Show that if ${\mathbf{\psi }}_{\mathbf{n}}$ is real, the geometric phase vanishes. (Problems 10.3 and 10.4 are examples of this.) You might try to beat the rap by tacking an unnecessary (but perfectly legal) phase factor onto the eigenfunctions: ${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\equiv }{\mathbf{e}}^{\mathbf{i\phi n}}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$, where${\mathbf{\varphi }}_{\mathbf{n}}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$ is an arbitrary (real) function. Try it. You'll get a nonzero geometric phase, all right, but note what happens when you put it back into Equation 10.23. And for a closed loop it gives zero. Moral: For nonzero Berry's phase, you need (i) more than one time-dependent parameter in the Hamiltonian, and (ii) a Hamiltonian that yields nontrivially complex eigenfunctions.

Geometric phase is a phase difference gained during the course of a cycle when a system is subjected to cyclic adiabatic processes, which derives from the geometrical features of the Hamiltonian's parameter space in both classical and quantum mechanics.

Let the wave function after a time be,

${\mathrm{\theta }}_{\mathrm{n}}{\left(\mathrm{t}\right)}^0-\frac{1}{\mathrm{h}}{\int }_0^{\mathrm{t}}{\mathrm{E}}_{\mathrm{n}}\left({\mathrm{t}}^{\text{'}}\right){\mathrm{dt}}^{\text{'}}...\left(2\right)$

The equation is given by, localid="1656051708104"

By using the normalization condition, So, by differentiating to get:

Multiply the real function by the phase factor to get the new wave function: ${\mathrm{\psi }}^{\text{'}}={\mathrm{e}}^{\mathrm{i\varphi }}\mathrm{\psi }$

It’s time derivative is ${\mathrm{\psi }}^{\text{'}}={\mathrm{i\varphi e}}^{\mathrm{i\varphi }}\mathrm{\psi }+{\mathrm{e}}^{\mathrm{i\varphi }}\dot{\mathrm{\psi }}$. Thus,

The geometric phase for the modified function is,

$$\begin{gathered} {\mathrm{Y}}^{\text{'}}=\mathrm{i}{\int }_0^{\mathrm{t}}\mathrm{i}\dot{{\mathrm{\varphi dt}}^{\text{'}}\mathrm{n}} \\ =-\left(\mathrm{\varphi }\right(\mathrm{t})-\mathrm{\varphi }(0\left)\right) \end{gathered}$$

Substitute the value in the equation (1).

$$\begin{gathered} {\psi }^{\text{'}}\left(t\right)=e^{ie\left(t\right)}{e}^{-i\left(\varphi \right(t)-\varphi (0\left)\right)}{\psi }^{\text{'}}\left(t\right) \\ =e^{ie\left(t\right)}{e}^{-i\left(\varphi \right(t)-\varphi (0\left)\right)}{e}^{ie\left(t\right)}\psi \left(t\right) \\ =e^{ie\left(t\right)}{e}^{ie\left(0\right)}\psi \left(t\right) \end{gathered}$$