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M07Q1 is a short name for the first problem in the Quantum Mechanics section of the May 2007 Princeton University Prelims. The problem statement can be found in the problems list. Here is the solution.

(a)

$\frac{d}{dt} F =-i [F,H] +\left(\frac{\partial{F}}{\partial{t}}\right)_{classical}$

In our case, the last term is 0, and because we are working with eigenstates:

$[F,H]=\psi_n|FH|\psi_n-\psi_n|HF|\psi_n=\psi_n|F|\psi_n E_n-E_n\psi_n|F|\psi_n=0$

(b)

Note that the result $\lang\psi_n|[F,H]|\psi_n\rang=0$ holds for any operator F, and so $\lang\psi_n|[\vec{r}\cdot\vec{p},H]|\psi_n\rang=0$. At the same time:

$\lang\psi_n|[\vec{r}\cdot\vec{p},H]|\psi_n\rang=\lang\psi_n|[xp_x,H]|\psi_n\rang+\lang\psi_n|[yp_y,H]|\psi_n\rang+\lang\psi_n|[zp_z,H]|\psi_n\rang$

$\lang[xp_x,H]\rang=\lang x[p_x,H]+[x,H]p_x\rang$

$[p_x,H]=-i\frac{\partial{V}}{\partial{x}}$

$[x,H]=\frac{ip_x}{m}$

$\lang x(-i)\frac{\partial{V}}{\partial{x}}+\frac{ip_x}{m}p\rang=0$

$\lang\frac{p_x^2}{m}\rang=\lang x\frac{\partial{V}}{\partial{x}}\rang$

which naturally generalizes to the vector case.

(c)

Since the force is independent of distance, so the potential must be something like:

$V=-\vec{r}\cdot\vec{F}$

$\vec{r}\cdot\nabla V=-\vec{r}\cdot\vec{F}=V$

$=\frac{1}{2}$

At rest, all the energy of the proton must be internal:

$E=+=\frac{3}{2}$

But also at rest we know that the energy of the proton is equal to its mass. Therefore:

$=\frac{2}{3}E=\frac{2}{3}m_P$