Virial theorem

In mechanics, a virial of a stable system of n particles is a quantity proposed by Rudolf Clausius in 1870. The virial is defined by

\tfrac{1}{2} \sum_{i=1}^n  \mathbf{r}_i \cdot \mathbf{F}_i , $$ where Fi is the total force acting on the i th particle and ri is the position of the i th particle; the dot stands for an inner product between the two 3-vectors. Indicate long-time averages by angular brackets. The importance of the virial arises from the virial theorem, which connects the  long-time average of the virial to the long-time average &lang; T &rang; of the total kinetic energy T of the n-particle system,

\tfrac{1}{2} \sum_{i=1}^n \langle   \mathbf{r}_i \cdot \mathbf{F}_i\rangle = - \langle T \rangle. $$

Proof of the virial theorem
Consider the quantity G defined by

G \equiv \sum_{i=1}^n  \mathbf{r}_i \cdot \mathbf{p}_i\quad \hbox{with}\quad \mathbf{p}_i = m_i \frac{d\mathbf{r}_i}{dt}. $$ The vector pi is the momentum of particle i. Differentiate G with respect to time:

\frac{dG}{dt} =  \sum_{i=1}^n\left[\frac{d\mathbf{r}_i}{dt} \cdot\mathbf{p}_i + \mathbf{r}_i  \cdot \frac{d\mathbf{p}_i}{dt}  \right] $$ Use Newtons's second law and the definition of kinetic energy:

\mathbf{F}_i = \frac{d\mathbf{p}_i}{dt}\quad \hbox{and}\quad 2 T_i = \frac{d\mathbf{r}_i}{dt}\cdot \mathbf{p}_i $$ and it follows that

\frac{dG}{dt} = 2 T + \sum_{i=1}^n \mathbf{r}_i\cdot\mathbf{F}_i   \quad\hbox{with}\quad T \equiv \sum_{i=1}^n T_i. $$ Averaging over time gives:

\left\langle \frac{dG}{dt} \right\rangle \equiv \frac{1}{T} \int_0^T \frac{dG}{dt} dt = \frac{1}{T}\left[ G(T) -G(0) \right]. $$ If the system is stable, G(t) at time t = 0  and at time t = T  is finite. Hence, if T goes to infinity, the quantity on the right hand side goes to zero. Alternatively, if the system is periodic with period T, G(T) = G(0) and the right hand side will also vanish. Whatever the cause, we assume that the time average of the time derivative of G is zero, and hence

2 \langle T \rangle + \sum_{i=1}^n \langle \mathbf{r}_i\cdot \mathbf{F}_i  \rangle = 0, $$ which proves the virial theorem.

Application
An interesting application arises when each particle experiences a potential V of the form

V = A r^k\quad\hbox{with}\quad r = (x^2+y^2+z^2)^{1/2}, $$ where A is some constant (independent of space and time).

An example of such potential is given by Hooke's law with k = 2 and Coulomb's law with k = &minus;1. The force derived from a potential is

\mathbf{F} = -\boldsymbol{\nabla}V \equiv -\left( \frac{ \partial V}{\partial x},\; \frac{ \partial V}{\partial y},\; \frac{ \partial V}{\partial z}\right) $$ Consider

\frac{ \partial V}{\partial x} = a \frac{ \partial r^k}{\partial x} = a k r^{k-1} \frac{ \partial r}{\partial x}= a k r^{k-1} (x/r) = k \frac{x}{r^2} V\quad \Longrightarrow \mathbf{F}\quad = - k V \frac{\mathbf{r}}{r^2}. $$ Then applying this for i = 1, &hellip; n,

2\langle T \rangle = k \sum_{i=1}^n \left \langle V(\mathbf{r}_i) \cdot \frac{\mathbf{r}_i\cdot \mathbf{r}_i}{r_i^2}\right\rangle = k\langle V\rangle \quad\hbox{with}\quad V = \sum_{i=1} V(\mathbf{r}_i). $$ For instance, for a system of charged particles interacting through a Coulomb interaction:

2\langle T \rangle = - \langle V \rangle. $$

Quantum mechanics
The virial theorem holds also in quantum mechanics. Quantum mechanically the angular brackets do not indicate a time-average, but an expectation value with respect to an exact stationary eigenstate of the Hamiltonian of the system. The theorem will be proved and applied to the special case of a potential that has a rk-like dependence. Everywhere Planck's constant ℏ is taken to be one.

Let us consider a n-particle Hamiltonian of the form

H = T + V(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_n)\quad\hbox{with}\quad T = \sum_{j=1}^n \frac{\mathbf{p}_j\cdot\mathbf{p}_j}{2m_j} $$ where mj is the mass of the j-th particle. The momentum operator is

\mathbf{p}_j = -i \boldsymbol{\nabla}_j, \qquad (\hbar =1). $$

Using the self-adjointness of H and the definition of a commutator one has for an arbitrary operator G,

0 = \langle \Psi | [G, H] | \Psi \rangle\quad\hbox{with}\quad H| \Psi \rangle = E | \Psi \rangle. $$

In order to obtain the virial theorem, we consider

G = \sum_{k=1}^n \mathbf{r}_k \cdot \mathbf{p}_k. $$ Use

[\mathbf{r}_k \cdot \mathbf{p}_k, H] = [\mathbf{r}_k, T] \cdot\mathbf{p}_k  +   \mathbf{r}_k \cdot[\mathbf{p}_k,V] $$ Define

\mathbf{F}_k \equiv -i [\mathbf{p}_k,V] = - [\boldsymbol{\nabla}_k, V]. $$ Use

[r_{k\alpha}, p_{j\beta}^2] = \delta_{kj}\delta_{\alpha \beta} 2 i p_{k \alpha}, \quad \alpha,\beta=x,y,z;\quad k,j=1,,2,\ldots,n, $$ and we find

[G, H] = i\big( 2T + \sum_{j=1}^n \mathbf{r}_j \cdot\mathbf{F}_j \big) $$ The quantum mechanical virial theorem follows

2\langle T\rangle = -\sum_{j=1}^n \langle \mathbf{r}_j \cdot\mathbf{F}_j \rangle $$ where &lang; &hellip; &rang; stands for an expectation value with respect to the exact eigenfunction &Psi; of H.

If V is of the form

V = \sum_{j=1}^n a_j (r_j)^k, $$ it follows that

\mathbf{F}_j = - [\boldsymbol{\nabla}_j, V] = - a_j\, k \mathbf{r}_j\, (r_j)^{k-2}. $$ From this:

2\langle T\rangle = k \sum_{j=1}^n a_j \langle   (\mathbf{r}_j\cdot\mathbf{r}_j)\, (r_j)^{k-2}\rangle = k \langle V \rangle $$

For instance, for a stable atom (consisting of charged particles with Coulomb interaction): k = &minus;1, and hence 2&lang;T &rang; = &minus;&lang;V &rang;.