Density Distribution: Solving for V=0 and V=mω²x²

[Hypnotoad]

Consider a system with one degree of freedom whose density distribution at time t=0 is given by:

$$D(x,p,t=0)=\frac{1}{\pi\sigma^2}exp[-\frac{m\omega^2}{2}x^2-\frac{1}{2m}p^2]$$

where $$x$$ is the generalized coordinate and $$p$$ the conjugate momentum. The Hamiltonian of the system is given by:

$$H=\frac{p^2}{2m}+V(x)$$

a) For $$V=\frac{1}{2}m\omega^2x^2$$ find the density distribution at time t. Choose a convenient area R in phase space and study the way it moves as a function of time.

b) Same question for $$V=0$$.


I don't even know how to start this problem. Any hints on where I can begin?

 

[member 553083]

Thanks.a) For V=\frac{1}{2}m\omega^2x^2, the Hamiltonian is given by H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2. The density distribution at time t is given by: D(x,p,t)=\frac{1}{2\pi h}exp[-\frac{iHt}{\hbar}]\times D(x,p,t=0)where h is Planck's constant. Choosing a convenient area R in phase space, the way it moves as a function of time can be studied by calculating the time evolution of the density distribution in this area: R(t)=\int_{R}\int_{x}D(x,p,t)dx dpb) For V=0, the Hamiltonian is given by H=\frac{p^2}{2m}. The density distribution at time t is given by: D(x,p,t)=\frac{1}{2\pi h}exp[-\frac{iHt}{\hbar}]\times D(x,p,t=0)As before, the way the area R in phase space moves as a function of time can be studied by calculating the time evolution of the density distribution in this area: R(t)=\int_{R}\int_{x}D(x,p,t)dx dp

 

[wais]

To solve this problem, we can use the Liouville's theorem which states that the density distribution in phase space is constant along the trajectory of the system. This means that the density distribution at time t will be the same as at time t=0, with the only difference being the position and momentum of the system.

a) For V=\frac{1}{2}m\omega^2x^2, the Hamiltonian becomes:

H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2

Using the given density distribution, we can write the Liouville equation as:

\frac{\partial D}{\partial t}+\{\{D,H\}\}=0

Substituting the values of D and H, we get:

\frac{\partial D}{\partial t}+\frac{1}{m}\frac{\partial}{\partial p}(pD)+\frac{m\omega^2}{m}\frac{\partial}{\partial x}(xD)=0

Simplifying this equation, we get:

\frac{\partial D}{\partial t}+\frac{\partial}{\partial p}(pD)+\frac{\partial}{\partial x}(xD)=0

This is a first-order partial differential equation which can be solved using the method of characteristics. The characteristic curves for this equation are given by:

\frac{dx}{dt}=x, \frac{dp}{dt}=p

Solving these equations, we get:

x(t)=x(0)e^t, p(t)=p(0)e^t

Substituting these values in the given density distribution, we get:

D(x,p,t)=\frac{1}{\pi\sigma^2}exp[-\frac{m\omega^2}{2}x(0)^2e^{2t}-\frac{1}{2m}p(0)^2e^{2t}]

We can see that the density distribution at time t is the same as the density distribution at time t=0, with the only difference being the scaling factor of e^{2t}. This means that the density distribution spreads out over time as the system evolves.

b) For V=0, the Hamiltonian becomes:

H=\frac{p^2}{2m}

Using the same method as above, we can write the Liouville equation