- #1
Bashyboy
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Homework Statement
Why can't you do integration-by-parts directly on the middle expression in equation 1.29--pull out the time derivative over onto x, note that [itex]\displaystyle \frac{\partial x}{\partial t} = 0[/itex], and conclude that [itex]\displaystyle \frac{d \langle x \rangle }{dt} = 0[/itex]
Homework Equations
Equation 1.29 [itex]\displaystyle \frac{d \langle x \rangle }{dt} = \int x \frac{\partial}{\partial t} | \psi |^2 dx = \frac{i \hbar}{2m} \int x \frac{\partial }{\partial x} \left( \psi^* \frac{\partial \psi}{\partial x} - \frac{\partial \psi^*}{\partial x} \right) dx[/itex]
The Attempt at a Solution
Here is my solution, which is not consistent with the one found in the answer key:
The notation [itex]\displaystyle \frac{\partial x}{\partial t} = 0 [/itex] expresses two things: one is that the position of the particle depends on things other than time; the second is that the position of the particle does not change with time, which is not true, in general. One could make such an assumption, but it would be a rather restricting one; the result you would derive from such an assumption would be that the particle never changes its position, and we would not even need such sophisticated mathematics and physics to predict/describe the behavior of the system.
According to the answer key:
For integration by parts, the differentiation has to be with respect to the integration variable – in this case the differentiation is with respect tot, but the integration variable is x. It’s true that
[itex]\displaystyle \frac{\partial }{\partial t} (x |\psi |^2 )= \frac{\partial x}{\partial t} |\psi |^2 + x \frac{\partial }{\partial t} |\psi |^2[/itex]
but this does not allow us to perform the integration:
[itex]\displaystyle \int_a^b x \frac{\partial }{\partial t} {\psi }^2 dx = \int_a^b (x |\psi |^2)dx = (x |\psi|^2)|_a^b[/itex]
Is my reasoning incorrect?