As far as I understand, the momentum operator is:(adsbygoogle = window.adsbygoogle || []).push({});

[itex] \hat{p} = -i \hbar \frac{\partial}{\partial \hat{q}} [/itex]

Where I'm not sure at this point if it's mathematically correct to talk about a derivative wrt. the position operator - but the point is, as far as I understand, that this equality is true in general, without taking the Schrodinger representation. It is not a derivative wrt the position eigenvalues.

Since [itex] \hat{p} [/itex] is a Hermitian operator, [itex] \frac{\partial}{\partial \hat{q}} [/itex] must be pure imaginary. So I think it's true that:

[itex] \left( \frac{\partial}{\partial \hat{q}} |ψ> \right)^{\dagger} =

<ψ|\left( \frac{\partial}{\partial \hat{q}} \right)^{\dagger} =

-<ψ| \frac{\partial}{\partial \hat{q}} [/itex]

Similarly:

[itex] \hat{H} |ψ> = i \hbar \frac{\partial |ψ>}{\partial t} [/itex]

Hence:

[itex] \left( \frac{\partial}{\partial t} |ψ> \right)^{\dagger} =

<ψ|\left( \frac{\partial}{\partial t} \right)^{\dagger} =

- \frac{\partial}{\partial t} <ψ| [/itex]

But that is wrong. The minus shouldn't be there. What mistake am I making?

Also, can we say: [itex] \hat{H} = i \hbar \frac{\partial }{\partial t} [/itex] ?

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# Confusion about derivative operators

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