Why is a proton assumed to be always at the center while applying the Schrödinger equation?

Why is a proton assumed to be always at the center while applying the Schrödinger equation? Isn't it a quantum particle?

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Self interactions are not considered in a non-relativistic quantum mechanical treatment and the Hydrogen atom is usually treated that way, in a first course. – Torsten Hĕrculĕ Cärlemän Dec 31 '13 at 8:51
@TorstenHĕrculĕCärlemän : What about proton being at the center? – Rajesh D Dec 31 '13 at 8:53
I don't get the fact about it being at the center of a coordinate frame, and it being a quantum particle. You can infact take any point as the origin, only to complicate the expressions further. It is most natural to hence take the nucleus at the center. – Torsten Hĕrculĕ Cärlemän Dec 31 '13 at 8:56
@RajeshD The assumption that the proton is stationary is just an approximation used since protons are about 2000 times as massive as the electrons and 2000 is approximately infinity. – David H Dec 31 '13 at 8:57
@DavidH : Thanks David. That seems very reasonable. – Rajesh D Dec 31 '13 at 9:00

There is a rigorous formal analysis which lets you do this. The true problem, of course allows both the proton and the electron to move. The corresponding Schrödinger equation thus has the coordinates of both as variables. To simplify things, one usually transforms those variables to the relative separation and the centre-of-mass position. It turns out that the problem then separates (for a central force) into a "stationary proton" equation and a free particle equation for the COM.

There is a small price to pay for this: the mass for the centre of mass motion is the total mass - as you'd expect - but the radial equation has a mass given by the reduced mass $$\mu=\frac {Mm}{M+m}=\frac{m}{1+m/M} ,$$ which is close to the electron mass $m$ since the proton mass $M$ is much greater.

It's important to note that an exactly analogous separation holds for the classical treatment of the Kepler problem.

Regarding self-interactions, these are very hard to deal with without invoking the full machinery of quantum electrodynamics. Fortunately, in the low-energy limits where hydrogen atoms can form, it turns out you can completely neglect them.

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I assume you're talking of the hydrogen atom; the hamiltonian of the nucleus + electron system is $$H = \frac{p_e^2}{2 m _e} + \frac{p_n^2}{2 m _n} - \frac{e^2}{|r_e - r_n|}.$$ You can do a change of coordinates (center of mass coordinates) $$\vec{R} = \frac{m_e \vec{r}_e + m_n \vec{r}_n}{m_e+m_n} \\ \vec{r} = r_e -r_n$$ and find the conjugate momenta to these coordinates: $$\vec{P} = \vec{p}_e + \vec{p}_n \\ \vec{p} = \frac{m_n \vec{p}_e - m_e \vec{p}_n}{m_e+m_n}.$$ Defining also the reduced mass $\mu$ such that $$\frac{1}{\mu} = \frac{1}{m_e} + \frac{1}{m_n}$$ and the total mass $M = m_e + m_n$, you can write the hydrogen atom hamiltonian as $$H = \frac{P^2}{2 M} + \frac{p^2}{2 \mu} - \frac{e^2}{r} = H_{CM} + H_{rel}.$$ In this calculations I always treated the nucleus as a quantum particle; but if you look at $H_{rel} = p^2/2\mu - e^2/r$ and let the mass of the nucleus tend to infinity, you obtain the hydrogen atom hamiltonian usually taught in basic QM courses Also, you don't have other terms like spin-orbit, j-j couplings etc. because they are relativistic effects that come out from the Dirac equation.

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thanks for the explanation @AlexA – Rajesh D Dec 31 '13 at 9:36

A similar (the same?) question you might reasonably ask is: how can we assume that the proton is stationary, at the centre of the problem, since it is surely going to be attracted by the electron and jiggle about a little? This is a question that would be just as valid directed at a classical system --- say, a planet orbiting a star --- as a quantum mechanical one.

The solution to this is as described above, by others: the fact that the star/proton is so much more massive than the planet/electron means that it is going to move very little (the acceleration of an object is inversely proportional to its mass, and hence with a large mass we have a very small acceleration i.e. very little motion), and so the stationary nature of the star/proton is a great approximation. And in fact, we can make the analysis completely rigorous by dealing with relative separations and reduced masses. But the finite mass of the proton means that indeed, the proton won't actually be stationary.

However, I'm not sure this is the question you're asking. Your concern was not "isn't the proton a particle of finite mass" but rather "isn't it a quantum particle". The suggestion is that you think the proton should jiggle due to its quantum mechanical nature --- that is, due to the uncertainty principle etc. --- irrespective of the mass of the proton (perhaps I am mistaken about this).

In the limit of the proton having infinitely more mass than the electron, the quantum mechanical nature of the proton won't force it to jiggle. In other words, the uncertainty in its position, $\Delta x$, can be made arbitrarily close to zero. This is consistent with the uncertainty principle since its momentum $p$ (mass x velocity) can tend to infinity in the limit of an infinitely massive proton. Hence we can still achieve

$$\Delta p \Delta x \geq \frac{\hbar}{2}$$

with an arbitrarily small velocity and positional uncertainty, if we make the mass arbitrarily large.

In other words, in the assumption that we're using to neglect motion of the proton due to it being attracted to the electron, we are also able to neglect the motion of the proton due to quantum mechanical effects.

The reality of course is that the proton will jiggle --- it will jiggle a bit due to its intrinsic quantum mechanical nature, and it will jiggle a bit more due to the attractive force on it of the electron. However, this can be dealt with rigorously just as before, using relative separations and reduced masses.

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protected by Qmechanic♦Mar 3 '14 at 16:12

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