MathematicalPhysicist
Aug30-09, 07:44 AM
1. The problem statement, all variables and given/known data
With the wave function \psi(r) of a particle, one forms the function:
D(R,P)=\frac{\int exp(-\frac{i}{\hbar} P\cdot r)\psi^*(R-r/2)\psi (R+r/2) dr}{(2\pi \hbar)^3}, which is the density in phase of a classical statistical mixture associated with this wave function, Show that:
1. \int D(R,P)dP=|\psi (R)|^2 \int D(R,P)dR=|\psi (P)|^2.
2. if the particle is free, the evolution in time of the mixture is strictly that of a statistical mixture of free classical particles of the same mass;
3. find the spreading law of a free wave packet.
2. Relevant equations
I think one of them is dirac delta function:
\int \frac{exp(-\frac{i}{\hbar} P\cdot r)}{(2\pi \hbar)^3}dr=\delta(r) \ \int \frac{exp(-\frac{i}{\hbar} P\cdot r)}{(2\pi \hbar)^3}dP=\delta(P)
3. The attempt at a solution
1. So far by using the above equality I get that:
\int D(R,P)dP=\int \frac{\int exp(-\frac{i}{\hbar} P\cdot r)\psi^*(R-r/2)\psi (R+r/2) dr}{(2\pi \hbar)^3} dP=\int \psi^*(R-r/2)\psi (R+r/2) dr, I don't see how this becomes the amplitude of the wave function in the R space squared?
2. Don't know what I need to show here.
3. The same as with 2.
Any hints?
Not answers!
With the wave function \psi(r) of a particle, one forms the function:
D(R,P)=\frac{\int exp(-\frac{i}{\hbar} P\cdot r)\psi^*(R-r/2)\psi (R+r/2) dr}{(2\pi \hbar)^3}, which is the density in phase of a classical statistical mixture associated with this wave function, Show that:
1. \int D(R,P)dP=|\psi (R)|^2 \int D(R,P)dR=|\psi (P)|^2.
2. if the particle is free, the evolution in time of the mixture is strictly that of a statistical mixture of free classical particles of the same mass;
3. find the spreading law of a free wave packet.
2. Relevant equations
I think one of them is dirac delta function:
\int \frac{exp(-\frac{i}{\hbar} P\cdot r)}{(2\pi \hbar)^3}dr=\delta(r) \ \int \frac{exp(-\frac{i}{\hbar} P\cdot r)}{(2\pi \hbar)^3}dP=\delta(P)
3. The attempt at a solution
1. So far by using the above equality I get that:
\int D(R,P)dP=\int \frac{\int exp(-\frac{i}{\hbar} P\cdot r)\psi^*(R-r/2)\psi (R+r/2) dr}{(2\pi \hbar)^3} dP=\int \psi^*(R-r/2)\psi (R+r/2) dr, I don't see how this becomes the amplitude of the wave function in the R space squared?
2. Don't know what I need to show here.
3. The same as with 2.
Any hints?
Not answers!