We now derive the virial theorem. Let $\hat{H}$ be the time-independent Hamiltonian of a system in the bound stationary state $\psi$ :
\(
\begin{equation}
\hat{H} \psi=E \psi \tag{14.59}
\end{equation}
\)
Let $\hat{A}$ be a linear, time-independent operator. Consider the integral
\(
\begin{equation}
\int \psi^{}[\hat{H}, \hat{A}] \psi d \tau=\langle\psi| \hat{H} \hat{A}-\hat{A} \hat{H}|\psi\rangle=\langle\psi| \hat{H}|\hat{A} \psi\rangle-E\langle\psi| \hat{A}|\psi\rangle \tag{14.60}
\end{}
\)
where (14.59) was used. Since $\hat{H}$ is Hermitian, we have
\(
\langle\psi| \hat{H}|\hat{A} \psi\rangle=\langle\hat{A} \psi| \hat{H}|\psi\rangle^{}=E^{}\langle\hat{A} \psi \mid \psi\rangle *=E\langle\psi \mid \hat{A} \psi\rangle=E\langle\psi| \hat{A}|\psi\rangle
\)
and Eq. (14.60) becomes
\(
\begin{equation}
\int \psi^{}[\hat{H}, \hat{A}] \psi d \tau=0 \tag{14.61}
\end{}
\)
Equation (14.61) is the hypervirial theorem. [For some of its applications, see J. O. Hirschfelder, J. Chem. Phys., 33, 1462 (1960); J. H. Epstein and S. T. Epstein, Am. J. Phys., 30, 266 (1962).] In deriving (14.61), we used the Hermitian property of $\hat{H}$. The proof that $\hat{p}{x}$ and $\hat{p}{x}^{2}$ are Hermitian, and hence that $\hat{H}$ is Hermitian, requires that $\psi$ vanish at $\pm \infty$ [see Eq. (7.17)]. Hence the hypervirial theorem does not apply to continuum stationary states, for which $\psi$ does not vanish at $\infty$.
We now derive the virial theorem from (14.61). We choose $\hat{A}$ to be
\(
\begin{equation}
\sum{i} \hat{q}{i} \hat{p}{i}=-i \hbar \sum{i} q{i} \frac{\partial}{\partial q{i}} \tag{14.62}
\end{equation}
\)
where the sum runs over the $3 n$ Cartesian coordinates of the $n$ particles. (Particle 1 has Cartesian coordinates $q{1}, q{2}, q{3}$ and linear-momentum components $p{1}, p{2}, p{3}$. In this chapter the symbol $q$ will indicate a Cartesian coordinate.) To evaluate $[\hat{H}, \hat{A}]$, we use (5.4), (5.5), (5.8), and (5.9) to get
\(
\begin{align}
& {\left[\hat{H}, \sum{i} \hat{q}{i} \hat{p}{i}\right]=\sum{i}\left[\hat{H}, \hat{q}{i} \hat{p}{i}\right]=\sum{i} \hat{q}{i}\left[\hat{H}, \hat{p}{i}\right]+\sum{i}\left[\hat{H}, \hat{q}{i}\right] \hat{p}{i}} \
& =i \hbar \sum{i} q{i} \frac{\partial V}{\partial q{i}}-i \hbar \sum{i} \frac{1}{m{i}} \hat{p}{i}^{2}=i \hbar \sum{i} q{i} \frac{\partial V}{\partial q_{i}}-2 i \hbar \hat{T} \tag{14.63}
\end{align}
\)
where $\hat{T}$ and $\hat{V}$ are the kinetic- and potential-energy operators for the system. Substitution of (14.63) into (14.61) gives
\(
\begin{equation}
\langle\psi| \sum{i} q{i} \frac{\partial V}{\partial q_{i}}|\psi\rangle=2\langle\psi| \hat{T}|\psi\rangle \tag{14.64}
\end{equation}
\)
Using $\langle B\rangle$ for the quantum-mechanical average of $B$, we write (14.64) as
\(
\begin{equation}
\left\langle\sum{i} q{i} \frac{\partial V}{\partial q_{i}}\right\rangle=2\langle T\rangle \tag{14.65}
\end{equation}
\)
Equation (14.65) is the quantum-mechanical virial theorem. Note that its validity is restricted to bound stationary states. (The word vires is Latin for "forces." In classical mechanics, the derivatives of the potential energy give the negatives of the force components. There is also a classical-mechanical virial theorem.)
For certain systems the virial theorem takes on a simple form. To discuss these systems, we introduce the idea of a homogeneous function. A function $f\left(x{1}, x{2}, \ldots, x_{j}\right)$ of several variables is homogeneous of degree $n$ if it satisfies
\(
\begin{equation}
f\left(s x{1}, s x{2}, \ldots, s x{j}\right)=s^{n} f\left(x{1}, x{2}, \ldots, x{j}\right) \tag{14.66}
\end{equation}
\)
where $s$ is an arbitrary parameter. For example, the function $g=1 / y^{3}+x / y^{2} z^{2}$ is homogeneous of degree -3 , since $g(s x, s y, s z)=1 / s^{3} y^{3}+s x / s^{2} y^{2} s^{2} z^{2}=s^{-3} g(x, y, z)$.
Euler's theorem on homogeneous functions states that, if $f\left(x{1}, \ldots, x{j}\right)$ is homogeneous of degree $n$, then
\(
\begin{equation}
\sum{k=1}^{j} x{k} \frac{\partial f}{\partial x_{k}}=n f \tag{14.67}
\end{equation}
\)
The theorem is proved as follows. Let
\(
u{1} \equiv s x{1}, \quad u{2} \equiv s x{2}, \quad \ldots, \quad u{j}=s x{j}
\)
Use of the chain rule gives for the partial derivative of the left side of (14.66) with respect to $s$
\(
\begin{aligned}
\frac{\partial f\left(u{1}, \ldots, u{j}\right)}{\partial s} & =\frac{\partial f}{\partial u{1}} \frac{\partial u{1}}{\partial s}+\frac{\partial f}{\partial u{2}} \frac{\partial u{2}}{\partial s}+\cdots+\frac{\partial f}{\partial u{j}} \frac{\partial u{j}}{\partial s} \
& =x{1} \frac{\partial f}{\partial u{1}}+x{2} \frac{\partial f}{\partial u{2}}+\cdots+x{j} \frac{\partial f}{\partial u{j}}=\sum{k=1}^{j} x{k} \frac{\partial f}{\partial u_{k}}
\end{aligned}
\)
The partial derivative of Eq. (14.66) with respect to $s$ is thus
\(
\begin{equation}
\sum{k=1}^{j} x{k} \frac{\partial f\left(u{1}, \ldots, u{j}\right)}{\partial u{k}}=n s^{n-1} f\left(x{1}, \ldots, x_{j}\right) \tag{14.68}
\end{equation}
\)
Let $s=1$, so that $u{i}=x{i}$; Eq. (14.68) then gives (14.67). This completes the proof.
Now we return to the virial theorem (14.65). If the potential energy $V$ is a homogeneous function of degree $n$ when expressed in Cartesian coordinates, Euler's theorem gives
\(
\begin{equation}
\sum{i} q{i} \frac{\partial V}{\partial q_{i}}=n V \tag{14.69}
\end{equation}
\)
and the virial theorem (14.65) simplifies to
\(
\begin{equation}
2\langle T\rangle=n\langle V\rangle \tag{14.70}
\end{equation}
\)
for a bound stationary state. Since (Prob. 6.35)
\(
\begin{equation}
\langle T\rangle+\langle V\rangle=E \tag{14.71}
\end{equation}
\)
we can write (14.70) in two other forms:
\(
\begin{align}
\langle V\rangle & =\frac{2 E}{n+2} \tag{14.72}\
\langle T\rangle & =\frac{n E}{n+2} \tag{14.73}
\end{align}
\)
EXAMPLE
Apply the virial theorem to (a) the one-dimensional harmonic oscillator; (b) the hydrogen atom; (c) a many-electron atom.
(a) For the one-dimensional harmonic oscillator, $V=\frac{1}{2} k x^{2}$, which is homogeneous of degree $n=2$. Equations (14.70) and (14.72) give
\(
\begin{equation}
\langle T\rangle=\langle V\rangle=\frac{1}{2} E=\frac{1}{2} h \nu\left(v+\frac{1}{2}\right) \tag{14.74}
\end{equation}
\)
This was verified for the ground state in Prob. 4.9.
(b) For the H atom, $V=-1 /\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}$ in Cartesian coordinates and atomic units. $V$ is a homogeneous function of degree -1 . Hence
\(
\begin{equation}
2\langle T\rangle=-\langle V\rangle \tag{14.75}
\end{equation}
\)
which was verified for the ground state in Prob. 6.36. For every hydrogen-atom bound stationary state,
\(
\begin{equation}
\langle V\rangle=2 E \quad \text { and } \quad\langle T\rangle=-E \tag{14.76}
\end{equation}
\)
(c) For a many-electron atom with spin-orbit interaction neglected,
\(
V=-Z \sum{i=1}^{n} \frac{1}{\left(x{i}^{2}+y{i}^{2}+z{i}^{2}\right)^{1 / 2}}+\sum{i} \sum{j>i} \frac{1}{\left[\left(x{i}-x{j}\right)^{2}+\left(y{i}-y{j}\right)^{2}+\left(z{i}-z{j}\right)^{2}\right]^{1 / 2}}
\)
Replacing each of the $3 n$ coordinates by $s$ times the coordinate, we find that $V$ is homogeneous of degree -1 . Hence Eqs. (14.75) and (14.76) hold for every atom.
Now consider molecules. In the Born-Oppenheimer approximation, the molecular wave function is [Eq. (13.12)]
\(
\psi=\psi{\mathrm{el}}\left(q{i} ; q{\alpha}\right) \psi{N}\left(q_{\alpha}\right)
\)
where $q{i}$ and $q{\alpha}$ symbolize the electronic and nuclear coordinates, respectively. $\psi_{\mathrm{el}}$ is found by solving the electronic Schrödinger equation (13.7):
\(
\hat{H}{\mathrm{el}} \psi{\mathrm{el}}\left(q{i} ; q{\alpha}\right)=E{\mathrm{el}}\left(q{\alpha}\right) \psi{\mathrm{el}}\left(q{i} ; q_{\alpha}\right)
\)
where $E_{\mathrm{el}}$ is the purely electronic energy and where (in atomic units)
\(
\begin{gather}
\hat{H}{\mathrm{el}}=\hat{T}{\mathrm{el}}+\hat{V}{\mathrm{el}} \tag{14.77}\
\hat{T}{\mathrm{el}}=-\frac{1}{2} \sum{i}\left(\frac{\partial^{2}}{\partial x{i}^{2}}+\frac{\partial^{2}}{\partial y{i}^{2}}+\frac{\partial^{2}}{\partial z{i}^{2}}\right) \tag{14.78}\
\hat{V}{\mathrm{el}}=-\sum{\alpha} \sum{i} \frac{Z{\alpha}}{\left[\left(x{i}-x{\alpha}\right)^{2}+\left(y{i}-y{\alpha}\right)^{2}+\left(z{i}-z{\alpha}\right)^{2}\right]^{1 / 2}} \
+\sum{i} \sum{j>i} \frac{1}{\left[\left(x{i}-x{j}\right)^{2}+\left(y{i}-y{j}\right)^{2}+\left(z{i}-z{j}\right)^{2}\right]^{1 / 2}} \tag{14.79}
\end{gather}
\)
Let the system be in the electronic stationary state $\psi{\mathrm{el}}$. If we put the subscript el on $\hat{H}$ and $\psi$ in (14.59) and regard the variables of $\psi{\mathrm{el}}$ to be the electronic coordinates $q{i}$ (with the nuclear coordinates $q{\alpha}$ being parameters), then the derivation of the virial theorem (14.65) is seen to be valid for the electronic kinetic- and potential-energy operators, and we have
\(
\begin{equation}
2\left\langle\psi{\mathrm{el}}\right| \hat{T}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle=\left\langle\psi{\mathrm{el}}\right| \sum{i} q{i} \frac{\partial V{\mathrm{el}}}{\partial q{i}}\left|\psi_{\mathrm{el}}\right\rangle \tag{14.80}
\end{equation}
\)
Viewed as a function of the electronic coordinates, $V_{\mathrm{el}}$ is $n o t$ a homogeneous function, since
\(
\begin{aligned}
{\left[\left(s x{i}-x{\alpha}\right)^{2}+\left(s y{i}-y{\alpha}\right)^{2}+\left(s z{i}-\right.\right.} & \left.\left.z{\alpha}\right)^{2}\right]^{-1 / 2} \
& \neq s^{-1}\left[\left(x{i}-x{\alpha}\right)^{2}+\left(y{i}-y{\alpha}\right)^{2}+\left(z{i}-z{\alpha}\right)^{2}\right]^{-1 / 2}
\end{aligned}
\)
Thus the virial theorem for the average electronic kinetic and potential energies of a molecule will not have the simple form (14.75), which holds for atoms. We can, however,
view $V{\text {el }}$ as a function of both the electronic and the nuclear Cartesian coordinates. From this viewpoint $V{\mathrm{el}}$ is a homogeneous function of degree -1 , since
\(
\begin{aligned}
{\left[\left(s x{i}-s x{\alpha}\right)^{2}+\left(s y{i}-s y{\alpha}\right)^{2}+\left(s z{i}\right.\right.} & \left.\left.-s z{\alpha}\right)^{2}\right]^{-1 / 2} \
& =s^{-1}\left[\left(x{i}-x{\alpha}\right)^{2}+\left(y{i}-y{\alpha}\right)^{2}+\left(z{i}-z{\alpha}\right)^{2}\right]^{-1 / 2}
\end{aligned}
\)
Therefore, considering $V_{\mathrm{el}}$ as a function of both electronic and nuclear coordinates and applying Euler's theorem (14.67), we have
\(
\sum{i} q{i} \frac{\partial V{\mathrm{el}}}{\partial q{i}}+\sum{\alpha} q{\alpha} \frac{\partial V{\mathrm{el}}}{\partial q{\alpha}}=-V_{\mathrm{el}}
\)
Using this equation in (14.80), we get
\(
\begin{equation}
2\left\langle\psi{\mathrm{el}}\right| \hat{T}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle=-\left\langle\psi{\mathrm{el}}\right| \hat{V}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle-\left\langle\psi{\mathrm{el}}\right| \sum{\alpha} q{a} \frac{\partial V{\mathrm{el}}}{\partial q{\alpha}}\left|\psi{\mathrm{el}}\right\rangle \tag{14.81}
\end{equation}
\)
which contains an additional term as compared with the atomic virial theorem (14.75). Consider this extra term. We have
\(
\left\langle\psi{\mathrm{el}}\right| \sum{\alpha} q{\alpha} \frac{\partial V{\mathrm{el}}}{\partial q{\alpha}}\left|\psi{\mathrm{el}}\right\rangle=\sum{\alpha} q{\alpha} \int \psi{\mathrm{el}}^{*} \frac{\partial V{\mathrm{el}}}{\partial q{\alpha}} \psi{\mathrm{el}} d \tau_{\mathrm{el}}
\)
where the nuclear coordinate $q_{\alpha}$ was taken outside the integral over electronic coordinates. In Section 14.7 we shall show that [see the bracketed sentence after Eq. (14.126)]
\(
\begin{equation}
\int \psi_{\mathrm{el}}^{} \frac{\partial V{\mathrm{el}}}{\partial q{\alpha}} \psi{\mathrm{el}} d \tau{\mathrm{el}}=\frac{\partial E{\mathrm{el}}}{\partial q{\alpha}} \tag{14.82}
\end{}
\)
\(
\begin{align}
2\left\langle\psi{\mathrm{el}}\right| \hat{T}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle & =-\left\langle\psi{\mathrm{el}}\right| \hat{V}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle-\sum{\alpha} q{\alpha} \frac{\partial E{\mathrm{el}}}{\partial q{\alpha}} \
2\left\langle T{\mathrm{el}}\right\rangle & =-\left\langle V{\mathrm{el}}\right\rangle-\sum{\alpha} q{\alpha} \frac{\partial E{\mathrm{el}}}{\partial q{\alpha}} \tag{14.83}
\end{align}
\)
where the $q_{\alpha}$ 's are the nuclear Cartesian coordinates. Using
\(
\begin{equation}
\left\langle T{\mathrm{el}}\right\rangle+\left\langle V{\mathrm{el}}\right\rangle=E_{\mathrm{el}} \tag{14.84}
\end{equation}
\)
we can eliminate either $\left\langle T{\mathrm{el}}\right\rangle$ or $\left\langle V{\mathrm{el}}\right\rangle$ from (14.83), which is the molecular form of the virial theorem.
Now consider a diatomic molecule. The electronic energy is a function of $R$, the internuclear distance: $E{\text {el }}=E{\text {el }}(R)$. The summation in (14.83) is over the nuclear Cartesian coordinates $x{a}, y{a}, z{a}, x{b}, y{b}, z{b}$. We have
\(
\begin{gather}
\frac{\partial E{\mathrm{el}}}{\partial x{a}}=\frac{d E{\mathrm{el}}}{d R} \frac{\partial R}{\partial x{a}}, \quad \frac{\partial E{\mathrm{el}}}{\partial x{b}}=\frac{d E{\mathrm{el}}}{d R} \frac{\partial R}{\partial x{b}} \
R=\left[\left(x{a}-x{b}\right)^{2}+\left(y{a}-y{b}\right)^{2}+\left(z{a}-z{b}\right)^{2}\right]^{1 / 2} \
\frac{\partial R}{\partial x{a}}=\frac{x{a}-x{b}}{R}, \quad \frac{\partial R}{\partial x{b}}=\frac{x{b}-x{a}}{R} \tag{14.85}
\end{gather}
\)
with similar equations for the $y$ and $z$ coordinates. The sum in (14.83) becomes
\(
\begin{aligned}
\sum{\alpha} q{\alpha} \frac{\partial E{\mathrm{el}}}{\partial q{\alpha}}=\frac{1}{R} \frac{d E{\mathrm{el}}}{d R}\left[x{a}\left(x{a}-x{b}\right)+x{b}\left(x{b}-x{a}\right)\right. & +y{a}\left(y{a}-y{b}\right) \
& \left.+y{b}\left(y{b}-y{a}\right)+z{a}\left(z{a}-z{b}\right)+z{b}\left(z{b}-z_{a}\right)\right]
\end{aligned}
\)
\(
\sum{\alpha} q{\alpha} \frac{\partial E{\mathrm{el}}}{\partial q{\alpha}}=R \frac{d E_{\mathrm{el}}}{d R}
\)
where $\left(x{a}-x{b}\right)^{2}+\left(y{a}-y{b}\right)^{2}+\left(z{a}-z{b}\right)^{2}=R^{2}$ was used. The virial theorem (14.83) for a diatomic molecule becomes
\(
\begin{equation}
2\left\langle T{\mathrm{el}}\right\rangle=-\left\langle V{\mathrm{el}}\right\rangle-R \frac{d E_{\mathrm{el}}}{d R} \tag{14.86}
\end{equation}
\)
Using (14.84), we have the two alternative forms
\(
\begin{align}
\left\langle T{\mathrm{el}}\right\rangle & =-E{\mathrm{el}}-R \frac{d E{\mathrm{el}}}{d R} \tag{14.87}\
\left\langle V{\mathrm{el}}\right\rangle & =2 E{\mathrm{el}}+R \frac{d E{\mathrm{el}}}{d R} \tag{14.88}
\end{align}
\)
In deriving the molecular electronic virial theorem (14.83), we omitted the internuclear repulsion
\(
\begin{equation}
V{N N}=\sum{\beta} \sum{\alpha>\beta} \frac{Z{\alpha} Z{\beta}}{\left[\left(x{\alpha}-x{\beta}\right)^{2}+\left(y{\alpha}-y{\beta}\right)^{2}+\left(z{\alpha}-z_{\beta}\right)^{2}\right]^{1 / 2}} \tag{14.89}
\end{equation}
\)
from the electronic Hamiltonian (14.77) to (14.79). Let
\(
V=V{\mathrm{el}}+V{N N}
\)
where $V{\text {el }}$ is given by (14.79). We can rewrite the electronic Schrödinger equation $\hat{H}{\mathrm{el}} \psi{\mathrm{el}}=E{\mathrm{el}} \psi_{\mathrm{el}}$ as [Eq. (13.4)]
\(
\left(\hat{T}{\mathrm{el}}+\hat{V}\right) \psi{\mathrm{el}}=U\left(q{\alpha}\right) \psi{\mathrm{el}}
\)
where
\(
U\left(q{\alpha}\right)=E{\mathrm{el}}\left(q{\alpha}\right)+V{N N}
\)
$U\left(q{\alpha}\right)$ is the potential-energy function for nuclear motion. Consider what happens to the right side of (14.83) when we add $V{N N}$ to $V{\mathrm{el}}$ and $E{\mathrm{el}}$. We have
\(
\begin{align}
&-\int \psi_{\mathrm{el}}^{}\left(\hat{V}{\mathrm{el}}+V{N N}\right) \psi{\mathrm{el}} d \tau{\mathrm{el}}-\sum{\alpha} q{\alpha} \frac{\partial U}{\partial q{\alpha}} \
&=-\left\langle\psi{\mathrm{el}}\right| \hat{V}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle-V{N N}-\sum{\alpha} q{\alpha} \frac{\partial E{\mathrm{el}}}{\partial q{\alpha}}-\sum{\alpha} q{\alpha} \frac{\partial V{N N}}{\partial q_{\alpha}} \tag{14.90}
\end{align*}
\)
Since $V_{N N}$ is a homogeneous function of the nuclear Cartesian coordinates of degree -1 , Euler's theorem gives
\(
\sum{\alpha} q{\alpha} \frac{\partial V{N N}}{\partial q{\alpha}}=-V_{N N}
\)
and (14.90) becomes
\(
\begin{equation}
-\left\langle\psi{\mathrm{el}}\right| \hat{V}{\mathrm{el}}+V{N N}\left|\psi{\mathrm{el}}\right\rangle-\sum{\alpha} q{\alpha} \frac{\partial U}{\partial q{\alpha}}=-\left\langle\psi{\mathrm{el}}\right| \hat{V}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle-\sum{\alpha} q{\alpha} \frac{\partial E{\mathrm{el}}}{\partial q{\alpha}} \tag{14.91}
\end{equation}
\)
Substitution of (14.91) into (14.83) gives
\(
\begin{align}
2\left\langle\psi{\mathrm{el}}\right| \hat{T}{\mathrm{el}}\left|\psi{\mathrm{el}}\right\rangle & =-\left\langle\psi{\mathrm{el}}\right| \hat{V}{\mathrm{el}}+V{N N}\left|\psi{\mathrm{el}}\right\rangle-\sum{\alpha} q{\alpha} \frac{\partial U}{\partial q{\alpha}} \
2\left\langle T{\mathrm{el}}\right\rangle & =-\langle V\rangle-\sum{\alpha} q{\alpha} \frac{\partial U}{\partial q{\alpha}} \tag{14.92}
\end{align}
\)
which has the same form as (14.83). Therefore the molecular electronic virial theorem holds whether or not we include the internuclear repulsion. Corresponding to Eqs. (14.86) to (14.88) for diatomic molecules, we have
\(
\begin{align}
2\left\langle T{\mathrm{el}}\right\rangle & =-\langle V\rangle-R(d U / d R) \tag{14.93}\
\left\langle T{\mathrm{el}}\right\rangle & =-U-R(d U / d R) \tag{14.94}\
\langle V\rangle & =2 U+R(d U / d R) \tag{14.95}
\end{align}
\)
The potential energy $V=V{\text {el }}+V{N N}$ takes the zero of energy with all particles (electrons and nuclei) at infinite separation from one another. Therefore, $U(R)$ in (14.93) to (14.95) does not go to zero at $R=\infty$ but goes to the sum of the energies of the separated atoms, which is negative.
The true (nonrelativistic) wave functions for a system with $V$ a homogeneous function of the coordinates must satisfy the form of the virial theorem (14.70). What determines whether an approximate wave function for such a system satisfies (14.70)? The answer is that, by inserting a variational parameter as a multiplier of each Cartesian coordinate and choosing this parameter to minimize the variational integral, we can make any trial variation function satisfy the virial theorem. (For the proof, see Kauzmann, page 229.) This process is called scaling, and the variational parameter multiplying each coordinate is called a scale factor. For a molecular trial function, the scaling parameter must be inserted in front of the nuclear Cartesian coordinates, as well as in front of the electronic coordinates.
Consider some examples. The zeroth-order perturbation wave function (9.49) for the heliumlike atom has no scale factor and so does not satisfy the virial theorem. If we were to calculate $\langle T\rangle$ and $\langle V\rangle$ for (9.49), we would find $2\langle T\rangle \neq-\langle V\rangle$; see Prob. 14.26. The Heitler-London trial function for $\mathrm{H}_{2}$, Eq. (13.100), has no scale factor and does not satisfy the virial theorem. The Heitler-London-Wang function, which uses a variationally determined orbital exponent, satisfies the virial theorem. Hartree-Fock wave functions satisfy the virial theorem; note the scale factor in the Slater basis functions (11.14).