Virial Theorem

Statement

For a bound stationary state of

$$H = \frac{\mathbf p^2}{2m}+V(\mathbf r),$$

the quantum virial theorem gives

$$2\langle T\rangle = \left\langle \mathbf r\cdot\nabla V(\mathbf r) \right\rangle,$$

where $T=\mathbf p^2/(2m)$.

In one dimension this becomes

$$2\langle T\rangle = \langle xV'(x)\rangle.$$

Homogeneous Potentials

If the potential is homogeneous of degree $k$,

$$V(\lambda\mathbf r)=\lambda^kV(\mathbf r),$$

then Euler’s theorem gives

$$\mathbf r\cdot\nabla V=kV,$$

and therefore

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

For the harmonic oscillator, $k=2$, so $\langle T\rangle=\langle V\rangle$. For a Coulomb potential, $k=-1$, so $2\langle T\rangle=-\langle V\rangle$.

Assumptions

• The state is stationary and bound, or a suitable time average exists.
• The expectation values are finite.
• Boundary terms vanish under integration by parts or the equivalent commutator argument is well-defined.
• Singular potentials require domain care.

Derivation Idea

Use the dilation generator

$$G = \frac12 \left( \mathbf r\cdot\mathbf p+\mathbf p\cdot\mathbf r \right).$$

For a stationary state, $\langle G\rangle$ is time independent, so the expectation of its commutator with $H$ vanishes. The commutator produces the kinetic term and the scale derivative of the potential.

Canonical Links

• Conservation Laws
• Heisenberg Equation
• Quantum Harmonic Oscillator
• Hydrogen Atom
• Variational Principle

Common Mistakes

• Applying the stationary bound-state form to scattering states.
• Forgetting boundary terms in finite boxes or singular potentials.
• Assuming $\langle T\rangle=\langle V\rangle$ for every potential.
• Using the Coulomb relation with the wrong sign.
• Treating the virial theorem as enough to determine the full spectrum.

Quick Check

For a Coulomb potential $V(r)=-\alpha/r$, what are $\langle T\rangle$ and $\langle V\rangle$ in terms of the total energy $E$?

Solution

The potential is homogeneous of degree $k=-1$, so $2\langle T\rangle=-\langle V\rangle$. Since $E=\langle T\rangle+\langle V\rangle$, we get $E=-\langle T\rangle$ and therefore

$$\langle T\rangle=-E, \qquad \langle V\rangle=2E.$$

For bound Coulomb states, $E<0$, so $\langle T\rangle>0$ and $\langle V\rangle<0$.

References

• R. Clausius, “On a mechanical theorem applicable to heat,” Philosophical Magazine 40, 122-127, 1870.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• A. Messiah, Quantum Mechanics, Dover, 1999.