0
$\begingroup$

I'm struggling with the concept of knowing simultaneously the eigenvalues of two different total angular momentum operators.

In the theory of angular momentum, the predominant choice when choosing a particular axe is of course the $z$ axes, with its well known operator $J_z$ and eigenvalues $\hbar m$. If we didn't define our $z$ axe as the one that has this eigenvalues for its total angular momentum operator, we could perfectly use $x$ instead of $z$ and having $\hbar m$ for the eigenvalues of $J_x$ instead of $J_z$. But if we have already used $z$ as our particular axe, then we can't know simultaneously the eigenvalues of $J_x$ since $\left[ J_x, J_z\right] \ne 0$, right?

I know this can seem pretty obvious, but in an exercise, I've seen that the argument I gave before (that we could perfectly use $x$ instead of $z$ as our "good" axe) was used despite having used already $z$ as the particular axe, and with the effects of its eigenvalues still playing a role in the exercise, so I'm a bit confused, and I would appreciate if someone could explain me if it's possible to do so or if otherwise the problem is wrong.

Edit:

Ok, so now I've got the general idea, but I think I haven't connected all the dots yet for my particular case, so I'm just going to write it here.

The thing is that I have a Hamiltonian of the form $H = \alpha (a_1^\dagger a_1 + a_2^\dagger a_2) + \beta (a_1^\dagger a_2 + a_2^\dagger a_1)$ that corresponds to a double well, where $\left[ a_1, a_2 \right] = 0$, and where $n_1$ represents excitations in the first well and $n_2$ excitations in the second one. I have the definitions $J_z = \frac{\hbar}{2} (N_1 - N_2)$, $J^2 = \hbar^2 \frac{N}{2}\left(\frac{N}{2} + 1\right)$, $J_x = \frac{J_+ + J_-}{2}= \frac{\hbar}{2} (a_1^\dagger a_2 + a_2^\dagger a_1)$, and so this way we can obtain the relations $j = \frac{n_1 + n_2}{2}$ and $m = \frac{n_1 - n_2}{2}$.

$N=N_1 + N_2$.

The Hamiltonian can be rewritten into $H = \alpha N + \beta \frac{2}{\hbar} J_x$. The thing is that if I have the state $\left|\Psi\right> = \frac{1}{\sqrt{2}} (\left|1, 0\right> + \left|0,1\right>)$, where inside the ket there's $\left|n_1, n_2\right>$, then, if we apply the hamiltonian to this state we'll see that it's an eigenvector with eigenvalue $\alpha + \beta$.

If we also say that $J_x$ can be substituted by $\hbar m$ in the hamiltonian we'll get that the eigenvalues of the hamiltonian are $E= \alpha n + \beta 2 m$. I say also, because I didn't explain it, but the relation of $j$ and $m$ was obtained by doing the same with $J_z$.

Okay, so back to our state, if we know substitute our values of $n$ and $m$ of the state in the formula, which are $1$ and $0$ respectively, we get that its eigenvalue doesn't have any $\beta$. There's obviously something wrong with this formula, which I think is the step of the eigenvalue for $J_z$.

So my question is, is the formula really wrong because of what I said about having $\hbar m$ for the eigenvalue of $J_z$, or is it something else that I'm missing?

I apologize for making it so long and poorly structured, I didn't expect this to be so long.

Thanks in advance.

$\endgroup$
  • $\begingroup$ what is the question? $\endgroup$ – ZeroTheHero Apr 3 '18 at 22:24
  • $\begingroup$ Your second paragraph is just fine. We get a lot of questions of the form "I think X is true... but I remember seeing somebody do something that might have suggested X was false!" for some clearly true thing X. We can't tell you what was wrong with their reasoning, or if you just misunderstood them, unless you write out what they said! $\endgroup$ – knzhou Apr 3 '18 at 22:31
  • $\begingroup$ If, when already knowing the eigenvalues of an angular momentum operator to be $\hbar m$, you can still say that the eigenvalues of another angular momentum operator is $\hbar m$ too. $\endgroup$ – Mr. Nobody Apr 3 '18 at 22:31
  • $\begingroup$ @knzhou Well I gave a bit more general question that's kind of what I wanted to know, since it's what I think that has flaws. Sorry for not being clear. $\endgroup$ – Mr. Nobody Apr 3 '18 at 22:32
  • $\begingroup$ added to my answer to deal with your edit. $\endgroup$ – ZeroTheHero Apr 4 '18 at 2:12
1
$\begingroup$

Mathematically, the eigenvalues are solutions of the determinental equation $$ \hbox{Det}\left( M-\lambda \boldsymbol{I}\right)=0 \tag{1} $$ Since the determinant is unchanged under a similarity transformation $U$: $$ \hbox{Det}(A)=\hbox{Det}(UAU^{-1}) $$ and such a similarity transformation implements a change of basis, the eigenvalue do not depend on the choice of basis used to construct the matrix $M$. Indeed, "eigen" is loosely translated from German as "characteristic" or "proper"; in French, eigenvalues are known as "valeurs propres", i.e. proper values, indicating they are enough to characterize the matrix $M$ without any specification of the basis used to obtain $M$.

Thus, there is no issue about knowing the eigenvalues of the angular momentum operators. The issue is about the eigenvectors, which are quite basis dependent. Basically, if you work in a basis where $J_z$ is diagonal, the eigenvectors form a basis, but they are not eigenvectors of $J_x$ or $J_y$. This is easily seen because $J_x$ or $J_y$ contain off-diagonal terms in their matrix representations.

Nevertheless, it must be that the eigenvalues of $J_x$ or $J_y$ are the same as those of $J_z$ on physical grounds because these eigenvalues are the possible outcomes of a measurement, and there is no preferred direction in space. Why would measuring along any otherwise equivalent direction produce different possible outcomes? If the outcomes were different, the directions would NOT be equivalent.

You can easily verify using (1) that, even if $J_x$ or $J_y$ are NOT diagonal, they will indeed have the same eigenvalues of $J_z$.


Edit to partially answer v2.

Your state $\vert\Psi\rangle$ is not an eigenstate of $J_z$ since the first part has $N_1-N_2=1$ while the second has $N_1-N_2=-1$. It is, however, an eigenstate of $J_x$. Note that you are using a fixed basis $\vert n_1,n_2\rangle$ in which $J_z$ is diagonal. Now, if you find the matrix realization of $J_x$ in the 2-dimensional space spanned by $\vert 1,0\rangle$ and $\vert 0,1\rangle$, you will find that the eigenvalues are the same as those of $J_z$ in that space. The eigenvectors are NOT the same: indeed $\vert\Psi\rangle$ is an eigenstate of $J_x$ as you can verify using your $J_x$ written in terms of h.o. creation and destruction operators. Knowing that $\vert\Psi\rangle$ is an eigenvector, just act with $J_x$ and you will get $\lambda \vert\Psi\rangle$, where $\lambda$ is the eigenvalue you are looking for.

$\endgroup$

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.