All exercise S.Qs of Chapter#2: Approximation Methods of Quantum Mechanics-II. We have gathered and arranged all S.Qs for BS/MSc Physics students.
Q.2.1 Differentiate between degenerate and non- degenerate states?
Answer: Degenerate state: If two or more unperturbed states have same energy, they said to be degenerate states.( One energy value and many wave functions ).
Non-degenerate: If two or more unperturbed states have different energy in different states, they are said to be non- degenerate states. (For every energy value, we have a corresponding wave functions).
Q.2.2 why λ is regarded as strength of perturbation?
Answer: A dimensionless real parameter λ which tells us about strength of perturbation and its value is bound between 0 and 1 ($0\ \le\ \lambda\ \le1$).
Q.2.3 Show that $\mathbf{ \sum_{\alpha=1}^{f}{|a\alpha}|2 = 1}$ in time independent degenerate perturbation.
Answer:
$$|\Psi n>=\sum_{\alpha = 1}^{f} a\alpha \ |\Psi n\alpha>$$
Taking complex conjugate;
$$(|\mathrm{\Psi n}> )* =\left ( \sum_{\alpha =1}^{f} \right )\alpha \alpha |\mathrm{\Psi n\alpha}> )^*$$
$$|{\Psi n|}=(\sum _{\alpha =1}^f*\alpha \left | \psi {n\alpha } \right)$$
$$\psi _n|\psi _n =1$$
$$\sum_{\alpha =1}^{f}\left | \psi {n\alpha } \right |\alpha ^*\alpha \alpha \alpha \left | \psi _{n\alpha } \right |=1$$
$$\sum_{\alpha=1}^{f}\left | \alpha \alpha \right |^2 <\psi {n\alpha }|\psi {n\alpha }>=1$$
$$\sum_{\alpha=1}^{f}{|a\alpha|}^2 =1$$
Q.2.4 why we use the Variational method for approximation?
Answer: That is used to approximately calculate the energy levels of difficult quantum systems. Variational method is used to find the approximation to the ground state.
Q.2.5 Write down the equation of wave function for WKB approximation?
Answer:
${\Psi}\left(x\right)=\begin{Bmatrix}
\frac{1}{\sqrt{p\left(x\right)}}\ [Be^{\frac{i}{\hbar}} \int_{x}^{0}p(x’)dx’] & x<0\\ \frac{1}{\sqrt{\left | p(x) \right |}}De^{-\frac{1}{\hbar}\int_{0}^{x}\left | P(x’)dx’ \right |} & x>0
\end{Bmatrix}$
Q.2.6 what is the validity of WKB approximation?
Answer: The validity of WKB approximation is;
The wavelength is bound states and much less than 1.
Q.2.7 Write about the quantum tunneling in quantum mechanics.
Answer: Tunneling is a quantum mechanical phenomenon when a particle is able to penetrate through a potential energy barrier that is higher in energy than the particles kinetic energy (where a subatomic particle passes through a potential barrier).
Q.2.8 Write down the condition for perturbation.
Answer: The conditions are varied slightly the solution to the problem will also vary slightly when deviation between two problems is small, Perturbation theory is good for calculation of these deviations.
Q.2.9 why we use the WKB approximation?
Answer: Time independent in one dimension and is useful for finding the energy Eigen values and wave functions of system for which the classical limit is valid. It is slowly varying potential.
Q.2.10 What is the value of Land splitting factor?
Answer: Value of Land splitting factor is,
L is orbital quantum number, s is spin quantum number and j is total angular momentum quantum number.
$$g=1+\frac{j\left(j+1\right)+s\left(s+1\right)-l\left(l+1\right)}{2j(j+1)}$$
Q.2.11 The unperturbed wave functions for infinite square well are,
$$\Psi_n^{\left(0\right)}\left(x\right)=\sqrt{\frac{2}{L}}\ sin\left(\frac{n\pi x}{L}\right)$$
Find 1st order correction in energy corresponding to perturbation. H’ = V0 .
Answer: First order correction to energy is,
$$E_n^{\left(1\right)}=\int_{0}^{L} \mathrm{\Psi}n^0 \left(x\right)V_0\mathrm{\Psi}_n^0\left(x\right)dx=V_0 \int{0}^{L} \mathrm{\Psi}_n^0 \left(x\right)\mathrm{\Psi}_n^0\left(x\right)dx$$
Using normalization condition, we have
$$E_n^{\left(1\right)}=V_0$$
Q.2.12 Show that variational equation $\mathbf{\delta E\left(\Psi\right)=0 }$ is equivalent to Schrodinger equation,
$$\hat{H}\left|\Psi\right.\left.\ \right\rangle=E\left|\Psi\right.\left.\ \right\rangle$$
Answer: From variational Principle,
$$E\left(\mathrm{\Psi}\right)=\frac{\left\langle\mathrm{\Psi}\middle|\hat{H}\middle|\mathrm{\Psi}\right\rangle}{\left\langle\mathrm{\Psi}\middle|\mathrm{\Psi}\right\rangle}$$
Using $\delta E\left(\mathrm{\Psi}\right)=0$ , we have
$$\delta\left(\left\langle\mathrm{\Psi}\middle| H-E\middle|\mathrm{\Psi}\right\rangle\right)=0$$
Carrying out variation $\left\langle\delta\Psi\right.\left.\ \right|$ we have
$$\Rightarrow\left\langle\delta\Psi\middle| H-E\middle|\mathrm{\Psi}\right\rangle=0$$
Since $\left|\mathrm{\Psi}\right.\left.\ \right\rangle$ is arbitrary, so
$$\hat{H}\left|\mathrm{\Psi}\right.\left.\ \right\rangle=E\left|\mathrm{\Psi}\right.\left.\ \right\rangle$$
Similarly carrying out variation $\left|\delta\Psi\right.\left.\ \right\rangle,$, we have
$$\left\langle\mathrm{\Psi}\right.\left.\ \right|\hat{H}=E\left\langle\mathrm{\Psi}\right.\left.\ \right|$$
Taking complex conjugate,
$$\hat{H}\left|\mathrm{\Psi}\right.\left.\ \right\rangle=E\left|\mathrm{\Psi}\right.\left.\ \right\rangle\$$
