I Expansion of position kets in terms of angular momentum

[hokhani]

TL;DR Summary

Why we cannot expand entirely the position kets in $|l,m\rangle$ ?

As far as I know, in QM, the eigenvectors of an observable form a complete base. However, in the expansion of $|x,y,x\rangle$ in terms of the eigenvectors of angular momentum operators $L^2$ and $L_z$ , i.e., $|l,m\rangle$, we can only expand the angular dependent part of $|x,y,x\rangle$ in terms of $|l,m\rangle$ and the radius part cannot be expanded! Any help is appreciated.

 

[div_grad]

If the symmetry of the Hamiltonian allows separating the radial and angular variables, then you can think of the wave function
$$
\psi(r, \vartheta, \varphi) = \sum_{l,m} c_l(r) Y_{lm}(\vartheta, \varphi)
$$
as being expanded in the $Y_{lm}(\vartheta,\varphi) \equiv \langle \vartheta, \varphi|l,m\rangle$ basis, with radial-dependent coefficients $c_l(r)$ (that have to be determined by solving the "leftover" radial Schrödinger equation). So the radius part

(...) the radius part cannot be expanded!

is just the radial-dependent expansion coefficient. That being said, the transformation from Cartesian to spherical coordinates (or to any other curvilinear coordinates, for that matter) is one of the examples in which the rigid use of the Dirac bra-ket notation leads to confusion.

 

[hokhani]

If the symmetry of the Hamiltonian allows

Do you mean that, in general, if the system doesn't have the required symmetry, then we cannot expand $|x,y,z\rangle$ in terms of $|l,m\rangle$?
Isn't it related to the number of degrees of freedom (DOF) so that for $|x,y,z\rangle$there are three DOF while two for $|l,m\rangle$?

 

[pines-demon]

TL;DR Summary: Why we cannot expand entirely the position kets in $|l,m\rangle$ ?

As far as I know, in QM, the eigenvectors of an observable form a complete base. However, in the expansion of $|x,y,x\rangle$ in terms of the eigenvectors of angular momentum operators $L^2$ and $L_z$ , i.e., $|l,m\rangle$, we can only expand the angular dependent part of $|x,y,x\rangle$ in terms of $|l,m\rangle$ and the radius part cannot be expanded! Any help is appreciated.

You can always do the expansion but you need to indicate a radial basis, angles are not enough.

 

[div_grad]

Do you mean that, in general, if the system doesn't have the required symmetry, then we cannot expand $|x,y,z\rangle$ in terms of $|l,m\rangle$?

The quantum system is defined by its Hamiltonian $H$, and the appropriate wave function satisfies the Schrödinger equation $\mathrm{i} \hbar \partial_t \Psi(q, t) = H \Psi(q, t)$ - I denote by $q$ the set of all coordinates on which $\Psi$ depend. Now, if the Hamiltonian does not depend explicitly on time, then the first separation-of-variables that you can do is put $\Psi(q, t) = \psi(q) \exp\left\{-\mathrm{i}Et/\hbar\right\}$ and solve the eigenvalue problem $H\psi(q) = E\psi(q)$. There are various ways in which you can solve such problems, for example using some variational principle or numerically-exact propagators. One effective way of simplifying the solution of the eigenvalue problem is to use basis expansions. In the case of the time-independent Schrödinger equation, you look for the operators $A_i$ which commute with the Hamiltonian, $\left[H, A_i\right]=0$, because then the eigenvectors of $A_i$ are simultaneously the eigenvectors of $H$ - and, in general, it is way easier to find the eigenvectors of the appropriate $A_i$ than to find the eigenvector $\psi(q)$ of the total Hamiltonian.

For example, if the symmetry of the Hamiltonian $H$ is such that the angular momentum operators $L^2$ and $L_z$ commute with $H$, then you can find the eigenvectors of $L^2$ and $L_z$ - which are the spherical harmonics $|l,m\rangle$ - and expand $|\psi\rangle$ in the basis of these eigenvectors (the expansion coefficients themselves depend on the radial variables). The problem is thus simplified because, in this case, the angular variables have been effectively separated-out from $H\psi(q) = E\psi(q)$. This is the case, e.g., for the hydrogen-like atoms, where you have the spherically-symmetric potential energy $V(r)$. But in the case of, say, diatomic molecules, the potential is no longer spherically-symmetric because of the existence of "privileged" direction in space, defined by the internuclear axis. In this case, $H$ commutes with $L_z$ - the angular momentum projection onto the internuclear "z-axis" - but it does not commute with $L^2$. So the basis in which you expand the wave function of a diatomic molecule is different from $|l, m\rangle$.

For spatial-homogeneous systems, the momentum operator commutes with $H$ and you can therefore expand the wave function in the basis of momentum eigenfunctions - which is effectively the Fourier transform of $\psi(q)$. And so on, and so forth. It all depends on the symmetry of the Hamiltonian and on the set of mutually commuting operators $A_i$ that also commute with $H$.

 

[div_grad]

Isn't it related to the number of degrees of freedom (DOF) so that for $|x,y,z\rangle$there are three DOF while two for $|l,m\rangle$?

Yes, it is. For a single free particle, the wave function in Cartesian variables would be written as a product $f_x(x) f_y(y) f_z(z)$ of functions depending solely on one of the three coordinates. And in the spherical coordinates, the free-particle wave function would likewise be a product $f_r(r) f_{\vartheta}(\vartheta) f_{\varphi}(\varphi)$. So indeed, if you "separate-out" the angular variables (by using spherical harmonics) then you're still left with the radial function $f_r(r)$. This is why the expansion coefficients depend on $r$ when you expand total wave function in the basis of solely-angle-dependent functions like $|l,m\rangle$.