Question:
In lecture 6, part III, we showed that the two-particle wave function for particles with center of mass position $\vec{R}$, relative separation $\vec{r}$, and central potential $V(\vec{r})$, in a state of energy $E$, can be written as $\Psi(\vec R, \vec r)=e^{\frac{i}{\hbar}\vec{P}\cdot \vec{R}} \psi(\vec r)$ where $\psi(\vec{r})$ obeys the single particle Schrödinger equation:
$\left(-\frac{\hbar^2 \nabla^2}{2\mu} + V(\vec{r})\right)\psi(\vec r) = E' \psi(\vec r)$
with $\mu = \frac{m_1 m_2}{m_1+m_2}$ being the reduced mass, and $E'=E-\frac{\vec P^2}{2(m_1+m_2)}$.
Suppose we have two identical, free ($V(\vec r) = 0$), fermions of spin $\frac{1}{2}$ (just for concreteness) and mass $m$. Which of the following is a valid two-particle wavefunction $\Psi(\vec R, \vec r)$ for a state of energy $E$, where $\vec p$ is a vector of magnitude $|\vec p| = \sqrt{m E'}$?
Assume that the spin part of the wavefunction is symmetric or more simply that the scalar wave function $\psi$ determines the symmetry properties under exchange.
Solution:
It appears the problem is quite complicated but it really isn't. The method in lecture already tell us the form of the solution, and the fact that the particles are fermions means they are bounded by Pauli exclusion principle that the probability of finding them with together at the same place is 0, so the only plausible solution is $\Psi(\vec R, \vec r) = \exp\left(\frac{i}{\hbar} \vec P \cdot \vec R\right) \sin(\frac{1}{\hbar} \vec p \cdot \vec r)$
In lecture 6, part III, we showed that the two-particle wave function for particles with center of mass position $\vec{R}$, relative separation $\vec{r}$, and central potential $V(\vec{r})$, in a state of energy $E$, can be written as $\Psi(\vec R, \vec r)=e^{\frac{i}{\hbar}\vec{P}\cdot \vec{R}} \psi(\vec r)$ where $\psi(\vec{r})$ obeys the single particle Schrödinger equation:
$\left(-\frac{\hbar^2 \nabla^2}{2\mu} + V(\vec{r})\right)\psi(\vec r) = E' \psi(\vec r)$
with $\mu = \frac{m_1 m_2}{m_1+m_2}$ being the reduced mass, and $E'=E-\frac{\vec P^2}{2(m_1+m_2)}$.
Suppose we have two identical, free ($V(\vec r) = 0$), fermions of spin $\frac{1}{2}$ (just for concreteness) and mass $m$. Which of the following is a valid two-particle wavefunction $\Psi(\vec R, \vec r)$ for a state of energy $E$, where $\vec p$ is a vector of magnitude $|\vec p| = \sqrt{m E'}$?
Assume that the spin part of the wavefunction is symmetric or more simply that the scalar wave function $\psi$ determines the symmetry properties under exchange.
Solution:
It appears the problem is quite complicated but it really isn't. The method in lecture already tell us the form of the solution, and the fact that the particles are fermions means they are bounded by Pauli exclusion principle that the probability of finding them with together at the same place is 0, so the only plausible solution is $\Psi(\vec R, \vec r) = \exp\left(\frac{i}{\hbar} \vec P \cdot \vec R\right) \sin(\frac{1}{\hbar} \vec p \cdot \vec r)$
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