2
$\begingroup$

When we say an particle is in the state: \begin{equation} |l,m\rangle, \end{equation} what is the underlying state space, as a vector space? Is it a tensor product vector space, of dimension: \begin{equation} l\times(2l+1)\ ? \end{equation} How can I find the matrix representation of the angular momentum operators that act on the $2l+1$ vector space in that tensor product? I am used to angular momentum operators taking the form of a cross-product: \begin{equation} x_ip_j - p_ix_j, \end{equation} but can we still do that for the $2l+1$ dimensional space corresponding to $m$?

$\endgroup$
4
  • $\begingroup$ $|l,m\rangle$ is normally for the eigenvectors of one of the angular momentum components operators and the total angular momentum operator. It does not refer to the Hydrogen Atoms eigenfunctions. The radial part and with that the Coulomb potential is missing. Still would like to know if you can write them as tensor prod. though. $\endgroup$ – pindakaas May 6 '15 at 20:07
  • $\begingroup$ pindakaas - thanks - I'm just learning QM, and angular momentum turns out to be very hard for me. No doubt am I making errors :-) I should probably edit the title. What would you suggest? $\endgroup$ – Frank May 6 '15 at 20:34
  • 1
    $\begingroup$ I don't believe it is a tensor product space. The states with a certain $l$ are $2l+1$-fold degenerate, so in this case I don't think $|l,m\rangle$ is shorthand for $|l\rangle\otimes|m\rangle$, but rather a convenient way of representing the $2l+1$ possible different eigenstates for each $l$. $\endgroup$ – mr blick May 6 '15 at 20:49
  • 1
    $\begingroup$ As for finding matrix elements, $L_z$ and $L^2$ should be easy since they are diagonal in the $|l,m\rangle$ basis. As for $L_x$ and $L_y$, try expressing those in terms of the ladder operators $L_{\pm}$, whose action you know on the $|l,m\rangle$ $\endgroup$ – mr blick May 6 '15 at 20:54
2
$\begingroup$

For orbital angular momentum, indeed, $L = x\times p$ even as a quantum operator, see this question.

When writing a ket $\lvert l,m \rangle$, this is meant to live in the $2l+1$-dimensional space $\mathcal{H}_l = \mathbb{C}^{2l+1}$ on which the representation of the angular momentum algebra labelled by $l$ exists ($m$ is the eigenvalue of the ket for $L_z$). The total space of (bound) states for your system is then the infinite sum of these spaces for all possible $l$, i.e. $$ \mathcal{H} = \bigoplus_l \mathcal{H}_l$$

$\endgroup$
1
  • $\begingroup$ Except, that if you go to high in energy, there won't be any bound state anymore - so the sum is not really infinite in practice :-) Thanks ACuriousMind, that's precisely the answer I was looking for. $\endgroup$ – Frank May 7 '15 at 0:03
0
$\begingroup$

My understanding of this limited, but this might help (too long for a comment):

The state space is spanned by the set of simultaneous eigenstates of the Hamiltonian, $ \hat L^2$, and $ L_z $. In fact, they form an orthonormal basis of a Hilbert space $ H $ which is the state space.

Out of convenience, we denote the eigenstates by the quantum numbers, indexing them with $ n, \ell$ and $m $ which correspond (though are not equal to) their respective eigenvalues for each of the operators.

I suppose that a state with only $\ell$ and $ m $ specified lies in the subspace of $ H $ with an orthonormal basis equal to the set of simultaneous eigenstates with those quantum numbers.

$\endgroup$

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.