In Sakurai's quantum mechanics, the derivation of momentum operator and Hamlitonian operator is based on spatial translation and time translation as below,
- for spatial translation and momentum operator, we have
$$\mathfrak{T}(d\pmb{x})|\alpha\rangle = \left(1 – \frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\right)|\alpha\rangle = \int{d\pmb{x}\ \mathfrak{T}(d\pmb{x})\ |\pmb{x}\rangle\langle\pmb{x}|\alpha\rangle} = \int{d\pmb{x}|\pmb{x}+d\pmb{x}\rangle\langle\pmb{x}|\alpha\rangle}$$
and if we change the variable
$$\left(1 – \frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\right)|\alpha\rangle = \int{d\pmb{x}|\pmb{x}\rangle}\langle\pmb{x}-d\pmb{x}|\alpha\rangle = \int{d\pmb{x}|\pmb{x}\rangle}\left(\langle\pmb{x}|\alpha\rangle-d\pmb{x}\frac{\partial }{\partial \pmb{x}}\langle\pmb{x}|\alpha\rangle\right)$$
and then we get the momentum operator is,
$$\pmb{p} = -i\hbar \frac{\partial}{\partial \pmb{x}}$$ - However, he derives Hamiltonian operator in a different manner.
$$\mathfrak{U}(t + dt) – \mathfrak{U}(t)= \left(1 – \frac{iHdt}{\hbar}\right)\mathfrak{U}(t) – \mathfrak{U}(t) = -\frac{iHdt}{\hbar}\mathfrak{U}(t)$$
and then we get,
$$\frac{\mathfrak{U}(t+dt)-\mathfrak{U}(t)}{dt} = \frac{d\mathfrak{U}(t)}{dt} = -i\frac{H}{\hbar}\mathfrak{U}(t)$$
and then we get Hamiltonian operator,
$$H = i\hbar\frac{\partial}{\partial t}$$ - Here comes a problem, if we derive the momentum operator in similar manner as 2., we get
$$\mathfrak{T}(\pmb{x} + d\pmb{x}) – \mathfrak{T}(\pmb{x})= \left(1 – \frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\right)\mathfrak{T}(\pmb{x}) – \mathfrak{T}(\pmb{x}) = -\frac{i\pmb{p}\cdot d\pmb{x}}{\hbar}\mathfrak{T}(\pmb{x})$$
and then we get,
$$\frac{\mathfrak{T}(\pmb{x}+d\pmb{x})-\mathfrak{T}(\pmb{x})}{d\pmb{x}} = \frac{d\mathfrak{T}(\pmb{x})}{d\pmb{x}} = -i\frac{\pmb{p}}{\hbar}\mathfrak{T}(\pmb{x})$$
and then we get,
$$\pmb{p} = i\hbar\frac{\partial}{\partial \pmb{x}}$$
There is a minus sign, and I don't know where is wrong , they should result in same result. Can someone tell me where is wrong?
Best Answer
The confused operators you wrote are undefined and sloppy. If only you wrote them in clean Dirac notation, there would be no ambiguity. Work in one dimension, w.l.o.g., so
Acting on bras, you get the conventional sign, which defines the coordinate representation realization. Acting on kets, unconventional, you get the opposite sign. $$\mathfrak{T}(\delta x )= \left(1 - \frac{i \hat p \delta x }{\hbar}\right) = \int d x ~\mathfrak{T}( \delta x )\ | x \rangle\langle x| = \int dx~| x +\delta x \rangle\langle x | \\ = \int d x~ | x \rangle \langle x-\delta x | = \int d x ~\ | x \rangle\langle x| \mathfrak{T}( \delta x )~~ \implies \\ \hat p =\int d x~ | x \rangle ( -i\hbar\partial_x )\langle x |=\int d x~ | x \rangle ( i\hbar \overset{_\gets}\partial_x )\langle x | .$$
Note the evolution operator $\mathfrak{U}(t)$ is defined on kets, not bras, $$\hat H = i\hbar\frac{\partial}{\partial t},$$ so it corresponds to the above, "unconventional", choice.
So the option you believe is opposite in sign, is really the same as in 1., except you have integrated by parts to be acting on kets.
After your additional comment. These time and space derivatives are meant to act on the Schroedinger wavefunction in the coordinate representation, $\phi(t,x)= \langle x| \phi(t)\rangle $, so $$i\hbar \partial_t \phi(t,x)= \langle x|\hat H |\phi(t)\rangle, \\ -i\hbar \partial_x \phi(t,x)= \langle x|\hat p|\phi(t)\rangle. $$