How do we know that the raising operator only raises the state by one step?

[Mayan Fung]

In the simple harmonic oscillator, I was told to use the raising and lowering operator to generate the excited states from the ground state. However, I am just thinking that how do we confirm that the raising operator doesn't miss some states in between.

For example, I can define a raising operator $b^\dagger = (a^\dagger)^2$, so that I can have
\begin{align*}
H|n\rangle &= E_n |n\rangle \\
H(b^\dagger|n\rangle) &= (E_n + 2\hbar \omega)(b^\dagger |n \rangle)
\end{align*}
If I use this $b^\dagger$ as the raising operator, I will miss one state for each raising operation. I have read some online resources and some books. None of them address this issue. I wonder how do we confirm that $a^\dagger$ is the raising operator which gives every eigenstates?

 

[Vanadium 50]

It's defined that way.

Now, a fair question is how one determines that an explicit expression is in fact the raising operator.

 

[Mayan Fung]

Yes, I agree. I should change my question to:
In a quantum harmonic oscillator, why is the operator
$$a^\dagger = N(\hat{x} - \frac{i}{m\omega}\hat{p})$$
, where N is some normalization constant, the raising operator of a harmonic oscillator which only raises the state by one level?

 

[Vanadium 50]

The proof is in Schiff, but it's not very enlightening: it's "plug it in and that's what you get".

 

[vanhees71]

Well you usually make your life as easy as possible and define the creation and annihilation operators such that
$$[\hat{a},\hat{a}^{\dagger}]=1.$$
The Hamiltonian of the harmonic oscillator is
$$\hat{H}=\hbar \omega (\hat{a}^{\dagger} \hat{a}+1/2).$$
So the eigenvalue problem is solved if you solve the eigenvalue problem of the number operator
$$\hat{N}=\hat{a}^{\dagger} \hat{a}.$$
Next we need the commutator
$$[\hat{N},\hat{a}]=[\hat{a}^{\dagger},\hat{a}]\hat{a}=-\hat{a}.$$
Adjoining this equation leads to
$$[\hat{N},\hat{a}^{\dagger}]=+\hat{a}^{\dagger}.$$
Now suppose $|u \rangle$ is an eigenvector of $\hat{N}$ to eigenvalue $n$ then
$$\hat{N} \hat{a} |u \rangle=([\hat{N},\hat{a}]+\hat{a} \hat{N})|u \rangle=(n-1) |u \rangle.$$
Thus $\hat{a}|u \rangle$ is either an eigenvector of $\hat{N}$ with eigenvalue $(n-1)$ or the null vector.

Now there must be an eigenvector of $\hat{N}$, $u_0 \rangle$ such that
$$\hat{a} |u_0 \rangle=0,$$
because otherwise you can get eigenvectors of $\hat{N}$ with arbitrarily small (negative) eigenvalue, which is impossible, because obviously $\hat{N}$ is positive semidefinite, because for all $|\psi \rangle$
$$\langle \psi|\hat{N} \psi \rangle=\langle \hat{a}^{\dagger} \psi|\hat{a}^{\dagger} \psi \rangle \geq 0.$$
In the same way you get
$$\hat{N} \hat{a}^{\dagger} |u \rangle=(n+1) \hat{a}^{\dagger} |u \rangle,$$
which implies that you get all eigenvectors starting from $|u_0 \rangle$ through successive application of $\hat{a}^{\dagger}$ to $|u_0 \rangle$. This can be written as the recursion relation
$$\hat{a}^{\dagger} |u_n \rangle=C_n |u_{n+1} \rangle.$$
Now we want to normalize the eigenvectors. Suppose all the $|u_n \rangle$ are normalized then we get
$$|C_n|^2=\langle \hat{a}^{\dagger}u_n|\hat{a}^{\dagger} u_n \rangle = \langle u_n |\hat{a} \hat{a}^{\dagger} u_n \rangle = \langle u_n |([\hat{a},\hat{a}^{\dagger}]+ \hat{N}) u_n \rangle=(n+1).$$
From this we get (up to an arbitrary phase factor) $C_n=\sqrt{n+1}$. So we get the recursion relation
$$|u_{\n+1} \rangle = \frac{1}{\sqrt{n+1}} \hat{a}^{\dagger} |u_n \rangle$$
which is solved by
$$|u_n \rangle=\frac{1}{\sqrt{n!}} (\hat{a}^{\dagger})^n |u_0 \rangle.$$
The eigenvalues are $n \in \mathbb{N}_0$.

 

[hilbert2]

Another way to check this is to solve the harmonic oscillator Schrodinger eqn and show that the solution isn't normalizable between the eigenvalues obtained with raising and lowering operators.

 

[PeroK]

In the simple harmonic oscillator, I was told to use the raising and lowering operator to generate the excited states from the ground state. However, I am just thinking that how do we confirm that the raising operator doesn't miss some states in between.

For example, I can define a raising operator $b^\dagger = (a^\dagger)^2$, so that I can have
\begin{align*}
H|n\rangle &= E_n |n\rangle \\
H(b^\dagger|n\rangle) &= (E_n + 2\hbar \omega)(b^\dagger |n \rangle)
\end{align*}
If I use this $b^\dagger$ as the raising operator, I will miss one state for each raising operation. I have read some online resources and some books. None of them address this issue. I wonder how do we confirm that $a^\dagger$ is the raising operator which gives every eigenstates?

The uniqueness of eigenvalues comes from the factorisation of the Hamiltonian: $$ \hat{H} = \hbar\omega\left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right)$$
Any ground state must have eigenvalue $\frac{\hbar\omega}{2}$. If there were any intermediate eigenvalues, you could repeatedly apply the lowering operator and reach a contradiction.

There cannot, therefore, be intermediate eigenvalues. The above factorisation is also what identifies $a^{\dagger}$ as more fundamental than your $b^{\dagger}$.

 

[andresB]

Another way to check this is to solve the harmonic oscillator Schrodinger eqn and show that the solution isn't normalizable between the eigenvalues obtained with raising and lowering operators.

what solutions are those?

 

[hilbert2]

what solutions are those?

If you set the eigenvalue $E$ in equation

$\displaystyle -\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2} + \frac{1}{2}m\omega^2 x^2 \psi (x) = E\psi (x)$

to a value that is not one of the eigenvalues

$\displaystyle E_n = \hbar\omega \left(n + \frac{1}{2}\right)$,

the solution $\psi (x)$ becomes something that approaches infinity when $x\rightarrow\pm\infty$, and therefore can't be a physically acceptable solution.