Problem set 61

1. The product of the uncertainties $\left(\Delta L_x\right)\left(\Delta L_y\right)$ for a particle in the state $a|1,1 > +b|1,-1 > $ (where $|l,m > $ denotes an eigenstate of $L^2$ and $L_z$) will be a minimum for
1. $a=\pm ib$
2. $a=0$ and $b=1$
3. $a=\frac{\sqrt{3}}{2}$ and $b=\frac{1}{2}$
4. $a=\pm b$

\begin{align*} L_z |l,m >&=m\hbar|l,m >\\ L^2 |l,m >&=m(m+1)\hbar|l,m >\\ L_+ |l,m >&={\scriptstyle\sqrt{(l-m)(l+m+1)}\hbar|l,m+1 >}\\ L_- |l,m >&={\scriptstyle\sqrt{(l+m)(l-m+1)}\hbar|l,m-1 >} \end{align*} Let $|\psi >=a|1,1 > +b|1,-1 > $ \begin{align*} L_z |\psi >&=a\hbar |1,1 >-b\hbar |1,-1 > \\ L^2 |\psi >&=2a\hbar |1,1 > \\ L_+ |\psi >&=\sqrt{2}b\hbar |1,0 > \\ L_- |\psi >&=\sqrt{2}a\hbar |1,0 > \\ L_+^2 |\psi >&=2b\hbar^2 |1,1 > \\ L_-^2 |\psi >&=2a\hbar^2 |1,-1 > \end{align*} $$\left(\Delta L_x\right)=\sqrt{< L_x^2 > -< L_x >^2 }$$ \begin{align*} &< L_x >=\left <\psi|L_x|\psi\right>\\ &=\frac{1}{2}\left <\psi|L_++L_-|\psi\right>\\ &=\frac{1}{2}\left <\psi|L_+|\psi\right>+\frac{1}{2}\left <\psi|L_-|\psi\right>\\ &=\frac{1}{2}\sqrt{2}b\left <\psi|1,0\right>+\frac{1}{2}\sqrt{2}a\left <\psi|1,0\right>\\ &=0 \end{align*} \begin{align*} &< L_x^2 >=\left <\psi|L_x^2|\psi\right>\\ &=\frac{1}{4}\left <\psi|(L_++L_-)(L_++L_-)|\psi\right>\\ &={\scriptstyle\frac{1}{4}\left <\psi|L_+^2+L_-^2+L_+L_-+L_-L_+|\psi\right>}\\ &={\scriptstyle\frac{1}{4}\left[2a^*b\hbar^2+2ab^*\hbar^2+2\hbar^2|a|^2+2\hbar^2|b|^2\right]}\\ &=\frac{\hbar^2}{2}\left[a^*b+ab^*+|a|^2+|b|^2\right]\\ &=\frac{\hbar^2}{2}\left[a^*b+ab^*+1\right] \end{align*} $$\left(\Delta L_x\right)=\sqrt{\frac{\hbar^2}{2}\left[a^*b+ab^*+1\right]}$$ Similarly $$\left(\Delta L_y\right)=\sqrt{\frac{\hbar^2}{2}\left[1-(a^*b+ab^*)\right]}$$ $$\left(\Delta L_x\right)\!\left(\Delta L_y\right)=\!\frac{\hbar^2}{2}\sqrt{1-(a^*b+ab^*)^2}$$ Clearly, for $a=\pm b$, we get,$\left(\Delta L_x\right)\left(\Delta L_y\right)=0$

Hence, answer is (D)

2. Of the nuclei of mass number $A=125$, the binding energy calculated from the liquid drop model (given that the coefficients for the Coulomb and the asymmetry energy are $a_c=0.7$ MeV and $a_{sym}=22.5$ MeV respectively) is a maximum for
1. $^{125}_{54}$Xe
2. $^{125}_{53}$I
3. $^{125}_{52}$Te
4. $^{125}_{51}$Sb

Using relation $$Z=\frac{4a_{sym}+a_cA^{-1/3}}{8a_{sym}A^{-1}+2a_cA^{-1/3}}$$ we get $Z=52$

Hence, answer is (C)

3. The value of $\oint\limits_C\frac{e^{2z}}{(z+1)^4}dz$, where $C$ is circle defined by $|z|=3$, is
1. $\frac{8\pi i}{3}e^{-2}$
2. $\frac{8\pi i}{3}e^{-1}$
3. $\frac{8\pi i}{3}e$
4. $\frac{8\pi i}{3}e^{2}$

Inside circle $|z|=3$, $z=-1$ is a pole of order 4. \begin{align*} &\oint\limits_Cf(z)dz\\ &=2\pi i\times&\text{sum of residues at}\\ &&\text{ poles inside }C \end{align*} In general the residue at a pole of order $m$ at $z=z_0$ is \begin{eqnarray} \begin{array}{c}R=\frac{1}{(m-1)!}\lim\limits_{z\to z_0}\left\{\frac{d^{m-1}\left((z-z_0)^mf(z)\right)}{dz^{m-1}}\right\}_{\!z=z_0}\end{array} \end{eqnarray} $$R=\frac{1}{3!}\lim\limits_{z\to -1}\left\{\frac{d^3\left((z+1)^4e^{2z}\right)}{dz^3}\right\}$$ $$R=\frac{8}{6}e^{-2}$$ $$\oint_Cf(z)dz=2\pi i\times\frac{8}{6}e^{-2}=\frac{8\pi i}{3}e^{-2} $$

Hence, answer is (A)

4. Consider the following processes involving free particles
   1. $\bar n\rightarrow \bar p+e^++\bar \nu_e$
   2. $\bar p+n\rightarrow \pi^-$
   3. $ p+n\rightarrow \pi^++\pi^0+\pi^0$
   4. $p+\bar \nu_e\rightarrow n+e^+$
   Which of the following statements is true?
1. Process (i) obeys all conservation laws
2. Process (ii) conserves baryon number, but violates energy-momentum conservation
3. Process (iii) is not allowed by strong interactions, but is allowed by weak interactions
4. Process (iv) conserves baryon number, but violates lepton number conservation

	$\bar n\rightarrow \bar p+e^++\bar \nu_e$
Q	$0\rightarrow -1+1+0$	Conserved
B	$-1\rightarrow -1+0+0$	Conserved
L	$0\rightarrow 0-1-1$	Not Conserved
$I_3$	$-\frac{1}{2}\rightarrow-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}$	Not Conserved

Similarly, one can check for other process. Process (ii) conserves baryon number, but violates energy-momentum conservation

Hence, answer is (B)

5. A one-dimensional harmonic oscillator is in the state $\psi(x)=\frac{1}{\sqrt{14}}\left[3\psi_0(x)-2\psi_1(x)-\psi_2(x)\right]$, where $\psi_0(x)$, $\psi_1(x)$ and $\psi_2(x)$ are the ground, first and second excited states respectively. The probability of finding the oscillator in the ground state is
1. 0
2. $\frac{3}{\sqrt{14}}$
3. $\frac{9}{14}$
4. 1

Any state of the harmonic oscillator can be expressed in terms of eigen states as follow $$\sum\limits_nc_n\psi_n(x)$$ Probability of finding the system in state $\psi_n(x)$ is given by $|c_n|^2$. Hence The probability of finding the oscillator in the ground state is $\frac{9}{14}$

Hence, answer is (C)