Consider a system with a time-independent Hamiltonian $\hat{H}$ that involves parameters. An obvious example is the molecular electronic Hamiltonian (13.5), which depends parametrically on the nuclear coordinates. However, the Hamiltonian of any system contains parameters. For example, in the one-dimensional harmonic-oscillator Hamiltonian operator $-\left(\hbar^{2} / 2 m\right)\left(d^{2} / d x^{2}\right)+\frac{1}{2} k x^{2}$, the force constant $k$ is a parameter, as is the mass $m$. Although $\hbar$ is a constant, we can consider it as a parameter also. The stationary-state energies $E_{n}$ are functions of the same parameters as $\hat{H}$. For example, for the harmonic oscillator
\(
\begin{equation}
E_{n}=\left(v+\frac{1}{2}\right) h \nu=\left(v+\frac{1}{2}\right) \hbar(k / m)^{1 / 2} \tag{14.110}
\end{equation}
\)
The stationary-state wave functions also depend on the parameters in $\hat{H}$. We now investigate how $E{n}$ varies with each of the parameters. More specifically, if $\lambda$ is one of these parameters, we ask for $\partial E{n} / \partial \lambda$, where the partial derivative is taken with all other parameters held constant.
We begin with the Schrödinger equation
\(
\begin{equation}
\hat{H} \psi{n}=E{n} \psi_{n} \tag{14.111}
\end{equation}
\)
where the $\psi_{n}$ 's are the normalized stationary-state eigenfunctions. Because of normalization, we have
\(
\begin{gather}
E{n}=\int \psi{n}^{} \hat{H} \psi{n} d \tau \tag{14.112}\
\frac{\partial E{n}}{\partial \lambda}=\frac{\partial}{\partial \lambda} \int \psi{n}^{*} \hat{H} \psi{n} d \tau \tag{14.113}
\end{gather*}
\)
The integral in (14.112) is a definite integral over all space, and its value depends parametrically on $\lambda$ since $\hat{H}$ and $\psi_{n}$ depend on $\lambda$. Provided the integrand is well behaved, we can find the integral's derivative with respect to a parameter by differentiating the integrand with respect to the parameter and then integrating. Thus
\(
\begin{equation}
\frac{\partial E{n}}{\partial \lambda}=\int \frac{\partial}{\partial \lambda}\left(\psi{n}^{} \hat{H} \psi{n}\right) d \tau=\int \frac{\partial \psi{n}^{}}{\partial \lambda} \hat{H} \psi{n} d \tau+\int \psi{n}^{} \frac{\partial}{\partial \lambda}\left(\hat{H} \psi_{n}\right) d \tau \tag{14.114}
\end{}
\)
We have
\(
\begin{equation}
\frac{\partial}{\partial \lambda}\left(\hat{H} \psi{n}\right)=\frac{\partial}{\partial \lambda}\left(\hat{T} \psi{n}\right)+\frac{\partial}{\partial \lambda}\left(\hat{V} \psi_{n}\right) \tag{14.115}
\end{equation}
\)
The potential-energy operator is just multiplication by $V$, so
\(
\begin{equation}
\frac{\partial}{\partial \lambda}\left(\hat{V} \psi{n}\right)=\frac{\partial V}{\partial \lambda} \psi{n}+V \frac{\partial \psi_{n}}{\partial \lambda} \tag{14.116}
\end{equation}
\)
The parameter $\lambda$ will occur in the kinetic-energy operator as part of the factor multiplying one or more of the derivatives with respect to the coordinates. For example, taking $\lambda$ as the mass of the particle, we have for a one-particle problem
\(
\begin{aligned}
\hat{T} & =-\frac{\hbar^{2}}{2 \lambda}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right) \
\frac{\partial}{\partial \lambda}(\hat{T} \psi) & =-\frac{\hbar^{2}}{2} \frac{\partial}{\partial \lambda}\left[\frac{1}{\lambda}\left(\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}\right)\right] \
& =\frac{\hbar^{2}}{2 \lambda^{2}}\left(\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}\right)-\frac{\hbar^{2}}{2 \lambda}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)\left(\frac{\partial \psi}{\partial \lambda}\right)
\end{aligned}
\)
since we can change the order of the partial differentiations without affecting the result. We can write this last equation as
\(
\begin{equation}
\frac{\partial}{\partial \lambda}\left(\hat{T} \psi{n}\right)=\left(\frac{\partial \hat{T}}{\partial \lambda}\right) \psi{n}+\hat{T}\left(\frac{\partial \psi_{n}}{\partial \lambda}\right) \tag{14.117}
\end{equation}
\)
where $\partial \hat{T} / \partial \lambda$ is found by differentiating $\hat{T}$ with respect to $\lambda$ just as if it were a function instead of an operator. Although we got (14.117) by considering a specific $\hat{T}$ and $\lambda$, the same arguments show it to be generally valid. Combining (14.116) and (14.117), we write
\(
\begin{equation}
\frac{\partial}{\partial \lambda}\left(\hat{H} \psi{n}\right)=\left(\frac{\partial \hat{H}}{\partial \lambda}\right) \psi{n}+\hat{H}\left(\frac{\partial \psi_{n}}{\partial \lambda}\right) \tag{14.118}
\end{equation}
\)
Equation (14.114) becomes
\(
\begin{equation}
\frac{\partial E{n}}{\partial \lambda}=\int \frac{\partial \psi{n}^{}}{\partial \lambda} \hat{H} \psi{n} d \tau+\int \psi{n}^{} \frac{\partial \hat{H}}{\partial \lambda} \psi{n} d \tau+\int \psi{n}^{} \hat{H} \frac{\partial \psi_{n}}{\partial \lambda} d \tau \tag{14.119}
\end{}
\)
For the first integral in (14.119), we have
\(
\begin{equation}
\int \frac{\partial \psi_{n}^{}}{\partial \lambda} \hat{H} \psi{n} d \tau=E{n} \int \frac{\partial \psi{n}^{*}}{\partial \lambda} \psi{n} d \tau \tag{14.120}
\end{}
\)
The Hermitian property of $\hat{H}$ and (14.111) give for the last integral in (14.119)
\(
\int \psi{n}^{*} \hat{H} \frac{\partial \psi{n}}{\partial \lambda} d \tau=\int \frac{\partial \psi{n}}{\partial \lambda}\left(\hat{H} \psi{n}\right)^{} d \tau=E{n} \int \psi{n}^{} \frac{\partial \psi_{n}}{\partial \lambda} d \tau
\)
Therefore,
\(
\begin{equation}
\frac{\partial E{n}}{\partial \lambda}=\int \psi{n}^{} \frac{\partial \hat{H}}{\partial \lambda} \psi{n} d \tau+E{n} \int \frac{\partial \psi{n}^{*}}{\partial \lambda} \psi{n} d \tau+E{n} \int \psi{n}^{} \frac{\partial \psi_{n}}{\partial \lambda} d \tau \tag{14.121}
\end{equation}
\)
The wave function is normalized, so
\(
\begin{gather}
\int \psi_{n}^{} \psi{n} d \tau=1, \quad \frac{\partial}{\partial \lambda} \int \psi{n}^{} \psi{n} d \tau=0 \
\int \frac{\partial \psi{n}^{}}{\partial \lambda} \psi{n} d \tau+\int \psi{n}^{} \frac{\partial \psi_{n}}{\partial \lambda} d \tau=0 \tag{14.122}
\end{gather}
\)
Using (14.122) in (14.121), we obtain
\(
\begin{equation}
\frac{\partial E{n}}{\partial \lambda}=\int \psi{n}^{} \frac{\partial \hat{H}}{\partial \lambda} \psi_{n} d \tau \tag{14.123}
\end{}
\)
Equation (14.123) is the (generalized) Hellmann-Feynman theorem. [For a discussion of the origin of the Hellmann-Feynman and related theorems, see J. I. Musher, Am. J. Phys., 34, 267 (1966).]
EXAMPLE
Apply the generalized Hellmann-Feynman theorem to the one-dimensional harmonic oscillator with $\lambda$ taken as the force constant.
For the harmonic oscillator, $\hat{H}=-\left(\hbar^{2} / 2 m\right)\left(d^{2} / d x^{2}\right)+\frac{1}{2} k x^{2}$ and $\partial \hat{H} / \partial k=\frac{1}{2} x^{2}$. The energy levels are $E_{v}=\left(v+\frac{1}{2}\right) h \nu=\left(v+\frac{1}{2}\right) h(k / m)^{1 / 2} / 2 \pi$. We have
\(
\partial E_{v} / \partial k=\frac{1}{2}\left(v+\frac{1}{2}\right) h k^{-1 / 2} m^{-1 / 2} / 2 \pi=\frac{1}{2}\left(v+\frac{1}{2}\right) h \nu / k
\)
Substitution in (14.123) gives
\(
\begin{equation}
\int{-\infty}^{\infty} \psi{v}^{} x^{2} \psi_{v} d x=\left(v+\frac{1}{2}\right) h \nu / k \tag{14.124}
\end{}
\)
We have found $\left\langle x^{2}\right\rangle$ for any harmonic-oscillator stationary state without evaluating any integrals. This result was also obtained from the virial theorem; see Eq. (14.74). For a third derivation, see Eyring, Walter, and Kimball, p. 79.
The derivation of the Hellmann-Feynman theorem assumes that $\partial \psi / \partial \lambda$ exists. For a state belonging to a degenerate energy level, this assumption may not be true and (14.123) need not hold. Changing the parameter's value from $\lambda$ to $\lambda+d \lambda$ amounts to applying a perturbation $\hat{H}^{\prime} \equiv \hat{H}(\lambda+d \lambda)-\hat{H}(\lambda) \approx(\partial \hat{H} / \partial \lambda) d \lambda$. This perturbation changes $\psi$ from $\psi(\lambda)$ to $\psi(\lambda+d \lambda)$. If $\psi(\lambda)$ is not one of the correct zeroth-order wave functions (9.73) for the perturbation $\hat{H}^{\prime}$, then $\psi(\lambda)$ need not equal $\lim {d \lambda \rightarrow 0} \psi(\lambda+d \lambda)$, so $\psi$ will make a discontinuous jump at $d \lambda=0$ and $\partial \psi / \partial \lambda$ will not exist at this point. The Hellmann-Feynman theorem (14.123) applies to the wave functions of a degenerate level only if we use the correct zeroth-order wave functions for the perturbation $\hat{H}^{\prime}$. These correct functions are found by solving (9.83) and (9.81) with $\hat{H}^{\prime} \equiv(\partial \hat{H} / \partial \lambda) d \lambda$. [Since $E{n}^{(1)}$ will be proportional to $d \lambda$, we can replace $\hat{H}^{\prime}$ with $\partial \hat{H} / \partial \lambda$ and $E{n}^{(1)}$ with $E{n}^{(1)} / d \lambda$ in (9.83) and (9.81).] For further details, see references 4-7 in G. P. Zhang and T. F. George, Phys. Rev. B, 69, 167102 (2004).
Application of the Hellmann-Feynman theorem to the hydrogenlike atom, with $Z$ as the parameter, gives (Prob. 14.37a)
\(
\begin{equation}
\int r^{-1}|\psi|^{2} d \tau=\left\langle\frac{1}{r}\right\rangle=\frac{Z}{n^{2}}\left(\frac{1}{a}\right) \tag{14.125}
\end{equation}
\)
This result was also obtained from the virial theorem; see Eq. (14.76).