# One body harmonic oscillator states expressed in terms of creation operators

I am reading trough chapter one of Moshinsky's "The harmonic Oscillator in Modern Physics". However i am having some trouble with the mathematics in section 8 of chapter 1. I will sketch a summary of what the author is trying to do and then point out my problem.

In the 3D quantum harmonic oscillator a general state may be constructed trough, $$|n_1 n_2 n_3 \rangle = [ n_1! n_2!n_3!]^{-\frac{1}{2}} \eta_1^{n_1} \eta_{2}^{n_2} \eta_{3}^{n_3} | 0 \rangle$$ with $\eta_j$ the creation operators, $$\eta_j = \frac{1}{\sqrt{2}} (x_j - i p_j)$$ The above state is valid for a Cartesian basis. If we want to characterize the state in function of the eigenvalues of $H, L^{2}$ and $L_z$, namely $N,l,m$, we must construct a homogeneous polynome of degree $N$ in the creation operators $\eta_i$'s. It turns out that the following state $$| n l m \rangle \equiv A_{nl} (\boldsymbol{\eta} \cdot \boldsymbol{\eta} )^{n} \mathcal{Y}_{lm}(\boldsymbol{\eta}) | 0 \rangle$$ with $$\mathcal{Y}_{lm}(\boldsymbol{r}) \equiv r^{l} Y_{lm} (\theta, \phi)$$ satisfies the following equations \begin{aligned} (H - \frac{3}{2}) | n l m \rangle & = \boldsymbol{\eta} \cdot \boldsymbol{\eta}^{\dagger} | n l m \rangle = N | n l m \rangle \\ L^{2} | n l m \rangle & = l(l+1)| n l m \rangle \\ L_z | n l m \rangle & = m | n l m \rangle \end{aligned}

The last two equations I can easily derive using that $$[\eta^{\dagger}_i, \eta_j] = \delta_{ij}, \hspace{20pt} [\eta^{\dagger}_i, \eta^{\dagger}_j ] = [\eta_i,\eta_j] = 0$$ we can derive, \begin{aligned} {} [L_k, \eta_m \eta_m] & = -i [ \varepsilon_{ijk} \eta_{i} \eta^{\dagger}_{j}, \eta_m \eta_m] \\ &\propto \varepsilon_{ijk} \eta_{i} \eta^{\dagger}_{j} \eta_m \eta_m - \varepsilon_{ijk} \eta_m \eta_m \eta_{i} \eta^{\dagger}_{j} \\ &= \varepsilon_{ijk} \eta_{i} \eta^{\dagger}_{j} \eta_m \eta_m + 2 \varepsilon_{ijk} \eta_i \eta_m \delta_{jm} - \varepsilon_{ijk} \eta_{i} \eta^{\dagger}_{j} \eta_m \eta_m \\ &= 2 \varepsilon_{ijk} \eta_i \eta_j = 0 \end{aligned} As we can see from the definition of the state, we can drag the vector operator $\vec{L}$ trough all the products $(\boldsymbol{\eta} \cdot \boldsymbol{\eta})$ up to the spherical harmonics $Y_{lm}(\theta, \phi)$. This gives us the last two eigenvalue equations.

In order to prove the first equation we need the commutator \begin{aligned} \left[ \eta_{i}^{\dagger}, \eta_{j} \eta_{j} \right] = 2 \delta_{ij} \eta_{j} \end{aligned} Hence dragging the operator $\eta_{i}$ trough the $n + \frac{l}{2}$ factors $(\boldsymbol{\eta} \cdot \boldsymbol{\eta} )$ will deliver a term $\propto 2(n+\frac{l}{2} ) \eta_i$, \begin{aligned} \boldsymbol{\eta} \cdot \boldsymbol{\eta}^{\dagger} | n l m \rangle & = \eta_{i} \eta_i^{\dagger} | n l m \rangle \\ & = \eta_{i} \eta_i^{\dagger} A_{nl} (\boldsymbol{\eta} \cdot \boldsymbol{\eta} )^{n} \mathcal{Y}_{lm}(\boldsymbol{\eta}) | 0 \rangle \\ &= A_{nl} \eta_{i} \eta_i^{\dagger} (\boldsymbol{\eta} \cdot \boldsymbol{\eta} )^{n} (\boldsymbol{\eta} \cdot \boldsymbol{\eta} )^{\frac{l}{2}} Y_{lm} | 0 \rangle \\ & = A_{nl} \eta_{i} (\boldsymbol{\eta} \cdot \boldsymbol{\eta} )^{n+\frac{l}{2}} \eta_i^{\dagger} Y_{lm} | 0 \rangle + (2n + l)A_{nl} \eta_{i} \eta_i (\boldsymbol{\eta} \cdot \boldsymbol{\eta} )^{n+\frac{l}{2}-1} Y_{lm} | 0 \rangle \\ & = A_{nl} \eta_{i} (\boldsymbol{\eta} \cdot \boldsymbol{\eta} )^{n+\frac{l}{2}} \eta_i^{\dagger} Y_{lm} | 0 \rangle + N | n l m \rangle, \end{aligned} with $N = 2n+l$. Now I really don't see why the first term here, $$\propto \eta_{i} \eta_i^{\dagger} Y_{lm} | 0 \rangle$$ should be zero. The author gives some handwaving arguments that $\eta_i^{\dagger}$ can be interpreted as $\frac{\partial}{\partial \eta_{i}}$ and thus this could be seen as $$\eta_{i} \frac{\partial}{\partial \eta_{i}} Y_{lm} = \boldsymbol{\eta} \cdot \boldsymbol {\nabla} Y_{lm} = 0$$ analogues to $$\mathbf{r} \cdot \boldsymbol{\nabla} Y_{lm} = 0.$$ However the author is very wary not to write $\eta^{\dagger} \equiv \frac{\partial}{\partial \eta_{i}}$ as the equivalence is only valid if the operators are acting on a polynome of $\eta_{j}$'s. I really don't see how a spherical harmonic could be seen as a polynome in $\eta_{j}$'s or how one could prove that $$\propto \eta_{i} \eta_i^{\dagger} Y_{lm} | 0 \rangle = 0$$ with ($\hbar = 1$) $$\eta_{i} = \frac{1}{\sqrt{2}} ( x_{i} - i p_{i} ) = \frac{1}{\sqrt{2}} ( x_{i} - \frac{\partial}{\partial x_{i}} )$$ rigourously without using (the somewhat, in my eyes, dirty trick) $\eta_{i}^{\dagger} \rightarrow \frac{\partial}{\partial \eta_{i}}$

I hope this post will not get flagged as too localized as the mathematics involved are extensively used in quantum mechanics.

Well, the $Y_{lm}$ in the derivation above is actually a $Y_{lm}(\eta)$, so it is a function of $\eta$ and as such can be expanded in powers of $\eta$.
The author might only give a handwaving argument, but you can very rigorously prove that operators satisfying bosonic commutation relations, e.g., $[a^\dagger,a] = 1$, satisfy $$[a^\dagger, f(a)] = \frac{\partial f(a)}{\partial a}$$
The proof goes via induction. If $f(a)$ is a constant, then the commutator with $a^\dagger$ is $0$ and obviously the derivative of a constant is zero too.
Next, if $f(a) = a^n$, then $[a^\dagger, a^n] = [a^\dagger, a^{n-1}] a + a^{n-1} [a^\dagger, a] = na^{n-1}$ using the induction hypothesis.
Thus, the statement is true if $f(a)$ is any power of $a$, and since commutators and derivatives are additive, the statement follows for all (well-behaved) functions $f(a)$.
• Ok, thank you for your answer. It is now more clear that the term is zero, using the commutation relation we get, $\eta_j \eta_{j}^{\dagger} Y_{lm} (\boldsymbol{\eta}) | 0 \rangle = \eta_j \frac{\partial}{\partial \eta_j} Y_{lm}(\boldsymbol{\eta}) | 0 \rangle + \eta_j Y_{lm} (\boldsymbol{\eta}) \eta_j^{\dagger} | 0 \rangle = \boldsymbol{\eta} \cdot \boldsymbol{\nabla}_{\boldsymbol{\eta}} Y_{lm}(\boldsymbol{\eta}) | 0 \rangle = 0$. I guess I was also confused by the fact that one is able put a vector operator rather than a vector as an argument in a function, as I have not seen it before. Commented May 29, 2013 at 20:33