <psi|A|psi> = 2<A> <A> = 2<A>²

 

[Leb]

Now I am confused even more... Where was my manipulation wrong ? I thought <A>=<psi|A|psi>... Or was your step simply correct in this particular situation due to insight ?

Anyway, thank you for your help! At least I know how to solve it. Still not sure how I can manipulate them and why I cannot always just factor out something from inside the bra-ket.

 

[dextercioby]

Yes, but A is an operator, you can't equate A to a number such <psi|A|psi>, nor take it out of the bra-ket.