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The variational principle

Suppose that we wish to solve the time-independent Schrödinger equation
\begin{displaymath}
H\,\psi = E\,\psi,
\end{displaymath} (1146)

where $H$ is a known (presumably complicated) time-independent Hamiltonian. Let $\psi$ be a normalized trial solution to the above equation. The variational principle 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 wave-function: i.e.,
\begin{displaymath}
E_0 \leq \langle\psi\vert H\vert\psi\rangle.
\end{displaymath} (1147)

Thus, by varying $\psi$ until the expectation value of $H$ is minimized, we can obtain an approximation to the wave-function and energy of the ground-state.

Let us prove the variational principle. Suppose that the $\psi_n$ and the $E_n$ are the true eigenstates and eigenvalues of $H$: i.e.,

\begin{displaymath}
H\,\psi_n = E_n\,\psi_n.
\end{displaymath} (1148)

Furthermore, let
\begin{displaymath}
E_0 < E_1 < E_2 < \cdots,
\end{displaymath} (1149)

so that $\psi_0$ is the ground-state, $\psi_1$ the first excited state, etc. The $\psi_n$ are assumed to be orthonormal: i.e.,
\begin{displaymath}
\langle \psi_n\vert\psi_m\rangle = \delta_{nm}.
\end{displaymath} (1150)

If our trial wave-function $\psi$ is properly normalized then we can write
\begin{displaymath}
\psi = \sum_n c_n\,\psi_n,
\end{displaymath} (1151)

where
\begin{displaymath}
\sum_n \vert c_n\vert^{\,2} = 1.
\end{displaymath} (1152)

Now, the expectation value of $H$, calculated with $\psi$, takes the form
$\displaystyle \langle\psi\vert H\vert\psi\rangle$ $\textstyle =$ $\displaystyle \left.\left\langle \sum_n c_n\,\psi_n\right\vert
H\left\vert\sum_...
...\right. = \sum_{n,m} c_n^{\,\ast}\,c_m\,\langle \psi_n\vert H\vert\psi_m\rangle$  
  $\textstyle =$ $\displaystyle \sum_n\,c_n^{\,\ast}\,c_m\,E_m\,\langle \psi_n\vert\psi_m\rangle=
\sum_n E_n\,\vert c_n\vert^{\,2},$ (1153)

where use has been made of Eqs. (1148) and (1150). So, we can write
\begin{displaymath}
\langle \psi\vert H\vert\psi\rangle = \vert c_0\vert^{\,2}\,E_0 + \sum_{n>0} \vert c_n\vert^{\,2}\,E_n.
\end{displaymath} (1154)

However, Eq. (1152) can be rearranged to give
\begin{displaymath}
\vert c_0\vert^{\,2} = 1-\sum_{n>0}\vert c_n\vert^{\,2}.
\end{displaymath} (1155)

Combining the previous two equations, we obtain
\begin{displaymath}
\langle \psi\vert H\vert\psi\rangle = E_0 + \sum_{n>0} \vert c_n\vert^{\,2}\,(E_n-E_0).
\end{displaymath} (1156)

Now, the second term on the right-hand side of the above expression is positive definite, since $E_n-E_0>0$ for all $n>0$ [see (1149)]. Hence, we obtain the desired result
\begin{displaymath}
\langle \psi\vert H\vert\psi\rangle \geq E_0.
\end{displaymath} (1157)

Suppose that we have found a good approximation, $\tilde{\psi}_0$, to the ground-state wave-function. If $\psi$ is a normalized trial wave-function which is orthogonal to $\tilde{\psi}_0$ (i.e., $\langle \psi\vert\tilde{\psi}_0\rangle=0$) then, by repeating the above analysis, we can easily demonstrate that

\begin{displaymath}
\langle \psi \vert H\vert\psi\rangle \geq E_1.
\end{displaymath} (1158)

Thus, by varying $\psi$ until the expectation value of $H$ is minimized, we can obtain an approximation to the wave-function and energy of the first excited state. 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 method, so that any approximations to highly excited states are unlikely to be very accurate. For this reason, the variational method is generally only used to calculate the ground-state and first few excited states of complicated quantum systems.


next up previous contents
Next: The helium atom Up: Variational methods Previous: Introduction   Contents
Richard Fitzpatrick 2006-12-12