- #1
snoopies622
- 852
- 29
Looking through this matrix approach to the quantum harmonic oscillator,
http://blogs.physics.unsw.edu.au/jcb/wp-content/uploads/2011/08/Oscillator.pdf
especially the equations
<br /> <br /> m \hat{ \ddot { x } } = \hat { \dot {p} } = \frac {i}{\hbar} [ \hat {H} , \hat {p} ]<br /> <br />
I'm getting the impression that there's no difference between the time derivative of an operator (of an observable) and the operator representing the time derivative of that observable. That is, between <br /> <br /> \dot { \hat {x } }<br /> <br />and<br /> <br /> \hat { \dot {x} } Is this true? If so it would mean that, once I knew the position operator matrix for a one-particle system, finding the momentum operator matrix would be simply taking the time derivative of the position matrix (replacing every element of the matrix with its time derivative) and multiplying the whole thing by the mass.
Something tells me that that's too easy and this is just a notation problem in this particular essay, but then I'm new to matrix mechanics.
http://blogs.physics.unsw.edu.au/jcb/wp-content/uploads/2011/08/Oscillator.pdf
especially the equations
<br /> <br /> m \hat{ \ddot { x } } = \hat { \dot {p} } = \frac {i}{\hbar} [ \hat {H} , \hat {p} ]<br /> <br />
I'm getting the impression that there's no difference between the time derivative of an operator (of an observable) and the operator representing the time derivative of that observable. That is, between <br /> <br /> \dot { \hat {x } }<br /> <br />and<br /> <br /> \hat { \dot {x} } Is this true? If so it would mean that, once I knew the position operator matrix for a one-particle system, finding the momentum operator matrix would be simply taking the time derivative of the position matrix (replacing every element of the matrix with its time derivative) and multiplying the whole thing by the mass.
Something tells me that that's too easy and this is just a notation problem in this particular essay, but then I'm new to matrix mechanics.
