2
$\begingroup$

If $H$ and $L_i$ are commuting ( $[H, L_i] = 0$ ) could we deduce that the eigenvalues of $H$ depend/ do not depend on $m$ and $\ell$ ( eigenvalue of $L_z, L^2$ )? I don't think so since it does not depend for the hydrogen atom but in a more general case could we deduce some relation between eigenvalues when two operators are commuting?

And in the particular case where $H = L^2/2\mu a$?

$\endgroup$
  • $\begingroup$ In your example the eigenvalues are prop. to $\ell(\ell+1)$ so they certainly depend on $\ell$, so I’m not sure what you mean. Moreover, have $H=L^2/2\mu a)+ k L_z$ and the eigenvalues are now $\ell(\ell+1)/(2\mu a)+km$, which depend on both $\ell$ and $m$. $\endgroup$ – ZeroTheHero Jan 1 at 8:53
1
$\begingroup$

If $\,H\,$ commutes with $\,A\,$ then it commutes with its square $\,A^2\,$ since
\begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!HA\boldsymbol{-}AH\boldsymbol{=}0\boldsymbol{\Longrightarrow}HA^2\boldsymbol{-}AHA\boldsymbol{=}0 \boldsymbol{\Longrightarrow} HA^2\boldsymbol{-}A(AH)\boldsymbol{=}0 \boldsymbol{\Longrightarrow}HA^2\boldsymbol{-}A^2H\boldsymbol{=}0 \tag{01}\label{01} \end{equation} and by induction with any power $\,A^n, n \in \mathbb{N}$.

Also if $\,H\,$ commutes with $\,A\,$ and $\,B\,$ then it commutes with their sum $\,A\boldsymbol{+}B\,$ since \begin{equation} \left. \begin{cases} \:\:HA\boldsymbol{-}AH\boldsymbol{=}0 \\ \:\:HB\boldsymbol{-}BH\boldsymbol{=}0 \end{cases} \right\} \boldsymbol{\Longrightarrow} H(A\boldsymbol{+}B)\boldsymbol{-}(A\boldsymbol{+}B)H\boldsymbol{=}0 \tag{02}\label{02} \end{equation}

Now, if $\,H\,$ commutes with the three components of angular momentum $\,L_1,L_2,L_3\,$ \begin{equation} HL_{\boldsymbol{i}}\boldsymbol{-}L_{\boldsymbol{i}}H\boldsymbol{=}0, \quad i=1,2,3 \tag{03}\label{03} \end{equation} then by equations \eqref{01}, \eqref{02} it commutes with the sum of their squares $\,L^2\boldsymbol{=}L_1^2\boldsymbol{+}L_2^2\boldsymbol{+}L_3^2\,$ \begin{equation} HL^2\boldsymbol{-}L^2H\boldsymbol{=}0 \tag{04}\label{04} \end{equation}

Suppose now that $\,H,A\,$ operate on the same (finite dimensional) linear space and commute \begin{equation} HA\boldsymbol{-}AH\boldsymbol{=}0 \tag{05}\label{05} \end{equation} Let $\,\psi^{(\lambda)}\,$ an eigenvector (or eigenfunction) of $\,A\,$ of eigenvalue $\,\lambda$ \begin{equation} A\psi^{(\lambda)}\boldsymbol{=}\lambda\psi^{(\lambda)} \tag{06}\label{06} \end{equation} Then from \eqref{05} \begin{equation} A\left[H\psi^{(\lambda)}\right]\boldsymbol{=}\lambda\left[H\psi^{(\lambda)}\right] \tag{07}\label{07} \end{equation}

If $\,H\psi^{(\lambda)}\boldsymbol{= 0}\,$ then $\,\psi^{(\lambda)}\,$ is an eigenvector of $\,H\,$ of eigenvalue $\,\mu\boldsymbol{=}0$. If $\,H\psi^{(\lambda)}\boldsymbol{\ne 0}\,$ then $\,H\psi^{(\lambda)}\,$ is also an eigenvector of $\,A\,$ of eigenvalue $\,\lambda$. Furthermore if $\,\lambda\,$ is a single eigenvalue of $\,A\,$ then its eigenspace is one-dimensional and so necessarily \begin{equation} H\psi^{(\lambda)}\boldsymbol{=}\mu\psi^{(\lambda)} \tag{08}\label{08} \end{equation} that is $\,\psi^{(\lambda)}\,$ is a common eigenvector of $\,A,H\,$ of eigenvalue $\,\lambda,\mu\,$ respectively. From this conclusion no relation could be derived between $\,\lambda\,$ and $\,\mu$. Now, if $\,\lambda\,$ is an eigenvalue of $\,A\,$ of multiplicity $\,\rho_{\lambda}\,$ then its eigenspace is $\,\rho_{\lambda}-$dimensional and under the condition $\,H\psi^{(\lambda)}\boldsymbol{\ne 0}\,$ equation \eqref{07} yields \begin{equation} H\psi^{(\lambda)}\boldsymbol{=}\sum\limits_{k\boldsymbol{=}1}^{k\boldsymbol{=}\rho_{\lambda}} a_{k}\psi^{(\lambda)}_{k} \tag{09}\label{09} \end{equation} where \begin{equation} \{\psi^{(\lambda)}_{k}\} \quad k\boldsymbol{=}1,2,3\cdots, \rho_{\lambda} \tag{10}\label{10} \end{equation} is a basis of the $\,\rho_{\lambda}-$dimensional eigenspace. From \eqref{09} no conclusion could be derived for the eigenvalues and eigenvectors of $\,H$. This is the case of equation \eqref{04}. The angular momentum operator $\,L^2\,$ has one eigenvalue $\,\lambda\boldsymbol{=}\ell(\ell\boldsymbol{+}1)\,$ of multiplicity $\,\rho_{\lambda}\boldsymbol{=}2\ell\boldsymbol{+}1\,$ so no conclusion could be derived for the eigenvalues and eigenvectors of $\,H$.

But from \eqref{03} for $i=3$ \begin{equation} HL_{\boldsymbol{3}}\boldsymbol{-}L_{\boldsymbol{3}}H\boldsymbol{=}0 \tag{11}\label{11} \end{equation} Since $\,L_{3}\,$ has $\,2\ell\boldsymbol{+}1\,$ single eigenvalues \begin{equation} L_{\boldsymbol{3}}\psi_{m}\boldsymbol{=}m\psi_{m} \qquad m\boldsymbol{=}\boldsymbol{-}\ell,\boldsymbol{-}\ell\boldsymbol{+}1,\cdots,\ell\boldsymbol{-}1,\boldsymbol{+}\ell \tag{12}\label{12} \end{equation} from the discussion after \eqref{07} we correspond to each pair of eigenvalue-eigevector $\,[m,\psi_{m}]\,$ of $\,L_{3}\,$ a pair of eigenvalue-eigevector $\,[\mu_{\ell m},\psi_{m}]\,$ of $\,H$ \begin{equation} H\psi_{m}\boldsymbol{=}\mu_{\ell m}\psi_{m} \tag{13}\label{13} \end{equation}

In case that for a $m'$ we have $\,H\psi_{m'}\boldsymbol{= 0}\,$ then $\,\mu_{\ell m'}\boldsymbol{=}0$.

$\endgroup$
0
$\begingroup$

If, by $L$, you mean the total angular momentum vector $\mathbf{L}$ i.e. $H$ commutes with all three components of the angular momentum, then your (energy) eigenvalues sure won't depend on $m$, but they might (and in general they will) depend on $\ell$.

From classical mechanics, we know that the angular momentum components $L_i$ are the generators of 3D rotations. This notion is carried over to quantum mechanics. The unitary operator $R_i(\phi)=e^{-i\phi L_i/\hbar}$, when acted by on some state $|\psi\rangle$, 'rotates' it by an angle $\phi$, in the counterclockwise sense, around the $i$-axis. This means that if you represent your state in the (normal) position bases as such: $\psi(x,y,z)$, then the coordinates transform as they would (classically) do under such rotation.

Now if $H$ commutes with all $L_i$, then it also commutes with the rotation operators $e^{-i\phi L_i/\hbar}$. What this physically means is if you consider the transformed Hamiltonian: $$H'=R_i(\phi)HR_i^\dagger(\phi)=HR_i(\phi)R_i^\dagger(\phi)=H.$$ In other words, $H$ is invariant under rotations; if you rotate the system, it stays the same. From this argument, we can deduce that the energies must not depend on $m$, because rotations in general do not preserve the value of $m$, and if the energies did depend on $m$, then rotations would change the energies, which we know is not true. However, $\ell$ remains unchanged under rotations, and so there is no need for the energy eigenvalues to be independent of $\ell$. You can visualize what I am saying pretty easily by imagining a classical system consisting of a bead moving around the surface of a sphere; rotating the system will mix the individual components of $\mathbf{L}$, but it will keep its magnitude $L^2$ the same. Be careful not to take the analogy too far, though.

*You can skip this paragraph, if you want. I included it just for the sake of completeness.* In the language of group theory, the 3D rotation group, dubbed SO(3), is a Lie group, with the algebra of the fundamental angular momentum commutation relations I suppose you know: $$[L_i,L_j]=i\hbar\varepsilon_{ijk}L_k.$$ The relation $[H,\mathbf{L}]=0$ implies that $H$ enjoys the full SO(3) symmetry. This is another way of saying $H$ is invariant under 3D rotations. This means that the spectrum of $H$ will fall under different representations of SO(3). Each representation is a (degenerate) subspace labeled by $\ell=0,1,2,\dots$. There is no need for states in different representations to be degenerate, and any such degeneracy would be the work of other symmetries.

For the case of hydrogen, it turns out there is an extra 'hidden' symmetry (which I will not discuss in detail here) due to the $\frac{1}{r}$ potential: an SO(4) symmetry, or rotations in 4D (don't try to visualize that). This symmetry explains the interesting spectrum of hydrogen.

To summarize, the absence of $m$ in the energy eigenvalues is explained by 3D rotational symmetry, and is a signature of any spherically symmetric potential $V(r)$. The absence of $\ell$ in the particular case of hydrogen is explained by 4D rotational symmetry and is found only in $\frac{1}{r}$ potentials.

PS If you're up to it, have a look at Sakurai's Modern Quantum Mechanics. It has a beautiful chapter on symmetries in quantum mechanics.

$\endgroup$
0
$\begingroup$

In general, you cannot say anything as two operators may be commuting simply because they act on different subspaces (i.e. momentum and isospin) so their eigenvalues need not be related in any way.

Another example of commutativity is when an operator is made out of another, as in $\hat H\propto\hat L^2$: in this particular case you will trivially have that the eigenvalues of $\hat H$ are a function of the eigenvalues of $\hat L^2$.

In the case of any central potential (i.e. the hydrogen atom), although $\hat H$ contains $\hat L_z$ so in principle one could imagine a relationship between $E_n$ and $m$, this doesn’t happen because $\hat H$ commutes with all $\hat L_i$, and in particular with the combinations $\hat L_\pm$ which alter the value of $m$ on a state $|l,m\rangle$.

So to answer your question: knowing commutativity alone doesn’t hint anything about eigenvalues, and you need to know the explicit form of the operators.

$\endgroup$

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.