naele
Nov4-10, 02:30 PM
1. The problem statement, all variables and given/known data
Show that, for all complexe numbers \alpha, a has a unique eigenvector |\alpha\rangle that is nothing else but the coherent state
\psi(x)=\frac{e^{-\frac{i}{2\hbar}\langle X\rangle\langle P\rangle}}{(\pi\ell^2)^{1/4}}e^{-\frac{(X-\langle X\rangle)^2}{2\ell^2}+\frac{i\langle P\rangle X}{\hbar}}
with
\alpha=\langle a \rangle=\frac{1}{\sqrt{2}}\left(\frac{\langle X\rangle}{\ell}+\frac{i\ell\langle P\rangle}{\hbar}\right)
2. Relevant equations
a=\frac{1}{\sqrt{2}}\left(\frac{X}{\ell}+\frac{i\e ll P}{\hbar}\right)
\ell=\sqrt{2}\Delta X
3. The attempt at a solution
Ok, so I think I have a game plan. Since a isn't Hermitian then its eigenvalues can be complex. So I should solve the eigenvalue problem for a|\alpha\rangle=\alpha|\alpha\rangle. But since I already showed that when the equality is valid in the Heisenberg inequality we get a gaussian like \psi(x) so if I can show that the eigenvalue problem admits a differential equation similar to what I already showed then that should be sufficient?
Show that, for all complexe numbers \alpha, a has a unique eigenvector |\alpha\rangle that is nothing else but the coherent state
\psi(x)=\frac{e^{-\frac{i}{2\hbar}\langle X\rangle\langle P\rangle}}{(\pi\ell^2)^{1/4}}e^{-\frac{(X-\langle X\rangle)^2}{2\ell^2}+\frac{i\langle P\rangle X}{\hbar}}
with
\alpha=\langle a \rangle=\frac{1}{\sqrt{2}}\left(\frac{\langle X\rangle}{\ell}+\frac{i\ell\langle P\rangle}{\hbar}\right)
2. Relevant equations
a=\frac{1}{\sqrt{2}}\left(\frac{X}{\ell}+\frac{i\e ll P}{\hbar}\right)
\ell=\sqrt{2}\Delta X
3. The attempt at a solution
Ok, so I think I have a game plan. Since a isn't Hermitian then its eigenvalues can be complex. So I should solve the eigenvalue problem for a|\alpha\rangle=\alpha|\alpha\rangle. But since I already showed that when the equality is valid in the Heisenberg inequality we get a gaussian like \psi(x) so if I can show that the eigenvalue problem admits a differential equation similar to what I already showed then that should be sufficient?