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gfd43tg
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Homework Statement
Homework Equations
The Attempt at a Solution
a) I am having some trouble understanding the notation. I'm uncertain whether it should be
$$ \langle {f} | \hat {O_{2}} | g \rangle = \int_{-\infty}^{\infty} f^{*}g \frac {dg}{dx} dx $$
or
$$ \langle {f} | \hat {O_{2}} | g \rangle = \int_{-\infty}^{\infty} f^{*}x \frac {dg}{dx} dx $$
b) Once again, the notation is really screwing with my head. I know to start,
$$ [\hat {O_{1}}, \hat {O_{2}}] \psi = (\hat {O_{1}} \hat {O_{2}} - \hat {O_{2}} \hat {O_{1}}) \psi $$
$$ = \hat {O_{1}}(\hat {O_{2}} \psi) - \hat {O_{2}}(\hat {O_{1}} \psi) $$
$$ = \hat {O_{1}}(x \frac {d \psi}{dx}) - \hat {O_{2}}(x^{3} \psi) $$
From here, I am unsure how to do this. My thoughts are either
$$ = x^{3} \Big (x \frac {d \psi}{dx} \Big ) - x \frac {d(x^{3} \psi)}{dx} $$
or
$$ = x^{3} \frac {d^{3} \psi}{dx^{3}} - ? $$
Wouldn't even know what to use for the second term.
I will assume it's the first one since at least I know what to do with that one. Using product rule,
$$ = x^{4} \frac {d \psi}{dx} - x \Big [ x^{3} \frac {d \psi}{dx} + \psi \frac {d (x^{3})}{dx} \Big ] $$
$$ = x^{4} \frac {d \psi}{dx} - x^{4} \frac {d \psi}{dx} - 3x^{3} \psi $$
Therefore, the commutator is ##-3x^{3}##
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