Question

What is the derivative of a differential operator?
d/dx{d/dx}=? eg. the hamiltonian operator is H=p^2/2m+V, where p=hbar/i*(d/dx) and V=V(x,y,z) and we are interested in in dH/dx.

Answer 1

is this what you mean ?\[\frac{\text d}{\text dx}\left(\frac{\text d}{\text dx}\right)=\frac{\text d^2}{\text dx^2}\]

Answer 2

shouldn't they be partial derivatives ?

Answer 3

well yes

Answer 4

I think, it is supposed to be zero, i mean in the case of the hamiltonian, my book says that dH/dx is equal to only dV/dx, and the

Answer 5

...and the d/dx ((1/2m)*(hbar/i)*d/x) part is left out.

Answer 6

but i just had hard time to figure out what it does it really mean to differentiate an operator with respect to a parameter (x here), when there is a differential operator (d/dx) of that said parameter inside the operator. btw, i dont know whether you are familiar with the noether theorem or continuous symmetry transformations, but this bit was need to it.

Answer 7

\[\frac{\text d\langle x\rangle}{\text d t}=\left\langle -\frac{\partial V}{\partial x}\right\rangle\] ?

Answer 8

can you right out the exact question, ?

Answer 9

okay, so here it is.
given the following \[\frac{ d }{ dx } \left\{ H f(x) \right\} =? \] , where \[H=\frac{ p^2 }{ 2m } + V(x), p=\frac{ \hbar }{ i } \frac {d} {dx}\] what happens if the the hamiltonian is independent of the variable x.
my book proceeds:
\[\frac{ d }{ dx } \left\{ H f(x) \right\} = \frac {dH }{ dx } f(x)+ H \frac {df }{ dx }\] and as\[\frac{ dH }{ dx } = 0 \rightarrow \frac{ dV }{ dx}=0\], and i was a little uneasy with the la implicationst

Answer 10

i was a little unease with the last row's implication*

Answer 11

\[\frac{ \text d }{ \text dx } \left\{ H f(x) \right\} = \frac {\text dH }{ \text dx } f(x)+ H \frac {\text df }{ \text dx }\]
with \(\frac {\text dH }{ \text dx }=0\)
\[\frac{ \text d }{ \text dx } \left\{ H f(x) \right\} = 0\cdot f(x)+ H \frac {\text df }{ \text dx }\]\[\frac{ \text d }{ \text dx } \left\{ H f(x) \right\} = H \frac {\text df }{ \text dx }\]

Answer 12

im not sure where we are going here?

Answer 13

i just sketched what was the thought process, but the real problem (at least for me) was that
\[if \frac{ dH }{ dx }=0 \it \implies \frac{ dV }{ dx }=0\],
whereas the H operator had the differential operator \[p = \frac{ \hbar }{ i } \frac{d }{ dx }\] in itself

Answer 14

how did you get that?