This question is about the Hamiltonian for more than one particle (non-relativistic).

Griffiths (Introduction to Quantum Mechanics, 2e) seems to imply that it is $\displaystyle H=-\frac{\hbar^2}{2}\left(\sum_{n=1}^N\frac{1}{m_n}\nabla_{\mathbf{r}_n}^2\right)+V(\mathbf{r}_1,\dots,\mathbf{r}_N,t)$, but wikipedia isn't so clear: https://en.wikipedia.org/wiki/Hamiltonian_%28quantum_mechanics%29#Many_particles.

Initially the article quotes that formula, but it quickly gets confusing:

However, complications can arise in the many-body problem. Since the potential energy depends on the spatial arrangement of the particles, the kinetic energy will also depend on the spatial configuration to conserve energy. The motion due to any one particle will vary due to the motion of all the other particles in the system. For this reason cross terms for kinetic energy may appear in the Hamiltonian; a mix of the gradients for two particles:
where M denotes the mass of the collection of particles resulting in this extra kinetic energy. Terms of this form are known as mass polarization terms, and appear in the Hamiltonian of many electron atoms (see below).

Unfortunately, the author wrote nothing below which might explain where the mass polarization terms come from.
Could I get some mathematical background, maybe a derivation, of why these terms are present in the Hamiltonian, and why they are sometimes omitted/forgotten?

  • $\begingroup$ Welcome on physics.SE, especially if you have good questions like this one. I believe there is a typo on your Hamiltonian: the kinetic energy should be squared. I do not know the answer to your question, but I guess a correct description of a many-body problem can only be done after second quantization is introduced. Then, because the fermions and bosons easilly "sum-up" to form a giant tensor product state (properly (anti-)symmetrised of course), the total kinetic energy is just the sum of the kinetic energy of the individual particles. $\endgroup$
    – FraSchelle
    Jul 26, 2013 at 8:07
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    $\begingroup$ @Oaoa There is no need to second-quantize a many body system except for the great convenience. But the ordinary many body Schrodinger equation is perfectly consistent... as long as relativistic effects don't matter of course. :) $\endgroup$
    – Michael
    Jul 26, 2013 at 8:46

2 Answers 2


I've never heard of this before and it seems rather weird, but I'll have a guess that seems like the only reasonable thing that could be going on. The kinetic energy of one particle definitely does not depend (fundamentally) on what another particle is doing. It only indirectly depends on the other particles through interactions (i.e. potentials or gauge fields). The author must have in mind some sort of Born-Oppenheimer-like approximation where some subset of interactions are taken care of or fixed before hand, leaving correlations between the remaining degrees of freedom. So here is my guess about what is happening.

If you have two equal mass particles with momenta $\vec{p}_1,\vec{p}_2$ then the total energy is

$$ H = \frac{\vec{p}_1^2}{2m} + \frac{\vec{p}_2^2}{2m} + V(\vec{r}_1,\vec{r}_2).$$

Now introduce the total $\vec{P}=\vec{p}_1+\vec{p}_2$ and relative momentum $\vec{p}=(\vec{p}_1-\vec{p}_2)/2$ then the Hamiltonian becomes

$$ H = \frac{\vec{P}^2}{2M} + \frac{\vec{p}^2}{2\mu} + V(\vec{r}_1,\vec{r}_2),$$

where $M=2m$ is the total mass and $\mu=m/2$ the reduced mass. Now if you neglect the relative momentum because the two particles are rigidly bound and move together always, then you get

$$ H \approx \frac{\vec{P}^2}{2M} + V(\vec{r}_1,\vec{r}_2). $$

Finally substituting back in you get a cross term $\propto \vec{p}_1\cdot\vec{p}_2$ of the kind mentioned in the article, and the mass $M$ indeed refers to the mass of a conglomerate of several particles. You can also do this if the potential is such that you can separate the centre of mass and relative motion problems and solve the relative motion problem on its own. And this idea should straightforwardly extend to the $N$ particle case. I leave that to you.

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    $\begingroup$ This is correct; in fact there are a handful of other wiki pages that refer to "mass polarization terms" that do make explicit that they come from this coordinate transformation. Why on earth anyone would do this for the many-body problem is beyond me. $\endgroup$
    – wsc
    Jul 26, 2013 at 4:21
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    $\begingroup$ @wsc Good to know. It was probably invented by the same sort of person who calls a function of temperature a "temperature dependent constant." $\endgroup$
    – Michael
    Jul 26, 2013 at 4:41
  • $\begingroup$ Thanks for the clarification. +1 to whomever fixes the wiki page. $\endgroup$
    – user86436
    Jul 26, 2013 at 5:46
  • $\begingroup$ @MichaelBrown Few remarks: the center-of-mass / relative-coordinate transformation should also apply to the potential. Then you need an argument like $V\left(R,r\right)\approx V\left(R\right)$. Then one understands that the logic of the approximation is to put $r_{1}\rightarrow r_{2}$. Then we should have $p_{1}\rightarrow p_{2}$ as well, and the mass polarization term vanishes out at the same time. This goes back to the comment I've done above: in second quantization these terms do not exist, since the ground-state is a tensor-product state correctly (anti-)symmetrised. $\endgroup$
    – FraSchelle
    Jul 26, 2013 at 8:19
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    $\begingroup$ @Oaoa Of course the potential should be transformed as well, I just didn't bother writing down. :) I don't think $r_1 \to r_2$ is necessary though (indeed, it incompatible with $p_1\to p_2$ by the uncertainty principle, strictly speaking). $V(R,r)\approx V_R (R) + V_r (r)$ is enough. $\endgroup$
    – Michael
    Jul 26, 2013 at 8:49

The mass polarization term, arises when we consider the motion of the nucleus as well(or in other words, give it a finite amount of mass). The kinetic energy of an atom with a nucleus of mass $M$ and charge $Ze$ and $N$ electrons of mass $m$ and charge $-e$ can be written as:

$$T=-\frac{\hbar^2}{2M}\nabla_{R_0}^2+\sum_{i=1}^N \left(-\frac{\hbar^2}{2m}\nabla_{R_i}^2\right) $$ where we have denoted the coordinates of the nucleus with $R_0$ and the electrons with $R_i$.

Now, moving to the center of mass frame; we define new coordinates $(\vec R,\vec r_1,\vec r_2,\dots,\vec r_N)$, where:

$$\vec R=\frac{1}{M+Nm}(M\vec{R_0}+m\vec{R_1}+\dots+m\vec {R_N})$$ is the center of mass, and $$\vec{r_i}=\vec{R_i}-\vec{R_0}$$ are the relative coordinates. From the above two equations, one can easily verify that: $$\nabla_{R_0}=\frac M{M+Nm}\nabla_R-\sum_{i=1}^N \nabla_{r_i}$$ $$\nabla_{R_i}= \frac{m}{M+Nm}\nabla_R+\nabla_{r_i} $$ Hence, we will have: $$\nabla_{R_0}^2=\left(\frac{M}{M+Nm}\right)^2\nabla_R^2-\frac{2M}{M+Nm}\sum_{i=1}^N \nabla_R . \nabla_{r_i}+\left(\sum_{i=0}^N\nabla_{r_i}\right)^2$$ $$\nabla_{R_i}^2=\left(\frac{m}{M+Nm}\right)^2\nabla_R^2+\frac{2m}{M+Nm} \nabla_R . \nabla_{r_i}+\nabla_{r_i}^2$$

Now, substituting these back into the initial equation we get: \begin{align} T=&-\frac{\hbar^2}{2M}\left(\left(\frac{M}{M+Nm}\right)^2\nabla_R^2-\frac{2M}{M+Nm}\sum_{i=1}^N \nabla_R \cdot \nabla_{r_i}+ \left(\sum_{i=0}^N\nabla_{r_i}\right)^2\right)+\\ &\quad\sum_{i=1}^N\left(-\frac{\hbar^2}{2m}\left(\left(\frac{m}{M+Nm}\right)^2\nabla_R^2+\frac{2m}{M+Nm} \nabla_R . \nabla_{r_i}+\nabla_{r_i}^2\right)\right)\\ =&-\frac{\hbar^2}{2(M+Nm)^2}(M+Nm)\nabla_R^2-\frac{\hbar^2}{2M}\sum_{i,j=1}^n\nabla_{r_i}\cdot\nabla_{r_j}-\frac{\hbar^2}{2m}\sum_{i=1}^n\nabla_{r_i}^2\\ =& -\frac{\hbar^2}{2(M_{tot})}\nabla_R^2-\frac{\hbar^2}{2m}\sum_{i=1}^n\nabla_{r_i}^2 -\frac{\hbar^2}{2M}\sum_{i,j=1}^n\nabla_{r_i}\cdot\nabla_{r_j} \end{align} where the last is the mass polarization term.

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    $\begingroup$ In your final formula, the coefficient of the $\del_{r_i}^2$ operator should be the reduced mass, not $m$. $\endgroup$ Oct 1, 2017 at 18:21
  • $\begingroup$ I think it is fine as it is. $\endgroup$ Feb 27, 2018 at 5:36
  • $\begingroup$ But $\vec{r}_i=\vec{R}_i-\vec{R}$ not $\vec{R}_i-\vec{R}_0$. Did you take the $\vec{R}_0\approx\vec{R}$? $\endgroup$
    – user279043
    Oct 6, 2018 at 0:53

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