User talk:RamanRana

Q2) Consider the potential V(x) = -g (x-a) –g (x+a), Prove that the ground state always exists and has even parity. By means of a sketch, also show that the first excited state exists for well separated values of 2a.

Ans : The potential is not affected by inversion. This implies that the wave functions have definite parity. The above potential is an attractive one. An attractive potential always has a ground state. The 1D space is divided into 3 regions. If we consider a particle with E<0, then in the leftmost and rightmost regions the wave function is of the type $Ae^{\left|-x\right|}$ and $De^{\left|-x\right|}$ respectively. The solution in the second region is $Be^{-\alpha x}+Ce^{\alpha x}$. For functions of even parity A=D, B=C. By continuity of wavefunction Ae^{-\alpha a}=Be^{-\alpha a}+Ce^{-\alpha a}. Integrating the Schrodinger equation from – ε to + ε and where ε tending to zero,

    ${\psi }'\left(a+\right)-{\psi }'\left(a-\right)={\frac {-2mg}{\hbar ^{2}}}\psi \left(a\right)$

Therefore,

               $\alpha e^{2\alpha a}={\frac {mg}{\hbar ^{2}}}\left(e^{2\alpha a}+1\right)$

For function of odd parity, A= -D and B= -C. Using the same method as we followed above,we get

                $\alpha e^{2\alpha a}={\frac {mg}{\hbar ^{2}}}\left(e^{2\alpha a}-1\right)$

The first equation always has a solution whereas the second does not always have one. This is seen easily from the fact that for positive values of α, the right hand side of the second equation is zero only at the origin and increases slower than the left hand side (except if the coefficient is adjusted so that it increases faster in the beginning).