Probability Density of finding a Particle with Generalized Coordinate

Question

The energy of a particle is given by $$E=|p|+|q|$$ where $p$ and $q$ are generalized momentum and generalized coordinate respectively. All the states with $$E\le E_0$$ are equally probable and states with $$E>E_0$$ are inaccessible. What is the probability density of finding the particle at coordinate $q$ with $q >0$?

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The probability density of finding the particle at coordinate $q$ is $${P_{q>0}}={\langle\psi (x)|q|\psi (x)\rangle}={{\langle\psi (x)|E-|p||\psi (x)\rangle}\over{\langle\psi (x)|E|\psi (x)\rangle}}$$ Considering only $$E\le E_0$$ The above expression gives $$P_{q>0}={{E_0-q}\over {E_0}^2}$$

Is the argument correct?

Answer 1

This does not appear to be a quantum mechanical problem, but a classical statistical problem. In particular, "all states with $E\leq E_0$ are equally probable" does not make sense quantum mechanically since most states do not have a well-defined energy.

Your $E$ is bounded from below by $0$. Since we are told that all states with $E\leq E_0$ are equally probable, the probability density that tells you how likely the state with momentum $p$ and position $q$ is given as the uniform probability density $$ f(p,q) = \left(\int_{E\leq E_0} \mathrm{d}p\mathrm{d}q\right)^{-1},$$ which is just the reciprocal of the volume of the region of phase space with $E\leq E_0$. Now, to get the probability of arbitrary $p$ and $q>0$, just compute it like from any other probability density by integrating it over the region of phase space with $q > 0$.