Ladder Operators acting upon N Ket

[Godmar02]

I can't seem to find information regarding this anywhere.

I understand why when the ladder operators act upon an energy eigenstate of energy E it produces another eigenstate of energy E \mp\hbar \omega. What I don't understand is why the following is true:

\ a \left| \psi _n \right\rangle &= \sqrt{n} \left| \psi _{n-1} \right\rangle
\ a^{\dagger} \left| \psi _n \right\rangle &= \sqrt{n+1} \left| \psi _{n+1} \right\rangle

I don't really even know what \left| \psi _n \right\rangle represents, though I think it is something to do with the state of a system. How can you derive the above property?

I am a bit of a beginner to SHO in quantum theory.

 

[Godmar02]

Any help would be greatly appreciated, I am sure I am missing something simple. Thanks

 

[Feldoh]

\psi_n is a stationary state of the harmonic oscillator with energy E = (n+1/2)\hbar \omega

Ladder operators a_{+}, a_{-} raise the energy of the state from n to n+1 or lower n to n-1.

Since the energy changes the state must change as well since the state is characterized by its energy. If you raise the energy by n+1 you raise the state to \psi_{n+1}

So you can sort of see that if a ladder operator acts on a stationary wave function in state n it will raise the state to n+1 by some proportionality constant:

\ a_+ \left| \psi _n \right\rangle \alpha \left| \psi _{n+1} \right\rangle

The two are proportional. Mathematically you can show the proportionality constants are \sqrt{n+1} and \sqrt{n} respectively.

 

[Godmar02]

So would I be right in saying that this works because any constant multiple of an eigenstate must also be an eigenstate.
i.e. if the constant of proportionality is beta then this works because

H (a^\dagger \left| \psi _n \right\rangle) = (E + \hbar\omega) (a^\dagger \left| \psi _n\right\rangle)
\beta \ H \left| \psi _{n+1} \right\rangle = \beta (E + \hbar\omega) \left| \psi _{n+1} \right\rangle

To find beta would I then have to take the modulus squared of <br /> \ a_+ \left| \psi _n \right\rangle<br /> and set it to be equal to 1(normalising it). If that is correct what do I use for <br /> \left| \psi _n \right\rangle<br /> ?

Sorry if I have misunderstood you.

 

[vela]

Have you shown that the energies of the states are given by E_n = (n+1/2)\hbar\omega and that the Hamitonian can be written as H=\hbar\omega(a^\dagger a+1/2)?

The eigenstates |\psi_n&gt; are assumed to be normalized so that &lt;\psi_n|\psi_n&gt;=1. With a little fiddling, you can calculate what a|\psi_n&gt; and a^\dagger|\psi_n&gt; are.

 

[Godmar02]

yes I have. I think I understand

If I write <br /> H=\hbar\omega(a^\dagger a\ +1/2) \<br /> or \ H= \hbar\omega(aa^\dagger\ -1/2)<br />

And then act upon an eigenvector <br /> |\psi_n&gt;<br />, which returns <br /> E_n = (n+1/2)\hbar\omega <br /> |\psi_n&gt;

I can show that
<br /> aa^\dagger|\psi_n&gt;=n|\psi_n&gt;
and
<br /> a^\dagger a|\psi_n&gt;=(n+1)|\psi_n&gt;


I know that the identities solve this but I cannot prove it from here, since the constants of proportionality are functions of n. What can I do?

 

[vela]

Try calculate the norms of a\left|\psi_n\right&gt; and a^\dagger\left|\psi_n\right&gt;.

 

[Godmar02]

ahhhhhh you legend. Makes so much sense now, sorry for being a bit dense. Thanks so much!