- #1
Sparky_
- 227
- 5
Hello,
In Griffiths (2nd edition) pgs 110-111 - deriving the uncertainty principle
I have 2 questions
1)
I am stuck on a point ...
(h = ^ hat )
<f | g > = < ( Ah - <A>) ψ | ( Bh - <B>) ψ >
= <Ψ | ( Ah - <A>) ( Bh - <B>) Ψ>
FOIL
= <ψ | AhBh ψ> - <B><ψ | Ah ψ> - <A>< ψ | Bh ψ> + <A><B>< ψ | ψ>
I do see where < ψ | Ahψ > = <A> so <B>< ψ | Bh ψ> = <A><B>
I don't understand how / why : < ψ | AhBh ψ> = <AhBh> the expectation of A hat times B hat = <AhBh>
why would it not be <AB> instead <AhBh>
like expectation of A hat = <A> not <Ah>
Meaning with the single operator A-hat or B-hat the result is <A> and <B> respectively
the double < ψ | AhBh ψ>, Griffiths has = <AhBh>2) I want to confirm I am correct with this ...
the book shows <B><A> - <A><B> + <A><B>
the result is <A><B>
(I want to say <B><A> instead)
Am I correct that <B><A> = <A><B>
(thinking of the expectations as a resulting number or average)
Thanks
-Sparky
In Griffiths (2nd edition) pgs 110-111 - deriving the uncertainty principle
I have 2 questions
1)
I am stuck on a point ...
(h = ^ hat )
<f | g > = < ( Ah - <A>) ψ | ( Bh - <B>) ψ >
= <Ψ | ( Ah - <A>) ( Bh - <B>) Ψ>
FOIL
= <ψ | AhBh ψ> - <B><ψ | Ah ψ> - <A>< ψ | Bh ψ> + <A><B>< ψ | ψ>
I do see where < ψ | Ahψ > = <A> so <B>< ψ | Bh ψ> = <A><B>
I don't understand how / why : < ψ | AhBh ψ> = <AhBh> the expectation of A hat times B hat = <AhBh>
why would it not be <AB> instead <AhBh>
like expectation of A hat = <A> not <Ah>
Meaning with the single operator A-hat or B-hat the result is <A> and <B> respectively
the double < ψ | AhBh ψ>, Griffiths has = <AhBh>2) I want to confirm I am correct with this ...
the book shows <B><A> - <A><B> + <A><B>
the result is <A><B>
(I want to say <B><A> instead)
Am I correct that <B><A> = <A><B>
(thinking of the expectations as a resulting number or average)
Thanks
-Sparky