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We want to find the spectrum of the Harmonic Oscillator Hamiltonian: $$H=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega ^2 \hat{x}^2$$ From what I have seen in many books the procedure is as follows: We can define a destruction operator $a$ and a creation operator $a^\dagger$, such as: $$H=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega ^2 \hat{x}^2=\hbar \omega \left(a^\dagger a +\frac{1}{2}\right)$$ We then call: $$N := a^\dagger a$$ Now: if we find the spectrum of $N$ we have automatically found the spectrum of $H$, obviously. So let's start digging for the spectrum of $N$: we can easily prove that: $$[a^\dagger , a]=-1 \ \ ; \ \ [N , a]=-a \ \ ; \ \ [N , a^\dagger]=a^\dagger$$ And now, using these commutators, we can show the following: calling an eigenstate of $N$: $|n\rangle$ (so: $|n\rangle \ | \ N|n\rangle = \lambda _n |n\rangle$) $$Na^\dagger|n\rangle=(a^\dagger N + [N , a^\dagger])|n\rangle=(\lambda _n +1)a^\dagger | n \rangle$$ $$Na|n\rangle=(a N + [N , a])|n\rangle=(\lambda _n -1)a | n \rangle$$ So now, with this, we have managed to prove two things:

  • The operators $a^\dagger , a$ by acting on an eigenstate of $N$ allow us to find another eigenstate of $N$ (Fact 1)
  • The newly found eigenstate has eigenvalue one unit higher or lower, with respect to $\lambda _n$, depending on which of the two operators we chose to use (Fact 2)

Wonderful. Another thing that we can prove is that it must exist an eigenstate $|0\rangle$ of $N$ such that: $$a|0\rangle=0 \ \ \ (Fact \ 3)$$ this can be easily proven: the eigenvalues of the Harmonic Oscillator Hamiltonian must be positive (we omit the proof of this), but the operator $a$ always lowers the value of $\lambda$ by one, so the only option is that, at some point; applying $a$ on an eigenstate of $N$ returns the null vector.

Now here is my problem: In all the sources that I have found the author then goes on to state that fact 1, 2 and 3 allow us to say that the complete collection of all the eigenstates of the Harmonic Oscillator Hamiltonian is

$$|n\rangle = K_n (a^\dagger)^n|0\rangle \ \ \ \ \ \ n \in \mathbb{N} \tag{1}$$ with $K_n$ some number. So essentially all the eigenstates can be found by applying the creation operator $a^\dagger$ to the lowest state $|0\rangle$. My problem with this is: surely all the vectors that we can obtain with (1) are eigenstates of $N$ but there could be other eigenstates hiding somewhere! We have never managed to prove, in all this, that all the eigenstates of $H$ are findable by the creation and destruction operators, there could be other eigenstates that we cannot reach by applying $a,a^\dagger$.

How can we show that this is not the case? How can we show that all the eigenstates are in (1)? Is there something that I am missing?

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You’ve already said that (1) applying the lowering operator to any eigenstate of $N$ generates another eigenstate of $N$ whose eigenvalue is one unit lower, and that (2) the spectrum of $N$ is bounded below.

Therefore, if you repeatedly apply the lowering operator to any eigenvector of $N$, then you’re eventually going to hit the ground state; otherwise, the eigenvalues would eventually be arbitrarily large and negative, contradicting (2).

It follows that you can reconstruct the original eigenvector by climbing back up the ladder by successive applications of $a^\dagger$.

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  • $\begingroup$ Is there a quick way to see that we can't have half levels which are "skipped" by raising and lowering operators, yet are still eigenvectors of the Hamiltonian? $\endgroup$ – John Jun 1 at 21:28
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    $\begingroup$ @John Well, as per the above argument, repeated applications of $a$ would lead to the sequence of eigenvalues $\ldots,\frac{3}{2},\frac{1}{2},-\frac{1}{2},-\frac{3}{2},\ldots$ which never terminates. The only way to have both (1) and (2) hold is if, for any eigenvector, successive applications of $a$ lead eventually to a state $|0\rangle$ such that $a|0\rangle = 0$, so the sequence terminates. $\endgroup$ – J. Murray Jun 1 at 21:34
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Let's suppose for the moment that there is a state $|\psi\rangle$ which is not in the span of the harmonic oscillator states, as OP worries. That means no linear combination of the $|n\rangle$ states will produce $|\psi\rangle$.

However, the Hilbert space is, in particular, a vector space. And that means there exists an operator which maps $|0\rangle$ to $|\psi\rangle$, call this operator $A$: $$ A|0\rangle = |\psi\rangle. $$ I will reiterate there is no loss in generality doing this as it's just a property of vector spaces.

So let's think about how we would build this operator $A$ out of the operators we know about. First of all, any operator which depends on the $x$ and $p$ operators can be rewritten in terms of the $a$ and $a^\dagger$ operators...but any operator build out of the $a,a^\dagger$ acting on $|0\rangle$ will just give us some linear combination of the $|n\rangle$ states by nature of the construction of those states.

Therefore, the only way we could have a state which lies outside the number operator tower is if we had an operator other than the $a$ and $a^\dagger$. This kind of problem starts with the assumption that these are the only operators we have.

This entire line of reasoning can be restated in terms of the operators themselves. This is essentially the statement that $N$ is a complete set of commuting, independent observables and that $a,a^\dagger$ are the generators of the operator algebra. It's sort of like them being a basis of the operator algebra (but not just linear combinations...we can obviously build products of them too).

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The argument, as you have presented it, shows that the eigenstates form ladders and at the lower end of each ladder there is a state with eigenenergy $\hbar\omega / 2$. You are right, we have not shown yet that there is only one such ladder; in fact, that is generally not the case. Consider, for example, a spin-1/2 particle and assume that the spin degree of freedom does not contribute to the Hamiltonian. Then, the eigenstates are $|n, \uparrow\rangle$ and $|n, \downarrow\rangle$, that is, each energy level is 2-fold degenerate (there are 2 ladders).

The assumption that there is no additional degree of freedom is usually more or less implicit. The argument normally continues by saying: now we determine the ground state $|0\rangle$ by solving $a|\psi\rangle = 0$, in position representation $$ \bigl( x + \frac{\hbar}{m\omega} \partial_x \bigr) \psi(x) = 0 . $$ As you see, we here assumed that the particle is fully described by the wave function $\psi(x)$ and there are no other degrees of freedom. Since we find only one solution to this differential equation, the calculation shows that the ground state is not degenerate and, thus, that there is only one ladder.

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