Question

Quantum mechanical operators question.
(see attached)

Answer 1

\[[(-\frac{ h_{b}^2 }{ 2m }\frac{ d^2 }{ dx^2 })(-ih_b \frac{ d }{ dx })]-[(-ih_b \frac{ d }{ dx })(-\frac{ h_b }{ 2m }\frac{ d^2 }{ dx^2 })]\]

Answer 2

where h_b is 'h - bar' ie plancks constant over 2pi

Answer 3

I want to know the answer too.

Answer 4

The first one is \([\hat T, \hat p] = \hat T \hat p - \hat p \hat T\)

I think you get this

\(\dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} \left( i \hbar \dfrac{d \Psi}{dx} \right) - i \hbar \dfrac{d }{dx} \left( \dfrac{\hbar^2}{2m} \dfrac{d^2 \Psi}{dx^2} \right) = 0\)

Maybe my calculus is all wrong but at this late hour to me that just all cancels out....

That should mean that there is no H uncertainty between properties T and p - @osprey ?? Does that sound right?!

Second, dunno what operator first one is but I get

\(i \hbar \dfrac{d}{dt} ( t \Psi) - t ~ i \hbar \dfrac{d}{dt} ( \Psi)\)

\(= i \hbar \left( \Psi + t \dfrac{d \Psi}{dt} \right) - t ~ i \hbar \dfrac{d \Psi}{dt}\)

\(= i \hbar \Psi\)

if i knew what the time deriv of the wave function was, i might have a comment to add.... but i don't :-(

Answer 5

@518nad

Answer 6

yup 0 and ih
http://prntscr.com/d1xdht

Answer 7

the H operator is P^2/2m + V(X) all operators, and in this case V(x) was so, so its technically just function of operator P and P operators commuting

Answer 8

thank you sir :-)

Answer 9

yw bby