1. Aug 19, 2014

### mcheung4

I am referring to the section The Harmonic Oscillator in Griffiths's introductino to quantum mechanics (the older edition with the black cover). I understand how it all works, however there is a part that I am not sure about. How do we know when we apply a- or a+ (the ladder operators) to a state (to get the immediately lower or higher stationary states, as Griffiths stated) we are not skipping some other possible stationary states? how do we know this method finds all stationary states? Thanks!

2. Aug 19, 2014

### stevendaryl

Staff Emeritus
The argument is this: Let $|\psi\rangle$ be a state with energy $E$. That means that $H|\psi \rangle = E |\psi\rangle$. Now, if $\hat{a}$ is the lowering operator, then we can prove that:

$H (\hat{a}|\psi\rangle) = (E - \hbar \omega) (\hat{a}|\psi\rangle)$

Since $\hat{a}$ lowers the energy every time you apply it, if you keep applying it then eventually you run into one of two possibilities:

1. For some value of $n$, $\hat{a}^n |\psi\rangle$ has a negative energy (namely, $E - n \hbar \omega$)
2. For some value of $n$, $\hat{a}^n |\psi\rangle = 0$

Since there are no normalizable states with negative energy (you can prove that), the first case is impossible. So for any normalizable state $\psi$, there must be a number $n>0$ such that $\hat{a}^n |\psi\rangle = 0$. The only such state is $|n-1\rangle$ for positive integer $n$.

3. Aug 19, 2014

### stevendaryl

Staff Emeritus
I don't know whether this is obvious, or not, but you can perfectly well solve Schrodinger's equation for a value of energy that is not equal to an eigenvalue. But if you do that, you will get an unnormalizable wave function. For example,

$\psi(x) = e^{+\frac{m \omega}{2 \hbar} x^2}$

solves the Schrodinger equation for the harmonic oscillator,
$\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi + \frac{1}{2} m \omega^2 \psi = E \psi$

with $E = -\frac{1}{2} \hbar \omega$.

On the other hand, that wave function is not normalizable, since it doesn't go to zero for large values of $x$. So if you had an intermediate energy level that is not equal to $(n+\frac{1}{2})\hbar \omega$, the corresponding wave function would not be normalizable.

4. Aug 19, 2014

### tom.stoer

This does not anwer the original question completely.

Let's assume we have a supersymmetric harmonic oscillator, but we forget about all fermionic states. By the method described above, we can construct all bosonic states. What tells you that you have missed the fermionic states?

In our case: what tells you that you have not missed any bosonic state?

I think one has to fix E and show that only for one single value E = 1/2 the ladder operator generates normalizable eigenstates (and that for all other E' the states are either not normalizable or are not eigenstates / eigenfunctions)

Last edited: Aug 20, 2014
5. Aug 20, 2014

### mcheung4

OK I get the part that we're only seeking normalizable states, and that we cannot have a state corresponding to some negative energy. thanks for all that.

But the bit I really dont understand is that how are we so sure that the ladder method generates all the possible states? why energy can only move up and down according to (n+1/2)ℏω? I know we always get quantised energy levels for bounded systems, but in this case it seems to me that we are using some operators to solve the system and found out (rather than deriving) that the energy can move in such way, while it doesnt say the energy cannot move in other way.

6. Aug 20, 2014

### tom.stoer

1) First based on the commutation relations you can show that

$$N|n\rangle = n|n\rangle$$
$$Na^\dagger|n\rangle = (n+1)a^\dagger|n\rangle$$
$$Na|n\rangle = (n-1)a|n\rangle$$

So if |n> is an eigenstate then the operators act as ladder operators on |n>.

2) We immediately see that if n is an integer then the series

$$|n\rangle, a|n\rangle \sim |n-1\rangle, a^2|n\rangle \sim |n-2\rangle,\ldots$$

eventually terminates with

$$\ldots, |0\rangle, 0$$

3) So we see that if the series does not terminate then we have to start with a non-integer n; and we see that if we start with a non-integer n then the series does not terminate.

4) Now we calculate the matrix elements of the number operator N. By definition of N we can re-write this as

$$\langle n|N|n\rangle = \langle n|(a^\dagger a)|n\rangle = (\langle n|a^\dagger)(a|n\rangle)$$

The last term is nothing else but the norm of the state a|n>, so we can conclude that it is positive semidefinite (the norm of every non-zero state in a Hilbert space is positive).

5) A state with n<0 would violate this positive-norm-condition. And by the above reasoning (3) we know that starting with non-integer n we could construct states with negative, non-integer n. But these states cannot be Hilbert space states with positive norm and must be excluded.

7. Aug 20, 2014

### stevendaryl

Staff Emeritus
Indirectly, it does. Suppose that there is some normalizable state $|\psi\rangle$ with energy $E$ greater than $0$ but less than $\hbar \omega$. Then consider the state $|\psi'\rangle$ defined by:

$|\psi'\rangle = \hat{a}|\psi\rangle$

Then:

$\hat{H} |\psi'\rangle = (E - \hbar \omega) |\psi'\rangle$

Note that $E - \hbar \omega$ is less than zero. So

There are only two possibilities:
1. $|\psi'\rangle$ is a normalizable state with energy less than zero.
2. $|\psi'\rangle = 0$

The first possibility is not really possible, since no normalizable state has negative energy. So it must be that

$|\psi'\rangle = 0$

So we know:
If $|\psi\rangle$ is a normalizable state with energy less than $|\hbar \omega$, then $\hat{a}|\psi\rangle = 0$

There is only one state satisfying $\hat{a}|\psi\rangle = 0$, namely the one with energy $1/2 \hbar \omega$. So we know that there is only one normalizable state with energy $E$ in the range $0 < E < \hbar \omega$.

We can similarly prove that there is only one normalizable state with energy $E$ in the range $\hbar \omega < E < 2 \hbar \omega$. In general, we can prove that there is only one state with energy $E$ in the range $n \hbar \omega < E < (n+1)\hbar \omega$