On section 7.3 of Ballentine's "Quantum Mechanics: A Modern Development" there is a really nice argument on why the eigenvalues of the total angular momentum operator must be integer, cf. e.g. this Phys.SE answer by NessunDorma. By defining the operators $$q_1=\frac{Q_x+P_y}{\sqrt{2}},\quad q_2=\frac{Q_x-P_y}{\sqrt{2}}, \quad p_1=\frac{P_x-Q_y}{\sqrt{2}}, \quad p_2=\frac{P_x+Q_y}{\sqrt{2}},$$ the projection of the orbital angular momentum can be written as $$L_z=Q_xP_y-Q_yP_x=\frac{1}{2}(p_1^2+q_1^2)-\frac{1}{2}(p_2^2+q_2^2),$$ i.e., a difference of harmonic oscillators. The reason we can say these are indeed independent harmonic oscillators is that the following commutation relations are satisfied: $$[q_1,q_2]=[p_1,p_2]=0, \quad [q_a,p_b]=i\delta_{ab}.$$

Expressing this operator in terms of ladder operators, we can easily see that its eigenvalues would be of the form $(n_1+1/2)-(n_2+1/2)=n_1-n_2$, which is always an integer. This means the quantum number $m$ is an integer, which in turn implies $l$ is an integer.

If we make the change $p_2,q_2 \rightarrow iq_2,ip_2$, then we have $$L_z=\frac{1}{2}(p_1^2+q_1^2)+\frac{1}{2}(p_2^2+q_2^2),$$ and the commutation relations are still satisfied (if we make instead $p_2,q_2 \rightarrow ip_2,iq_2$, then there is an unwanted minus sign in the expression for $[q_a,p_b]$). As a footnote, I think this is unphysical since $ip_2$ and $iq_2$ are not hermitian, but probably the following still makes mathematical sense:

We know that the eigenfunctions of $L_z$ are spherical harmonics $Y_l^m(\theta,\phi)$. On the other hand, the eigenfunctions of the harmonic oscillator are Hermite polynomials (with some factors).

My questions are: If we write $L_z$ in terms of the operators $\{q_i,p_i\}$, can we say the eigenfunctions are products of Hermite polynomials (2D harmonic oscillator)? And if so, would this give a relation between the product of Hermite polynomials and spherical harmonics by means of a change of variables?

I hope there are no trivial mistakes in my reasoning.

  • 2
    $\begingroup$ Hmm... In order to construct representations of $SO(3)$, you define the raising/lowering operators $L_\pm = L_1 \pm i L_2.$ You then solve for a highest weight state that satisfies $L_3 | l, l \rangle = l |l, l \rangle$ and $L_+ | l, l \rangle = 0$. You then lower that state $l$ times using $L_-$ to get the rest of the vectors in the representation. What happens when you try to do that here? The special property of spherical harmonics is that they satisfy $L^2 Y_{lm} = l(l+1) Y_{lm}. $ Is that the case here? $\endgroup$ Apr 12 at 2:20
  • $\begingroup$ Linked. $\endgroup$ Apr 12 at 2:51
  • $\begingroup$ @user1379857 Do you mean $n_1-n_2$ can be any integer since $n_i \in \mathbb{N}$? Would that imply there is no highest (or lowest) weight state as in the case $Y_l^m$ where $m\leq l$ (and $m\geq -l$)? $\endgroup$
    – DanGVel
    Apr 12 at 14:14
  • $\begingroup$ The Hermite polynomial is starting from the ground state of oscillator: $a | 0 \rangle=0$, translating into r-space differential equation $\ x \psi_0 + \frac{\partial \psi_0}{\partial x} = 0$ with proper scaling. The r-space representation of the $\frac{1}{2}(p_1^2 + q_1^2)$ is different from the harmonic oscillator, therefore, it eigen function in r-space will not be Hermite polynomial. $\endgroup$
    – ytlu
    Apr 12 at 16:39
  • $\begingroup$ The algebraic structure of $L_+$ and $L_-$ is similar to that of $a^\dagger$ and $a$ in the harmonic oscillator. This is a smart device to put this two systems into a similar form of Hamiltonian. But it doesn't imply they have same eigen function in r-space. $\endgroup$
    – ytlu
    Apr 12 at 16:58

Ach, yes & no, but... this is the most ponderous summary of the classic Jordan map construction of SU(2) matrix generators there is.

Most of the difficulty and confusion here lies in the change of bases, which runaway abstraction of notation fosters. The notional spaces of the two oscillators are not the spacetime indexed by the generators of rotations, so our everyday $\theta, \phi$ space angles! They are oscillating in auxiliary spaces similar to those of QFT oscillators, likewise unconnected to our spacetime!

I will thus illustrate all this with a specific example: let's look at the $\ell =2$, spin two representation consisting of the three $2\ell +1=5$ -dimensional matrices, 5×5, satisfying $[L^j,L^k]= i \epsilon^{jkr}L^r$, where I've nondimensionalized $\hbar$ for sanity.

In Fock space, life is simple: spin two is but the symmetric spin addition of four spin-1/2s, two $a_1$ and two $a_2$ oscillators, so $n_1=2, ~ n_2=2$. The building block of the 5×5 matrices is the image of the three 2×2 Pauli matrices in this map,
$${\vec L} \equiv {\mathbf a}^\dagger \cdot\frac{ {\vec \sigma } } {2} \cdot {\mathbf a}^{\,} ~,$$ for two-vectors ${\mathbf a},{ \mathbf a}^\dagger$, the starting point of Schwinger’s 1952 treatment of the theory of quantum angular momentum, predicated on the action of these operators on Fock states built of arbitrary higher powers of such operators.

Here the two oscillators are tensor-squared to $$L^2\equiv {\vec L} \cdot {\vec L} = \frac{n_1+n_2}{2} \left ( \frac{n_1+n_2}{2}+1\right )I_5~, $$ the eigenvalue here being 6, as expected for spin 2, of course.

For instance, acting on an (unnormalized) Fock eigenstate, recalling $L_+=a_1^\dagger a_2$ and $L_-=a_2^\dagger a_1$, observe
$$L^2~ a^{\dagger k}_1 a^{\dagger n}_2 |0\rangle= \frac{k+n}{2} \left ( \frac{k+n}{2}+1\right ) ~ a^{\dagger k}_1 a^{\dagger n}_2 |0\rangle ~,$$ while $$L_z ~ a^{\dagger k}_1 a^{\dagger n}_2 |0\rangle= \frac{1}{2} \left ( k-n\right ) ~ a^{\dagger k}_1 a^{\dagger n}_2 |0\rangle ~,$$ so that, for $l=(k+n)/2$ , $m= (k−n)/2$, this is proportional to the eigenstate $ |l,m\rangle$,
$$|l,m\rangle= \frac{a_1^{\dagger ~(l+m)} a_2^{\dagger ~ (l-m)} }{\sqrt{(l+m)!~(l-m)!}}|0\rangle~ . $$

In our example, $l=2$, so $$|2,m\rangle= \frac{a_1^{\dagger ~2+m} a_2^{\dagger ~ 2-m} }{\sqrt{(2+m)!~(2-m)!}}|0\rangle~ . $$

Now recall the standard bases connection to now two disjoint one dimensional spaces (!) $$ \langle x_j|a_i^{\dagger ~~n}|0\rangle = \delta_{ij}\psi_n^{(i)}(x_i), $$ where the $\psi_n (x)$ are the Hermite functions related to the Hermite polynomials in the customary way, n indicates the energy excitation index, and (i) labels the oscillator 1,2, they came from, but now they are the same functions. So, e.g., take m=2, $$ (\langle x_1| \otimes \langle x_2 |) |2,2\rangle= \psi_4 (x_1)\psi_0(x_2), $$ etc.

Likewise, see here $$ \langle \theta,\phi| l,m\rangle= Y_l^m(\theta, \phi), $$ with this explicit form, for the rest

Consider $Y_2^2(\theta,\phi)= \frac{1}{4} \sqrt{\frac{15}{2\pi} } \sin^2\theta ~ e^{2i\phi}$. This is a function with arguments on the celestial sphere, so to speak, and having little to do with the two notional spaces in which the Hermite functions of the two Schwinger oscillators live.

  • Indeed, the two very different realizations describe the same states and the same 5×5 matrices (in our illustration), through the employment of motions on two very different manifolds; but these hardly induce a natural connection between associated Legendre polynomials and Hermite polynomials (and hence functions, above).

NB. Still, there is an important connection between Laguerre and Hermite polynomials, but that's yet another construction... phase space, where the Laguerres of Wigner functions are bilinears of the Hermites of the wave functions. Totally outranges your focus here.

Note in response to comment by @ytlu

The above review/deconstruction was meant to highlight the tenuousness of the connection between Hermite polynomials and spherical harmonics. The five states I mapped above, ($y\equiv \cos\theta$), $$ Y^2_2 \sim e^{i2\phi} P_2^2 (y) \leftrightarrow \psi_4(x_1)\psi_0(x_2), \\ Y^1_2 \sim e^{i\phi} P_2^1 (y) \leftrightarrow \psi_3(x_1)\psi_1(x_2), \\ Y^0_2 \sim P_2^0 (y) \leftrightarrow \psi_2(x_1)\psi_2(x_2), \\ ..., $$ indeed provide equivalent realizations, in two variables, of the very same matrix of derivatives.

I provide the nontrivial quantities with flat measure, $\int dx ~\psi_n(x) \psi_k(x)=\delta_{nk}$ (not polynomials!), and $\int dy ~P_l^m(y)~ P_k^m(y)\propto \delta_{lk}$, to compare apples with apples: so they provide an obvious correspondence dictionary, with evident correspondence rules. But I don't think that was the change of variables in the sky the OP was asking for.

  • $\begingroup$ This is rather irrelavant to the OP's question. $\endgroup$
    – ytlu
    Apr 12 at 16:47
  • $\begingroup$ As I pointed out, the operators $iq_2$ and $ip_2$ are not physical (hermitian). This discards any physical connection between the angular variables $\theta$, $\phi$ and $\{q_1,p_1,iq_2,ip_2\}$. What I am asking is if this (complex) change of variables would be mathematically correct (I'm not sure how to write everything in coordinate space). I know it has no relevance in physical terms. Nonetheless, I find your answer very interesting and I'll think about how it relates to what I'm looking for. Maybe I'm not getting the full picture. $\endgroup$
    – DanGVel
    Apr 12 at 18:53
  • $\begingroup$ I edited the answer to make the coordinatization of Schwinger's model more conventional. You really are talking about the Schwinger two-oscillator model, except in weird superfluous coordinate-dependent terms, something that QFT with its infinity of oscillators studiously avoids! I disagree with @ytlu that this is irrelevant to the question. Is my bridge to coordinates, a tensor product, not transparent? You are conversant with the standard bridge between Fock space and coordinates, no? $\endgroup$ Apr 12 at 19:14
  • $\begingroup$ @CosmasZachos Your general transformation may cover the transformation in this post. But the specific question is about the connection between Hermite polynomial and spherical harmonic funnction, arising from the similar algebraic relation between the two Hamiltonians . That is rather unlikely to be true. $\endgroup$
    – ytlu
    Apr 13 at 3:43
  • $\begingroup$ I did not disagree that the connection is tenuous, although cf. Bailey, W. N. (1939), On Hermite polynomials and associated Legendre functions, Journal of the London Mathematical Society 1 (4), 281-286. Perhaps I'll add an appendix on the tenuous links one is allowed to see. $\endgroup$ Apr 13 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.