# Angular momentum - proof for integer or half-integer eigenvalues

I am confused about a proof my Quantum Mechanics textbook has left "as an exercise for the reader".

So, we've got the angular momentum operator $\hat{L}$. We've also got the generalized angular momentum $\hat{J}: \hat{L}=\hbar\hat{J}$. We've got the commutation relations $[\hat{J_k},\hat{J_l}]$ and $[\hat{J^2},\hat{J_k}]$.

We've introduced the "ladder operators" $\hat{J_+}=\frac{1}{\sqrt{2}}(\hat{J_1}+i\hat{J_2})$ and $\hat{J_-}=\frac{1}{\sqrt{2}}(\hat{J_1}-i\hat{J_2})$.

Then, we went on to prove three properties for the eigenvalues and eigenvectors of $\hat{J^2}$ and $\hat{J_3}$: $\hat{J^2}\left|J,m\right\rangle=J^2\left|J,m\right\rangle$, $\hat{J_3}\left|J,m\right\rangle=m\left|J,m\right\rangle$:

1. $m^2\leq J^2$ (so there are minimal and maximal $m$s).

2. $J_+$ "raises" $m$ to $m+1$, $J_-$ "lowers" $m$ to $m-1$.

3. $j$ (which comes from $J^2 \rightarrow j(j+1)$) is an integer or half-integer number.

The question my textbook asks is: Why is $\Delta m$ an integer number?

I thought it was because of the second property but when I asked my professor, he told me this was not a good proof. "$J_+$ changing $m$ from 0 to 1 does not prove that $\Delta m = 1/3$ is impossible".

So, how do I prove this? I thought it was quite trivial, but it turned out it is not.

P.S.: I already viewed this question but it doesn't help me much.

Edit: I may have got a little "lost in translation". The real question my textbook asks is Why is $\Delta m$ an integer number?

• I'm confused, how did you prove that j is a half-integer number? I thought it could either be an integer or a half integer (see my response here: physics.stackexchange.com/q/27899) . Once you've proven that for j, then I agree with you that the conclusion for m follows automatically from the 2nd statement in your question – kleingordon May 20 '15 at 1:03
• Sorry, I meant "integer or half-integer number". I was regarding integers as composed of an even number of half-integers. Also, I meant "Why is $\Delta m$ an integer number?" I know $m$ may have any value, but $\Delta m$ (the step from one $m$ to the next) may only be an integer value. – iordan_93 May 20 '15 at 20:41

## 2 Answers

Your points ,1-3 are fine. There are is a maximal and a minimal value of $m$. Call the maximal value $M$ (we have to call it something). Now we can apply the lower operator any number of times, each time it lowers the value of $m$ by a full integer amount. The maximum and minimum value have a a finite difference $d$. So if you round $d$ up to the nearest integer $n$ you see that applying the lowering operator $n$ times must yield the state of lowest $m$ (or else hit a zero magnitude state first). So a finite number of applications of the lowering operator sent the maximum value $M$ to the minimum value, so they differ by an integer amount (each time you lowered, $m$ went down by 1). So the maximum and the minimum values of $m$ differ by an integer.

To me, this is the proof that $j=M$ is an integer or half integer value ($n=j-(-j)=2j$). It sounds like your proofs are backwards and you are also trying to prove an untrue claim (that $m$ must be integer when for instance the spin of a spin 1/2 particle can have $m=1/2$).

To explicitly show that m=1/2 is possible, let $J_x=\hbar\sqrt{3/4} \sigma_x$, $J_y=\hbar\sqrt{3/4} \sigma_y$, $J_z=\hbar\sqrt{3/4} \sigma_z$ and $J^2=\hbar^23/4\left( \sigma_x^2+\sigma_y^2+\sigma_z^2\right)$. Then note that they satisfy the commutation relations. Then note that the eigenvalues of $J_z$ are $\pm \hbar/2$ hence $m=\pm 1/2$ by definition.

Thus it is impossible to prove your desired claim that $m$ is an integer from the hyopthesi since the above paragraph satisfies the hypothesi and yet the conclusion is false as $m=1/2$ is not an integer but is a perfectly fine value.

Response to the edited question

If you have two values of $m$ that differ by a noninteger then the lowering operator applied many times to each can't both stop at one and the very same lowest $m$ state. So there would have to be a state besides the lowest $m$ state that is sent to zero by the lowering operator.

Show (or assume) that can't happen and you are pretty much done.

• Thank you for your answer, but I made a mistake. My textbook asked why is $\Delta m$ an integer value and I think you answered my question. P.S.: I know about spin one-half and its great importance in quantum mechanics :) – iordan_93 May 20 '15 at 20:46

Your point 1. show that if $j$ (assumed $>0$) is the max value of $m$, then $-j$ is the lowest value, i.e the conditions is symmetric on $m$.

Your point 2. shows that you must be able to reach $j$ from $-j$ using an integer number of steps, which is same as saying $2j$ must be an integer.

As to your final question: since you proved that $J_\pm$ raise or lower by 1, start with the maximal value of $m$, which is $j$ and "crank down" using $J_-$. You can only reach states with $m$ values given by $j,j-1,\ldots,-j$.

Assume, for the sake of discussion, that your $j=4/3$, so that $2j$ is not an integer. Applying $J_-$ repeatedly produces the sequence of $m$ values $4/3,1/3,-2/3$. It is easy to see that the smallest $m$ is not the negative of the largest $m$; this sequence of $m$'s has no physical meaning since reversing the $z$ axis should just reverse the sign of the projection $m$, justifying the symmetry in the sign of $m$ encapsulated in your point 1. Moreover you never get anything but $\Delta m=\pm 1$.