- #1

George Keeling

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- When and why can I avoid expanding momentum as derivative operator (in position space) and treat it as an independent variable? So ∂p/∂t=0

I'm doing Griffiths & Schroeter Problem 3.37 and I would like to prove that the expectation value of a rather peculiar observable vanishes:

$$\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0$$My cheat sheet expands the LHS to an integral and continues …

$$\int{\Psi^\ast\frac{\partial}{\partial t}\left(xp\right)\Psi d x}=\int{\Psi^\ast\left(0\right)\Psi d x}=0$$but, along with ##\Psi\left(x,t\right)##, we are in position space so should be setting

$$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$so the interesting term becomes

$$-i\hbar\frac{\partial}{\partial t}\left(x\frac{\partial\Psi}{\partial x}\right)=-i\hbar\left(\frac{\partial x}{\partial t}\frac{\partial\Psi}{\partial x}+x\frac{\partial^2\Psi}{\partial t\partial x}\right)$$

The first part vanishes because ##\frac{\partial x}{\partial t}=0## but the second doesn't.

It's a bit tempting to say that ##x,p,t## are all independent variables so that

$$\frac{\partial}{\partial t}\left(xp\right)=\frac{\partial x}{\partial t}p+x\frac{\partial p}{\partial t}=0+0$$but I don't see why you can do that in this case. Or can you? If so why?

$$\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0$$My cheat sheet expands the LHS to an integral and continues …

$$\int{\Psi^\ast\frac{\partial}{\partial t}\left(xp\right)\Psi d x}=\int{\Psi^\ast\left(0\right)\Psi d x}=0$$but, along with ##\Psi\left(x,t\right)##, we are in position space so should be setting

$$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$so the interesting term becomes

$$-i\hbar\frac{\partial}{\partial t}\left(x\frac{\partial\Psi}{\partial x}\right)=-i\hbar\left(\frac{\partial x}{\partial t}\frac{\partial\Psi}{\partial x}+x\frac{\partial^2\Psi}{\partial t\partial x}\right)$$

The first part vanishes because ##\frac{\partial x}{\partial t}=0## but the second doesn't.

It's a bit tempting to say that ##x,p,t## are all independent variables so that

$$\frac{\partial}{\partial t}\left(xp\right)=\frac{\partial x}{\partial t}p+x\frac{\partial p}{\partial t}=0+0$$but I don't see why you can do that in this case. Or can you? If so why?