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Let us consider the Hamiltonian for the isotropic three dimensional harmonic oscilator:

$$H = \dfrac{\mathbf{P}^2}{2m}+\dfrac{m\omega^2\mathbf{R}^2}{2},$$

where $\mathbf{P}$ and $\mathbf{R}$ are the usual momentum and position operators in three dimensions. I want to show that if we define

$$\mathbf{T} = \dfrac{1}{\sqrt{2\hbar}}\left(\sqrt{m\omega}\ \mathbf{R}-\dfrac{i}{\sqrt{m\omega}}\mathbf{P}\right),$$

we will have $H = \hbar\omega(\mathbf{T}\mathbf{T}^{\dagger}+3/2)$.

To do that I just computed $\mathbf{T}^{\dagger}$ using the fact that $\mathbf{R}$ and $\mathbf{P}$ are hermitian and computed the product:

$$\mathbf{T}\mathbf{T}^\dagger = \dfrac{1}{2\hbar}\left(\sqrt{m\omega}\ \mathbf{R}-\dfrac{i}{\sqrt{m\omega}}\mathbf{P}\right)\left(\sqrt{m\omega}\ \mathbf{R}+\dfrac{i}{\sqrt{m\omega}}\mathbf{P}\right),$$

that is:

$$\mathbf{T}\mathbf{T}^\dagger = \dfrac{1}{2\hbar}\left(m\omega \mathbf{R}^2+i\mathbf{R}\mathbf{P}-i\mathbf{P}\mathbf{R}+\dfrac{\mathbf{P}^2}{m\omega}\right),$$

and using the fact that $[\mathbf{R},\mathbf{P}]=i\hbar$ we get

$$\mathbf{T}\mathbf{T}^\dagger = \dfrac{1}{2\hbar}\left(m\omega\mathbf{R}^2-\hbar+\dfrac{\mathbf{P}^2}{m\omega}\right)=\dfrac{1}{\hbar\omega}\left(H-\dfrac{\hbar \omega}{2}\right),$$

in other words we have $H = \hbar\omega\mathbf{T}\mathbf{T}^\dagger +\dfrac{\hbar \omega}{2} = \hbar\omega\left(\mathbf{T}\mathbf{T}^\dagger+\dfrac{1}{2}\right).$

In other words there's something quite wrong here. I tried the same calculation again some times but I always get the same. What am I missing here? How does one end up with $H = \hbar\omega(\mathbf{T}\mathbf{T}^\dagger + 3/2)$?

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  • $\begingroup$ I originally thought this was off topic, but after reviewing the homework policy carefully, I changed my mind. I could see the argument going either way, though. $\endgroup$ – David Z Apr 11 '16 at 10:02
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The answer is given by Prahar in his comments :

\begin{equation} {\bf T} \cdot {\bf T}^\dagger= {\rm T}_{1}{\rm T}^\dagger_{1}+{\rm T}_{2}{\rm T}^\dagger_{2}+{\rm T}_{3}{\rm T}^\dagger_{3} \tag{01} \end{equation} For $k=1,2,3$(1) \begin{equation} {\rm T}_{k}{\rm T}^\dagger_{k}=\dfrac{1}{2\hbar}\left(\sqrt{m\omega}\ {\rm R}_{k}-\dfrac{i}{\sqrt{m\omega}} {\rm P}_{k}\right)\left(\sqrt{m\omega}\ {\rm R}_{k}+\dfrac{i}{\sqrt{m\omega}} {\rm P}_{k}\right) \tag{02} \end{equation}

\begin{equation} {\rm T}_{k}{\rm T}^\dagger_{k}= \dfrac{1}{2\hbar}\left(m\omega {\rm R}_{k}^2+\underbrace{i{\rm R}_{k} {\rm P}_{k}-i{\rm P}_{k}{\rm R}_{k}}_{-\hbar}+\dfrac{ {\rm P}_{k}^2}{m\omega}\right) \tag{03} \end{equation} etc..... \begin{equation} \mathbf{T}\mathbf{T}^\dagger = \dfrac{1}{2\hbar}\left(m\omega\mathbf{R}^2-3\hbar+\dfrac{\mathbf{P}^2}{m\omega}\right)=\dfrac{1}{\hbar\omega}\left(H-\dfrac{3\hbar \omega}{2}\right) \tag{04} \end{equation}


(1) In the following equations (02) & (03) it's supposed that we don't make use of the Einstein summation convention for repeated indices.

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