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. The theorem will be proved and 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 ⟩.