- #1
rem1618
- 14
- 0
(First post, hi everyone.)
I'm reading Griffith's textbook on intro to quantum mech right now. After establishing the following equation
$$\left\langle f(x) \right\rangle = \int f(x)p(x)\,dx$$
And thus
$$\left\langle x \right\rangle = \int x|\Psi|^2\,dx$$
He goes on to find ##\left\langle v \right\rangle## by taking the derivative with respect to time, and he writes
$$\frac{d\left\langle x \right\rangle}{dt} = \int x\frac{\delta}{\delta t}|\Phi|^2\,dx$$
which I don't completely understand. Why isn't the x inside the partial derivative? I'm not sure if it's a physical or mathematical concept I'm not grasping here.
I first thought x is the position, so it could be a function of time and written as x(t), but then I figure maybe x(t) doesn't really have any meaning from the statistical perspective. If it can be pulled out of that derivative, then it's just a...constant? Is there a physical meaning to that x or is it just an operator tool that makes the math work out?
I'm reading Griffith's textbook on intro to quantum mech right now. After establishing the following equation
$$\left\langle f(x) \right\rangle = \int f(x)p(x)\,dx$$
And thus
$$\left\langle x \right\rangle = \int x|\Psi|^2\,dx$$
He goes on to find ##\left\langle v \right\rangle## by taking the derivative with respect to time, and he writes
$$\frac{d\left\langle x \right\rangle}{dt} = \int x\frac{\delta}{\delta t}|\Phi|^2\,dx$$
which I don't completely understand. Why isn't the x inside the partial derivative? I'm not sure if it's a physical or mathematical concept I'm not grasping here.
I first thought x is the position, so it could be a function of time and written as x(t), but then I figure maybe x(t) doesn't really have any meaning from the statistical perspective. If it can be pulled out of that derivative, then it's just a...constant? Is there a physical meaning to that x or is it just an operator tool that makes the math work out?