# Why do the ladder operators in harmonic oscillators work?

The Hamiltonian can be diagonalized by transforming $x$ and $p$ to $a$ and $a^\dagger$. I understand how one proceeds from there to find the spectrum of $a^\dagger a$, the ground state $|0\rangle$ and so on. But I have trouble understanding why the simple choice $[a, a^\dagger] = 1$ is everything one needs in order to diagonalize the Hamiltonian.

In SU(2) groups one can perform the highest-weight construction for the $2j+1$ dimensional irreducible representation (spin $j$ irrep). But there one has the Cartan-Weyl basis consisting of $\sigma_3$ and then uses $\sigma_1$ and $\sigma_2$ to find $\sigma_\pm$ such that this is special with $\sigma_3$ such that $\sigma_\pm$ raises and lowers the eigenvalue of $\sigma_3$.

The harmonic oscillator feels simpler than the SU(2) group as we only have excitations of one kind. With angular momentum or spin, there seem to be much more degrees of freedom. On the other hand the basis for the harmonic oscillator is infinite and that makes all the matrix representations of $a$ and $a^\dagger$ a tad more complicated.

Why does the algebraic method work for the harmonic oscillator?

Similarly to AccidentalFourierTransform I am not sure to understand well your issue.

However there is a crucial missed point in your argument, usually absent in many textbooks on these topics.

It is true that decomposing $H$ as $H= \hbar\omega( a^\dagger a + \frac{1}{2})$ and taking the relations (following from CCR) $[a, a^\dagger] =I$ into account one finds a set of vectors labeled by $n \in \mathbb R$, $|n\rangle$, such that $\langle n|m \rangle = \delta_{nm}$, but it is by no means sufficient to prove that the spectrum of $H$ is discrete with $$\sigma(H)= \{ \hbar \omega (n + 1/2)\:|\: n \in \mathbb N\}\:.\tag{1}$$

The missed fact is that the vectors $|n\rangle$ form a complete set of orthonormal vectors.

Investigating this issue is necessary since the Hilbert space is $L^2(\mathbb R)$ thus infinite-dimensional.

Completeness does not arise from algebraic arguments and has to be established separately focusing on the explicit form of wavefunctions $\psi_n(x) = \langle x|n\rangle$. These are the basis of Hermite functions whose finite span is dense in $L^2(\mathbb R)$, assuring in turn that the orthonormal set $\{|n\rangle\}_{n \in \mathbb N}$ is complete as wanted. As a corollary of spectral theorem, $H$ is essentially selfadjoint on the span of the $|n\rangle$s and the spectrum of its unique self-adjoint extension is just (1).

Concerning the evident similarities with the analogous constructions related with $SU(2)$ and the theory of angular momentum, it is in fact possible to prove that $L^2(\mathbb R)$ is the carrier space of a (strongly continuous) irreducible representation of Weyl-Heisenberg Lie group, and the algebraic procedure based on the algebraic manipulations of $a$ and $a^\dagger$ to construct this representation is strictly analogous to the one employed to construct the corresponding irreducible unitary representations of $SU(2)$ dealing with the ladder operators $J_-$ and $J_+= (J_-)^\dagger$.

The situation of the compact Lie group $SU(2)$ is however different, due to the known fact that all strongly-continuous irreducible unitary representations of a compact topological group are necessarily finite dimensional in view of Peter-Weyl's theorem. This feature guarantees that pure algebraic manipulations are enough to find a (necessarily finite!) orthogonal basis of the angular momentum, for instance. The argument cannot be used for the harmonic oscillator because the Weyl-Heisenberg group is not compact and admits infinite dimensional representations.

• Impeccable, as usual. Do you know where I can find information about the completeness of the basis of Fock spaces in QFT (that is, about the completeness of the multiparticle basis $|\boldsymbol p_1,\boldsymbol p_2,\cdots,\boldsymbol p_n\rangle$). It's always bugged me that the completeness of free states in QFT is never discussed (at least in the books I've read). – AccidentalFourierTransform May 16 '16 at 11:47
• It is easy, as the basis $|p_1\rangle$ is complete and as the Fock space is the Hilbertian direct sum of symmetric tensor products of one-particles space the multiparicle set of vectors is complete as well. – Valter Moretti May 16 '16 at 13:52
• Hmm to me it's not that easy :-P I'm not even sure how to prove that $|p_1\rangle$ is complete in the one-particle space! (I mean, in standard QM I know how to deal with this: as $\langle x|p\rangle\sim \exp(ipx)$, completeness follows, but in QFT, what does $|x\rangle$ even mean?) Anyway, I wouldn't like to waste your time: if you ever feel like writing about this, ping me and I'll ask a formal question in the main page. – AccidentalFourierTransform May 18 '16 at 21:02
• I see. So you are interested in a rigorous approach. In this case $|k\rangle$ does not exist in the Hilbert space and the completeness problem does not make sense. One should exploit the approach based on rigged Hilbert spaces to say something about completeness in another sense. Otherwise a proper approach based on standard spectral theory would be more suitable, but the language has to be made more proper as the momentum has a continuous spectrum. The issue regarding the position operator in QFT is in a sense "orthogonal" to your main question and would deserve a separate discussion. – Valter Moretti May 18 '16 at 21:15
• Well, I do not know if I feel like writing about this presently :) I am really submersed by several duties especially regarding the doctoral school of my dept in addition to my teaching load...I guess that I could not afford a discussion about your mathematically delicate issues... no spare time enough... – Valter Moretti May 18 '16 at 21:22

Well, I'm not sure I understood your question so I'm going to write what I think and let's see if it's useful :-)

The algebra $[a,a^\dagger]=1$ is all you need to diagonalise $H$, but this is because what $H$ looks like: $$H=\omega a^\dagger a$$

The important observables, namely $H,P,X$, can be written as polynomials in $a,a^\dagger$: \begin{aligned} X&=a+a^\dagger\\ P&=i(a-a^\dagger)\\ H&=\omega a^\dagger a \end{aligned} and, of course, we can show that any observable $\mathcal O(P,X)$ can be written as a linear combination of monomials in $a,a^\dagger$.

Now, diagonalising $H$ is the same as solving for the time evolution of operators, because in the basis where $H$ is diagonal time evolution is trivial. But, time evolution is given by the commutator $[\mathcal O,H]$, and using the product rule and the linearity of $[\cdot,\cdot]$, it's easy to see that if we know $[a,a]$, $[a^\dagger,a^\dagger]$ and $[a,a^\dagger]$, we know the commutator of any observable $\mathcal O$ with $H$, that is, we know the time evolution of any observable.

For example, \begin{aligned} i\dot X&=[X,H]=\omega[a+a^\dagger,a^\dagger a]=[a,a^\dagger a]+[a^\dagger,a^\dagger a]=\\ &=\omega\left(a^\dagger[a, a]+[a,a^\dagger ]a+a^\dagger[a^\dagger,a]+[a^\dagger,a^\dagger ]a\right) \end{aligned} where I only used the algebraic properties of $[\cdot,\cdot]$.

With this, if we know the individual commutators $[a,a]=[a^\dagger,a^\dagger]=0$ and $[a,a^\dagger]=1$ we can write \begin{aligned} i\dot X&=\omega(a-a^\dagger) \end{aligned} and, by taking a second derivative we get $\ddot X+\omega^2X=0$, that is, we find the explicit form (ODE) of the time evolution of $X(t)$.

Conclusion: the algebra of $a,a^\dagger$ is enough to completely specify the commutator of $H=\omega a^\dagger a$ with any operator $\mathcal O(X,P)$, and therefore it's enough to determine the time evolution of any observable. This in turns means that, once we know, $[a,a]$, $[a^\dagger,a^\dagger]$ and $[a^\dagger,a]$, we know the eigenvalues of $H$.

EDIT

There are two ways to introduce the $a,a^\dagger$ operators.

1) Dirac's way (as can be found on most books on QM): We assume that there exist two operators $X,P$ that we take as fundamental, and define $$H=\frac12P^2+\frac12X^2$$ together with $[X,P]=i$. From this, the usual analysis follows (see e.g., here where they motivate the definition of $a$ and diagonalise $H$).

In this method, all the observables can be written as polynomials in $X$ and $P$, that is, as polynomials in $a,a^\dagger$.

2) Weinberg's method (see Weinberg I. for more details): We assume that there exists a discrete basis $|n\rangle$ $n=0,1,2,\dots$ such that any $\psi$ can be written as $|\psi\rangle= c_n|n\rangle$ (implicit sum). Then we can write $$a|n\rangle=|n-1\rangle\quad a^\dagger|n\rangle=|n+1\rangle$$ up to a normalisation, and this defines the operator $a$ and its commutation relations. With this, we can prove that any operator $\mathcal O$ can be written as $$\mathcal O=o_0 \mathbb 1+o_i a^i+o_{ij}a^ia^j+o_{ijk}a^ia^ja^k$$ where $\{a^i\}=\{a,a^\dagger\}$ and there are implicit sums over repeated indices. The proof of this theorem can be found in W. I, but the meaning is very simple: any operator can be written as a linear combination of $a,a^\dagger$.

In this picture, the operators $a,a^\dagger$ are "fundamental", and we can define, for example, $X=a+a^\dagger$. Now, how do we know that $H\propto a^\dagger a$? well, we don't. But WLOG we can write $$H=h_1a+h_1^*a^\dagger+h_2 a^\dagger a+\text{cubic terms}+\dots$$ but the terms with $h_1$ would make $H$ unbounded (as can be seen by evaluating $\langle 1|H|0\rangle$), so we must take $h_1=0$. This means that $H=\omega a^\dagger a$ plus higher order terms. These higher order terms would make the EoM for $a$ non-linear, which means that we must neglect them if we want a harmonic oscillator (which is linear, by definition).

This analysis shows how we can derive the usual harmonic oscillator if we assume that $a,a^\dagger$ are the fundamental operators. In any case, it should be clear that, whether we treat $a$ as fundamental or derived, the commutator $[a,a^\dagger]$ is all we need to find the eigenvalues of $H$, because diagonalising $H$ is the same as solving the time-evolution, which in turns is given by $[\mathcal O,H]$. As in both 1) and 2) we can write any $\mathcal O$ as a polynomial in $a,a^\dagger$, once we know $[a,a^\dagger]$ we know $[\mathcal O,H]$ for any $\mathcal O$.

• @MartinUeding please let me know if this is more or less what you had in mind (and, of course, if you need me to add more details just say so) – AccidentalFourierTransform May 15 '16 at 11:34
• My questions are: “Why can $H$ be written as $a^\dagger a$ (up to constants)?” and probably equivalent “Why can $X$ and $P$ be written in terms of $a$ and $a^\dagger$? Right now I know that there is one transformation from the space $(x, p)$ to the space $(a, a^\dagger)$. If one uses that, all the rest follows. But why can one choose it this way? Could one choose something different for $a$ and $a^\dagger$? – Martin Ueding May 15 '16 at 11:54
• @MartinUeding well, how do you define $a,a^\dagger$ to begin with? are they derived from $X,P$? or do you want to treat them as fundamental? in the case of the latter, what properties do you define $a,a^\dagger$ to have? and what would your definition for $H$ be? would you accept that there exists a discrete basis $|n\rangle$ such that $a|n\rangle\propto |n-1\rangle$ and $a^\dagger|n\rangle\propto |n+1\rangle$? or would this be a derived property? – AccidentalFourierTransform May 15 '16 at 12:05
• Perhaps there is a connection via $[x, p] = \mathrm i$? I would think that the analytic method is the default. There one obtains the energy quantization after a tedious calculation. Honestly, I do not really know where to start and what to called derived. Perhaps I am wondering how one came up with the definitions of $a$ and $a^\dagger$ in the first place. Did one just try some variants until one diagonalized $H$? – Martin Ueding May 15 '16 at 13:05