For a quantum mechanical harmonic oscillator of constant real mass m and frequency ω, define the following operators on the Hilbert space:
$ h = a^{+}a $
$e = [\sqrt{-1 + a^{+}a}]a^{+} $
$ f = -a\sqrt{-1 + a^{+}a} $
Derive the action of the three operators defined above on a generic state |n>
I dont understand how to obtain the following answers
$ h|n> = n|n> $
e|n> = $\sqrt{(n(n+1)} | n+1 >$
f| n > = $ -\sqrt{n(n-1)} | n-1> $
Can someone please explain how to obtain these answers? ,cause am looking through my notes and seem pretty confused...
Beginning with the premise that
$$a\left|n\right>=\sqrt{n}\left|n-1\right>\\ a^\dagger\left|n\right>=\sqrt{n+1}\left|n+1\right>$$
we have that the first relation you have is
$$h\left|n\right>=a^\dagger\,a\left|n\right>=\sqrt{n}a\left|n-1\right>=\sqrt{n}\sqrt{(n-1)+1}\left|n-1+1\right>=n\left|n\right>$$
For the second case, you can proceed in the same way
$$ e\left|n\right>=\sqrt{a^\dagger a-1} a^\dagger\left|n\right>=\sqrt{a^\dagger a-1} \sqrt{n+1}\left|n+1\right>$$
Now there comes a kind of tricky part, the square-rooted operators. Luckily (otherwise if this were not the case I'm not sure how to actually proceed), $\left|n+1\right>$ is an eigenvalue of both operators inside the square root, so we have
$$ e\left|n\right>= \sqrt{n+1}\sqrt{a^\dagger a-1}\left|n+1\right>= \sqrt{n+1}\sqrt{n+1-1}\left|n+1\right>=\sqrt{n(n+1)}\left|n+1\right>$$
Using the same reasoning as above you get the final one, namely
$$ f\left|n\right>=-a\sqrt{a^\dagger a-1}\left|n\right>=-\sqrt{n-1}a\left|n\right>=-\sqrt{n(n+1)}\left|n-1\right>$$