Problem set 92

1. Given $[x_i,P_j]=i\hbar\delta_{ij}$, $i,j=1,2,3$. $[x_1,P_2^2]$ is:
1. 0
2. $i\hbar P_2$
3. $2x_1$
4. $2P_2$

$$\begin{align*} [x_1,P_2^2]&=P_2[x_1,P_2]+[x_1,P_2]P_2\\ &=0 \end{align*}$$

Hence, answer is (A)

2. Which of the following is an eigen state of square of linear momentum operator $P_x^2$?
1. $Ax^2$
2. $A\left(\sin{kx}+\cos{kx}\right)$
3. $Ae^{-\alpha x^2}$
4. $A\sin^2{kx}$

$P_x=-i\hbar\frac{\partial}{\partial x}$. Hnece, $P_x^2=-\hbar^2\frac{\partial^2}{\partial x^2}$.
Clearly, $$\begin{align*} {\scriptstyle P_x^2\left(A\left(\sin{kx}+\cos{kx}\right)\right)}&={\scriptstyle-\hbar^2\frac{\partial^2}{\partial x^2}\left(A\left(\sin{kx}+\cos{kx}\right)\right)}\\ &=k^2A\left(\sin{kx}+\cos{kx}\right) \end{align*}$$

Hence, answer is (B)

3. The electron in a hydrogen atom is in a superposition state described by the wavefunction $\psi(\vec r)=A\left[4\psi_{100}(\vec r)-2\psi_{211}(\vec r)+\sqrt{6}\psi_{210}(\vec r)-\sqrt{10}\psi_{21-1}(\vec r)\right]$, $\psi_{nlm}(\vec r)$ normalized wave function. The value of normalization constant, $A$, is:
1. $\frac{1}{3}$
2. $\frac{1}{6}$
3. $6$
4. $36$

Normalization condition is $\int\psi^*(\vec r)\psi(\vec r)dr=1$. As $\psi_{nlm}(\vec r)$ are normalized wave functions, we have, $\int\psi^*_{nlm}(\vec r)\psi_{n'l'm'}(\vec r)dr=\delta_{n'n}\delta_{l'l}\delta_{m'm}$ $$\int\psi^*(\vec r)\psi(\vec r)dr=|A|^2\left(16+4+6+10\right)=1$$ $$A=\frac{1}{6}$$

Hence, answer is (B)

4. Two coherent light sources of intensities I and 9I are used in an interference experiment. The resultant intensity at points where the waves from the two sources with phase difference $\pi$ is :
1. 16I
2. 9I
3. 4I
4. zero

Resultant intensity is given by $$I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\delta$$ $$I_R=I+9I+2\sqrt{9I}\cos\pi$$ $$I_R=10I-6I=4I$$

Hence, answer is (C)

5. Non-relativistic hydrogen atom spectrum is proportional to $-1/n^2$. The degeneracy of $n^{th}$ level is:
1. $n$
2. $2n+1$
3. $n^2$
4. $1/n^2$

The energy levels in the hydrogen atom depend only on the principal quantum number $n$. For given values of $n$ and $l$, the $( 2 l + 1 )$, states with $m_{l}=-l \rightarrow l$ are degenerate. The degree of degeneracy of the energy level $E_n$ is therefore : $\sum _{l=0}^{n-1}(2l+1)=n^{2}$, which is doubled if the spin degeneracy is included.

Hence, answer is (C)