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  Linear superposition and expectation values

+ 0 like - 0 dislike
When you make a linear superposition like this $\psi(\boldsymbol{r},t)=h^{-\frac{2}{3}}\int [C^{+}(\boldsymbol{p})e^{-i\omega t}+C^{-}(\boldsymbol{p})e^{i\omega t}]e^{(i\boldsymbol{p}\cdot\boldsymbol{r}/\hbar)}\, d^{3}\boldsymbol{p}$ where $\omega=\frac{1}{\hbar}(c^{2}p^{2}+m^{2}c^{4})^{\frac{1}{2}}$, the exponential factors are transparent to the integration?. And if I calculate the expectation value of the momentum gradient operator like this $<\hat{\boldsymbol{\nabla_{p}}}>=\int\int\,d^{3}\boldsymbol{p}\,d^{3}\boldsymbol{q}\,(\phi(\boldsymbol{q},t)\hat{\boldsymbol{\nabla_{p}}}\phi(\boldsymbol{p},t))$ where $\phi(\boldsymbol{p},t)=C^{+}(\boldsymbol{p})e^{-i\omega t}+C^{-}(\boldsymbol{p})e^{i\omega t}$ then, the gradient operator acts on the exponentials, being $\omega$ a function of $p$?
Closed as per community consensus as the post is Not graduate-level upwards
asked Jan 15, 2015 in Closed Questions by Calogero (0 points) [ revision history ]
recategorized Jan 17, 2015 by dimension10
Most voted comments show all comments
Nothing can be pulled out of integral because of p-dependence. Vote to close.
The $i \omega t$ factors can be brought in front of the integral when integrating over p and pulled through the p-gradient.
No, it cannot - $\omega$ is explicitly p-dependent.
The quantum field operators is Fourier-decomposed in 4D $\omega - k$ space, when you integrate or differentiate only with respect to 3D momentum (or k respectively as $p=\hbar k$), the frequency part can be treated as constant.
Replace $\omega$ with its explicit -p-dependent expression. Generally, a function of four-coordinate $x$ can be represented as a four-dimensional Fourier integral where all variables $(\omega,\mathbf{p})$ are independent. But the formula above is not of this kind. Here the $\omega$-integration is already done - with help of a delta-function, so no omega exists anymore but a function of $p$.

By the way, the second expression (with gradient) is a symbolic one since the gradient should act on the operators $C^{\pm} (\mathbf{p})$. The whole expression makes sense when it is sandwiched between states and then the action of $C^{\pm} (\mathbf{p})$ gives functions of $\mathbf{p}$.
Most recent comments show all comments
@JiaYiyang I suppose he means "does the gradient operator act on the exponentials?". The answer seems to obviously be "yes".

This doesn't seem to be graduate-level+. Voting to close.

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