# A Highest state of SU(2) algebra

1. Oct 6, 2016

### spaghetti3451

Let the generators of the SU(2) algebra be $\tau_{1}$, $\tau_{2}$ and $\tau_{3}$.

Consider an $N$ dimensional representation, which means that the $\tau_{i}$ are $N \times N$ matrices which act on some $N$-dimensional vector space.

Consider the ladder operators $\tau_{\pm}=\tau_{1}\pm i\tau_{2}$. Let's work in a basis where $\tau_{3}$ is diagonal, and let $|m\rangle$ denote a unit normalized eigenstate of $\tau_{3}$ with eigenvalue $m$.

Now, I can show that $\tau_{\pm}$ can raise/lower the eigenvalue of $|m\rangle$ by 1, since $\tau_{3}(\tau_{\pm}|m\rangle)=(m\pm 1)(\tau_{\pm}|m\rangle)$.

I can also show that $\tau_{-}$ can annihilate the state $|1\rangle$, since $\tau_{3}(\tau_{-}|1\rangle)=(1-1)(\tau_{-}|1\rangle)=0$

How do I show that $\tau_{+}$ also annihilates some state?

2. Oct 6, 2016

### Fightfish

This sounds a bit handwavy, but you have a finite-dimensional vector space - so the number of eigenstates of $\tau_{3}$ is finite. So it is not possible to keep applying $\tau_{+}$ to raise the state indefinitely; it must terminate at some maximum $|m\rangle$.

3. Oct 6, 2016

### spaghetti3451

Consider the highest state $|j\rangle$ labelled by its eigenvalue $j$. Then, $\tau_{3}(\tau_{+}|j\rangle)=(j+1)(\tau_{+}|j\rangle)$.

But $|j\rangle$ is the highest state, and $(j+1) \neq 0$. Therefore, $\tau_{+}|j\rangle = 0$.

Is this sound?

4. Oct 6, 2016

### Fightfish

Sounds good to me (that's how I would do it too if I had to write it down mathematically)

5. Oct 6, 2016

### spaghetti3451

How do I argue that there must be exactly one state annihilated by $\tau_{+}$ and that this is a $\tau_{3}$ eigenstate?

I have already shown that $\tau_{+}$ annihilates some highest state $|j\rangle$. All I can do now is to argue that $\tau_{+}$ does not annihilate the lowest state - $\tau_{+}$ cannot annihilate the lowest state, because it can only raise a state, not lower it.

6. Oct 6, 2016

### Fightfish

Well, if the set of $\tau_3$ eigenstates forms a complete basis of your vector space (as it should), then any state can be represented in terms of these $\tau_3$ eigenstates, and we know that $\tau_+$ only annihilates $|j\rangle$ (among the set of $\{|m\rangle\}$), so if $\tau_{+} |\psi\rangle = \sum c_{n} \tau_{+} |n\rangle$ vanishes, it must be that all the $c_n$s vanish for $n \neq j$.

7. Oct 7, 2016

### stevendaryl

Staff Emeritus
I think that a way to get both an upper and lower bound on $\tau_3$ is by realizing that $\tau_1, \tau_3, \tau_3$ commute with $S^2 \equiv (\tau_1)^2 + (\tau_2)^2 + (\tau_3)^2$. So we can find states that are simultaneously eigenstates of $S^2$ and $\tau_3$. So let's restrict attention to eigenstates of $S^2$ with eigenvalue $\lambda$. Then we can write:

$\tau_+ \tau_- = (\tau_1 + i \tau_2)(\tau_1 - i \tau_2) =$ (using commutation relations and definitions) $S^2 - \tau_3(\tau_3 - 1)$

If we let $|m\rangle$ be a state where $\tau_3 |m\rangle = m |m \rangle$, then we know that $\tau_- |m\rangle$ returns a state that is proportional to $|m-1\rangle$. Let $\alpha_m$ be the constant of proportionality:

$\tau_- |m\rangle = \alpha_m |m-1\rangle$

We can show that $\tau_+ |m-1\rangle = \alpha_m |m\rangle$. So putting the two together, we have:

$\tau_+ \tau_- |m \rangle = (\alpha_m)^2 |m\rangle$

But we also know that $\tau_+ \tau_- = S^2 - \tau_3(\tau_3 - 1)$. So we have:

$(S^2 - \tau_3(\tau_3 -1)) |m\rangle = (\alpha_m)^2 |m\rangle$

Since we assumed we're dealing with eigenstates of $S^2$ with eigenvalue $\lambda$, and by definition, $\tau_3 |m\rangle = m |m \rangle$, we have:

$(\lambda - m(m-1)) |m\rangle = (\alpha_m)^2 |m\rangle$

So $\alpha_m = \sqrt{\lambda - m(m-1)}$

So we have:
$\tau_- |m\rangle = \sqrt{\lambda - m(m-1)} |m-1\rangle$

$\tau_+ |m-1\rangle = \sqrt{\lambda - m(m-1)}|m\rangle \Rightarrow \tau_+ |m\rangle = \sqrt{\lambda - m(m+1)}|m+1\rangle$

If we require that $\alpha_m$ is real (and why is that?), then it must be that $\lambda \geq m(m-1)$. If $m$ keeps getting larger, than eventually that constraint will be violated. If $m$ keeps getting smaller, than eventually that constraint will be violated. So if you keep acting with $\tau_-$, eventually you get to zero: For the smallest possible value of $m$, it must be that:

$\tau_- |m_{min}\rangle = 0 = \sqrt{\lambda - m_{min} (m_{min}-1)} |m_{min}-1\rangle$

So $\lambda = m_{min}(m_{min}-1)$

Similarly, if you keep acting with $\tau_+$, you will eventually violate the constraint. So there must also be an $m_{max}$ such that

$\tau_+ |m_{max}\rangle = 0 = \sqrt{\lambda - m_{max}(m_{max} + 1)} |m_{max} + 1\rangle$

So $\lambda = m_{max}(m_{max} + 1)$

So if we define $m_{max} = j$, and $m_{min} = -j$, then we have:

$\lambda = j(j+1)$
$\tau_+ |j\rangle = 0$
$\tau_- |-j\rangle = 0$