# Why is the action of lowering operator on the ground state of a harmonic oscillator to give a 0 wave function?

In quantum mechanics of the harmonic oscillator, when we use the operator method to find out the solutions, we find that the action of $$\hat{a}$$ is to lower the energy of a state by $$\hbar\omega$$ and the action of $$\hat{a}^\dagger$$ is to raise the energy by $$\hbar\omega$$. And so we claim that the action of the lowering operator on the ground state $$\phi_0$$ would be $$\hat{a}\phi_0=0$$ and we use this to calculate the ground-state wave function and so on. But I have neither found not been able to construct a good enough argument as to why $$\hat{a}\phi_0=0$$. Why not something other than 0? or maybe some other wave function entirely?

It would be very nice if you could provide arguments in terms of wave functions and not the vectors as the Dirac notation confuses me as I am taking a quantum mechanics course for the first time and we have not covered that yet.

• If your understand that $\phi(x)=\langle x|\phi\rangle$, I believe that your can easily understand this point: $a|\phi\rangle=0\rightarrow \langle x|a|\phi\rangle\rightarrow a\phi=0$. Mar 17, 2020 at 20:11
• How did you get $a|\phi\rangle=0$? Mar 17, 2020 at 20:13
• Is the question here why there isn't an infinite descending chain of energy states rather than stopping at some value? Mar 17, 2020 at 20:14
• Yes. I am actually confused as to why does the lowering operator anihillate the ground state entirely and not give any other wave function. Although, i do understand that the energy can't be less than 0 as the expectation value of hamilltonian is strictly positive. Mar 17, 2020 at 20:16
• The point is that $[H,a]=-\omega a$. Which means if $\psi$ is the ground state, with energy $E_0$, it has to be killed by $a$, for otherwise $a\psi$ would have energy $E_0-\omega$ contradicting the assumption that $\psi$ was the ground state. This is most likely explained by whatever source you are following. Mar 18, 2020 at 19:29

The eigenstates of the operator $$N = a^\dagger a$$ can be labeled by their eigenvalues, i.e. $$N \phi_n = n \phi_n$$, where $$n$$ is an integer. Note that $$n = \langle \phi_n,N\phi_n\rangle = \underbrace{\langle\phi_n,a^\dagger a \phi_n\rangle = \langle a \phi_n,a\phi_n\rangle}_{a \text{ and } a^\dagger\text{ are mutually adjoint}} = \Vert a \phi_n\Vert^2 \geq 0$$

In particular, we have that $$0 = \Vert a\phi_0 \Vert^2$$ Since the only element of a Hilbert space with zero norm is the zero vector itself, we have that $$a \phi_0 = 0$$

• @Tachyon209 Just let $n=0$ in that first expression I wrote. Mar 17, 2020 at 21:09
• Sorry to ask if this is trivial, but how do you write n as expectation of N for $\phi_n$ ? Mar 17, 2020 at 21:11
• @Tachyon209 If you understand that $N\phi_n = n \phi_n$, then $\langle \phi_n, N\phi_n\rangle = \langle \phi_n, n \phi_n\rangle = n \langle \phi_n,\phi_n\rangle = n$. Mar 17, 2020 at 21:12
• But you are taking another assumption that n is an integer. How do you ensure there are no states woth negative n? Mar 17, 2020 at 21:17
• Or not doing all of this, I meant to ask in this question whether there is any argument by which you can say that setting $a\phi=0$ makes sense? Mar 17, 2020 at 21:20

Expanding on J. Murray's answer (which basically is the answer), we can start from the so-called number operator and have $$\hat{N} = a^{\dagger}a$$. As it was explained Murray's answer, its spectrum is non-negative, because $$\langle \psi | \hat{N} | \psi \rangle = \langle \psi | a^{\dagger}a | \psi\rangle = || a|\psi\rangle ||^2 \geq 0$$ so every eigenvector $$|n\rangle$$ must have non-negative eigenvalue $$n$$ (currently no assumption on $$n$$. Can be fractional, can be integer, who knows?)

Now an important point is that $$[\hat{N},a]=-a$$, which means that if $$|n\rangle$$ is an eigenvector with eigenvalue $$n$$, then, applying this commutator we get $$\hat{N} a |n\rangle = a(\hat{N}-1)|n\rangle = (n-1)a|n\rangle$$ so $$a|n\rangle$$ is also an eigenvector of $$\hat{N}$$ with an eigenvalue $$n-1$$! This means that if we start with some arbitrary eigenvector, we can create a series of eigenvectors with decreasing eigenvalues, just by repeatedly applying $$a$$, hurray! The only problem is that we know that this series cannot go on endlessly. It must stop otherwise we will get to negative eigenvalues territory, which cannot be true. So the only way for this series of lowering eigenvalues to stop is if it exactly hits zero. Then we don't have a state any more, and acting with $$a$$ on the zero vector just remains zero. So the eigenvalues $$n$$ have to be the non-negative integers. The state with the lowest possible eigenvalue has to be $$a|0\rangle = 0$$.

Edit: I just now noticed your last paragraph where you asked for the answer to be in terms of wave-functions. So you can start with $$\int\! dx \psi^*(x) \hat{N} \psi(x) = \int\! dx |a\psi(x)|^2 \geq 0$$ ensuring that the eigenvalues of $$\hat{N}$$ are nonnegative, and then have $$\hat{N} a \psi_n(x) = a(\hat{N}-1)\psi_n(x) = (n-1)a\psi_n(x)$$ telling us that $$a\psi_n(x)$$ is a wavefunction which is an eigenfunction of $$\hat{N}$$ with eigenvalue $$n-1$$. Other than that all progresses as before

• I would be very glad if you could answer these two questions- $\\$(1) How do infer from $\langle\hat{N}\rangle\geq0$ that $\hat{N}$ has non-negative eigenvalues? $\\$(2) How did you arrive at the conclusion that states of nth energy level $(\phi_n)$ have eigenvalues of $\hat{N}$ as n? As far as I know, it comes from the structure of the Hamiltonian and knowing that $E_n=(n+\frac{1}{2})\hbar\omega$ but this result comes from taking $\hat{a}\phi_0=0$ and then using the equal spacing to derive this result. Mar 18, 2020 at 21:43
• (1) since $\langle \psi| \hat{N}|\psi\rangle \geq 0$ for any state, it should be true also for an eigenstate $|n\rangle$ for which $\langle \psi| \hat{N}|\psi\rangle=n$. So any eigenstate must satisfy this. (2) I didn't talk about the energy, as this is a property of the operator $\hat{N} = a^{'dagger}a$ as can be inferred no matter how we reach it. In the context of the Harmonic oscillator Hamiltonian, indeed we have $H = \hbar\omega (\hat{N}+1/2)$ so up to a constant $H$ and $N$ are the same, so we can apply what we know of $\hat{N}$ to the energy eigenvalues
– user245141
Mar 18, 2020 at 22:10
• note that we know that $H=\hbar\omega(\hat{N}+1/2)$ by construction and independently of the spectrum of $\hat{N}$ and its eigenvectors.
– user245141
Mar 18, 2020 at 22:11
• I know that $\hat{N}$ has eigenfunctions (because it is a Hermitian operator), and therefore has some eigenvalues which I denote by $n$ (not restricting it to any specific value). Then, from the arguments in my answer, I discover that these $n$ must be the non-negative integers $0,1,2,\ldots$. Then, using the identity $H=\hbar\omega(\hat{N}+1/2)$ I conclude that the same eigenfunctions of $\hat{N}$, must be also eigenfunctions of the Hamiltonian, with energies $E_n = \hbar\omega(n+1/2)$.
– user245141
Mar 18, 2020 at 22:18
• The operators $H$ and $\hat{N}$ are basically the same operators. Every eigenfunction of one of them is an eigenfunction of the other. You can question how do we know that we got all eigenfunctions of $\hat{N}$, and you're right - they might be more. However this mean that $N$ is degenerate, as all these functions must terminate with the eigenvalue $n=0$ upon enough applications of $a$, and we know that in 1d there is no degeneracy of bound-state solutions. So we found all eigenfunctions.
– user245141
Mar 19, 2020 at 7:02

You can do this by direct substitution in the position or momentum basis. Here it is for $$x$$:

\begin{align} \hat a|0\rangle &= \left[i\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x - \frac i {m\omega}\hat p \right) \right] \left[\left(\frac{m\omega}{\pi\hbar}\right) e^{-mwx^2/2\hbar} \right]\\ &\propto \left[x+\frac {\hbar}{m\omega}\frac d{dx} \right]e^{-mwx^2/2\hbar}\\ &\propto xe^{-mwx^2/2\hbar}+\frac {\hbar}{m\omega}(-m\omega x/\hbar )e^{-mwx^2/2\hbar} \\ &= 0 \, . \end{align}

• Actually, we derive the ground state wave function by assuming that $\hat{a}\phi=0$. So, this is kind of a circular argument. I mean to ask why did we take the particular assumption that $\hat{a}\phi=0$? Mar 17, 2020 at 20:23
• That is one way to get the ground state wavefunction, but there are others. One way is just to write down all the energy eigenstates and find the one with the lowest energy.
– d_b
Mar 17, 2020 at 22:14
• @d_b OP asked why you can assume $\hat{a} \phi=0$, from which you can derive the wave function of above of the ground state. You can brute force solve the Schroedinger equation, but it is not what the OP asked. Mar 18, 2020 at 7:44