knightpraetor
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So i was working on a homework problem to calculate energy, and i can calculate it directly. But i was thinking it should be easier to use the number operator on the wave function and find out what it's energy should be. The function involved is the initial wave state of a quantum harmonic oscillator. Anyways, it's getting late but i thought i'd ask to see if I'm thinking the right way, and try to solve it more tomorrow. so
ANyways, Asub+ * Asub- *Psi(x) = n * Psi(x)
where Psi is the wave function.
by the definitions of Asub+ and Asub-, i will end up getting
(2m *(Hamiltonian operator ) -m*omega* hbar) / (2m*omega*hbar)) * Psi = n *Psi
i then insert the hamiltonian operator, which is
- hbar^2/ 2m * d^2/dx^2 + V(x), where V(x) for a harmonic oscillator is
kx^2/2
if i do this and multiply by the wave function should it come out to N * the wave function so that i can learn what energy state the function is in?
the wave function at issue is
Psi (x,0) = A (1-2 *sqrt (mw/hbar)*x)^2 * e^ (-mwx^2/2hbar)
from griffith 2.41
i'm pretty sure i can solve it just by calculating the integral manually, but that seems inefficient..i thought since that in my notes it gave me a "number operator" and I have the energy for any given N, then that should be faster
ANyways, Asub+ * Asub- *Psi(x) = n * Psi(x)
where Psi is the wave function.
by the definitions of Asub+ and Asub-, i will end up getting
(2m *(Hamiltonian operator ) -m*omega* hbar) / (2m*omega*hbar)) * Psi = n *Psi
i then insert the hamiltonian operator, which is
- hbar^2/ 2m * d^2/dx^2 + V(x), where V(x) for a harmonic oscillator is
kx^2/2
if i do this and multiply by the wave function should it come out to N * the wave function so that i can learn what energy state the function is in?
the wave function at issue is
Psi (x,0) = A (1-2 *sqrt (mw/hbar)*x)^2 * e^ (-mwx^2/2hbar)
from griffith 2.41
i'm pretty sure i can solve it just by calculating the integral manually, but that seems inefficient..i thought since that in my notes it gave me a "number operator" and I have the energy for any given N, then that should be faster