Variational Principle

Suppose that we wish to find an approximate solution to the time-independent Schrödinger equation,

$$H\,\vert\psi\rangle = E\,\vert\psi\rangle,$$ (9.68)

where $H$ is a known (presumably complicated) time-independent Hamiltonian, and $E$ the energy eigenvalue. Let $\vert\psi\rangle$ be a properly normalized trial solution to the above equation that corresponds to the trial wavefunction $\psi({\bf x}')\equiv \langle{\bf x}'\vert\psi\rangle$ . The so-called variational principle [94] states, quite simply, that the ground-state energy, $E_0$ , is always less than, or equal to, the expectation value of $H$ calculated with the trial solution: that is,

$$E_0 \leq \langle\psi\vert\,H\,\vert\psi\rangle.$$ (9.69)

Thus, by varying $\vert\psi\rangle$ until the expectation value of $H$ is minimized, we can obtain approximations to both the energy and the wavefunction of the ground state. (Incidentally, if $\vert\psi\rangle$ is not properly normalized then we must minimize $\langle\psi\vert\,H\,\vert\psi\rangle/\langle\psi\vert\psi\rangle$ . See Exercise 16.)

Let us prove the variational principle. Suppose that the $E_n$ and the $\vert n\rangle$ are the true eigenvalues and eigenkets of $H$ : that is,

$$H\,\vert n\rangle = E_n\,\vert n\rangle.$$ (9.70)

Furthermore, let

$$E_0 < E_1 < E_2 < \cdots,$$ (9.71)

so that $\vert\rangle$ is the ground state, $\vert 1\rangle$ the first excited state, et cetera. The $\vert n\rangle$ are assumed to be orthonormal: that is,

$$\langle n\vert m\rangle = \delta_{nm}.$$ (9.72)

If our trial ket $\vert\psi\rangle$ is properly normalized then we can write

$$\vert\psi\rangle = \sum_{n} c_n\,\vert n\rangle,$$ (9.73)

where

$$\sum_{n} \vert c_n\vert^{\,2} = 1,$$ (9.74)

and the $c_n$ are complex numbers. Here, $n$ is summed from 0 to $\infty$ . Now, the expectation value of $H$ , calculated with $\vert\psi\rangle$ , takes the form

$$\langle\psi\vert\,H\,\vert\psi\rangle = \sum_{n,m} c_n^{\,\ast}\,... ...n^{\,\ast}\,c_m\,E_m\,\langle n\vert m\rangle=\sum_n E_n\,\vert c_n\vert^{\,2},$$ (9.75)

where use has been made of Equations (9.71), (9.73), and (9.74). So, we can write

$$\langle \psi\vert\,H\,\vert\psi\rangle = \vert c_0\vert^{\,2}\,E_0 + \sum_{n>0} \vert c_n\vert^{\,2}\,E_n.$$ (9.76)

However, Equation (9.75) can be rearranged to give

$$\vert c_0\vert^{\,2} = 1-\sum_{n>0}\vert c_n\vert^{\,2}.$$ (9.77)

Combining the previous two equations, we obtain

$$\langle \psi\vert\,H\,\vert\psi\rangle = E_0 + \sum_{n>0} \vert c_n\vert^{\,2}\,(E_n-E_0).$$ (9.78)

But, the second term on the right-hand side of the previous expression is positive definite, because $E_n-E_0>0$ for all $n>0$ . [See Equation (9.72).] Hence, we obtain the desired result:

$$\langle \psi\vert\,H\,\vert\psi\rangle \geq E_0.$$ (9.79)

If $\vert\psi\rangle$ is a properly normalized trial ket that is orthogonal to the true ground-state of the system (i.e., $\langle 0\vert\psi\rangle=0$ ) then, by repeating the previous analysis, we can easily demonstrate that

$$\langle \psi \vert\,H\,\vert\psi\rangle \geq E_1.$$ (9.80)

Thus, by varying $\vert\psi\rangle$ until the expectation value of $H$ is minimized, we can obtain approximations to both the energy and wavefunction of the first excited state. In reality, because we do not generally know the exact ground state, we have to use a trial ket that is orthogonal to the approximate ground state obtained via the variational method described in the preceding paragraph. Obviously, we can continue this process until we have approximations to all of the stationary eigenstates. Note, however, that the errors are clearly cumulative in this approach, so that any approximations to highly excited states are likely to inaccurate. For this reason, the variational method is generally used only to calculate the ground state, and the first few excited states, of complicated quantum systems.

We can employ the variational principle to obtain an improved estimate for the ground-state energy of a helium atom. The Hamiltonian is written [see Equation (9.52)]

$$H = \frac{{\bf p}_1^{\,2}}{2\,m_e} + \frac{{\bf p}_2^{\,2}}{2\,m_... ...{2\,e^{\,2}}{4\pi\,\epsilon_0\,r_2} + \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{12}},$$ (9.81)

where we have now explicitly taken into account the fact that the nuclear charge is $2\,e$ . Let us take [see Equation (9.54)]

$$\phi({\bf x}_1,{\bf x}_2)= \psi({\bf x}_1)\,\psi({\bf x}_2)$$ (9.82)

as our properly normalized trial wavefunction, where [see Equation (9.56)]

$$\psi({\bf x})= \frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\exp\left(\frac{-Z\,r}{a_0}\right),$$ (9.83)

and $a_0$ is the Bohr radius. In the following, we shall treat $Z$ as a variable parameter. In fact, $Z\,e$ is the nuclear charge experienced by each electron. We would expect this quantity to be somewhat less that the true nuclear charge, $2\,e$ , because the electrons partially shield one another from the nucleus.

It is convenient to rewrite the Hamiltonian in the form

$$H = H_0 +H_1,$$ (9.84)

where

$$H_0 = \frac{{\bf p}_1^{\,2}}{2\,m_e} + \frac{{\bf p}_2^{\,2}}{2\,... ...c{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_1} -\frac{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_2},$$ (9.85)

and

$$H_1 = \frac{(Z-2)\,e^{\,2}}{4\pi\,\epsilon_0\,r_1}+ \frac{(Z-2)\,e^{\,2}}{4\pi\,\epsilon_0\,r_2}+ \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{12}}.$$ (9.86)

The trial wavefunction, (9.83), is an exact eigenstate of $H_0$ belonging to the eigenvalue $2\,Z^{\,2}\,E_0$ , where $E_0$ is the ground-state energy of hydrogen. [See Equation (9.57).] If we treat $H_1$ as a perturbation (see Chapter 7) then we obtain the estimate

$$E = 2\,Z^{\,2}\,E_0 + \langle H_1\rangle,$$ (9.87)

for the ground-state energy of helium. Here, the expectation value is calculated using the trial wavefunction.

Given that $E_0=-e^{\,2}/(8\pi\,\epsilon_0\,a_0)$ , we get

$$\langle H_1\rangle= \left[2\,(Z-2)\left\langle\frac{a_0}{r_1}\rig... ...ght\rangle + 2\left\langle\frac{a_0}{r_{12}}\right\rangle\right]\vert E_0\vert.$$ (9.88)

However, we already proved in Section 9.5 that

$$2\left\langle \frac{a_0}{r_{12}}\right\rangle = \frac{5\,Z}{4}.$$ (9.89)

[See Equations (9.58) and (9.63).] Furthermore,

$$\left\langle \frac{a_0}{r_1}\right\rangle = \left\langle \frac{a_0}{r_2}\right\rangle =\frac{1}{\pi}\left(\... ...right)^3\int d^{\,3}{\bf x}\,\exp\left(-\frac{2\,Z\,r}{a_0}\right)\frac{a_0}{r}$$

$$=4\left(\frac{Z}{a_0}\right)^3\int_0^\infty dr\,r^{\,2}\exp\left(-\frac{2\,Z\,r}{a_0}\right)\frac{a_0}{r}=Z\int_0^\infty dx\,x\,{\rm e}^{-x} = Z,$$ (9.90)

where $x=2\,Z\,r/a_0$ , and use has been made of Exercise 11. Thus,

$$\frac{\langle H_1\rangle}{\vert E_0\vert} = 4\,Z\,(Z-2) + \frac{5\,Z}{4} = 4\,Z^{\,2} - \frac{27\,Z}{4}.$$ (9.91)

According to the previous analysis, when expressed as a function of $Z$ , the ground-state energy of a helium atom takes the form

$$E(Z) = \left(2\,Z^{\,2} - 4\,Z^{\,2}+\frac{27\,Z}{4}\right)E_0 =\left(\frac{27\,Z}{4}-2\,Z^{\,2}\right)E_0.$$ (9.92)

We must now minimize this expression with respect to $Z$ to obtain our new estimate for the actual ground-state energy. It is easily seen that

$$\frac{dE}{dZ} = \left(\frac{27}{4}-4\,Z\right) E_0,$$ (9.93)

and

$$\frac{d^{\,2}E}{dZ^{\,2}} = -4\,E_0 > 0.$$ (9.94)

Thus, $E$ clearly attains a minimum value when

$$Z = \frac{27}{16} = 1.6875.$$ (9.95)

The fact that $Z<2$ confirms our earlier conjecture that the electrons partially shield the nuclear charge from one another. Our new estimate for the ground-state energy of helium is [68]

$$E(Z=27/4) = \frac{27^{\,2}}{2^{\,7}}\,E_0 = 5.695\,E_0=-77.5\,{\rm eV}.$$ (9.96)

This is clearly an improvement on our previous estimate, $-74.8\,{\rm eV}$ [see Equation (9.64)], recalling that the correct result is $-78.98\,{\rm eV}$ [75].