Understanding Cartan subalgebra applied to the n-harmonic oscillator

[JD_PM]

TL;DR

I want to understand the usefulness of obtaining the Cartan subalgebra

I was studying the $n$-dimensional harmonic oscillator, whose Hamiltonian is

$$\hat H = \sum_{j=1}^{n} \Big( \frac{1}{2m} \hat p_j^2 + \frac{\omega^2 m}{2} \hat q_j^2 \Big)$$

The ladder operators are

$$a_{\pm} = \frac{1}{\sqrt{2 \hbar m \omega}} ( \mp ip + m \omega q)$$

And came across an exercise related to it that states the following:

'Determine the Cartan subalgebra (i.e. the maximal abelian subset of these operators; the maximal number of operators $a_i^{\dagger} a_j$ which mutually commute). Hint: there are $n$ of them. Show also that the energy eigenstates are also eigenstates of the elements of this abelian subset.'

Before even attempting the exercise I need to understand the usefulness of determining the Cartan subalgebra.

Could you please shed some light on why is useful to determine it (i.e. why is it useful to determine the maximal number of operators $a_i^{\dagger} a_j$ which mutually commute)?

Thank you.

 

[vanhees71]

It's a great tool to deal with the groups SU(n), which are important in physics in various places, particularly in HEP physics, where it's used as gauge group (for $n=3$ in QCD and in chiral form in the electroweak standard model) or to classify the hadrons in terms of flavor SU(2) (isospin) or SU(3) (including also strangeness) as a global symmetry. There's also chiral symmetry used as low-energy effective theory of QCD, leading to all kinds of models, including hadronic models or various Quark-Meson models.

The idea behind the harmonic-oscillator treatment is pretty simple. You can calculate easily, using the creation and annihilation operators (for "phonons" in this case) the energy eigenvectors. It's clear that they are completely specified by the common eigenvectors of the $n$ number operators $\hat{N}_j=\hat{a}_j^{\dagger} \hat{a}_j$. The energy eigenvalue for given numbers $n_j \in \mathbb{N}_0$ is given by
$$E(n_1,\ldots,n_n)=\hbar \omega \left (\frac{n}{2} + \sum_{j=1}^n n_j \right).$$
Now think about what the operators $\hat{a}_i^{\dagger} \hat{a}_j$ do concerning the energy eigenvalue.

 

[JD_PM]

Hi vanhees71.

The idea behind the harmonic-oscillator treatment is pretty simple. You can calculate easily, using the creation and annihilation operators (for "phonons" in this case) the energy eigenvectors.

Alright, what I know so far is that the energy must be bounded below. Thus there's a ground state $| \ 0 >$ which, by definition, satisfies $a_{-} | \ 0 > = 0$

The energy of the ground state is $E_{o} = \frac 1 2 \hbar \omega$; another way of seeing it is as follows

$$H| \ 0 > = \frac 1 2 \hbar \omega | \ 0 >$$

I also know that we can get the excited states by repeatedly applying the creation operator to the $| \ 0 >$ state as follows: $(a^{\dagger})^n | \ 0 > = | \ n >$. The eigenvalue equation is then:

$$H| \ n > = (n + \frac 1 2 ) \hbar \omega | \ n >$$

The energy eigenvectors are $| \ 0 >$ (ground state) and $| \ n >$ (excited states).

Now think about what the operators $\hat{a}_i^{\dagger} \hat{a}_j$ do concerning the energy eigenvalue.

This is how I interpreted what you wanted me to do:

I'd proceed as follows

Now we have

$$(a^{\dagger} a_{-})^{2n} | \ 0 > = | \ n^2 >$$

The eigenvalue equation is then:

$$H| \ n ^2> = (2n + \frac 1 2 ) \hbar \omega | \ n^2 >$$

Do you agree with the following energy eigenvalue?

$$E_{n^2} = (2n + \frac 1 2 ) \hbar \omega$$

Thank you.

 

[vanhees71]

Well, in the $d$-dimensional harmonic oscillator you have $d$ independent creation and annhilation operators fulfilling
$$[\hat{a}_j,\hat{a}_k]=0, \quad [\hat{a}_j,\hat{a}_k^{\dagger}]=\delta_{jk}.$$
For a complete set of basis vectors you also need $d$ independent compatible observables to define a unique common eigenbasis.

The Hamiltonian is given by
$$\hat{H}=\frac{d}{2} \hbar \omega + \bar \Omega \sum_{j=1}^d \hat{N}_j$$
with the phonon-number operators
$$\hat{N}_{j}=\hat{a}_j^{\dagger} \hat{a}_j.$$
Now you can show that these number operators provide a complete set of compatible observables, and you can express the Hamiltonian with it. The corresponding basis is given by the common eigenvectors, $|n_1,\ldots,n_d \rangle$ with $n_j \in \{0,1,2,\ldots\}$.

Now think about what the operators $\hat{a}_j^{\dagger} \hat{a}_k$ do when applied to the corresponding basis vectors.

 

[JD_PM]

Now think about what the operators $\hat{a}_j^{\dagger} \hat{a}_k$ do when applied to the corresponding basis vectors.

Mmm I was thinking about it but the bulb did not light...

I may need to learn more before trying again. I will think more about it and post what I get.

 

[JD_PM]

Mmm I was thinking about it but the bulb did not light...

I may need to learn more before trying again. I will think more about it and post what I get.

Old thread but I've just recalled I did not end up getting it right. May someone please share bibliography on Cartan Subalgebra? Maybe reading more about it the bulb shines. My main sources at the moment on QFT are: Mandl & Shaw, Peskin & Schroeder and Schwartz . I did not find it in there.

 

[dextercioby]

A good text is Howard Georgi ”Lie Algebras in Particle Physics”.