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Sorry to ask, but I'm struggling all week with the following question, so I was hoping if you could orient me.

Suppose we have an orthonormal basis ${|n>}$, and we can define some operators which will call $s_+$, $s_-$, $s_z$ (which are the spin operators). For further detail, we can define them as:

$s_+=\sum_{n=0}^{+\infty}|2n+1\rangle\langle2n|$

$s_-=\sum_{n=0}^{+\infty}|2n\rangle\langle2n+1|$

$s_z=\sum_{n=0}^{+\infty}(|2n+1\rangle\langle2n+1|-|2n\rangle\langle2n|)$

Now if we define a new operator as the sum of the other operators, say, $n_{tot}=s_z+s_++s_z$, how can we find the eigenvalues of $n_{tot}$.

I tried using an infinite matrix and checking if the eigenvalues converge to something. I also tried applying directly the operator to the state, but I don't know how to proceed after that.

Edit: Following the tips, I tried to calculate by brute force:

$s_{tot}|v\rangle=(s_z|v\rangle+s_+|v\rangle+s_-|v\rangle)=\sum_{n=0}^{+\infty}(|2n+1\rangle\langle2n+1|v\rangle-|2n\rangle\langle2n|v\rangle+|2n+1\rangle\langle2n|v\rangle+|2n\rangle\langle2n+1|v\rangle=\sum_{n=0}^{+\infty}(|2n+1\rangle\delta_{2n+1,v}-|2n\rangle\delta_{2n,v})+|2n+1\rangle\delta_{2n,v}+|2n\rangle\delta_{2n+1,v})=E_v|v\rangle$

Now let's apply another vector on the LHS:

$\langle w|s_{tot}|v\rangle=\sum_{n=0}^{+\infty}(\langle w|2n+1\rangle\delta_{2n+1,v}-\langle w|2n\rangle\delta_{2n,v})+\langle w|2n+1\rangle\delta_{2n,v}+\langle w|2n\rangle\delta_{2n+1,v})=\sum_{n=0}^{+\infty}(\delta_{2n+1,w}\delta_{2n+1,v}-\delta_{2n,w}\delta_{2n,v})+\delta_{2n+1,w}\delta_{2n,v}+\delta_{2n,w}\delta_{2n+1,v})=E_v\langle w|v\rangle=E_v \delta_{w,v}$

And the only term that survives is E_v=+1 for odd n, and E_v=-1 for even n. However, this doesn't seem correct.

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  • $\begingroup$ If you have an orthonormal basis $\langle 2n+1 | 2n \rangle =0 \Rightarrow s_z=0$ $\endgroup$ – InertialObserver Aug 23 '18 at 2:48
  • $\begingroup$ My bad, I made a little mistake on the definition of $s_z$. If you apply them to the state $|n\rangle$ you will get deltas. Still, from there I don't know how to move forward. $\endgroup$ – Charlie Aug 23 '18 at 2:51
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Just take the sum and do it by brute force. Just apply to some vector $| v \rangle$

$$\sum_n |2n+1\rangle \langle 2n|v\rangle + |2n\rangle \langle 2n+1| v\rangle+ |2n+1\rangle \langle 2n+1| v\rangle-|2n\rangle\langle2n|v\rangle = v|v\rangle $$

and then exploit orthonormality to solve for $v$. (e.g. take the both sides with, say $\langle 2n^* |$). Do this with $2n^*+1$ as well. This will give you a system of equations that will let you solve the eigensystem. Let me know if you need more hints.

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  • $\begingroup$ Thanks, I never thought about having a system of equations. I'm going to try what you said and tell you if I need an additional hint. Indeed, in the example you show, we can clearly see the apparition of delta dues to orthonormality. $\endgroup$ – Charlie Aug 23 '18 at 3:29
  • $\begingroup$ I tried it and updated in my original post. However, all the terms vanish and I cannot build a system of equations. $\endgroup$ – Charlie Aug 23 '18 at 4:14
  • $\begingroup$ Why are you saying that $\langle 2n+ 1 | v \rangle $ is proportional to a kronecker delta? You cannot assume this. For example, apply $\langle 2n^* |$ to both sides. On the LHS you conclude that, for example, $\langle 2n^* | 2n+1 \rangle = \delta_{2n^*, 2n+1} = 0 $ for all integer $n$ leaving behind only the terms like $\langle 2n^* | 2n\rangle$ when $n = n^*$ $\endgroup$ – InertialObserver Aug 23 '18 at 4:20
  • $\begingroup$ I see, I was saying it's proportional due to orthogonality, but I suppose here we're talking about a general state. I'll try what you showed, I got a better idea now. $\endgroup$ – Charlie Aug 23 '18 at 5:00
  • $\begingroup$ Exactly. We can't make any assumptions about the general eigenstate. $\endgroup$ – InertialObserver Aug 23 '18 at 5:02

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