Raising and lowering operators on a simple harmonic oscillator

[qualal]

Homework Statement

Hi, I'm currently studying for a quantum mechanics exam but I am stuck on a line in my notes:

Ha\left|\Psi\right\rangle =\hbar\omega\left(a^{t}a a + \frac{a}{2}\right)\left|\Psi\right\rangleHa\left|\Psi\right\rangle =\hbar\omega\left(\left(a a^{t} - 1\right)a + \frac{a}{2}\right)\left|\Psi\right\rangle
Where H is the harmonic oscillator hamiltonian

H = \frac{P^{2}}{2m} + \frac{1}{2}m\omega^{2} x^{2}and a, a^{t} are the lowering and raising operators respectively.

a^{t} = \frac{1}{\sqrt{2\hbar m \omega}}\left(m \omega x - i p\right)

a= \frac{1}{\sqrt{2\hbar m \omega}}\left(m \omega x + i p\right)

The part I don't understand is how they got from a^{t}a a to \left(a a^{t} - 1\right)a

I'm not afraid of doing a bit of calculating but I really need a hint as to what method I should use.

 

[kuruman]

Are you sure your first equation is correct?

The Hamiltonian has one of two forms

H=\hbar\omega(aa^t-1/2)

H=\hbar\omega(a^ta+1/2)

Whichever form you use, if you tack on an extra a^t on the right, you don't end up with what you have. Check your notes farther up.

 

[qualal]

Apologies I mistyped when I wrote

Ha^{t}\left|\Psi\right\rangle

I meant to say Ha\left|\Psi\right\rangle

does this make more sense now?

It now looks just like your second hamiltonian multiplied by "a" to me.

 

[kuruman]

Now that you got the first line right, you can get from the first to the second line using the commutator

[a, a^t]=1

 

[qualal]

thanks so much! this has been bothering me all day!

I used \left[ a,a^{t}\right]= a a^{t}-a^{t} a =1

and re-arranged to get

a a^{t}-1=a^{t} a

which I then sub straight into the first line :-)