Virial theorem

In mechanics, a virial of a stable system of n particles is a quantity proposed by Rudolf Clausius in 1870. The virial (from the Latin vis, force) is defined by

$-{\tfrac {1}{2}}\sum _{i=1}^{n}\mathbf {r} _{i}\cdot \mathbf {F} _{i},$

where F_i is the total force acting on the i th particle and r_i 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 ⟨ T ⟩ of the total kinetic energy T of the n-particle system,

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

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 p_i 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:

$\mathbf {F} _{i}={\frac {d\mathbf {p} _{i}}{dt}}\quad {\hbox{and}}\quad 2T_{i}={\frac {d\mathbf {r} _{i}}{dt}}\cdot \mathbf {p} _{i}$

and it follows that

${\frac {dG}{dt}}=2T+\sum _{i=1}^{n}\mathbf {r} _{i}\cdot \mathbf {F} _{i}\quad {\hbox{with}}\quad T\equiv \sum _{i=1}^{n}T_{i}.$

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, being divided by infinite T, 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 the potential V is of the form

$V=\sum _{i=1}^{n}V(\mathbf {r} _{i})\quad {\hbox{with}}\quad V(\mathbf {r} _{i})=a_{i}r_{i}^{k}\quad {\hbox{and}}\quad r_{i}=(x_{i}^{2}+y_{i}^{2}+z_{i}^{2})^{1/2},$

where a_i 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 = −1. The force derived from a potential is

$\mathbf {F} _{i}=-{\boldsymbol {\nabla }}_{i}V\equiv -\left({\frac {\partial V}{\partial x_{i}}},\;{\frac {\partial V}{\partial y_{i}}},\;{\frac {\partial V}{\partial z_{i}}}\right)$

Consider

${\frac {\partial V}{\partial x_{i}}}=a_{i}{\frac {\partial (r_{i})^{k}}{\partial x_{i}}}=a_{i}k(r_{i})^{k-1}{\frac {\partial r_{i}}{\partial x_{i}}}=a_{i}k(r_{i})^{k-1}{\frac {x_{i}}{r_{i}}}=k{\frac {x_{i}}{r_{i}^{2}}}V(\mathbf {r} _{i}).$

Hence

$\mathbf {F} _{i}=-kV(\mathbf {r} _{i}){\frac {\mathbf {r} _{i}}{r_{i}^{2}}}.$

Then applying this for i = 1, … n,

$2\langle T\rangle =k\sum _{i=1}^{n}\left\langle V(\mathbf {r} _{i})\cdot {\frac {\mathbf {r} _{i}\cdot \mathbf {r} _{i}}{r_{i}^{2}}}\right\rangle =k\langle V\rangle \quad {\hbox{where}}\quad V=\sum _{i=1}^{n}V(\mathbf {r} _{i}).$

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. First the theorem will be proved and then it is applied to the special case of a potential that has a r^k-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 m_j 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

$[r_{k\alpha },p_{j\beta }^{2}]=\delta _{kj}\delta _{\alpha \beta }2ip_{k\alpha },\quad \alpha ,\beta =x,y,z;\quad k,j=1,,2,\ldots ,n,$

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

$\langle T\rangle =-{\tfrac {1}{2}}\sum _{j=1}^{n}\langle \mathbf {r} _{j}\cdot \mathbf {F} _{j}\rangle$

where ⟨ … ⟩ stands for an expectation value with respect to the exact eigenfunction Ψ 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 = −1, and hence 2⟨T ⟩ = −⟨V ⟩.