Uncertainty principle in terms of expectations values in Dirac notatio

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Homework Help Overview

The discussion revolves around the uncertainty principle in quantum mechanics, specifically in the context of expectation values expressed in Dirac notation. The original poster attempts to show a relationship involving the uncertainty of an operator and its expectation values.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definitions and properties of operators and expectation values, with some attempting to manipulate expressions involving these concepts. Questions arise regarding the completeness relation and the mathematical definitions of the terms involved.

Discussion Status

Participants are actively engaging with the problem, offering insights and corrections. Some guidance has been provided regarding the manipulation of terms, but confusion remains about the application of definitions and the properties of operators versus numbers.

Contextual Notes

There are indications of uncertainty regarding the definitions of certain terms and the mathematical framework being used. Participants express varying levels of familiarity with the concepts, and some mention the time spent trying to understand the problem.

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Homework Statement


Show that
[itex](\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 ,[/itex]
where [itex]\Delta A[/itex] is the uncertainty of an operator A and [itex]\langle A \rangle[/itex] is the expectation value of A.

Homework Equations


[itex]\langle \psi |A| \psi \rangle = \langle A \rangle[/itex]
completeness relation: [itex]|\psi \rangle = \sum_{n}b_{n}|\psi_{n}\rangle[/itex]. 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 [itex]\sum_{n}\left( b_{n}^{2}a_{n}^{2} - b_{n}^{2}a_{n} \right)[/itex]. To be honest, I am just guessing the left result with the square term.
 
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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'.
 


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

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

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


The 3rd term is (-2) times the second one. Can you see why ?
 
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Because [itex]\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[/itex] ?
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.
 


Not really <A> is a number, so

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


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.
 
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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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