Skip to main content

Virial theorem

Consider a quantum particle of mass \(m\) moving in a \(D\)-dimensional space under the action of a Hamiltonian \[H = T + V \,,\] where \[T = \frac{\boldsymbol{p} \cdot \boldsymbol{p}}{2 m} = \frac{p_1^2 + p_2^2 + \ldots + p_D^2}{2 m}\] is the kinetic energy, and \[V = V(\boldsymbol{x}) = V(x_1, x_2, \ldots, x_D)\] is the potential energy. Here, \(x_j\) and \(p_j\), with \(j = 1, 2, \ldots, D\), are Cartesian components of the particle's position and momentum vectors, respectively, satisfying the standard commutation relation: \[[x_j, p_k] = i \hbar \delta_{jk} \,.\] Suppose further that the potential \(V\) is confining, and that the particle is in a bound state \(| \psi \rangle\) of energy \(E\), i.e. \[H | \psi \rangle = E | \psi \rangle \,.\] The virial theorem states that the expectation value of the kinetic energy is given by \[\boxed{ \langle T \rangle = \tfrac{1}{2} \langle \boldsymbol{x} \cdot \boldsymbol{\nabla} V \rangle }\] Here, \[\langle \cdot \rangle \equiv \langle \psi | \cdot | \psi \rangle\] and \(\boldsymbol{\nabla} = (\partial / \partial x_1, \, \partial / \partial x_2, \, \ldots, \, \partial / \partial x_D)^{\text{T}}\).

Proof: For any observable \(A\), \[\langle [H, A] \rangle = 0 \,.\] Indeed, \[\langle [H, A] \rangle = \langle \psi | H A - A H | \psi \rangle = \langle \psi | E A - A E | \psi \rangle = 0 \,.\] Choosing \(A = \boldsymbol{x} \cdot \boldsymbol{p}\), we get \[\langle [H, \boldsymbol{x} \cdot \boldsymbol{p}] \rangle = 0 \,.\] Evaluating the commutator, \[\begin{align} [H, \boldsymbol{x} \cdot \boldsymbol{p}] &= \boldsymbol{x} \cdot [H, \boldsymbol{p}] + [H, \boldsymbol{x}] \cdot \boldsymbol{p} \\ &= \boldsymbol{x} \cdot [V, \boldsymbol{p}] + \tfrac{1}{2 m} [\boldsymbol{p} \cdot \boldsymbol{p}, \boldsymbol{x}] \cdot \boldsymbol{p} \\ &= \boldsymbol{x} \cdot (i \hbar \boldsymbol{\nabla} V) + \tfrac{1}{2 m} (-2 i \hbar \boldsymbol{p}) \cdot \boldsymbol{p} \\ &= i \hbar \left( \boldsymbol{x} \cdot \boldsymbol{\nabla} V - \tfrac{1}{m} \boldsymbol{p} \cdot \boldsymbol{p} \right) \,, \end{align}\] we obtain \[\langle \boldsymbol{x} \cdot \boldsymbol{\nabla} V - \tfrac{1}{m} \boldsymbol{p} \cdot \boldsymbol{p} \rangle = 0 \,.\] This result is equivalent to the statement of the virial theorem.

Special case: Spherically symmetric power-law potentials

Let \[V(\boldsymbol{x}) = \alpha |\boldsymbol{x}|^{\beta} \,,\] where \(\alpha\) and \(\beta\) are some constants (whose values are such that \(V\) supports at least one bound state). Then, \[\boldsymbol{\nabla} V = \alpha \beta |\boldsymbol{x}|^{\beta-1} \boldsymbol{\nabla} |\boldsymbol{x}| = \alpha \beta |\boldsymbol{x}|^{\beta-1} \frac{\boldsymbol{x}}{|\boldsymbol{x}|} = \alpha \beta |\boldsymbol{x}|^{\beta-2} \boldsymbol{x} \,,\] \[\boldsymbol{x} \cdot \boldsymbol{\nabla} V = \alpha \beta |\boldsymbol{x}|^{\beta-2} \boldsymbol{x} \cdot \boldsymbol{x} = \alpha \beta |\boldsymbol{x}|^{\beta} = \beta V \,,\] and the virial theorem reads \[\langle T \rangle = \frac{\beta}{2} \langle V \rangle \,.\] For example, \[\langle T \rangle = \langle V \rangle \qquad \text{if} \qquad \beta = 2 \quad (\text{harmonic trap}) \,,\] \[\langle T \rangle = -\tfrac{1}{2} \langle V \rangle \qquad \text{if} \qquad \beta = -1 \quad (\text{Coulomb potential}) \,.\]