# Proof that energy states of a harmonic oscillator given by ladder operator include all states

In quantum mechanics, while studying the harmonic oscillator, I learnt about ladder operators. And I realised that if you are able to find or determine any energy state of the quantum harmonic oscillator then, using the ladder operators, you can determine the other energy states as well. However in none of the texts I read I found the following fact:

The energy states determined by the above procedure, that is, using the ladder operator are the only possible energy states of the harmonic oscillator. There exists no energy state that is not given by the ladder operator.

So this is my question:

Are the energy states determined by the ladder operator in case of a harmonic oscillator, the only possible energy states? Is any other energy state possible? And what is the proof?

It is well known that the above-mentioned states are the only possible energy states but I want a rigorous proof that no other state is possible. However the analytical procedure mentioned in different books show an approximate solution of the Schrodinger equation which can be considered to be rigorous in the sense that it solves the equation to derive the solutions although it considers certain approximations.

I have searched for this answer in books related to quantum mechanics written by DJ Griffiths, Gasiorowicz, Dirac and online resources like OCW, University of Columbia's courses, caltech.edu, but I couldn't find a proper answer.

• I'm not sure what you mean when you say that the analytical procedure is an approximate solution. Are you referring to the asymptotic analysis, where you write your wavefunction as $\psi(x)\sim e^{-ax^2}$ at large $x$ and then define $\psi(x)=f(x)e^{-ax^2}$? This isn't an approximation, it's just a way of redefining your equation in terms of a nicer function $f(x)$. – Bob Knighton Jun 27 '17 at 17:26
• @BobKnighton yes I am talking about the asymptotic analysis and as much as I could learn from griffiths, it looks a bit like an approximation. It's not exactly what you call mathematically rigorous. – SchrodingersCat Jun 27 '17 at 17:28
• It's very much not an approximation. The solution given is exact, and the way to get there is rigorous. The part that looks like an approximation (the asymptotic analysis) is simply a way to motivate the following results. We could do the same thing without the analysis by saying: "Define a new function $f(x)=\psi(x)/e^{-ax^2}$ and write the Schrodinger equation in terms of $f$. Then we can show that $f$ must be a terminating polynomial if $\psi$ is to be bounded." It is perfectly rigorous, and equivalent to (but just a bit less elegant than) the operator method. – Bob Knighton Jun 27 '17 at 17:31
• @BobKnighton What is the need of defining a new function while solving a differential equation? So when we define a new function we must state that the solutions of the original equation are all contained in the solutions to the modified equation. Hence there is a certain lack of rigor in the solution of differential equation by asymptotic analysis. The given function has an exponential form and although that is physically justified, from the mathematical point of view a lot more rigor is required. – SchrodingersCat Jun 27 '17 at 17:40
• I'm not quite sure I understand what you're asking. We define a new function to make the equation easier to solve. Namely, it makes the new equation solvable by power series methods (while the original equation is not, since the solutions are required to decay). And the solutions to the original equation are certainly contained in the solutions to the new equation. If we find a solution for $f$, then we automatically have $\psi(x)=f(x)e^{-ax^2}$. No more rigor is needed, as every operation made is well defined. – Bob Knighton Jun 27 '17 at 17:43

This is a fantastic question! Let's get started.

I will assume that we already have defined the ladder operators $a$ and $a^{\dagger}$ and have defined a "ground state" $|0\rangle$ (we still have not proved it is the ground state) such that $a|0\rangle=0$. We will also assume that we already know that the Hamiltonian of the Harmonic oscillator can be written in the form

$$H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right).$$

(Note that the ground state $|0\rangle$ is trivially an eigenstate with $E_0=\hbar\omega/2$.) Finally, I will assume that we have already shown the commutation relations of the ladder operators. Namely,

$$[a,a^{\dagger}]=1.$$

With this, we have enough for a proof.

We can define a state $|n\rangle$ (let's forget about normalization for now) as

$$|n\rangle=(a^{\dagger})^n|0\rangle,$$

where $n$ is a nonnegative integer. The state $|n\rangle$ is an eigenstate of the Hamiltonian with energy $E_n=\hbar\omega(n+1/2)$. We wish to show that the set $\{|n\rangle\}_{n\in\mathbb{Z}^+}$ are all of the possible normalizable eigenstates of the Hamiltonian.

Recall that in the position representation, if we have a potential $V(x)$, then we cannot have a normalizable eigenstate $|\psi\rangle$ whose energy satisfies $E_{\psi}\leq\min V(x)$. That is, we can not have an energy less than the minimum potential energy of the system (ie the kinetic energy must be positive).

Now, we finish off with a proof by contradiction. Consider an eigenstate $|\psi\rangle$ whose energy is given by $E_{\psi}=\hbar\omega(n+1/2+\epsilon)$, with $\epsilon\in(0,1)$. Such a state would essentially describe any of the "other" states that $H$ could permit. Now, consider the state $|\psi^{(1)}\rangle=a|\psi\rangle$. By the commutator algebra, it is not hard to show that $|\psi^{(1)}\rangle$ has energy

$$E_{\psi^{(1)}}=\hbar\omega\left((n-1)+\frac{1}{2}+\epsilon\right).$$

Now, we can induct and define a state $|\psi^{(m)}\rangle\equiv(a^m)|\psi\rangle$. Clearly, its energy is given by

$$E_{\psi^{(m)}}=\hbar\omega\left((n-m)+\frac{1}{2}+\epsilon\right).$$

Thus, unless this process terminates at some point (that is, $a|\psi^{(m)}\rangle=0$ for some $m$), we can achieve an arbitrarily low energy. However, this process could never terminate, since the ground state $|0\rangle$ is unique (it's defined in terms of a position operator and a single derivative operator, so $a|0\rangle=0$ simply defines a first order differential equation in position space) and has energy $\hbar\omega/2$, this cannot be achieved for any $\epsilon$ in the given range. Thus, no such state $|\psi\rangle$ can occur. Similarly, we cannot have a state with energy $E_{\psi}\in(0,\hbar\omega/2)$ by the same logic.

Thus, we have (very rigorously) shown that the only normalizable of $H$ are those with energy $\hbar\omega(n+1/2)$, which are uniquely made from the action of ladder operators on the ground state.

I hope this helped!

(TL;DR -- If another state did exist, it would have an energy not of the form of those given by ladder operators. However, acting on this state many times with $a$ would produce an arbitrarily low energy, and is thus such a state could not exist.)

• I like this proof since it makes intuitive sense and answers the question. However, isn't this proof technically unnecessary given that we know the spectrum of H consists of $\hbar \omega(n+1/2)$ for n>=0, and show how we reach each eigenstate using the creation operator acting on the ground state? I don't want to diminish the question or answer, I think they're both very good. – QtizedQ Jun 27 '17 at 21:54
• The point is to prove that the energies you gave are the only ones possible for this problem. – Bob Knighton Jun 27 '17 at 22:31
• @BobKnighton (1/3) I think that one of the weak points in this proof is that you have not proved general uniqueness of the ground state here, except for (again) the traditional ladder operator ($a$) and its powers ($a^{m}$), whose uniquness is in question to start with. Yes, $a|0>=0$ defines a 1st-order differential equation, but for operators $a$. Another point is that, even after proving the uniqueness of the ground state for such operators (not difficult to do, see proof here: physics.stackexchange.com/questions/380655/…), … – user135626 Jan 29 '18 at 17:49
• @BobKnighton (2/3) … there is a second inherent assumption in your proof that the $a$ operators are the ones that would always produce the smallest steps ($\hbar\omega$), and hence act as means for comparison. Therefore, making steps up/down with the same resolution defined by $a$ is bound to reject any other levels. But if a finer resolution could be generated by … – user135626 Jan 29 '18 at 17:50
• @BobKnighton (3/3) … another ladder operator, the traditional operators $a$ would not be compatible with them. (Notice that instead of proving the point for additive $\epsilon$, one need to prove it for multiplicative factor such as $(1\pm\epsilon)\hbar\omega$ instead, to address the operators). What would be great to see, is a rigorous proof that no other operators could given smaller step. This has been asked recently in this post, and it would nice if you could add your insight to it: physics.stackexchange.com/questions/380655/… – user135626 Jan 29 '18 at 17:51

To prove this, you need to know the following theorem: There are no degenerate bound states in 1D. The proof is outlined in the link. In the case of the oscillator potential, every eigenstate is a bound state, so there is no degeneracy in the spectrum.

Once we know that, the rest of the proof is fairly straightforward. In what follows, I'll ignore normalization and set $\hbar\omega=1$ to make the notation easier. We have a unique ground state, given by $|0\rangle$, which satisfies $a|0\rangle=0$, and $H|0\rangle=\frac{1}{2}|0\rangle$. On top of that ground state, we define a tower of states $|n\rangle=(a^\dagger)^n|0\rangle$, which satisfy $H|n\rangle=(n+\frac{1}{2})|n\rangle$. I claim that these are all the eigenstates. We'll prove it by contradiction.

Say there's some other eigenstate $|\phi\rangle$ which is not one of the $|n\rangle$. Then $H|\phi\rangle=E_\phi|\phi\rangle$. Using the commutation relation of $a$ and $H$, you can show that $a^k|\phi\rangle$ is also an eigenstate of the Hamiltonian with eigenvalue $E_\phi-k$. Since we know the eigenvalues can't be negative, that means that for some $k$ this process must terminate. In other words, for some k, $a^k|\phi\rangle\neq 0$ but $a^{k+1}|\phi\rangle=0$. But if that is true, $a^k|\phi\rangle$ necessarily has energy $\frac{1}{2}$, since $Ha^{k}|\phi\rangle=(a^\dagger a+\frac{1}{2})a^{k}|\phi\rangle=\frac{1}{2}|\phi\rangle$. But we also knew that this state had energy $E_\phi-k$. Thus, $E_\phi-k=\frac{1}{2}$, or $E_\phi=k+\frac{1}{2}$.

But that is a contradiction! That means that $|\phi\rangle$ and $|k\rangle$ have the same energy. Thus, since there is no degeneracy in 1D, they must be the same state. This contradicts our assumption that $|\phi\rangle$ was not in our initial list of states.

EDIT: In response to your confusion in the comments, perhaps a concrete example would help. Let's say there were a state $|\frac{1}{2}\rangle$. Then $a|\frac{1}{2}\rangle$ would be a state $|-\frac{1}{2}\rangle$, and $a^2|\frac{1}{2}\rangle$ would be a state $|-\frac{3}{2}\rangle$, etc. This is not allowed, because we know our Hamiltonian has only non-negative eigenvectors, yet e.g. $H|-\frac{3}{2}\rangle=(-1)|-\frac{3}{2}\rangle$.

The only way to not get arbitrarily negative eigenvectors is if at some point, $a^k|\phi\rangle$ is no longer a valid state of your system. What is an invalid state of the system? The zero vector is the only vector in Hilbert space that is not an allowed state of your system. So at some point, you need to have $a^{k+1}|\phi\rangle=0$. But, as I argued above, having $a^{k+1}|\phi\rangle=0$ implies that $a^k|\phi\rangle$ necessarily has energy $\frac{1}{2}$. That means that $|\phi\rangle$ had energy $k+\frac{1}{2}$, thus is $|k\rangle$. It also means that $a^k|\phi\rangle$ is actually just $|0\rangle$, since $|0\rangle$ is the unique vector with energy $\frac{1}{2}$

Nowhere did we assume that $|\phi\rangle$ was not a state like $|1.41\rangle$. But we proved that if such a state existed, we could use ladder operators to generate a whole series of states with arbitrarily negative energies. Since we know our Hamiltonian does not have negative energy states, that implies that an eigenstate like $|1.41\rangle$ cannot exist.

• When you write 'In other words, for some k, $a^k|\phi\rangle\neq 0$ but $a^{k+1}|\phi\rangle=0$.' , you are assuming that $a^{k+1}|\phi\rangle$ is the ground state. It may not be so. I mean this assumption is not justified. When you are considering that state as the ground state, then the problem itself becomes solved. It may be so that $a^{k+1}|\phi\rangle$ is a state that lies between the ground state and the first excited state, like say $|\frac12\rangle$. – SchrodingersCat Jun 27 '17 at 17:19
• This is the main crux of my question. I want to know if there exist energy levels or energy states like $|\pi\rangle$, $|1.57\rangle$, $|2.31\rangle$ in between states like $|1\rangle$,$|2\rangle$, $|3\rangle$etc. And the name of these states is not just meant for some funny nomenclature, but by this naming I want to denote states which actually lie between $|n\rangle$ where n is an integer. Or better said, the energy states which cannot be found using the ladder operator. – SchrodingersCat Jun 27 '17 at 17:26
• @SchrodingersCat I'm not assuming that at all! $a^{k+1}|\phi\rangle$ is not a ground state. It is not a state at all. It is $0$. Not $|0\rangle$, but actually $0$, as in the null vector. – Jahan Claes Jun 27 '17 at 19:42
• The only thing I'm assuming (and this is also something you can prove) is that there are no eigenstates with negative energy. That means that not every $a^k|\phi\rangle$ can be an eigenstate, since for sufficiently large $k$, $E_\phi-k$ will be negative. That means that at some point $a^{k+1}|\phi\rangle$ must be identically zero for some $k$, since otherwise it would be an eigenstate. – Jahan Claes Jun 27 '17 at 19:45
• Keep in mind that $0$ is not a valid state for a quantum system. It's just the null vector in our Hilbert space. – Jahan Claes Jun 27 '17 at 19:45
1. Here we will assume that the quantum harmonic oscillator is given in algebraic (as opposed to a geometric) setting. Say that we only know that
$$\tag{1}\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1}, \qquad\qquad \nu\in\mathbb{R},$$ $$\tag{2} \hat{N}~:=~\hat{a}^{\dagger}\hat{a},$$ $$\tag{3} [\hat{a},\hat{a}^{\dagger}]~=~{\bf 1}, \qquad\qquad[{\bf 1}, \cdot]~=~0.$$ Also we assume that the physical states live in an inner product space $(V,\langle \cdot,\cdot \rangle )$.

2. In my Phys.SE answer here it was with these assumptions then shown that the point spectrum of the number operator $\hat{N}$ is precisely all the non-negative integers $${\rm Spec}_p(\hat{N})~=~ \mathbb{N}_0 .\tag{4}$$

3. In particular, it is possible to get to all (possible degenerate) energy-levels by acting on a vacuum state with the creation operator $\hat{a}^{\dagger}$.

4. However, in the algebraic (as opposed to a geometric) setting there is a caveat: The vacuum state need not be unique!

5. Example: $$V~=~F_A\oplus F_B\tag{5}$$ could be a direct sum of two Fock spaces with vacuum states $|0\rangle_A$ and $|0\rangle_B$, respectively. In this system a general vacuum state is a linear combination of $|0\rangle_A$ and $|0\rangle_B$. Note that it would then not be possible to transform a given fixed vacuum state into an arbitrary energy state by only acting with the creation operator $\hat{a}^{\dagger}$. We need the other vacuum states as well.

Let us assume that $| m \rangle$ is an eigenstate of $N=a^\dagger a$, i.e. $N | m \rangle = m | m \rangle$, and that $m \notin \mathbb{N}$.

Then, since

$$a | m\rangle = c | m-1\rangle$$

with $c \in \mathbb C$, for some $k \in \mathbb N$ we will have

$$a^k | m\rangle = c' | m-k\rangle$$

with

$$m-k<0$$

We will now prove that this is an absurdum. Indeed, given an eigenvalue $n$ of $N$, we have

$$n = \langle n | N | n \rangle = \langle n | a^\dagger a |n \rangle = \| a | n \rangle \|^2\geq 0$$

We then conclude that out hypothesis $m \notin \mathbb N$ must be wrong, and that $m$ must be a non-negative integer.

The eigenfunctions produced by the ladder operators are the Hermite functions, which are a complete basis for $L^2(R)$. Thus there are no states orthogonal to them.