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I want to express the degenerate eigenstates of the three-dimensional isotropic harmonic oscillator written as eigenstates of $\mathbf{L}^2$ and $L_z$, in terms of the Cartesian eigenstates $|n_x\, n_y\, n_z\rangle$ for the first excited state, but I'm not sure how.

I know that the first excited state is threefold degenerate: $n_x=1$ or $n_y=1$ or $n_z=1$, so $n\equiv n_x+n_y+n_z=1$. Label the spherical state $|n\, \ell\, m\rangle$, since we know that $L_z=i\hbar[a_x a_y^\dagger-a_x^\dagger a_y]$, we have $$ \begin{align*} \langle n_x\, n_y\, n_z|L_z|n\, \ell\, m\rangle &= m\hbar\langle n_x\, n_y\, n_z|n\, \ell\, m\rangle\\ \langle n_x\, n_y\, n_z|L_z|n\, \ell\, m\rangle &= i\hbar\langle n_x\, n_y\, n_z|[a_x a_y^\dagger-a_x^\dagger a_y]|n\, \ell\, m\rangle \end{align*} $$ which leads to $$ \begin{align*} m\langle n_x\, n_y\, n_z|n\, \ell\, m\rangle &= i\sqrt{(n_x+1)n_y}\langle n_x+1\, n_y-1\, n_z|n\, \ell\, m\rangle\\ &- i\sqrt{n_x(n_y+1)}\langle n_x-1\, n_y+1\, n_z|n\, \ell\, m\rangle \end{align*} $$ Therefore, $$ \begin{align*} m\langle 1\,0\,0|n\,l\,m\rangle &= -i\langle 0\,1\,0|n\,l\,m\rangle \\ m\langle 0\,1\,0|n\,l\,m\rangle &= +i\langle 1\,0\,0|n\,l\,m\rangle \\ m\langle 0\,0\,1|n\,l\,m\rangle &= 0 \end{align*} $$ In addition, we can decompose $|n\,l\,m\rangle$ in $|n_x\, n_y\, n_z\rangle$ basis $$ |n\,l\,m\rangle=\sum_{n_x\, n_y\, n_z}|n_x\, n_y\, n_z\rangle\langle n_x\, n_y\, n_z|n\,l\,m\rangle $$ So, for example, I want to decompose $|1\,1\,1\rangle_{n\ell m}$ $$ \begin{align*} |1\,1\,1\rangle &= |1\,0\,0\rangle\langle1\,0\,0|1\,1\,1\rangle+|0\,1\,0\rangle\langle0\,1\,0|1\,1\,1\rangle+|0\,0\,1\rangle\langle0\,0\,1|1\,1\,1\rangle \\ &=|1\,0\,0\rangle\langle1\,0\,0|1\,1\,1\rangle+|0\,1\,0\rangle(i\langle1\,0\,0|1\,1\,1\rangle) \\ &=(|1\,0\,0\rangle+i|0\,1\,0\rangle)\langle1\,0\,0|1\,1\,1\rangle \end{align*} $$

but I got stuck here. I don't know how to get to the result $$ |1\,1\,1\rangle=\frac{1}{\sqrt{2}}|1\,0\,0\rangle+\frac{i}{\sqrt{2}}|0\,1\,0\rangle $$ Somehow $\langle1\,0\,0|1\,1\,1\rangle$ evaluates to $1/\sqrt{2}$.

Thank you very much in advance.

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You know that the normalisation of the inner product is 1, that is, $$ \langle n\, l\, m\ |\ n\, l\, m\rangle = 1 $$

you can use this information to find the value of $\langle n_x=1\, n_y=0\, n_z=0\ |\ n=1\, l=1\, m=1\rangle$ as,

$$ \langle 1\,1\,1|1\,1\,1\rangle= 1 $$ leaving some of the algebra for you as part of the exercise*, you will obtain, $$ 1 = \left( 1 + 1 \right) \Big|\langle 1\, 0\, 0\ |\ 1\, 1\, 1\rangle\Big|^2 $$

then solving gives you the transition element for $\langle 1\, 0\, 0\ |\ 1\, 1\, 1\rangle = 1/\sqrt{2}$

remember $|\phi\rangle^{\dagger} = \langle\phi|$ and also the orthogonality conditions when both "bra" and "ket" are in the same basis.

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