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I've gone through an intermediate classical mechanics course, and in solving the two-body problem, we reduce it to a one-body between a larger stationary mass, and a smaller reduced mass.

Most solutions I've seen of the hydrogen atom simply neglect the movement of the nucleus. Some solutions replace the electron mass with the reduced mass, for a more accurate answer.

Since the reduced mass is a result derived in classical mechanics, is it valid to apply it in quantum mechanics? Is it possible to derive the reduced mass within a quantum mechanical framework?

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Related: and links therein. – Qmechanic Apr 15 '14 at 8:12
up vote 4 down vote accepted

When you use the reduced mass, what you have first done is to go from the variables $(r_1,p_1)$ and $(r_2,p_2)$ to $(r=r_1-r_2, p/\mu=p_1/m_1-p_2/m_2)$ and $(M R=m_1 r_1+m_2 r_2,P=p_1+p_2)$, where $M=m_1+m_2$ is the total mass and $1/\mu=1/m_1+1/m_2$ is the (inverse of the) reduced mass. As you said, this is usually introduced in classical mechanics to simplify the two-body problem, and it is not a priori valid in quantum mechanics.

But in fact, it is. You just have to show that $\hat r,\hat R,\hat p,\hat P$ have all the expected commutation relations of independent position and momentum operators. This then allows to separate the two-body hamiltonian in two parts $\hat H(\hat r_1,\hat r_2,\hat p_1,\hat p_2)=\hat H_{red}(\hat r,\hat p)+\hat H_{CM}(\hat R,\hat P)$.

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Minor notational suggestion to the answer (v1): Since OP presumably is aware of the $2\to 1$ reduction in classical (Hamiltonian) mechanics, it would be good to notationally stress that $\hat{\bf r}_1$, $\hat{\bf p}_1$, etc, in the answer are vector-valued operators. – Qmechanic Sep 26 '13 at 17:11

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