Example 2: Relaxation method.

The operator $f\left( \widehat{H'}\right)$ to approximate is $\widehat{U}=e^{-\widehat{H'}\tau }$ with the maximal and minimal eigenvectors $\lambda _{min}=E_{min}$, $\lambda _{max}=E_{max}$. This is often referred to as propagation in imaginary time as it results from the above propagation operator by setting

$$t=-i\hbar \tau .$$

(Note that the propagation is actually in negative imaginary time.)

If the operator is applied to a mixed state and the result renormalised, the state with the lowest energy in the mixed state is produced and all other states are filtered out at the rate $e^{-(E_{1}-E_{0})\tau }$, where $E_{0}$ and $E_{1}$ are the lowest eigenvalues. If a higher state is sought, the ground state has to be first produced and in the second run of the method projected out before every renormalization step. There are operators $e^{-(\widehat{H'}-E)^{m}\tau },\, m$ an even number, which produce the eigenstate closest to the energy $E$ directly without the need of producing the lower states first but I shall not go into any details about these [8].

The factor $\Phi =e^{\left( \Delta E/2+E_{min}\right) \tau }$ for the relaxation operator $\widehat{U}$ is not a phase factor anymore, $f(z')=e^{-z'}$ and hence

$$b_{n}=(-1)^{n}2I_{n}\left( \alpha \right) ,\, \alpha =\frac{\Delta E\tau }{2\hbar }$$

with $I_{n}$ the modified Bessel function of first kind of order n.

This time, the coefficients go to zero exponentially as $n$ becomes greater than $\sqrt{\alpha }$ so the method is very fast except for very short times where this exponentially decaying tail becomes dominant in the calculation time.

The same result is produced by

$$f(z')=\frac{a_{0}}{2}+\sum ^{\infty }_{n=1}a_{n}\chi _{n}(z')$$

with the recursion

$$\chi _{0} = \psi ,$$

$$\chi _{1} = -\widehat{H'}\psi ,$$

$$\chi _{n+1} = -2\widehat{H'}\chi _{n}-\chi _{n-1}$$

(note the changed sign of $\chi _{n-1}$ compared to the second algorithm in Ex. 1) and the coefficients

$$a_{n}=2I_{n}\left( \alpha \right) .$$

This can be shown by proving inductively that $\chi _{n}=(-1)^{n}\phi _{n}$, as obviously $b_{n}=(-1)^{n}a_{n}$:

$$\chi _{n+1} = -2\widehat{H'}\chi _{n}-\chi _{n-1}$$

$$= (-1)^{n+1}2\widehat{H'}\phi _{n}-(-1)^{n-1}\phi _{n-1}$$

$$= (-1)^{n+1}\left( 2\widehat{H'}\phi _{n}-\phi _{n-1}\right)$$

$$= (-1)^{n+1}\phi _{n+1}.$$