# What does $X|n\rangle \propto |n+c\rangle$ mean?

$$\renewcommand{\ket}{\left \lvert #1 \right\rangle}$$

I'm transcribing below (but see edit history for a scan) a calculation from pg 17 of this article on Lie groups and Lie algebras

$$[N,X]=cX. \tag{6.1.1}$$ If $$N$$ has an eigenvector $$\ket{n}$$ which satisfies $$N \ket{n} = n \ket{n}$$, then we can apply $$NX$$ to $$\ket{n}$$ and use (6.1.1) to obtain \begin{align} NX \ket{n} &= \{XN + [N, X]\}\ket{n} \\ &= XN\ket{n} + [N, X] \ket{n} \\ &= Xn \ket{n} + cX \ket{n} \\ &= (n + c) X \ket{n} \end{align} Hence, \begin{align} N \ket{n} &= n \ket{n}\tag{6.1.2}\\ X \ket{n} &\propto \ket{n + c}\tag{6.1.3} \\ N \ket{n + c} &= (n + c)\ket{n + c}\tag{6.1.4} \end{align} So the action of $$X$$ on $$\ket{n}$$ is to create a new eigenvector of $$N$$, $$\ket{n + c}$$ which eigenvalue $$n+c$$. If $$c$$ is positive, then $$X$$ is a raising operator and if $$c$$ is negative it is a lowering operator.

What does $$X|n\rangle \propto |n+c\rangle$$ mean? They both seem to be vectors. Are they proportional as vectors? What about the previous calculation suggests this?

• They are proportional as vectors. – Prahar May 11 '19 at 0:33
• Minor comment to the post (v4): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic May 11 '19 at 8:09

It just means that $$X|n\rangle$$ is proportional to $$| n + c \rangle$$.

You can see this from the top half of the derivation: $$N\left( X |n\rangle \right)$$ results in some prefactor ($$n+c$$) times $$X|n \rangle$$. This means that we may not know what $$X$$ exactly does to $$|n\rangle$$, but it does produce an eigenstate of $$N$$ with eigenvalue $$n+c$$. From the definition of $$N$$ you can then see that $$X$$ must map $$|n\rangle$$ onto $$|n+c\rangle$$; we just do not know if there's some additional factor involved, hence the $$\propto$$ symbol.

• Why does $X$ have to map $|n\rangle$ to $|n+c\rangle$? Is it obvious that if $N|n\rangle = n|n\rangle$, then $N|n+c\rangle= (n+c)|n+c\rangle$? – Anju George May 10 '19 at 23:34
• @AnjuGeorge yes, that's pretty much the definition of $N$. These can be number states, so if $c$ is just a number, $|n+c\rangle$ is in that number state basis as well. – Julius May 10 '19 at 23:40

$$\renewcommand\ket{\left|{#1}\right>} \renewcommand\braket{\left<{#1}\middle|{#2}\right>}$$

It means that the operator $$X$$ changes the state $$\ket{n}$$ into a different state, $$\ket{n+c}$$. However, if $$\ket{n}$$ is normalized so that $$\braket nn = 1$$, the new state $$X\ket n$$ isn't necessarily normalized the same way. If you care you have to figure out what the new normalization constant is, but if you don't care (for now) you can leave it alone.

For instance, in the spherical harmonic ladder operators, the normalization is

$$L_\pm \ket{\ell,m} = \sqrt{\ell(\ell+1) - m(m\pm 1)} \ket{\ell, m\pm1}$$

There the normalization is important, because that's what establishes the limit that the projection quantum number $$m$$ has to have magnitude $$|m| \leq \ell$$. Remembering that choice is enough information to construct a table of Clebsch-Gordan coefficients. But lots of times it's helpful to discuss raising and lowering operators without worrying about their normalizations, so people tend to leave them off. Other systems that admit ladder operators (such as the simple harmonic oscillator) have different normalizations. The proof you quote describes the most general type of "ladder" operator.

In this answer we assume that OP already knows what the proprotionality symbol $$\propto$$ means, and is instead trying to understand why eqs. (6.1.3) & (6.1.4) should hold.

Strictly speaking, assuming that $$\hat{N}$$ is a diagonalizable operator, so that the Hilbert space $${\cal H}~=~\oplus_n {\cal H}_n \tag{A}$$ decomposes into a sum of eigenspaces $${\cal H}_n~:=~{\rm ker}(\hat{N}-n\hat{\bf 1}), \tag{B}$$ then the commutation relation (6.1.1) only shows that the operator $$\hat{X}$$ maps the eigenspace $${\cal H}_n$$ into the eigenspace $${\cal H}_{n+c}$$.

We can now proceed in two ways:

1. If $$\hat{X}|n\rangle \neq 0$$, we could define $$|n+c\rangle~:=~\frac{\hat{X}|n\rangle}{||\hat{X}|n\rangle||}. \tag{C}$$ Then eqs. (6.1.3) & (6.1.4) would hold.

2. We could define $$|n+c\rangle$$ to be some normalized eigenvector in $${\cal H}_{n+c}$$. If furthermore $${\cal H}_{n+c}$$ is a non-degenerate eigenspace, then this definition is unique modulo a phase factor, and eqs. (6.1.3) & (6.1.4) would hold.