# Homework Help: Uncertainty principle in terms of expectations values in Dirac notatio

1. Feb 25, 2013

### Leb

1. The problem statement, all variables and given/known data
Show that
$(\Delta A)^{2} = \langle \psi |A^{2}| \psi \rangle - \langle \psi |A| \psi \rangle ^{2}\\ \phantom{(\Delta A)^{2} }=\langle \psi | (A - \langle A \rangle )^{2} | \psi \rangle ,$
where $\Delta A$ is the uncertainty of an operator A and $\langle A \rangle$ is the expectation value of A.
2. Relevant 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.

3. 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 jsut guessing the left result with the square term.

2. Feb 25, 2013

### dextercioby

Re: Uncertainty principle in terms of expectations values in Dirac not

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'.

3. Feb 25, 2013

### Leb

Re: Uncertainty principle in terms of expectations values in Dirac not

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 +?

Last edited: Feb 25, 2013
4. Feb 25, 2013

### dextercioby

Re: Uncertainty principle in terms of expectations values in Dirac not

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

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

5. Feb 25, 2013

### dextercioby

Re: Uncertainty principle in terms of expectations values in Dirac not

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

Last edited: Feb 25, 2013
6. Feb 25, 2013

### Leb

Re: Uncertainty principle in terms of expectations values in Dirac not

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.

7. Feb 25, 2013

### dextercioby

Re: Uncertainty principle in terms of expectations values in Dirac not

Not really <A> is a number, so

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

8. Feb 25, 2013

### Leb

Re: Uncertainty principle in terms of expectations values in Dirac not

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.

Last edited: Feb 25, 2013
9. Feb 25, 2013

### dextercioby

Re: Uncertainty principle in terms of expectations values in Dirac not

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.

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