# Ladder Operators acting upon N Ket

1. Jan 9, 2010

### 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.

Last edited: Jan 9, 2010
2. Jan 9, 2010

### Godmar02

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

3. Jan 9, 2010

### 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.

4. Jan 9, 2010

### 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 $$\ a_+ \left| \psi _n \right\rangle$$ and set it to be equal to 1(normalising it). If that is correct what do I use for $$\left| \psi _n \right\rangle$$ ?

Sorry if I have misunderstood you.

5. Jan 9, 2010

### vela

Staff Emeritus
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 $$|\psi_n>$$ are assumed to be normalized so that $$<\psi_n|\psi_n>=1$$. With a little fiddling, you can calculate what $$a|\psi_n>$$ and $$a^\dagger|\psi_n>$$ are.

6. Jan 10, 2010

### Godmar02

yes I have. I think I understand

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

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

I can show that
$$aa^\dagger|\psi_n>=n|\psi_n>$$
and
$$a^\dagger a|\psi_n>=(n+1)|\psi_n>$$

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?

7. Jan 10, 2010

### vela

Staff Emeritus
Try calculate the norms of $a\left|\psi_n\right>$ and $a^\dagger\left|\psi_n\right>$.

8. Jan 10, 2010

### Godmar02

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