# Why does the raising operator, when acting on a ket with a maximum second quantum number, yield zero?

I'm studying the general formalism of angular momentum in quantum mechanics from Zettili's "Quantum Mechanics: Concepts and Applications", and came across the following equation (labeled 5.37) on page 287:

$$\hat{J}_+|\alpha,\beta_{max}\rangle = 0$$

The book states that this is because $\beta$ has an upper limit of $\beta_{max}$. This part makes sense to me; if this weren't true, then there would exist an eigenvalue for operator $\hat{J}_z$ which is larger than an eigenvalue for $\hat{J}^2$, which isn't physical.

My question is: why zero? Why not:

$$\hat{J}_+|\alpha,\beta_{max}\rangle = |\alpha,\beta_{max}\rangle$$

I don't get why another raising operator destroys the ket altogether. In my head, this seems like reaching the top rung of the ladder, trying to go up another rung, and ending up at zero. Also, is this just a scalar zero? If that's the case, I'm even more confused, because I didn't think operators could convert vectors into scalars.

If someone could help me understand this concept, I'd really appreciate it.

You say you get the physical justification for having some ladder operator(s) give $\hat L_+ \left|\text{state}\right> = 0$. For angular momentum both the raising and lowering operators eventually terminate; for the harmonic oscillator only the lowering operator terminates, at the ground state. Here's a mathematical argument that the termination must end in zero, rather than getting idempotently "stuck" at some $\left|\alpha,\beta_\text{max}\right>$.

1. The parameter $\beta$ is the eigenvalue of the state under $J_z$; that is, $$J_z \left|\alpha,\beta\right> = \beta\hbar \left|\alpha,\beta\right>.$$

2. The operators $J_z$ and $J_+$ don't commute. Instead, $$[J_z,J_+] = +\hbar J_+.$$ (This commutation relation is the justifies the statement that $J_+\left|\alpha,\beta\right>$ is an eigenstate of $J_z$ with eigenvalue $(\beta+1)\hbar$.)

3. If $J_+\left|\alpha,\beta_\text{max}\right> = \gamma \left|\alpha,\beta_\text{max}\right>$ for some constant $\gamma$, then $\left|\alpha,\beta_\text{max}\right>$ is also an eigenstate of $J_+$.

4. What is the eigenvalue of $J_+ \left|\alpha,\beta_\text{max}\right>$ under $J_z$? The first statement says we can do \begin{align} J_z J_+ \left|\alpha,\beta_\text{max}\right> &= J_z \gamma \left|\alpha,\beta_\text{max}\right> \\ &= \beta\hbar \gamma \left|\alpha,\beta_\text{max}\right>, \end{align} which gives the expected eigenvalue of $\beta\hbar$. The second statement permits us to write \begin{align} J_z J_+ \left|\alpha,\beta_\text{max}\right> &= \big( [J_z, J_+] + J_+ J_z \big) \left|\alpha,\beta_\text{max}\right> \\ &= \big( \hbar J_+ + J_+ \beta\hbar \big) \left|\alpha,\beta_\text{max}\right> \\ &= \big(1+\beta\big)\hbar J_+ \left|\alpha,\beta_\text{max}\right> \\ &= \big(1+\beta\big)\hbar \gamma \left|\alpha,\beta_\text{max}\right> \end{align} which suggests the eigenvalue is $(\beta+1)\hbar$. We have a contradiction, unless $\gamma=0$.

A state cannot simultaneously be an eigenstate of two operators which do not commute, for reasons buried deep down in linear algebra. The raising operator $J_+$ must change the eigenvalue of $J_z$.

• I really appreciate this answer. I now see why supposing an eigenvalue of $\gamma=1$ would create a contradiction. I also see that this is due to the fact that $J_+$ and $J_z$ don't commute. I also learned what idempotently means. Nice work. – kingofharts May 2 '15 at 21:20

The raising operator has to raise something. It can't operate on an eigenstate of $J_z$,i.e. $|\alpha,\beta_{max}\rangle$, and give the same thing back. Think of it as raising the eigenstate to a new eigenstate, but with a 0 multiplying the resulting eigenstate to ensure an unphysical value of angular momentum doesn't exist. And this is not a scalar 0. It is the 0 vector, defined by a norm of 0.

• Thanks for the answer. I understand that $|\alpha,\beta_{max+1}\rangle$ is strictly forbidden. Can this zero result be observed in an experiment, if you try and raise the eigenstate beyond what is possible? I guess I don't understand how multiplying the state by zero doesn't have repercussions as well. – kingofharts May 1 '15 at 19:55
• The 0 state is a legitimate state (actually it's the additive identity for the Hilbert space). We can't have something like $|\alpha,\beta_{max+1}\rangle$ floating around because this would suggest a possibility of measuring something that is impossible. That is why we multiply by 0, to ensure what we get back is actually something allowed (the zero vector). And I don't really know if there is an experiment to determine this, but my guess would be no, since there isn't a state with z-component $\beta_{max+1}$ to observe. – mr blick May 1 '15 at 21:20