1. Aug 20, 2013

cooper607

The total energy of a particle in a harmonic oscillator is found to be 5/2
~!. To change the energy,
if i applied the lowering operator 4 times and then the raising operator 1 times successively. What
will be the new total energy?
i want the calculation plz

2. Aug 20, 2013

CompuChip

The energy is the eigenvalue of the Hamiltonian operator. So you have some state $\phi$ satisfying $\hat H \phi = E \phi$ with $E = \tfrac52$.
For brevity let me write the ladder operators as $\hat A$ (so $\hat A = a^\dagger a^\dagger \cdots a$); then you can work out what $\hat H(\hat A \phi)$ is by commuting it to the front, i.e. rewriting it as
$$\hat H \hat A \phi = \hat A \hat H \phi + [\hat H, \hat A] \phi = \tfrac52 \phi' + \Delta E \phi'$$
where $\phi' = A \phi$ is your new state. You can then read off your new energy $E = \tfrac52 + \Delta E$.
The trick, of course, is calculating $[\hat H, \hat A]$.